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# Demidovich Problems In Mathematical Analysis

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C.

T. C. BapaneHKoe. B. 77 AeMudoeuH, B. A M. Koean, r Jl JJyHit, E noptuneea, C. B. P. fl. UlocmaK, A. P.

E

H. Ctweea,

SAflAMM H VnPA)KHEHHfl

no

MATEMATM H ECKOMV AHAJ1H3V

I7od B.

H.

AE

rocydapcmeeHHoe a

M

G. Baranenkov* B. Drmidovich V. Efimenko, S. Kogan, G. Lunts>> E. Porshncva, E. bychfia, S. frolov, /?. bhostak, A. Yanpolsky

PROBLEMS IN

MATHEMATICAL ANALYSIS Under B.

the editorship of

DEMIDOVICH

Translated from the Russian by G.

YANKOVSKV

MIR PUBLISHERS Moscow

TO THE READER MIR opinion

of

Publishers would be the translation and

to

have your

design

of

this

book.

Please send your suggestions to 2, Pervy Rtzhtky Pereulok, Moscow, U. S. S. R.

Second Printing

Printed

in

the

Union

of

Soviet Socialist

Republic*

CONTENTS 9

Preface

Chapter Sec.

I.

INTRODUCTION TO ANALYSIS

1.

Functions

Sec. 2

Graphs

Sec. 3

Limits

of

11

Elementary Functions

16

22

Infinitely Small and Large Quantities Sec. 5. Continuity of Functions

33

Sec. 4

Chapter II Sec Sec

1.

2

36

DIFFERENTIATION OF FUNCTIONS Calculating Derivatives Directly Tabular Differentiation

42 46

Sec. 3 The Derivatwes of Functions Not Represented Explicitly Sec. 4. Geometrical and Mechanical Applications of the Derivative Sec 5 Derivatives of Higier Orders

Sec

6

Sec

7

Sec. 8

.

.

56

.

60

66

and Higher Orders Mean Value Theorems Taylor's Formula Differentials of First

9 The L'Hospital-Bernoulli Rule Forms

Sec

for

Evaluating

71

75

77

Indeterminate

78

THE EXTREMA OF A FUNCTION AND THE GEOMETRIC APPLICATIONS OF A DERIVATIVE

Chapter III

Sec.

1.

Sec. 2

Sec

3

Sec

4.

Sec. 5.

Chapter IV Sec.

1

Sec

2

Sec

3

Sec. 4

Sec. 5.

The Extrema of a Function The Direction of Concavity

of

One Argument

83

Points of Inflection

Asymptotes Graphing Functions by Characteristic Points Differential of an Arc Curvature

91

93

.

96 .

.

101

INDEFINITE INTEGRALS Direct Integration Integration by Substitution Integration by Parts

Standard Integrals Containing a Quadratic Trinomial Integration of Rational Functions

107

113 116

....

118 121

Contents

Sec. 6.

Integrating Certain Irrational Functions

125

Sec

Integrating Trigoncrretric Functions Integration of Hyperbolic Functions

128

7.

Sec. 8

Sec

9.

133

Using Ingonometric and Hyperbolic Substitutions

integrals of the

Form

f

R

(x,

^a^ + bx + c) dx,

for

Where R

Finding is

Ra-

a

tional Function

Sec

10

Sec

11

Sec.

12.

133

Integration of

Vanou* Transcendental Functions

135

Using Reduction Formulas Miscellaneous Examples on Integration

135 136

DEFINITE INTEGRALS

Chapter V Sec.

1.

Sec

2

The Definite Integral

as the Limit of a

Sum

138

Evaluating Ccfirite Integrals by Means of Indefinite Integrals 140

Sec. 3

Improper Integrals

Sec

Charge

4

of

143

Variable in a Definite Integral

146

Integration by Parts

Sec. 5.

149

Mean-Value Theorem Sec. 7. The Areas of Plane Figures Sec 8. The Arc Length of a Curve Sec 9 Volumes of Solids Sec 10 The Area of a Surface of Revolution

Sec

6

Sec

11

torrents

Sec

12.

Applying Definite Integrals

Centres

of

Gravity

150

153 158 161

166

Guldin's Theorems

to the Solution of Physical

168

Prob-

lems

Chapter VI. Sec.

1.

173

FUNCTIONS OF SEVERAL VARIABLES Basic Notions

180

Sec. 2. Continuity

184

Sec

3

Partial Derivatives

185

Sec

4

Total Differential of a Function

187

Sec

5

Differentiation of Composite Functions 190 Given Direction and the Gradient of a Function 193

Sec. 6. Derivative in a Sec. 7

Sec Sec Sec

HigKei -Order Derivatives and Differentials

197

8

Integration of Total Differentials 9 Differentiation of Implicit Functions 10 Change of Variables

202 205

.211 217

15

The Tangent Plane and the Normal to a Surface for a Function of Several Variables . The Extremum of a Function of Several Variables .... * Firdirg the Greatest and tallest Values of Functions Smcular Points of Plane Curves

16

Envelope

232

Sec.

11.

Sec

12

Sec.

13

Sec

14

Sec

Sec

Taylor's Formula

.

.

.

.

.

Sec. 17. Arc Length o! a Space

Curve

.

220 222 227

230 234

Contents Sec.

18.

Sec.

19

The Vector Function of a Scalar Argument The Natural Trihedron of a Space Curve Curvature and Torsion of a Space Curve

Sec. 20.

Chapter VII. Sec.

The Double

1

Integral in Rectangular Coordinates

Sec. Sec.

7.

Triple Integrals

Sec.

8.

Improper Integrals

Sec. Sec.

Sec.

238

242

MULTIPLE AND LINE INTEGRALS

Change of Variables in a Double Integral 3. Computing Areas 4. Computing Volumes 5. Computing the Areas of Surfaces 6 Applications of the Double Integral in Mechanics

Sec.

235

246 252

2

256 258

259 230 262

Dependent

on

a

Parameter.

Improper 269

Multifle Integrals 9 Line Integrals

273

Sec.

Sec.

10.

Surface Integrals

284

Sec.

11.

Sec.

12.

The Ostrogradsky-Gauss Formula Fundamentals of Field Theory

288

286

Chapter VIII. SERIES Sec.

Number

1.

293

Series

Sec. 2. Functional Series

304

Sec. 3. Taylor's Series Sec. 4. Fourier's Series

318

Chapter Sec.

311

IX DIFFERENTIAL EQUATIONS 1.

Verifying Solutions. Forming Differential Equations Curves. Initial Conditions

of

Fami-

lies of

Sec. 2

Sec.

3.

327

Orthogonal Trajectories Sec. 4

322

324 First-Order Differential Equations First-Order Diflerential Equations with Variables Separable. First-Order

Sec. 5. First-Order

Homogeneous Linear

Differential

Differential

330

Equations Equations.

Bernoulli's

332 Equation 335 Sec. 6 Exact Differential Equations. Integrating Factor Sec 7 First-Order Differential Equations not Solved for the Derivative 337 339 Sec. 8. The Lagrange and Clairaut Equations Sec. 9. Miscellaneous Exercises on First-Order Differential Equations 340 345 Sec. 10. Higher-Order Differential Equations 349 Sec. 11. Linear Differential Equations Sec.

12.

Linear Differential Equations of Second Order with Constant

Coefficients

351

Contents

8 Sec. 13. Linear

Differential

Equations of Order

Higher

than

Two 356

with Constant Coefficients Sec

14.

Euler's Equations

Sec

15.

Systems

of

Differential

357

Integration of Differential Equations by

Sec. 16.

359

Equations

Means

of

Power

Se-

361

ries

Sec

17.

Chapter X.

Problems on Fourier's Method

363

APPROXIMATE CALCULATIONS 367

Sec. 2.

Operations on Approximate Numbers Interpolation of Functions

Sec.

Computing the^Rcal Roots

376

Sec.

1

3.

Sec. 4

Numerical, Integration of

Nun

Equations Functions

Integration of Ordinary DilUrtntial Equations Sec. 6. Approximating Ftuncr's Coefficients Sec. 5.

372

of

er:ca1

382 .

.

384

3>3

396

APPENDIX

475

I.

II.

Greek Alphabet Some Constants

Inverse Quantities, Powers, Roots, Logarithms Trigonometric Functions V. Exponential, Hyperbolic and Trigonometric Functions VI. Some Curves

475

475

III.

476

IV

478 479 480

PREFACE This collection of problems and exercises in mathematical analcovers the maximum requirements of general courses in ysis higher mathematics for higher technical schools. It contains over 3,000 problems sequentially arranged in Chapters I to X covering branches of higher mathematics (with the exception of analytical geometry) given in college courses. Particular attention is given to the most important sections of the course that require established skills (the finding of limits, differentiation techniques, the graphing of functions, integration techniques, the applications all

of definite integrals, series, the solution of differential equations). Since some institutes have extended courses of mathematics,

the authors have included problems on field theory,

method,

and

the

Fourier

approximate calculaiions. Experience shows that problems given in this book not only fully satisfies

the number of the requireiren s of the student, as far as practical mas!ering of the various sections of the course goes, but also enables the instructor to supply a varied choice of problems in each section to select problems for tests and examinations. Each chap.er begins with a brief theoretical introduction that covers the basic definitions and formulas of that section of the course. Here the most important typical problems are worked out in full. We believe that this will greatly simplify the work of the student. Answers are given to all computational problems; one asterisk indicates that hints to the solution are given in the answers, two asterisks, that the solution is given. The are frequently illustrated by drawings. problems This collection of problems is the result of many years of teaching higher mathematics in the technical schools of the Soviet Union. It includes, in addition to original problems and examples, a large number of commonly used problems.

and

Chapter I

INTRODUCTION TO ANALYSIS

Sec.

1.

Functions

1. Real nurrbers. Rational and irrational numbers are collectively known numbers The absolute value of a real number a is understood to be the nonnegative number \a\ defined by the conditions' \a\=a if a^O, and = a if a < 0. The following inequality holds for all real numbers a |aj as real

ana

b:

2. Definition of a function. If to every value*) of a variable x, which belongs to son.e collection (set) E, there corresponds one and only one finite value of the quantity /, then y is said to be a function (single-valued) of x or a dependent tariable defined on the set E. x is the a r gument or independent variable The fact that y is a Junction of x is expressed in brief form by the notation y~l(x) or y = F (A), and the 1'ke If to every value of x belonging to some set E there corresponds one or several values of the variable /y, then y is called a multiple- valued function of x defined on E. From now on we shall use the word "function" only in the meaning of a single-valued function, if not otherwise stated 3 The domain of definition of a function. The collection of values of x for which the given function is defined is called the domain of definition (or the domain) of this function. In the simplest cases, the domain of a function is either a closed interval [a.b\, which is the set of real numbers x that satisfy the inequalities or an open intenal (a.b), which :s the set of real numbers that satisfy the inequalities a a more comx b. Also possible is plex structure of the domain of definition of a function (see, for instance, Prob-

a^^^b,

< <

lem 21)

Example

1.

Determine the domain

of definition of the function 1

Solution.

The function

is

defined

x that

is,

if

|x|>

1.

oo<x<

vals:

Thus, the 1

and

1

2

if

-l>0,

domain

<x<

-\-

of the function is a set

of

two

inter-

oo

=

4. Inverse functions. If the equation t/ /(x) may be solved uniquely for the variable x, that is, if there is a function x g(y) such that y^*

*)

stated.

Hencetorth

all

values will be

considered

as

real,

if

not

otherwise

Introduction to Analysis

12

\Ch.

I

then the function x = g(y), or, in standard notation, y=g(x), is the inverse of y = f(x). Obviously, g[f(x)]s&x, that is, the function f (x) is the inverse of g(x) (and vice ve^sa). In He fereia! case, the equation y f(x) defines a multiple-valued inverse function x f~ (y) such that y ==[[(-* (y)\ for all y that are values of the function f (x) Lxanple 2. Determine the inverse of the function

=

}

y=l-2-*. Solution. Solving equation

(1) for x,

(1)

we have

2-*=l

y

and

log(l-y)

*

j

log 2

w

'

of the function (2) is the domain of oo
Obviously, ries

or a function of a function. A function defined by an equation not solved for the dependent variable the equation x*+i/*=l defines is called an implicit (unction. For example, y as an implicit function of x. 6. The graph of a function. A set of points (x, y) in an ;o/-plane, whose coordinates are connected by the equation y f(x), is called the graph of the given function.

1**.

Prove that

if

a and b are real numbers then

\\a\-\b\\<\a-b\<\a\ + \b\. Prove the following equalities:

2.

a)

|ab|Ha|.|b|; 2

b)

|a|

d)

;

Solve the inequalities:

3.

1|<3;

a)

|x

b)

|x+l|>2;

*

Find /(-1),/(0),/(1),

4.

4- llx

c)

d) |jt-

/(2), /(3), /(4),

if

/(*)

= *'- 6*' -f

6.

5.

Find

6.

/ (x)

7.

The function

and /(2) *)

=a

c)

2

/(O),

-arc cos (log x). Find

=

Log x

/(-4). /(-^/(T) f(x)

is

linear.

/!, Find

7^.

if/

/(I), /(10).

this function, if/(

3.

is

the logarithm of the

number x

to the base 10.

1)

=2

Sec

Functions

1]

13

Find the rational integral function f(x) of degree two, = and /(3) = 5. 9. Given that f(4) = 2, /(5) = 6. Approximate the value if we consider the function / (x) on the interval 3) /(4, linear (linear interpolation of a function). 10. Write the function 8.

0,

if

of

if

as a single formula using the absolute-value sign. Determine the domains oi definition of the following functions: 11.

a)

y=

x+\;

y=

16. 17.

a),= ?E2L = *VV-2. 14**. =1/2 + * **.

x

= lo

-

/

13.

b)

21.

19.

(/

-

Determine the domain

22. f(jc)

= 2A:

SA;'

5x

8

+ 6A:

(-*)l

A

function

Find

10.

and

^(^)

=

defined in a symmetric region / and orfd if x) f(x). f( f(x) /( x) Determine which of the following functions are even and which

23. is

of definition of the function

= |/sin 2x.

y 4

t/=

called euen

f (x)

if

=

=

are odd:

e)

24.

Prove

/<;*
that

may

any function be represented

defined in the interval the form of the sum of an

f(x) in

even function and an odd function.

In t roduction to Analysis

14

[Ch.

1

25. Prove that the product of two even functions or of two odd functions is an even function, and that the product of an even function by an odd function is an odd function. 26. A function f (x) is called periodic if there exists a positive numter T (the period of the function) such that f(x+ T)^f(x) for all valves of x within the dcmain of definition of f(x). Determine uhich of the following functions are periodic, and for the periodic functions find their least period T:

= 10 sin 3 *, = a sin \K + b cos tar; (*)

= sin *; = sin (J/*). / (x) 1

a) / (x)

d) / (x)

b)

e)

/

c)

and the area S Express the length of the segment y = as a function of Construct (Fig 1). the graphs of these functions.

27.

x=AM

AMN

of the figure

28. The linear density (that is, mass per unit length) of a rod AB = l (Fig. 2) on the segments AC l^ CD = 1 2 and DB = l\ (/ + l t + / 3 =- 1) to
M

AfsM"

MN

f

I

\

B

B Fig.

Fig. 2

1

AM =

m

x of this of a variable segment tively. Express the mass rod as a function of x. Construct the graph of this function. 29.

Find cp|\M*)| and

30.

Find

31.

Find

H/Um, /U+l),

Let f(n) be the that

32.

Show

33.

Show

that

i|?

jtp(x)J,

if

v(x)

= x?

and q(x)

= 2*.

if

f(x

if

sum

l)

of

= x\

n terms

of

an arithmetic progression.

if

f(x)

=

and the numbers jc lf xt x t form an arithmetic progression, then the numbers J (x ) f (x t ) and / (xj likewise form such a pro,

l

gression.

9

Sec.

__ Functions

1]

15

Prove that if f(x) is an exponential function, that is, and the numbers x v *,, x t form an arithmetic progression, then the numbers /(*,), f (* 2 ) and /(jcj form a geo34.

/ (x)

= a x (a >0),

metric progression. 35. Let

Show

that

Let

36.

Show

(*)

= !(* + a-*)

that f

p

(*+
=

and ty

37.

(x

+ y)

and

W

(AT)

t|)

(y)

= 1 (a*

+ * (*)

^>

a-*).

(y)

Find /(-I), /(O), /(I) arc sin arc tan

if

^ ^ < # c + oo.

x for x for

1

r

0,

38. Determine the roots (zeros) of the rrgion of positivity of the region of negativity of the function y if: a)

b) c)

39.

In 40.

r/=l

-f-x;

y=2+x 0=1

*

d)

y = x*

^

ff

2 ;

and

3x;

_ ioo

2jc

-

Find the inverse of the function y

if:

what regions will these inverse functions be defined? Find the inverse of the function /

y

\

x, x*,

if

if

Write the given functions as a series of equalities each of which contains a simple elementary function (poweri exponential, trigonometric, and the like): 41.

member

a)

b)

= (2*-5r; * y=2 i/

c)

y

d)

y

COS

;

= arc sin (3-*

3

).

Introduction to Analysts

16

[Ch.

1

Write as a single equation the composite functions repre-

42.

sented as a series of equalities: a)

b)

y=

= sin#; # = arctan, u = Yv, w

u*>

w, ,

if

y

= log#;

t/<0,

ifu>0;

*_!. 43. Write, explicitly, functions of a)

x

2

b) 10* c)

*

arc cos y

y defined by the equations:

= n;

+10' =10;

+ \y\ = 2y.

Find the domains of definition of the given implicit functions. Sec, 2. Graphs of Elementary Functions

=

Graphs of functions # /(*) are mainly constructed by marking a suffiand / (*,-) (/ = 0, 1, 2, ...) ciently dense net of points Ai /(*,-, //), where */, by connecting the points with a line that takes account of intermediate points. Calculations are best done by a slide rule.

=

Fig. 3

Graphs of the basic elementary functions (see Ap pendix VI) are learned through their construction. Proceeding from the graph of y

we

= f(x),

(T)

get the graphs of the following functions by means of simple geometric constructions: M*) js th * mirror image of the graph T about the *-axis; 1) 0i 2) 0i=/( *) is the mirror image of the graph F about the #-axis;

=

Sec. 2]

3) #i 4) 1/4

Graphs

= /(*)

= & + /(*)

the

is is

F graph displaced along

the F

graph

The

to the right

desired line

is

17

th? x-axis by an Uve (/-axis by

along

displaced

(Fg. 3). Example. Construct the graph

Solution.

Elementary functions

of

amount a; an amount

of the function

y = sinx displaced along the *-axis

a sine curve

by an amount -j

(Fig. 4)

Y

Fig. 4

Construct

the

(straight lines): 44. y kx, if 45. 46.

i/

= = x+

6,

fc

if

= 0, 6 = 0,

1,

the

of

graphs

2,

1/2, 2,

1,

following

functions

-1, -2. 1,

2.

0=1. 5* +2.

Construct the graphs of rational integral

two

linear

functions of degree

(parabolas). 3

47. y=--ax , if a=l, 2, 48. // *'-{- c, if c=0, 49. ,/=(*-*)', if *.

1/2,

=

2,

1,

1,

2, 0.

1.

= 1, 2, -1. = y, 4 (x-l)\ = 0, 1, 2, -1. = 2, c = 3; 2) a = 51*. y = ax* + bx + c, 1) a=l, b 6 = 6, c = 0. = 2 f x x*. Find the points ol intersection of this 52. 50. y

if

f

if:

t/

rabola with the

2,

pa-

Ac-axis.

Construct the graphs of

the

following rational integral func-

tions of degree above two: x* (cubic parabola). 53*. y 2 +(*-!)'. 54.
= = = = =

Construct the graphs of tions (hyperbolas): 68*.

0=4.

the

following

linear

fractional func-

_ _ Introduction to Analysis

18

-.

59.

y= *-

61*.

62 *-

Construct the graphs of the fractional rational functions: 63.

=*+

-

-.

65*.

y=.

66.

67*.

=r (WtYc/i of Agnesl). ,^rj 2x =x (Newton's serpentine).

i/

68.

t

*/

.

t

69.

y=*-h^.

=^

70. y

2

(trident of Newton).

H

Construct the graphs of the irrational functions: 71*. */=]/*: 72.

73*.

y=t/x._2 y=-*/x

74.

y=x\fx

(semicubical parabola).

y=j/25

75*. 76.

(Niele's parabola).

f/=

Vx

x

2

l

(hyperbola).

y- >=,. J^l-A __ 78*. y = + x y ^^

77.

^

(ellipse).

2

2

(cissoid of Diodes).

2

==

x 1/25 x 79. r/ Construct the graphs of the trigonometric functions: 83*. */=-cotjc. 80*. y sinx. 84*. y sec x. 81*. y cosx. 85*. y = cosec x. 82*. /=-tanx. .

=

=

86.

87*.

=

{/-/4sinx

f

y^smnx, /

if if

y= sin(x 89*. y = 5sin(2x

88.

\

cp),

/4

= 1,

n=l, *

if

3).

10, 2,

9 = 0, rv

2.

1/2, 3,

1/2.

Jl

3ll

-J-,

-j-

n

ft

~T*

[Ch. 1

Sec. 2]

Graphs

of

Elementary functions

19

Construct the graphs of the exponential and logarithmic functions:

101.

= ax a = 2, l (?(e = 2, 718 ...)*). a =10, 2, 1, y = \oga x, -x x = where e sinhx, y sinhx=l/2(e = coshx, where coshx = = tanhx, where

/

102*.

if

,

f

*.

if

103*. 104*.

).

f/

105*.

f/

106. 0=10*. 2 107*. y=-e~* (probability curve).

= 2""

108. ^

*3

109. //-logx 2 110. y-=log A:. 111. //=-log(logx).

114.

112.

117

.

the

graphs

118*. y--=arc sin*. 119*.

(/=--

= = log"(cosx). ^ = 2-^ sin*.

115. 116.

/y==rV-log X

Construct

y=

113.

.

2

{/

t/

-

of the

inverse trigonometric functions?

= arcsin~. x = arc cos--. # = + arc cot x. ^

122. #

= arccosx. # = arc tan*.

123.

j/

120*. 124. A: 121*, (/= arc cot x. Construct the graphs of the functions: 125.

y=\x\.

127.

= ^(x + \x\). = x\x\\ a) y

128.

a)

126. y

129.

*)

b)

t/= sinA:+ sin jt|; b) 2 x when |jc|< 1. 3

^-<

|

y = log^^l x f/= sin x sinx|. \-

|

_l. whcn

* see p. 22 for

more

details.

Introduction to Analysis

20

[C/i.

/

= #=[*], b) y x[x], where [x] is the integral part number x, that is, the greatest in.eger less than or equal

130. a) of the

to

x.

Construct the graphs of the following coordinate system (r, cp) (r^O): 131.

r

132*.

133*.

= l. f = 7r = r = <*>

/-

134*.

in the polar

functions

(spiral of Archimedes).

(logarithmic spiral).

(hyperbolic spiral).

= 2cosip (circle). ' = -^- (straight line). = sec*y (parabola).

135. r 136. 137.

/-

138*. r=10sin3(p (three-leafed rose) 139*. r a(l fcoscp) (a>0) (cardioid). 2 I 143*. r a cos2(p (a>0) (lemniscate). Cjnstruct the graphs of the functions represented parametri-

= =

cally:

=

=

t* (semicubical parabola). 141*. x t\ y 142*. *=10 cos/, y=sin/ (ellipse). 3 1 143*. *=10cos /, 10 sin / (astroid). 144*. jc a(cos/-f / sin/), t/ a(sm / /cos/)

y=

=

=

(involute of a

circle).

=

^'

145*. ^

146

'

^3,

= rTT'

^0//wm ^

Descartes).

/==

2- (branch of y=2 = # = 2 sin (segment of *-/t\ y=t x^a (2 cos/ cos2/), = a(2sin/

147. xasfc'-t^143. jc 2cos f f

149. 150.

J/

t

t

,

2

/

f

a hyperbola). a straight line).

2

2

/

1

,

*/

t

sin 2/) (cardioid).

Cjnstruct 'the graphs of the following functions defined implicitly:

151*.x 152.

2

+

2

=

25 (circle). xy--= 12 (hyperbola). */

= 2jc (parabola). 154. ^1 + ^! = = jc'(10 155. 156*. x T + y T =;aT (astroid). 153*.

2

i/

j/*

t

157*. x 158. *'

=

2

Graphs

Sec. 2]

2

of

Elementary Functions

21

a"

=e =

* 159*. |/V y (logarithmic spiral). 8 160*. x* 3x// (folium of Descartes). y 161. Derive the conversion formula Irom the Celsius scale (Q to the Fahrenheit scale (F) if it is known that corresponds

+

+

0C

to

32F

and 100C corresponds to 212F. Construct the graph of the function obtained. 162. Inscribed in a triangle (base 6^=10, altitude h

rectangle (Fig. 5). tion of the base x.

Fig. 5

6

Fig

Construct the graph of this function and value. 163. Given a triangle ACB with BC a, AC

=

ACB = x (Fig. 6). angle area ABC as a function of Express # of this function and find its greatest value.

$= Give 164. a) 2x' b) x* c) + A find =b x. a graphic solution of the equations: 5x + 2 = x 1=0; 0; d) e) = 0.1jc; logJt f) I0' x x=l cot = x\ 4 x^x xy=10, x b) xr/-6, c) x d) * e) #=sinx, f y 5sin;c; (0<jc<jt). = 7\ 2 2 + j/ = cos# (0 <x< 2jt). its greatest and a variable Plot the graph 165. Solve tjie systems of equations graphically: a) = 6) is a as a of the functhe area y rectangle Express 22 Introdnction to Analysis (C/t. / Sec. 3. Lfmits xn lim n > if for a \xn e>0 any | <e there when Example 1. the limit of a sequence is a, such that (e) n> N. Show that 5L + 1.2. Urn n Solution. a oo N=N number a is number The 1. The limit of a sequence. x lt .... X0, .... or *! Form -* rt-r (1) 1 the difference 2* +1 Evaluating the absolute value 1 of this difference, -2 1 < we have: e, (2) if n>-\ = N Thus, for every number positive (e). there will be a number Af= 1 such > N we will have inequality (2) Consequently, the number 2 is that for n the limit of the sequence x n (2n-\- l)/(n-fl), hence, formula (1) is true. 2. The limit of a function. We say that a function / (x) -*- A as x -+ a (A and a are numbers), or lim f(x) x -a if for every 8 > \f(x)A <e we have < for | =6 6 () > such that f(jO = 4, a|<6. |x Similarly', lim * \f( X if -> 00 )A\<* The following conventional notation lim *-*a which means that f (x) \ > E for /(x) < | for |x|> /V(e). also used: is = oo, x a \ < 6 (E), where E number positive 3. One-sided x -- a | = A, limits. 0; similarly, f(a ii x 0)= * x If >a a ~ and x -+ a, -* a, then f(x) and /(a lim - and x o we is an arbitrary we write conventionally write * -^ a-f-0. The numbers then + 0)= lim / (x) *-*a + o *re called, respectively, the limit on the left of the function f (x} at the point a and the //mi/ on the right of the function / (x) at the point a (if these numbers exist). Limits Sec. 3] 23 For the existence of the limit of a function and sufficient to have the following equality: I the limits as jc-^o, it is necessary O)-/ /(a If / (x) lim /, (x) and lim f 2 (x) exist, then the x - rel="nofollow">a x -> a following theorems. old: 1) x 2) 3) lim -+ a [/, (*) + /, (*)] = x lima [f, (x) f 2 (jc)J = lim [f , x- o W/^ (JK)J = lim x-*a + xlimaf /, (x) -+ (x); lim f, (x). lim f t (x); x -* a x -* a lim * The following two , -* lim /, (x)l a -# -* jc a ^ (x) (lim f , (x) Jt -^ a ^ 0). limits are frequently used: ILi=i lim AP ^ ->-0 and lim Example 2. [ 1-J--L x ) = lim (l Find the limits on the / (x) + a a) right =* = 2 and left 71828 . . . of the function = arc tan- as x ->-0. Solution. We have lim x .+ arc tan )= 2 x J +o fa \ and f(-0)= lim - x->. Obviously, the function 166. is Prove that as n / (x) faictanlW-42 / A: o \ in this case has no limit as x--0. *oo the limit of the sequence equal to zero. For which values of n will we have the inequal- = 0.01; c) 8 ity an arbitrary positive number)? Calcula e numerically for a) e 0.1; b) e 167. Prove that the limit of the sequence (e is = = 0.001 24 _ _ as Introduction to Analysis >oo rt is which For unity. the inequality (e n>N values of an arbitrary positive number)? Find N for a) e = 0.1; b) e = 0.01; 168. Prove that liin x*=:4. X -> How we have will K-l|<e is = 0.001. e c) 2 should one choose, for a given positive number number positive [CH. 1 e, some 6 so that the inequality 2 -4|<8 |* should follow from Compute 6 e for a) = 0.1; = 0.01; 8 b) e c) Give the exact meaning of the Hoi log* = oo; b) lim 2*= +00; c) 169. a) * -> +0 170. c) _ 2*3' 1 _ I a) *} b) V) X + + Find the limits i, 1 1 1 3 ' 1/2", ' ' (- * 1 )"" 1 _2n_ ' ... . 4 2/i~l ' ' ' 1/21/21/2", . . ' . . . ; . limits: a+4 + l,+ ... *\* lirn fl ' ' Hm n 173 5 -> 00 of the sequences: 0.23, 0.233, 0.2333, Find the 172. 1 1/2 1/2; d) 0.2, 171. ' liai/(x) X 00 = 0.001. following -oo Hm * C+D ( "* + )(>. + 3) n l) 2n+11 2 178*. lim n -* CD J' notations: = oo. 6 Limits Sec. 3] Hm (Vn + 179. n -+ -o/% 1 \f~n). < / i 180. lim When x -+ the seeking is it oo, useful of a ratio of two integral polynomials in n to divide both terms of the ratio by x , where limit first the highest decree of these polynomials. A similar procedure is also possible in ing irrational terms. Example lim _ 2. * lim lim 181. r 182. cases for fractions contain- 1. J2^-3)(3t-f^)(4A'-6) Example many -. or lim =. . ^rrr. * *86. ^^. 187. ~' = J lim 1 lim * + ., 3* O ^2 184. 185. lim V J" . __ +5* 8v lim 1/ -11- . jc lim ->* h 3 -\- \ 10-j- A: ^ Y L 4- 188. . +7 X* lim * lim 1. ^~~^=J. <x> 00 183. * as n is 189. lirn -r-r~c * - ^5 190. lim Vx + Vx If P(A-) and Q (x) are integral polynomials and P (u) + or Q (a) then the limit of the rational fraction lim is obtained directly. But if P(a) Q(a)=0, then = out of the fraction Example P Q (x) it is advisable to camel the binomial * once or several times. 3. lim /'T 4 ^ lim !*""!!) f xf ?? Hm ^^4. a Introduction to Analysis 26 101. ^{. lim *-. 192. lim * _. *-+> The expressions containing irrational ized by introducing a new variable. Example I 196. lim ^ * | \Ch. Um ^ 198. * _ terms are fl in cases rational- many Find 4. lim Solution. Putting we !+* = have E=1 lim 199. lim X - Mm ^ * -4^-. *~ 200. lim T = lim " 2 ,'~ 201. ,,. lim l 1 */', 3/ t/x ~ , ] x . Another way of finding the limit of an irrational expression is to transfer the irrational term from the numerator to the denominator, or vice versa, from the denominator to the numerator. Example 5. = = lim lim x -+a(X a)(Vx _^ + V a) ! lim *-> a 203. 204. -49Q-. lim li.n *-* j-^= / x . ^ 206. 207. 2 jc -f lim lim *-+< 1 205. lim *-+' ^L"" / * . 1 208. lim ^-*o ! V a 2\f~i -=f. __ Limits Sec. #] _ 209. 27 212. K lim 213 [/*(* ^i 214. Hm(]/xfa + a) xj. -6* 4 6-*). ' 210. lim 211. lim X- li. | Jf--fCO 215. The formula llm X- X -i r frequently used when solving the following examples. cos a. sin a and lim cos granted that lim sin * = Example a) !!! lim; lim 227. a) lim X-* -> CO sill ,. 3x 228. xsinl; lim x sin b) b)li.n^. X 217. lim 00 . * x) tan (1 -~- lim X -0 sin 5* sin 2* 229. ' . ^ Jt-M 218. taken is 6. lira 216. It *= lim cot 2x cot f-^ * * -+0 x). / \ sin JTX 219. lim M ^ = . sin BJIJC 230. l lim *- 220. lim ( n n-*cc 221. lim 222. lim 223. lim 224. lim 225. lim 226. lim \ sin-). n I Jt 231. lim 232. lim cosmx - arc sin crs^ tan* 236. ' sui A: lim JC I - cosn V* tan 233. * ji 1-2 lim ^ ' sin six \ * for "28 _ _ Introduction to Analysis m. 24 .=T. ta - !!?. one should bear mind in C=4"; lim (p(x) if = ^ /l 1 cp A and (x) and if (x) where e \|? lim ty(x)^= is = B, (x) oo, then the problem of finding = we Hm a (x) ^ Hm (x) a = 2.718 Example 7. ... is Napier's number. Find lim Solution. Here, lim (5111=2 X and lim Jf-^O \ hence, lim x-*o Example Solution. put q>(x)= 1 +a(x), as x -+ a and, lien^e, -* 1 - lim solved in straightforward fashion; lim \|) (x) then co, lini(pU)=l and the limit of (3) x (3) that: lim where a -=C there are final limits if 3) '""" taking limits of the form lim l 2) *""r Jt p I -* 1 When then n nx f 1) [Ch. 1 8. Find We have lim 1 r-^( end Hm *2 = 2 J^ x 2 [(p (x) - ij ty (x) ^Sec Limits 3] Therefore, lim =0. / j Example Find 9. lim Solution. We f x ~~ \ t have ^11= lim X-+ 06 X -4- 1 lim (- CO i . = 1 +T , Transforming, as indicated above, we have In this case easier to find the limit without resorting to the general is it procedure: it Generally, is useful to remember that lim 250. 244. lim(* X "" 2 *" /^i 245> ]. i Hmfl ... V 251. lirn(l 1 252**. a) lim (cos x) Iim(l J_ * ; X ~* \ -)/ . . b) 247 + sinjc) *. -.o 2 \*a Jill ( 2?+T ) / 246. M 3x4-2/ X-K>\ li f I)*. H 30 _ _ Introduction to Analysis When solving the problems that follow, limit lim/(x) exists and is positive, then lim [In /(*)] x-+a Solution. lim X-*0 Formula We that if f (x)]. X-+Q have X X-+Q frequently used in the solution of problems. 253. lim [In (2*+!) *- 254. li 255. . X - " limfjlnl/J-i^). lX/ 260*. ,_*<> \ llmn(^/a ^ n V) 256. lim *[ln(jt+l) Inx]. 261. lim- <-*--- 00 - 257. lim. 258*. Hm=. 259*. ital! 262. -^o (a lim b) lim >0). (see Find the* 264. a) sin 263. a) *_^ . Hm *" + b)Jirn* p===. 267 a/lLutanh*; *-*-* b) limtanh*, *->+ where tanh^ = 266. a) . x* Problems 103 and fa 265. * following limits that occur on one side: lira lira - a ) lim *-b) Hm *-*+ ^^~. V ; (a>0). . * -- li \) ptX pCLX 268. a) lim b) |im i 1+ ' T ! the ln is (*) [Hm know Prove that tO. Example = In useful to is it [Ch. 104). Limits Sec. 31 269. a) 31 270. a) lim-^4i; ' Hm-^-; x~* Construct the graphs of the following functions: 2 \im (cos "*). 271**. y = n->oo * 272*. y=lim n-*c = \im 273. y * i (x^O). xn J/V-t-a 2 . a->o 274. t/ 275. = li;n| t/ = li -*< 276. Transform a common the following fraction: a Regard 277. it What mixed periodic fraction into = 0.13555... as the limit of the corresponding finite fraction. will happen to the roots of the quadratic equation if the coefficient a approaches zero while the coefficients b and c are constant, and fc^=0? 278. Find the limit of the interior angle of a regular n-gon > oo. as n 279. Find the limit of the perimeters of regular n-gons inscribed oo. in a circle of radius R and circumscribed about it as n 20. Find the limit of the sum of the lengths of the ordinates of the curve y = e~*cos nx, = x 0, 1, 2, ..., n, as n *oo. 281. Find the limit of the sum of the areas of the squares constructed on the ordinates of the curve drawn as at the points on bases, where x=^l, 2, 3, ..., n, provided that n 282. Find the limit of the perimeter of a broken line *oo. M^.. .Mn inscribed in a logarithmic spiral Introduction to Analysis [Ch. I oo), if the vertices of this broken line have, respectively, (as n the polar angles == 0, =y , . . . , q>rt =y A segment AB = a (Fig. 7) is divided into n equal parts, each pnrt serving as the base of an isoscelos triangle with base angles u ^45. Show that the limit of the perimeter of the broken line thus formed dilTers from the 283. length of AB despite the fact that in the limit the broken line "geometrically merges with the segment AB". Fig 7 Fig. 8 in half; the point C, divides a segment AB---1 the in AC a half; divides point C, divides a segment point C 2 C C 3 in half, and so C divides the in half; 2 point 4 segment C,C on. Determine the limiting position of the point C n when /i--oo. 285. The side a of a right triangle is divided into n equal which is constructed an inscribed rectangle parts, on each of the limit of the area of the step-like figure Determine 8). (Fig. *ou. thus formed if n 286. Find the constants k and b from the equation The 284. l l = 0. (1) What is the geometric meaning of (1)? 287*. A certain chemical process proceeds in such fashion that the increase in quantity of a substance during each interval of time r out of the infinite sequence of intervals (tr, (i -f l)t) substance (/~0, 1, 2, ...) is proportional to the quantity of the available at the commencement of each interval and to the length of the interval. Assuming that the quantity of substance at the n} after determine the quantity of substance Q nth each takes if the increase part of the elapse of time t place initial time is Q the time interval Find Q^lhi ( , t *= Sec. 4] 33 Small and Large Quantities Infinitely Sec. 4. Infinitely Small and Large Quantities 1. Infinitely small quantities (infinitesimals). If lim a (x) = 0, x->a i.e., < |a(x)|<e when if as infinitesimal * oo. as x x < then the function a (x) is an a fi(e), similar fashion we define the infinitesimal a (x) In a. The sum and product of x-+a. a K \ | limited number infinitesimals as x of +a are also infinitesimals as If a(x) and p *a and x (x) are infinitesimals as lim SlJfUc, x-+a P (x) where C is some number different from zero, then the functions a(x) and p(x) are called infinitesimals of the same order; but if C 0, then we say that the function a (x) is an infinitesimal of Higher order than p (x). The function u (x) is called an infinitesimal of order n compared with the function p (x) if = lim where < J C| < Q(x) -C " ' -f oo. If then the functions a For example, for (x) x and p are called equivalent functions as x *a: we have > sinx~x; and so (A*) tanx~ x; ln(l-fx)~ x forth. The sum two infinitesimals of different orders is equivalent to the lower. The limit of a ratio of two infinitesimals remains unchanged if the terms of the ratio are replaced by equivalent quantities. By virtue of this theorem, when taking the limit of a fraction of term whose order is lim !> aPW , as x *a we can subtract from (or add to) where a (x) >.0 and p (x) the numerator or denominator infinitesimals of higher orders chosen so that the resultant quantities should be equivalent to the original quantities. > t Example 1. j/?T2? = lim a/7 ' - ,. lim i- ,. *-*o 2. ber Af 2 2x Infinitely large quantities (infinites). If for an arbitrarily large numa x exists a 6(N) such that when 6(N) we have the < there | inequality lfMI>tf. then the function f(x) 2-1900 is called an infinite as x >a. | < _ _ Introduction to Analysis 34 > co The definition of an infinite f (x) as x of infinitesimals, we introduce the concept is [Ch. 1 analogous. As in the case of infinites of different orders. 288. Prove that the function is an infinitesimal as x *oo. For what values of x is the ine- quality l/WI<e fulfilled if e is an arbitrary number? = Calculate for: a) e 0.1; b) e-0.01; the function 289. Prove that is an infinitesimal for x >1. c) e-0.001. For what values of x is the ine- quality !/(*)!< fulfilled if e is an arbitrary positive number? Calculate numeri- = 0.01; c) e a) e-0.1; b) e 290. Prove that the function cally for: = 0.001. ~ x is an infinite for x *2. In 2 what neighbourhoods 2|<8 of |x is the inequality lf(x)\>N fulfilled if N is an arbitrary positive number? Find 5 if a) #=10; J2^ c) b) #=100; #=1000. 291. Determine the order of smallness the surface of a sphere, b) the volume of a sphere if the radius of the sphere r is an infinitesimal of order one. What will the orders be of the radius of the sphere and the volume of the sphere with respect to its surface? 292. Let the central angle a of a circular sector ABO (Fig. 9) with radius R tend to zero. Determine the orders of the infinitesimals relative to the infinitesimal a: a) of the chord AB\ b) of the line CD; c) of the area of A/4BD. o of: a) Infinitely Small Sec. 4] *0 determine the orders x 293. For x and Large Quantities of 35 smallness relative to of the functions: *\ *) ^* cos 1 d) +x \ tan e) *'* sin A: A:. b) c)$/*'-

294. Prove that the length of an infinitesimal arc of a circle constant radius is equivalent to the length of its chord. 295. Can we say that an infinitesimally small segment and an infinitesimally small semicircle constructed on this segment as a diameter are equivalent? of

Using the theorem si "

296. lim

!"

arc sin

x

^o

,

infinitesimals, find

lim^

298.

.

= _^-*

f

297. lim

two

of the ratio of

5*

3*' s

.

299. lim

~

1

ln(l--*)

300. Prove that

when x

*0

~

the quantities

VT+T1 + Applying formula a)

1/L06;

b)

+xl \x\ is

(1)

.

approximate the following:

(1),

1/0^7;

Y\

when

and

are equivalent. Using this result, demonstrate that small we have the approximate equality

c)

/lO;

d)

/T20

and compare the values obtained with tabular data. we have the following approxi301. Prove that when x mate equalities accurate to terms of order x 2 :

b) n +x) &\

c)

(1

d)

log(l+x)

where

+ nx

(n

a positive integer);

is

= Afx, Af = log e = 0.43429...

Using these formulas, approximate: *>

02

Compare

;

2>

0^7

;

3>

I<55

;

4)

^16;

5)

1.04'; 6) 0.93*; 7) log 1.1.

the values obtained with tabular data.

_ _ Introduction to Analysis

36

302.

Show

X+OQ the ~ P (x) = a.x n + a,x n + that for

[Ch.

1

rational integral function

l

.

.

.

-f

an

(

is an infinitely large quantity equivalent to the term of highest n degree a x 303. Let x*oo. Taking x to bean infinite of the first order, determine the order of growth of the functions: .

a)

b)

*>-

100* -1,000;

c)

7+2-

Sec. 5. Continuity of Functions

1. Definition of continuity.

A

function

continuous

is

/ (x)

when x =

(or "at the point g"), if: 1) this function is defined at the point g, that is, there exists a number / (g); 2) there exists a finite limit lim f (x); 3) this lim-

x-4 is

it

equal to the value of the function at the point

llmf

(*)

g,

i.e.,

= /().

(1)

*-*fc

Putting

where Ag

^0, condition lim

(1)

may be

A/(g) =

lim

rewritten as

l/(g+ Ag)-f

(g)]

= 0.

(2)

or the function / (x) is continuous at the point g if (and only if) at this point to an infinitesimal increment in the argument there corresponds an infinitesimal increment in the function. If a function is continuous at every point of some region (interval, etc.), then it is said to be continuous in this region. Example 1. Prove that the function

y fs

continuous Solution.

Ay = sin

<*

for every

We

= sin x

value of the argument

x.

have

+ A*)-sin x = 2 sin

cos

=

x+

_

sin

.

cos f

Since si

lim

T

T=

1

and

2 it

follows that for any x

we have lim A(/

Hence, the function sin*

is

=

(

continuous when

oo<x< +

x+

^

Sec. 5]

Continuity of Functions

37

2. Points of discontinuity of a function. We say that a function /(x)has discontinuity 'at (or at the point XQ) within the domain of definition of the function or on the boundary of this domain if there is a break in the continuity of the function at this point.

x=*

a

Example when x=l.

2.

The

This

function

function

1

f(x)=

is

10 a)

(Fig.

not defined at the point x

is 1,

discontinuous

and no matter

2

1-2

how we

choose the number /(I), the redefined tinuous for *=1. If the function f (x) has finite limits:

Hm and not

all

three

/(*)

= f(*

numbers f(x

-0) ),

a discontinuity of the first kind.

then *

is

Urn

/(*)

+ Q)

) /(* f (x In particular,

/ (x)

will not be con-

= /( are equal, the nxQ

is

called

if

called a removable discontinuity. of a function f(x) at a point

For continuity ficient that

and

function

JC

Q,

it

is

necessary and suf-

= 0.

*

The function

3.

Example at

[Ch. I

In t reduction to Analysis

38

/(jc)=j-y

has a discontinuity of the

kind

first

Indeed, here, /

(

+ 0)=

5!!L

lim

==+ i

and

/(_0)=

lim

x

jc-*-o

Example 4. The iunction y = E(x), where E(x) denotes the integral number x [i.e., E (x) is an integer that satisfies the equality x = E(x)

of the

where 1,

0<<7<1], i2,

and

...,

discontinuous

(Fig. 106) at every integral all the discontinuities are of the first kind.

is

point: x

part

+ q. = 0,

+ =

Indeed, if n is an integer, then /i. At all (/i 0)-=/il and (/i 0) other points this function is, obviously, continuous. Discontinuities of a function that are not of the first kind are called discontinuities of the second kind. Infinite discontinuities also belong to discontinuities of the second kind. These ane points * such that at least one of the one-sided limits, /(*<> 0) or /(*o 0) is equal to oo (see Example 2).

+

Example

The function # = cos

5.

lim cos 5L X X-+-0

the

lOc) at

(Fig.

discontinuity of the second kind, since here:

both one-sided

and

lim

cosi

+

X

Jt->

point

x=0

has a

limits are nonexistent

.

3. Properties of continuous functions. When testing functions for continuity, bear in mind the following theorems: of a limited number of functions continuous in 1) the sum and product some region is a function that is continuous in this region; 2) the quotient of two functions continuous in some region is a continuous function for all values of the argument of this region that do not make the divisor zero; 3) if a function

f (x) is continuous in an interval (a, b), and a set of its values is contained in the interval (A, B), and a function cp (x) is continuous in (A B), then the composite function cp[/(*)J is continuous in (a, b). A function f (x) continuous in an interval [a, b] has the following propert

ties: is

1) f (x)

boanded on

|/(*)|<M when 2) / (x) has a 3) / (x) takes

a<*<6;

[a,

minimum and

6J,

a

there

i.e.,

maximum

some number

is

value on

[a,

M

such that

b]\

between the two given values; matter what the if and fl (a
f(a) = A

has

at least

304. of the

one

Show

all

intermediate

= /(P)

values

=

real root in the interval

that the function y

argument

x.

(a,

=x

2

p).

is

continuous for any value

_ _ Continuity of Functions

Sec. 5]

39

305. Prove that the rational integral tunction

is

continuous for any value of x. 306. Prove that the rational fractional function

is continuous for all values of x except those that make the denominator zero. 307*. Prove that the function y = Yx is continuous for x&zQ. 308. Prove that if the function f (x) is continuous and non-

negative in the interval

is

then the function

(a, 6),

likewise continuous in this interval. 309*. Prove that the function y cos x is continuous for any x. 310. For what values of x are the functions a) tan* and

b) cotjc

continuous?

=

Show

that the function # |#| is continuous. graph of this function. 312. Prove that the absolute value of a continuous is a continuous function. 313. A function is defined by the formulas

311*.

How

Plot the

function

should one choose the value of the function A=f(2) so 2? the thus redefined function f(x) is continuous for # Plot the graph of the function y f(x). 314. The right side of the equation

=

that

=

f(x) is

meaningless for x =

0.

= lx sin

How

so that f(x) is continuous for 315. The function f(*) is

in for

should one choose the value /(O) jc

= 0?

= arctan--^

meaningless for x=--2. Is it possible to define the value of /(2) such a way that the redefined function should be continuous jc

= 2?

Introduction to Analysis

40

316. The so that fix)

function f(x)

-

f(x)

= ^ y\__1 y x

b)

/(*)

=

c)

/(*)

= ln(\+x)

d)

/(

f)

/(*)

I

/]

a)

is

not

continuous for x =

is

l

;

1

cos*

[Ch.

defined 0,

for

x=

0.

1

Define /(O)

if:

(n is a positive

integer);

.

A* 111(1

= * cot*.

Investigate the following functions for continuity: 317.

y=

318.

y=

3,9.

yssi

320.

^=. b)

322.

*/

-.

324.

-. =

y=

xs\n-.

=

l

325. y

= arc tan

326.

= (!+

329.

y

A:)

.

arc tan ,

=

= -/

323.

y=\n(cosx).

330.

y=J

** \

.

2#+l

for

'"T x rel="nofollow">3.

P'ot the graph of this function.

331. Prove that the Dirichlet function %(x) which is zero for and unity for rational x, is discontinuous for every t

irrational x value of x.

Investigate the following functions for continuity and construct their graphs:

= \in\ y = lim (x arc tan nx).

332. y 333.

Continuity of Functions

Sec. 5]

334. a) y

= sgnx,

function sgn x

is

b)

y

=x

sgnx,

c)

i/

= sgn(sinjt),

where the

defined by the formulas:

+

I

sgn x

=

1,

if

*>0,

0,

if

x=

-1,

if

*<0.

{

[

0,

=

=

335. a) y xE(x), b) y xE(x), where E (x) is the integral part of the number x. 336. Give an example to show that the sum of two discontinuous functions may be a continuous function. 337*. Let a be a regular positive fraction tending to zero (0
= which

is

338.

(l

a)

+

l,

true for all values of a? that the equation

Show

has a real root in the interval 339. Prove

that

any

(1,2).

polynomial

one real root. 340. Prove that the equation

least

has an infinite number of real roots.

Approximate P (x) of odd

this root.

power has

at

Chapter II

DIFFERENTIATION OF FUNCTIONS

Sec. 1. Calculating Derivatives Directly

1. Increment of the argument and increment of the function. If x and x l are values of the argument x, and y f(x) and t/ 1 /(jc 1 ) are corresponding values of the function y f(x), then

^x~x is

=

=

=

x

l

called the increment of the argument x in the interval

A0=0i

(x,

xj, and

y

or /

= f (x,) -f (x) = f (x + A*) -

r

Fig. is

11

called the increment of the function y in the and by AN). The ratio

where

&x=MA

MN

same

interval

(jc,

*,) (Fig.

11,

the slope of the secant of the graph of the function y=*f(x) (Fig. 11) is called the mean rate of change of the function y over the interval (x, *-f Ax). Example t. For the function

5s

and

Calculating Derivatives Directly

Sec. 1]

calculate

fromx=l

b)

from

a)

Ax=l. 1

x=3 Solution. We

to x=l.l; to x 2.

=

have

1=0.1,

= (l.l 5-1.1 + 6) Ax = 2 3 = At/ = (2* 5-2-1-6) (3* 2

Ai/

b)

to a change in the argument:

Ax and A#, corresponding

a)

2

1+6) =

5-

(I

2.

5-3 -f- 6)--=0.

In the case of the hyperbola y

M

secant passing through the points

3,

(

--

Solution. AJ/

Ax~~

2. The

Ax=10

Here,

3

=

and

)

V

N

the

find

,

10,

{

-r^

the

of

.

J

'

=7

)

slope

Ay = ^ 1U 1

and

7

1

4= o

5*5U

Hence,

1 30'

The derivative

derivative.

spect to the argument x that

0.29;

1,

Example

,

43

is

y'=j-

of a function

the limit of the ratio -r^

y-=f(x) with

when Ax approaches

re-

zero;

is.

y rel="nofollow">=

lim AJC -> o

The magnitude

>.

A*

of the derivative yields the slope of the f(x) at the point x (Fig. 11):

graph of the function y

=

tan

y'

tangent

MT

to

the

q>.

usually called differentiation of the function. The derivative y'=f' (x) is the rate of change of the function at the point x. Example 3. Find the derivative of the function

Finding the derivative

/'

is

y Solution.

From formula

= x*.

we have

(1)

Ay = (*+

xi

A*)*

2*Ax+ (Ax) 1

and

Hence,

'=

L^

lim

Ax 5*.

lim AJC->O

One-sided derivatives. The expressions

/'_(*)= lim AJ:-*--O

f

(*+**)-/(*) Ax

and

/(x)=

lim

Ax

'

44

[Ch. 2

Differentiation of Functions

are called, respectively, the left-hand or right-hand derivative of the function f(x) at the point x. For /' (x) to exist, it is necessary and sufficient that /'.(*)

Example 4 Find

By the

Solution.

and

/'_ (0)

f^

4.

Infinite derivative.

(0)

the function

we have

=

L

lim

=

lim A*--t-o

Ax

some point we have

at

If

(0) of

+

/'

definition /'_ (0)

= /+(*).

/(*+**)-/(*)_. Uoo,

lim

we say that the continuous function / (x) has an infinite derivative at x. In this case, the tangent to the graph of the function y f(x) is perpendicular to the x-axis. Example 5. Find /' (0) of the function

then

=

Solution.

We

V=V*

have

/'0)=llm *~ Ax

=

]im

-=-=<-

341. Find the increment of the function to a

change

a)

b) c)

in

from x= from x = from A: =

argument: 1

to

1

to

1

to

xt xl A:,

= 2; = = + h. 1

.

1

;

1

342. Find A// of the function a)

b) c)

*=

0,

^=8,

^=

a,

2 y = x that corresponds

y=-i/xil:

AA:- 0.001;

^ ==

AA:

9; /Z.

343. Why can we, for the function y = 2;c 3, determine the increment Ay if all we know is the corresponding increment Ax = 5, while for the function y x 2 this cannot be done?

+

=

344. Find

the

increment

by and

the ratio

tions:

a)

y

= ( Jg

f

f 01

"*^

1

2)

b)

y=l/"x

forx-0

c)

y

for

= \ogx

x -100,000

andAjc =

^ A *

for the func-

0.4;

and AJC = 0.0001; and A*90,000.

Sec.

Calculating Derivatives Directly

1]

which correspond Ay and ment fromx to x-(- Ax for the functions: 345. Find

-

a)

y-ax +

b)

y-x';

d) y

6;

y

e)

45

a

to

change in argu-

= /x; = 2*\

346. Find the slope of the secant to the parabola

y

==

x

~x

t

the abscissas of the points of intersection are equal:

if

a)

x,-l, x a -2;

c)

x^l,' x 2 =l+fc.

2

To what if

limit does the slope of the secant tend in the latter case

/i->0?

What

347. in the

is

mean

the

l^x^4?

interval

rate of change of the

=

2

function y

+

= x*

+

2/ 348. The law of motion of a point is s 3/ 5, where the distance s is given in centimetres and the time t is in seconds. What is the average velocity of the point over the interval of time from t~\ to ^ 5? 2* in the interval 349. Find the mean rise of the curve y

=

=

mean

350. Find the [x,

x+Ax]. 351. What

rise of the

curve

j/

= /(x)

in the interval

=

f(x) by the rise of the curve y given point x? 352. Define: a) the mean rate of rotation; b) the instantaneous rate of rotation. 353. A hot body placed in a medium of lower temperature cools off. What is to be understood by: a) the mean rate of cooling; b) the rate of cooling at a given instant? 354. What is to be understood by the rate of reaction of a substance in a chemical reaction? 355. Let /(X) be the mass of a non- homogeneous rod over the interval [0, x]. What is to be understood by: a) the mean linear density of the rod on the interval [x, x+Ax]; b) the linear density of the rod at a point x? is

to be understood

at a

M=

356.

x=

2,

if:

ative y'

Find

the

a)

Ax-1;

ratio

when x^2?

b)

of

the

Ax = 0.1;

c)

function

*/

=

Ax -0.01. What

at the point is

the deriv-

_ _ Differentiation of Functions

46

357**. Find the derivative of the function y 358. Find {/'=

a)

=x

t/

f

c)

;

^

lirn

y

[C/t.

2

= ianx.

of the functions:

=

359. Calculate f'(8), if 1 360. Find /'(0), /'(I), /'(2), if /(*) *(*- 1) (x-2)V 361. At what points does the derivative of the function #* coincide numerically with the value of the function itself, /(#) that is, /(*) /'(*)? 362. The law of motion of a point is s 5/*, where the distance s is in metres and the time t is in seconds. Find the speed

=

=

=

at

*

=

= 3.

Find

363.

the

of

slope

the

tangent

=

drawn

at a point with abscissa x 2. 364. Find the slope of the tangent

the point

the

to

to the

curve y = Q.lx*

curve y=sinjt at

0).

(ji,

f (*) \ /

= -i

are the slopes of the tangents to the curves y

=~

365. Find the value of the derivative of the function at the point

366*.

x = X Q (x

What

of their intersection?

Find the

367**. Show that the following functions derivatives at the indicated points:

y=^?_

b) y c)

y

=l/xl

at

x x

= |cosx|

at

*=

Sec. 2. Tabular

1. Basic v

ty(x) are

at

5)

(*)'=,;

6)

3)

(

)'-' r

4)

fc

= 0,

1,

(cu)'=cu

have

finite

2,

rules for finding a derivative. If c is a constant functions that have derivatives, then

2)

(c)'

jt,

do not

angle be-

Differentiation

= 0;

1)

x

+ 0).

and y = x* at the point tween these tangents.

a)

i

;

t;';

7)-

==

(v

* 0).

and

w

=

o>(jc) '

Sec. 2]

__

2. Table I.

of derivatives of basic functions n n- 1

(x )'

III.

= nx

(sinx)'

IV. (cosx)'

V1T

VIII.

.

47

= cos*. = sin*.

(arrdn*)':=

(

ZL.

(arccos*)'=

IX. (arc

_

Tabular Differentiation

|

*

|

<

1).

<|*|<1).

i*.

jo'^y-pj.

X. XI. XII.

XIII.

XIV.

XV. XVI. XVII. XVIII. (cothx)'

XIX.

"

cosh 1 x

=

(arcsinhjt)'

."""

.

= -^J==r. V\+x* 1

XXI.

(arc

tanh

x)'

(|jc|
;

XXII.

3. Rule that

is,

/==/

composite function. If y f(u) and u where the functions y and u have derivatives, then

for differentiating a [

y or in other notations

=

^/ dx~~ du dx' This rule extends to a series of any finite

number

of differentiate functions.

48

_ _ Differentiation of Functions

Example

Find the derivative

1.

=a

Solution. Putting #

5 ,

[Ch. 2

of the function

where w =

2

2jc

(*

+ 3),

by

formula

we

(1)

will

have

= (u*)'u (* -2x + 3); = 5u 2

y'

Example

4

Find the derivative

2.

(/

we

=

u

*;

2 (jt

-

of the function

y = sin Solution. Putting

= 10 (x-1)

(2x-2)

8

4*.

= sinu;

u

=

4jc,

find >-4

Find the derivatives a

differentiating 368-408).

= 12sin

2

4xcos4jt.

of the

composite

following functions (the rule for function is not used in problems

A. Algebraic Functions s

QO

AV

,. ouo. y

Q5Q oOy.

<>

i/

5

A V8 ~~ TEA

^

==:

373.

v

I

v^

-f-

&A:

O

Q7 i/. Of

*j. 4 A U.OAV

.

.

if

2

+6

y==+n n

Q7IS4T o/O

QJA v3

11 t^

f

.

y

= Xv*

5

Ov ~~~ ^A.

2

I

~f~

Av~'

.

*

/" +f* y X

.

+ C.

(,-a(-H-W". V^a

374.

~]~

Ov AA

;r-Jt-pJT

'":

370. y--=ax*

372

|

2

2.

379. il-

=

380.

.

2jc

1

JC

=i

381.

B. Inverse Circular and Trigonometric Functions 382.

383.

(/

t/

= 5 sin ^ + 3 cos x. = tanx cotx.

386. 387.

388

S85.j/-2/sin(-(''-2)cos(.

-

389.

y=arctan^-h f/

arc cot x.

= -

_('+"')' '"'-'. ^

Sec. 2]

_

Tabular Differentiation

_

49

C. Exponential and Logarithmic Functions

390.

= e*arc sin x.

396. y

y^K*-e*.

391. y

= (x-l)e*.

397.

y^ ~.

392.

= 5-

398

y

r/

393. (/==

-

399.

J.

=

394. / (x) e* cos 2 395. #=:(A; 2

400.

jc.

=

* .

\j

=7 y = \nx\ogx

*/

In

a log a

jc.

D. Hyperbolic and Inverse Hyperbolic Functions 401.

t/

= Jtsinhjt.

y=-Vcosh x

402. y 403.

404

//

t/ ^

405. 406.

.

= tanhA:

407.

*.

= ^iiL Inx

= arctanx = arc J (/

arctanh

t/

(/

408. ^ //

==

-

i

x-

E. Composite Functions In problems 409 to 466, use the rule for differentiating a composite function with one intermediate argument. Find the derivatives of the following functions:

40Q** "v u

H i \

a

y

.

Solution.

2 ~~~- ^r \* -i_^v *jAi i~ \jAt

j

Denote

1

+

= w; = ^ 30w M

3jt

5jc*

;

i^

- 30a 29 -(3

/

L. \

I

\0x)

then

t/

=w

=3 u'^

j<) .

We

10*;

=30 (1 + 3jc

1

410. i/=| 411. 412.

410 ,._ 10> f/

1

>

-

"~56(2*

1)'

414. t/=J/T^J?". 415.

y=^/

416. w=(a''.

24(2^1)'

'

40(2x

I)''

have:

Differentiation of Functions

50

417.

t/

= (3

2 sin*)

Solution.

2 sin x) 4

4 Jt)

2 sin x)'

-(3

419. y r/=J/coU

/coU.

423. y j/

427.

428

y

426.

-

429.

430. 431.

(

2

3

= |/ 3sin*--2cos* y=

f

/.

9

.

t/

y=/2ex + tart y = sin 3* + cos -|-

Solution

^ ^

1

sin Sm

5

f^

,

= cos 3^3*)' -sin 4 f 4V + 5

\

5 /

]

,

cos 2

x Y/-

x 5 "^2

432. t/=sin(x

f

=

433. /(x) cos(ct; 434. /(0=sin/si ,._

"

l+cos2*

= acot~ = ~ y = arc sin 2x.

436. /(x) 437.

438.

t/

Solution, y'

439. y

=

= arcsin^.

440. /(x)

2 cos x)

-- CGSJC = 30-^-3 cos *

424. y

= cosec ^+sec 425. f f(x) = 6(1 _ 3cosx) {/= 1/1 + arc sin x. (arc sin x) y = J/arc tan *

422.

2 sin x)*

t

= 2x + 5 cos' *.

421*. x

= 5 (3

.

-

418. j/=tanjc

[Ch. 2

.

= 5 (32 sin y'

- 10 cos x (3

420. y

5

_

= arccosJ/7.

441. y

= arc tan.

442. y

=

=

_

Tabular Differentiation

Sec. 2]

443.

AAA 444.

= 5e~*. 447. y = arc cose*. 448 0=1 = X = logsinjc. 5 449. = x 10'*. 450. y= ln(l *'). = = \n* * 451. ts'm2 y In(lnjc). f(t) 5 4 arc x sin sinx). y== \n(e* + = arctan (lnA:) + ln(arctan^). y = /In x+l + In (1/7+1). t/

>

-

.

t/

j/

445.

446. 452.

453. 454.

2

j/

t

.

t/

F. Miscellaneous Functions

455**. y=sin'5jccos*y.

15

10 3)'

458.

j/=

460.

az

461. y 462.

463.

f/

^-i-jc 2 x*

= :

3

= |-

y=44

465.

t/

=x

4

(a

__J "2 |/

=

470. z

=

468.

|

469.

471.

/(0=(2/-

(Jt-i-2)

1 '

51

52

Differentiation of Functions

[C/t.

2

--

= ln(]/l+e*-l)-ln(/l # = ^ cos'x (3 cos * 5).

473. y

2

474. ...

475

= (tan

-

2

4

*

l)(tan

x-HOtan 2 *-fl)

476. y=-ian*5x.

= ^ sin (x

477. y

478. j/=sin 479.

=

493. 494.

2

A;+sin'x.

ianx + x.

tan *

486. y

= arc sin

487. y

=

488. y

= 4~- af c sin fx

2

491.

jc

* .

cos *

^ V

.

\

^

= K^

2

490. t/=jt/a

y=/a sin + p cos x. y = arc sinjc + arccosA; 2

= arc sin

489. y

^f +cotx.

=

^-T +a 8

y=arcsin(l

.

= -^ (arc sin*)

2

arc cos jt.

=jc-I y = ln(arcsin5x). y = arc sin (Inx). =

495.

496.

497.

t/

498.

t/

499. 500.

501.

502.

/ i/

=

arc tan

= 36 = = =

F( F

-

5tan-i -

2

arc tan

}

1/2 arc cot

~

(36

=x.

I/ *

-) a /

+ a arc sin

x*

a

2

484. y

492.

(O-

5

-o-

O

481. y

483.

).

= 3sinA:cos

*/

480. w

482.

2

2

485. #

+ 2*) Vbxx*.

2

arc

. CL

sin-.

Tabular Differentiation

Sec. 2]

504.

y = ~e- (3sm3x x

cos 3*).

507

= x"a-*". 508. y= K r = = I/cos *a 509. 506. 510. y = x 2 1/7+2 In (!+/*) = ln(a + *+/2ax + *'). 514*. 511. 5I2 ^ = !^' 515. y y ^_ = 513. y _,_

505. y

<

*.

/

/

l

l

^

j/

1

i

.

516.

=

517. y 518.

y=lnln(3

=5 ^=

519. y 520.

522.

l

t/

/

=

.

x-sin(lnx

^-) 1

524. /(*)

=

525. y

= \l

526.

=2

i/

arc sln sin

527.

528.

*

+ (!

ux

u^?^^-

;/

COS X

= -Uln

arc cos 3x)

.

'

'

sm 8

1 .

m '

3 cos' bx

tan^+2-: = n

529. y

530. 531.

= arc

tan In x.

y^lnarc sinx + f/

= arctanln

.

In

2

jc-|

arc sin In x.

l

n

z x

3

_

Differentiation of Functions

54

633. t/=

[C/t.

1Vsin*

1 arc tan x. = 1 In Ji- + 1 In J^| + 1 In (1 + *)- In (*'-*+ 1) + /(*) =

534. y 535.

536. f(*)

=

= s'mh*2x. = e * cosh px. y = ianh*2x. y = Insinh2jc.

545.

= arcsinh~.

546.

537. y 538. y 539.

540. 541.

t/

542. y=r. arc cosh In*. arc tanh (tan x).

a

= # = arc cot h (sec*). y = arc tanh y^

543. 544.

.

t/

= -2-(

A:a

~

-^+--

547. r/=

548. Find y', a)

arc tan

if:

y=\x\\

b)

= *|*|.

Construct the graphs of the functions y and 549. Find #' 550. Find

f

(x)

if

x

1

551. Calculate

/'

(0)

f /'

x<0,

for

if

/(*) Solution,

y'.

if

=

= e~ x (3 sin 3x) = e (3 sin 0) e (0)

-* cos 3*. e~* cos

(x)

=

cos

3jc; 1

.

*

552. 553.

f(x)-ln(l+x) 1

y-tan ^.

554. Find

b) / (x)

+ (0)

/'

= arc sin Y

l+e

T

+ arcsin.

Find

2 ;

/'(I).

(-gj^.

and /1(0)

^^

Find

of the functions:

e)

f(x)

= x sin

-^

2

Tabular Differentiation

Sec. 2]

55

555. Find / (0) -f xf (0) of the function f(x)=e-*. 556. Find /(3) (x 3)/' (3) of the function f(x) 557. Given the functions f(x) tar\x and (p(*)

+

= Y\ + x. = ln(l x)>

=

n0)

find tind

q/(or

Given the functions /(x)=l

558.

2^

find nna

ff

x

and

cp(jc)

=

l

sin^

r

.

(1)

559. Prove that the derivative of an even function is an odd function, and the derivative of an odd function is an even function.

560. Prove that the

derivative of a periodic

a periodic function.

561.

Show

that

562.

Show

that the

563.

Show

that the function

the

function

xy' = d-x)y-

xy'

function

tion

xy' = y(y\i\x

is

also

y = xe~*

satisfies

the

equation

= xe~?

satisfies

the

e
y

= (\-x*)y.

function

y=

x

,

1

,

satisfies lrlx

n

the equa-

1).

G. Logarithmic Derivative

A logarithmic derivative of a function logarithm of this function; that is,

is

= f(x)

is

the

derivative

sometimes simplified by

first

of the

exponential function

=

where u = y(x) and v ty(x). Solution. Taking logarithms we get In

y

v In u.

Differentiate both sides of this equation with respect to x: (In y)'

= v' In u + v (In a)',

or 1

y

whence

~~

the

taking logs of the func-

tion.

Example. Find the derivative

of

fM

~y Finding the derivative

y

1

u

_ _

56 or

[Ch. 2

Differentiation of Functions

564. Find y'

Solution.

In

1

y

,

y'

if

,

2 = ^In x + In (1

x)

21, --ui(1) = _-

_-2x i-4-32

u

y

whence

y'

=y

/

2

^

565. Find

y\

Solution.

y

In

/'

if

x

1

x In sinx;

2x

1

-/'

)

1

cosx

sin x

1-j-x

-^

y = (smx)

(1

,

i

x

3

+ x + 3 In sin x +2 In cos x; 2

In

-~pj

-2smx cosx

,

\

2

+ 3cotx--2 tan xj

.

x .

= In sin x + x cot x;

= (sin x) x (In sin x + x cot x).

,Ln the following problems find y' after function y f(x):

=

first

taking logs of the

Sec. 3. The Derivatives of Functions Not Represented Explicitly 1. The derivative of an derivative

y'x

^ 0,

then

the

inverse

function.

I!

a

derivative of the inverse

*~7

function

function

y=f(x) has

x=/-

a

1

(t/)

is

The Derivatives

Sec. 3]

_

Functions Not Represented Explicitly

of

57

or

=

_ dy'

dy

Tx Example

1.

Solution.

\s

Find the derivative x

We

if

,

y

=1+1=^1 x x

have y x

=

x

hence,

;

*>

-7-. x-\- 1

2. The derivatives of functions represented parametrically. related to an argument x by means of a parameter t

If

a function

t

t

*-

I

=

then

*t or, in other notation, *JL

t^dx' dt

Example

2.

Find

^, dx

if

x

a cos

y = a sin Solution.

We

=

find

t, /

a sin/ and -r-

d\

dt

_

_ dx

3. The derivative and y

is

of an

= acosf.

a

implicit

sin

Whence

>

/

function.

If

the

relationship between

x

given in implicit form,

F(x,y) = Q,

(I)

then to find the derivative

y'x y' in the simplest cases it is sufficient: 1) to calculate the derivative, with respect to x, of the left side of equation (1), taking y as a function of x\ 2) to equate this derivative to zero, that is, to put

~F(A:,f/)

and

= 0,

3) to solve the resulting equation for

Example

3.

Find the derivative

yx

(2)

/'.

if

0.

Solution. ito

zero,

we

Forming the derivative

get 3*'

of the left

(3)

side

+ 3y V -3a (y + xy') = 0,

of (3)

and equating

it

58

Differentiation of Functions

[Ch. 2

whence 2

y

581. Find the derivative

,_* ay ~~axy*' xy

a)

f/

c)

y

if

=

= 0. .dy

In the following problems, find the derivative

#'=^

*

the

functions y represented parametrically: 582.

589.

x = acos*f,

(

= b sin* x = acos* y=b sin

\ y

590.

583.

t.

t,

8

cos

3

1.

/

T^nr 591.

584.

sin

8

/

?=

V coslr

585.

x

= arc cos

y

arc sin

592.

__

(

586.

'

y~=e:

{

= a( In tan + cos = a(sin + cosO. -2-

587.

t

588. /

t

~

595. Calculate

cos/).

when

f

///i

/t cn a sin

= 4= a(t = a(l

^

:

Solution.

*~ ^

593.

y

sin/

-r-~

a(l

cosO

1

cos/

if

sin

/),

cos/).

^

sin

^)

>

The Derivatives

Sec. 3]

Functions Not Represented Explicitly

of

59

and S1

fdy\

=

^ **

when

"T I

596.

Find

=

/

if<

!

i

= tlnt, in/ y=^ x = e cosf, = ^ sm^./ \ x

'

/

~.

597.

dv dx

^

i

Find

i

when

,

f

= 44 ji

-r

f

if

<

<

.

f/

698.

Prove that a function # represented parametrically by the

equations

satisfies

599.

the equation

When x = 2

the following equation jc

Does

when

it

_

x = 2?

= Va*

x*.

(x*)'

Is

x

= :r

5//+10 = 0.

2x

602.

5 + ? =1 s

8

8

x

604.

x'-

605.

l/^ +

606.

l/S + /~* =

607.

/'= y

--=a

f

possible to perform term-by-term

+ y*^0'? it

required to find the deriva-

is

609.

a cos

6I0

'

tan//

611.

xy-

613.

^=

-

603.

-t-y

= (2x)

implicit functions y.

of the

601.

608.

*

it

examples that follow

In the y'

true:

follow from this that

600. Let y differentiation of

tive

is

= 2x.

2

.

K^ = /"a.

0.3 sin y

'/a*.

= *.

2

614.

616.

arctan

^-

= -^

l

60

Differentiation of Functions

2

+ y = care tan 2

617.

1/x

619.

Find

Solution.

#=1, we

at the point

y'

Differentiating,

obtain

2*/'

= l+3f/',

620. Find the indicated points: a)

b)

for for

2

c)

we get whence

for

if

==y* + 3xy*y'.

x

Putting

2y' 0'

derivatives y' of

y)

x+l ye = e # = y

A! (1,1),

=

x*

618.

.

(Ch.

=

l

and

1.

functions y at the

specified

x=

*= x=

and and

y=l;

and

r/=l.

*/=!;

Sec. 4. Geometrical and Mechanical Applications of the Derivative

1. Equations of the tangent and the normal. From the geometric significance of a derivative it follows that the equation of the tangent to a curve will be t/ ) (* y = f(x) or F(x,y)=Q at a point

M

where

,

M

value of the derivative y' at the point (X Q y Q ). The straight through the point of tangency perpendicularly to the tangent is called the normal to the curve. For the normal we have the equation

y'Q is the

line passing

,

Y\ 2. The angle between angle between the curves

curves.

The

d and

at their

10 {Z

the

is

M

Q

common

A and

M

M

point

B

point Using a familiar formula of analytic geometry,

M

to

these

y Q ) (Fig. 12) tangents curves at the

(*

between

co

angle

,

the

.

we

get

3. Segments associated with the tangent and the normal in a rectangular coordinate system. The tangent and the normal determine the following four

Sec

Geometrical and Mechanical Applications of the Deriiative

4]

segments (Fig.

13): t

= TM

is

NM

n

the so-called

is

S t = TK

is

S n = KN

is

segment

and tan

\

y = y'Q

Sn N

/f

Fig.

KM = \y

of the tangent,

the subtangent, the segment of the normal, the subnormal.

St

Since

61

,

X

13

it

follows that

j/o

4. Segments

associated with the tangent

tern of coordinates. If a en in polar coordinates

tion

r

= /(q>),

then

formed by the tangent

OM

the

MT

u.

angle

and the

radius vector r (Fig. 14), defined by the following formula:

The tangent

and the normal

curve is givby the equa-

is

.

MN

MT

and the normal together with the radithe point of tangency and with the perpendicular to the radius vector drawn through the pole determine the following four segat the point us vector of

ments

M

(see Fig.

\Af

Fig.

14

14): t

= MT

n=

MN

S t = OT S n = ON

is is is

is

the segment of the polar tangent, the segment of the polar normal, the polar subtangent,

the polar subnormal.

in a polar sys-

[Ch. 2

Differentiation of Functions

62

These segments are expressed by the following formulas:

to

621. What angles cp are formed with the x-axis by the tangents 2 x x at points with abscissas: the curve y

=

x = 0;

a)

x=l/2;

b)

x=l?

c)

2x. Whence Solution. We have y'^\ 45; b) tan 9 0, q>=0; l,

q>

=

= =

622. At

N

2

the

sect

=

=

what angles do the sine and y= s'm2x inter-

axis

of

abscissas

at

the

origin? 623. At what angle does the tanianx intersect the gent curve y

Fig. 15

=

axis of abscissas at the origin? tx 624. At what angle does the curve y e*' intersect the line x 2? straight 625. Find the points at which the tangents to the curve 4 20 are parallel to the jc-axis. 12x* y =z 3* -f. 4x* 626. At what point is the tangent to the parabola

=

=

+

=

+ =

3 0? 5x y 627. Find the equation of the parabola y~x*-}-bx-\-c that is tangent to the straight line x y at the point (1,1). 628. Determine the slope of the tangent to the curve x*+y* Q at the point (1,2). 2 629. At what point of the curve y 2x* is the tangent perto 2 the line 0? straight pendicular 630. Write the equation of the tangent and the normal to the parallel to the straight line

xy7 =

= + 4x3y =

parabola

y=

at the point Solution.

We

= [y'] x= i = -T-

K

x

with abscissa x = 4. have y

f

2

1

k

,/-

whence

7=;

V

the

x

slope

the

of

Since the point of tangency has coordinates

*

#2 =

= 4,

follows that the equation of the tangent is 1/4 (* 4) or x Since the slope of the normal must be perpendicular, *,

whence the equation

of the

normal:

tangent

y = 2,

4 (x

4) or

4x

+y

It

4# + 4 = 0.

= -4; t/2 =

is

18

0.

_

Geometrical and Mechanical Applications of the Derivative

Sec. 4\

63

631. Write the equations of the tangent and the normal to the 4# 3 at the point (2,5). 2x* curve y = x' 632. Find the equations of the tangent and the normal to the curve

+

at the point (1,0).

633. Form the equations of the tangent and the normal to the curves at the indicated points: tan2x at the origin; a) y b)

= y = arc

sin

^^

at

the

point

of

intersection

with

the

A:-axis; c)

d)

y = arc cos 3x at the point of intersection with the y-axis; at the point of intersection with the #-axis; y = ln* ~ x* at the points of intersection with the straight y=e }

e)

line

y=

1.

634. Write the equations of the tangent point (2,2) to the curve t*

and the normal

at the

'

635. Write the equations of the tangent to the curve

at the origin

and

at the

point

^

= j-

636. Write the equations of the tangent and the normal to the x* 6=0 at the point with ordinate y 3. 2x y* 637. Write the equation of the tangent to the curve x* y*

curve

=

+ +

+

2xy = Q at the point (1,1). 638. Write the equations of the tangents and the normals to the curve y = (x 3) at the points of its intersection 2)(x l)(jt with the #-axis. 639. Write the equations of the tangent and the normal to the 4 curve y* = 4x 6xy at the point (1,2). 640*. Show that the segment of the tangent to the hyperbola xy = a* (the segment lies between the coordinate axes) is divided in two at the point of tangency. 641. Show that in the case of the astroid x 2 8 + y*t* = a*/ J the segment of the tangent between the coordinate axes has a constant value equal to a.

+

/

_

___

_

[Ch. 2

Differentiation of Functions

64

642.

Show

that the normals to the involute of the circle

x = a(cost

+

t

y = a(sinf

sin/),

t

cost)

are tangents to the circle 643. Find the angle at which the parabolas y (x 2 _. x i ntersect. 4 _|_ 6* y 2 x and y 644. At what angle do the parabolas y

=

=

Sect? 645.

=

Show

2

+

that the curves y 4x 2x 8 and y are tangent to each other at the point (3,34). Will

same thing 646.

at

Show

2

2)

and

= x*

inter-

= x*

x

we have

-\-

10

the

(2,4)? that the hyperbolas

intersect at a right angle. 647. Given a parabola y* 4x. At the point (1,2) evaluate the lengths of the segments of the subtangent, subnormal, tangent,

=

and normal. 648. Find the length of the segment 2* at any point of it. curve y 649.

Show

that in the

equilateral

of the

subtangent of the 2

2

=

2

a the y hyperbola x to the radius vector equal

length of the normal at any point is of this point. 650. Show that the length of the segment of the subnormal 2 2 2 a at any point is equal to the abscissa in the hyperbola x y

=

of this point.

651. x*

y

Show

that the segments of the subtangents of the ellipse

2

jjr+frl

an d the circle

2

x -+y

2

= a*

at

points

with the same

abscissas are equal. What procedure of construction of the tangent to the ellipse follows from this? 652. Find the length of the segment ol the tangent, the normal, the subtangent, and the subnormal of the cycloid ( \

x = a(t a(ts'mt), y = a(l

an arbitrary point t~t 653. Find the angle between the tangent and the radius vector of the point of tangency in the case of the logarithmic spiral at

.

Find the angle between the tangent and the radius vecthe point of tangency in the case of the lemniscate 1 a cos 2q>.

654. tor r*

=

of

Sec. 4]

Geometrical and Mechanical Applications of the Derivative

65

655. Find the lengths of the segments of the polar subtangent, subnormal, tangent and normal, and also the angle between the tangent and the radius vector of the point of tangency in the case of the spiral of Archimedes

at a

point with polar angle

= 2jt.

Find the lengths of the segments of the polar subtangent, subnormal, tangent, and normal, and also the angle between the tanat an gent and the radius vector in the hyperbolic spiral r= 656.

=

=

r rQ cp arbitrary point cp 657. The law of motion of a point ;

.

Find the velocity of the point at / in centimetres and / is in seconds).

on the *-axis

= 0, ^

and

is

t

2

=

(x

is

658. Moving along the #-axis are two points that have the 2 5t and #=l/2/ where t^O. following laws of motion: x=\00 With what speed are these points receding from each other at the time of encounter (x is in centimetres and / is in seconds)?

+

659. The end-points of a segment AB the coordinate axes OX and OY (Fig. 16).

,

^5 m A

is

are sliding along at 2 m/sec.

moving

A Fig.

17

is the rate of motion of B when A is at a distance OA = 3 m from the origin? 660*. The law of motion of a material point thrown up at an angle a to the horizon with initial velocity V Q (in the vertical plane OXY in Fig. 17) is given by the formulas (air resistance is

What

3-1900

66

_ _ Differentiation of Functions

[Ch. 2

disregarded):

#=i; /cosa, where

trajectory of motion speed of motion and

A

661.

i>

/sin a

^-,

point

g is the acceleration of gravity. Find the and the distance covered. Also determine the

is in

its

direction.

motion along a hyperbola

r/

=

so that its

x increases uniformly at a rate of 1 unit per second. the rate of change of its ordinate when the point passes

abscissa

What

=

the time and

is

/

*/

is

through (5,2)? 662. At what point of the parabola

y*=\8x does the ordinate increase at twice the rate of the abscissa? 663. One side of a rectangle, a 10 cm, is of constant length, while the other side, b, increases at a constant rate of 4 cm,'sec. At what rate are the diagonal of the rectangle and its area increas30 cm? ing when 6 664. The radius of a sphere is increasing at a uniform rate of 5 cm/sec. At what rate are the area of the surface of the sphere and the volume of the sphere increasing when the radius

=

=

becomes 50 cm? 665.

=10

A

point

is

in

motion along the spiral

of

Archimedes

so that the angular velocity of rotation of its radius constant and equal to 6 per second. Determine the rate of elongation of the radius vector r when r 25 cm. 666. A nonhomogeneous rod AB is 12 cm long. The mass of a part of it, AM, increases with the square of the distance of the from the end A and is 10 gm when 2 cm. moving point, Find the mass of the entire rod AB and the linear density at any point M. What is the linear density of the rod at A and S?

(a

vector

cm)

is

=

M

AM =

Sec. 5. Derivatives of Higher Orders

1. Definition of higher derivatives. A derivative of the second order, or Ihe second derivative, of the function y=f(x) is the derivative of its derivative; that is,
=

(
The second derivative may be denoted
If

* = /(/)

is

or

^.

as or

f"(x).

the law of rectilinear motion of a point, then

eration of this motion.

^

is

the accel-

Sec. 5]

_

Derivatives of Higher Orders

Generally, the ith

a derivative of order (n

derivative of a function y f(x) is the derivative 1). For the nth derivative we use the notation

(v

y

Example

t

or

~^,

or

f

Find the second derivative

t.

_

y

= \n(\

(n)

67 of

(x).

function

of the

x).

/.JZL; /

Solution.

=

2. Leibniz rule. If the functions u q>(x) and v=ty(x) have derivatives to the nth order inclusive, then to evaluate the nth derivative of a product of these functions we can use the Leibniz rule (or formula): up

(uv)

<">

= u<"

3. Higher-order derivatives

= -r

then the derivatives y'x

functions

of (

*

I

i^

== f/^jc

represented

can

^2

successively

xt

we have

For a second derivative

Example

Solution.

2.

F^nd

We

the formula

w /

if

,

have -

If

.

.

_ /

(a cos*),

& cos .

&

f

"~~ "~~~ .

asm*

a

LUl

I

and .

*

(acosO

If

= q>(0, = *(0,

by the formulas

xt

parametricaKy.

-asln<

osln

be

calculated

68

Differentiation of Functions

[Ch.

A. Higher-Order Derivatives of Explicit Functions In the examples that follow, find the second derivative of given function. 667. y 668. y

669.

= x* + 7x' = e*

5x

+ 4.

671.

672.

.

= \n t/\+x

2

v

Show

y= (arc sin x) = acosh u

673. 674.

.

V^

675.

=

2

y=sm*x.

670. j y

//

th<

OY

I

.

.

*/

y=

that the function

2

a

-\

satisfies the

2

differ

ential equation

676. tial

Show

that the function y

2y'+y = e

equation y" 677.

Show

equation y" 4-3y' 678. tion y" 679. 680. 681. 682. 683.

Show

= -|-2y

the

satisfies

differen

.

y=-C

l

e"

for all constants

the

that

2

x

function

the

that

= -^-x e x

function y

x

+ C 2 e' 2x C and C l

= e 2x s'm5x

satisfies th 2

.

satisfies the equa 4y s 2 5x + 7x 2. Find y"' if y = x 5 Find /'"(3) if /(*) - (2^: 3) Find y v of the function # = ln(l+x). Find t/ VI of the function y==sin2x. Show that the function y = e~ x cosx satisfies the differ ential equation y lv + 4y = Q. 684. Find /(O), f (0), T(0) and /'"(O; f

+29y = 0. ,

.

f

if

= e x sinx.

f(x) 685.

The equation of motion of a poin along the jc-axis is

O.OOU 8

X-100-H5/

.

Find the velocity and the acceleration the f

t

point

for

times

=10.

A

/

= 0,

t

M

l

=\

y

c

an

is in motion around with constant anguls Fig- 18 velocity CD. Find the law of motion of i1 projection M, on the x-axis if at time / = the point is at Q (a, 0) (Fig. 18). Find the velocity and the ac celeration of motion of M,. at the in What is the velocity and the acceleration of tial time and when it passes through the origin? What are the maximum values of the absolute velocity and th

686.

circle

x

2

point 2

+y = a

2

M

M

absolute acceleration of Ai,?

l

Sec. 5]

_

_

69

Derivatives of Higher Orders

687. Find

the nth derivative of the function 3 natural number. 688. Find the nth derivatives of the functions:

where n

y=

(

js

a)

and

y^T^x*

y^^**'

b)

689. Find the /zth derivative of the functions:

j/=sinx;

a)

e)

b)4, = cos2*; = e~ *; c) y

f)

y=^j\ = yJ; J/

9

g)

|/=ln(l+x);

d)

h)

y=sin*jr,

y=

690. Using the Leibniz rule, find y a)

y = x.f\

b)

x y = jc .e-*

{n

l

\

d)y =

2

c)

691. Find

//

= (!

(n)

/

(0),

if:

2 A:

)

e)

;

y

= x*

~

cos x\

if

B. Higher-Order Derivatives of Functions Represented Parametrically and of Implicit Functions d^u

In the following problems find

K

692. a)

693. _,

= \nt,

'

b)

,

:asin/; :

= 0cos'/,

.

= arc tan/, \0-l x

J

"'

iv

f

c)

* = arc sin/

= a(l -cos/); x = a (sin/- /cos/), = a(cos/-f-/ sin/).

\ y

f

696. Find

^

[Ch. 2

Differentiation of Functions

70

d*u

^

697. Find 698. tions x

ax 2

Show

== sin

if

y=

{ \

that y (as a function

^ + be~ iV y = ae' 2

t,

* = ln(l-f-/

I

/=0,

for

satisfies the differential

*

= sec, { = tan/. \ y = ^"'cos/, {xy-e-<*tnt. x

699.

703. jt'"

701.

Solution.

By

/,

= y'

/

r/

,

.

^

y = f(x), x = f~*(y).

/=-

and

if

'

find

y

~

a

composite

^^

+ **_ *

=

function

we have

-2^;^.

1

if

if

required to determine represented implicitly.

is

it

= f(x)

=

the

|y

l.

707.

# = *-}- arc tan

708.

Having the equation y = x + \ny

709. Find

/

|/.

at the point (1,1)

x

2

5xy +

710. Find y" at (0,1)

711. a)

*

2x

find

t

+

6

-j

= 0.

if

The function y

is

defined

implicitly

= 0.

at the point (1,1). b)

and

if

2

Find

the derivatives x",

finally get:

In the following examples derivative y" of the function y 705. y*

^+f=

^.

s

differentiating

-jwe ^

706.

the equaand 6

a

+ y*=*l.

Substituting the value of #',

y

=

x = e -1 = ^

function

the rule for

2* + 2i0'=0; whence

by

any constants

7 2 ' Find

x'

if

< 1

of the inverse function

704. Find

defined

x)

y'" f

the

.

of

for

'

Knowing

),

equation

In the following examples find (

t

2

Find

,

if

x

2

+

2

i/

==a 8 .

by the equation

Sec. 6]

Differentials of First

and Higher Orders

71

Sec. 6. Differentials of First and Higher Orders 1. First-order

differential. The differential (first-order) of a function the principal part of its increment, which part is linear relative to the increment Ax dx of the independent variable x. The differential of a

y

= f(x)

is

=

19

Fig.

function

is

equal to the

product of its

derivative

by the differential of the

independent variable dy--=y'dx,

whence u y

MN

If is an arc of the graph tangent at M. (x, y) and

dy dx

,

'

the function

of

= f(x)

y

(Fig.

MT

19),

is

the

PQ = Ax-=dx,

then the increment in the ordinate of the tangent

and the segment AN by. Example 1. Find the increment 2 = 3x x. y Solution. First method: A//

= 3 (x + Ax)

and

2

(x

the

differential

+ Ax)

3x 2

of

the

function

+x

or At/

Hence,

= (6*

dy = (6x

Ax + 3 (Ax) 2

1)

1)

Ax = (6x

1)

.

dx.

Second method: t/'

Example

2.

and Ax = 0.01. Solution.

= 6x

Calculate

A/ =

At/

1;

df/

= j/'dx = (6x

and dy

of

the

1)

dx.

function

y = 3x

l)-Ax

(6x

=

(6jt

x

+ 3 (Ax) = 5- 0.0 + 3- (0.01 = 0.0503 Ax = 5- 0.01 = 0.0500. 2

and

2

1

1)

2

)

for

x=l

72

Differentiation of Functions

[C/t.

2

2. Principal properties of differentials. = 0, where c = const. 1) dc 2) d*- Ax, where x is an independent variable. 3)

4) 5)

= cdu. dv. v) = du d (uv) udv + v du. d(cu)

d(u

7)

differential to approximate calculations. If the increment argument x is small in absolute value, then the differential dy of the function y = f(x) and the increment At/ of the function are approximately

3. Applying the

A*

of the

equal: that

A # =^ dy,

is,

whence 3. By how much (approximately) does the side of a square change 2 to 9.1 m 2 ? area increases from 9 Solution. If x is the area of the square and y is its side, then

Example

if

m

its

It

is

given that #

The increment mately as follows:

At/

=9 in

and A*

0.1.

the side of the square

ky^zdy--=y' Ax

4. Higher-order

differentials.

=

A

j=z

-0.1

may

be calculated approxi-

= 0. 016m.

second-order differential

is

the differential

of a first-order differential:

We

similarly define the differentials of the third and higher orders. If y f(x) and x is an independent variable, then

=

But

if

y

= /(),

where w

d*y

and so

forth.

= cp(x),

then

= y"' (du) + 3y" du 9

d*u

+ y' d'u

(Here the primes denote derivatives with respect to

M).

712. Find the increment Ay and the differentia! dy of the func2 # 5* -f x for x 2 and A# 0.001.

tion

=

=

=

_

Differentials of First

Sec. 6}

713.

_

and Higher Orders

Without calculating the derivative,

find

73

d(l-x') for

x=\ 714.

and Ax =

The area

.

square S with

of a

Find the increment and the

side x

differential of

is

this

=

x*. given by S function and ex-

plain the geometric significance of the latter. 715. Give a geometric interpretation of the increment and differential of the following functions: nx*\ a) the area of a circle, S b) the volume of a cube, v=^x\ 716. Show that when Ax *0, the increment in the function

=

= 2X

corresponding to an increment Ax in x, is, for any x, equivalent to the expression 2* In 2 A*. 717. For what value of x is the differential of the function 2 y = x not equivalent to the increment in this function as Ax >0? 718. Has the function y = \x\ a differential for x = 0? 719. Using the derivative, find the differential of the function //

y

,

cos x for x

=y

and Ax

--=

~

.

720. Find the differential of the

for

for

=

function

Ax-

0.01. 9 and x 721. Calculate the differential of the function

x-^-J

Ax^.

and

problems find the differentials of the given functions for arbitrary values of the argument and its increment. In the following

727. y

722.

y^'-m-

723.

<,=

724.

#= arc sin

= cot s = arc

729. r

.

730.

725. //--=arctan~.

=

= x\nx

q>

x.

-f

cosec

(p.

lane*.

x

726. y e~ \ 731 Find d//

if

x*

+ 2xy

y*

= a*.

Solution. Taking advantage of the invariancy of the form of a differential, x dy) 2y dy (y dx

we obtain 2x dx + 2 Whence

+

=

74

Differentiation of Functions

[Ch. 2

In the following examples find the differentials of the functions defined implicitly. 732.

=

733.

734. X

y'y = 6x*.

735. Find dy at the point (1,2), if 736. Find the approximate value of sin Solution. Putting

3) we have

(see

sin

*

=

arc30=-jr-

and Ax = arc

+ ^~ cos

31=^ sin 30

31.

1=^, from formula

(1)

30=0.500+0.017-J~^=0.515.

737. Replacing the increment of the function by the differencalculate approximately:

tial,

a) cos

b) tan c)

61; 44;

d) In 0.9; e) arc

tan 1.05.

e*\

738. What will be the approximate increase in the volume of a sphere if its radius 15 cm increases by 2 739. Derive the approximate formula (for \&x\ that are small compared to x)

#=

mm?

A*

Using

it,

V5

approximate

Y\7, /70, /640.

,

740. Derive the approximate formula

and find approximate values 741.

for

for

*=1.03; * = 0.2;

for

* = 0.1;

for __

c)

d)

j!/TO,

j/70, jI/200.

Approximate the functions:

/(x)- !/"}= y=e

l

~ x*

x =1.05.

for /

742. Approximate tan 453 20". 743. Find the approximate value of arc sin 0.54. 744.

Approximate

\

Sec. 7]

_ _ Mean-Value Theorems

745. Using

Ohm's

/

law,

= --,

show

the current, due to a small change found approximately by the formula

2/

the

in

resistance,

in

may be

A.

A/ 746. Show that, relative error of 1/

that a small change

75

of the radius, a error of approximately in calculating the area of a circle and the surface of a sphere. 747. Compute d*y, if y co$5x. determining the in length results in a relative = Solut ion. 748. u 749. 750. // d y = y" (dx = 2 2 25 cos 5* (dx) 2 ) = x\ = arccosx, find d*u. {/^sinxlnx, find d*y. find d*y. = ^, find d'z. z = **-*, find d'z. z = flnd d4 2=TJ, M = 3sin(2jt-f 5), find = e* cosa sin(;t sin u), 751. . e 752. 753. *' 754. 755. // d"w. n find d y. Sec. 7. Mean-Value Theorems 1. Rolle's b, then the theorem. If has a derivative argument x has at the interval is continuous on f (x) every interior point of this interval, and function a at /' (x) least one value , where a < 5< b, such that 2. Lagrange's theorem. If a function f (*) is continuous on the interval and has a derivative at every interior point of this interval, then where a <5< ft. 3. Cauchy's theorem. If the functions a<x a^x^b and for simultaneously, and F(b)^F(a) interval -f (a) 756. Show l<x<0 that and the g. and F (x) are derivatives continuous on the do not vanish that then _ function 0<x appropriate values of t f (x) have f(x) = xx* satisfies the Rolle on the intervals the theorem. Find _ 76 _ Differentiation of Functions The function / (x) is continuous and different! able for all values = /(0) = /(1)=0. Hence, the Rolle theorem is applicable on 1) intervals and 0<*<_1, To find form the equation _ we Solution. of x, the and /( Kx<0 n*) = l-3* = 0. Whence 2 0<E < and [Ch. 2 1 757. The function = fc^-J/l; 2 = J/l , -i< g, where 1. = = 2 takes on f(x) 2) \/(x equal values the end-points of the interval [0.4]. Does /(4) j/4 at the Rolle theorem hold for this function on [0.4]? 758. Does the Rolle theorem hold for the function /(0) on the interval 759. Show [0, JT]? Let that the equation /'(*) = has three real roots. 760. The equation obviously has a root x any other real = 0. Show that this equation cannot have root. 761. Test whether the Lagrange theorem holds for the function on the interval value of [2,1] and the find appropriate intermediate . = xx* all Solution. The function f(x) values of A:, and /' (x)= 1 3x 2 h3ve /(l)-/(-2y = 0-6 = 2 2 and 3^ 2 inequality 1 continuous and difTerentiable for by the Lagrange formula, we = [l-(-2)]/ (E), that is, Hence, the only suitable value is 1, for which the is Whence, / /'(E)2 = g=l; < < holds 1 762. Test the validity of the Lagrange theorem and find the intermediate point for the function f(x) x 4/s on the interval [ 1,1]. 2 763. Given a segment of the parabola y x lying between two points A (1,1) and 3(3,9), find a point the tangent to which is parallel to the chord AB. 764. Using the Lagrange theorem, prove the formula = appropriate = sin (x 4- h) where sin x = h cos , Taylor's Formula Sec. 8] = 2 77 +2 * 765. a) For the functions /(x) whether the Cauchy theorem holds on find E; b) do same with the on the interval ~1 To, and F(x) the respect to /(*) = x' interval = sin* test 1 and [1,2] and F(x) = cosx . Sec. 8. Taylor's Formula If (n is function a f (x) is continuous and has continuous derivatives up to the l)th order inclusive on the a finite derivative f (n} (x) at lor's a<x<6 interval each interior point &<* (or of the interval, and there then Tay- formula fw = f ( -V(a) + -V w+ . . . . nl where = + a 0(jc In particular, f W =/ (0) a) and when +xf (0) a 0<6<1, = holds true on the interval. we have (Maclaurin's formula) r (0) + + . . . + ( "-' rel="nofollow"> / (0) + /<> (I), ( where = ? 0<9<1. 0jc, = 766. Expand the polynomial /(A:) tive integral powers of the binomial x n*)=3jt Solution. for 2 4A- = 6^ f 8 2;c A: + 3; /' f'(2) = 7;r(2) = 8;r(2) 7 (jc) + 3^ + 5 in posi- 2. 4; /'" (x)=6; n^4. Whence H; =6. Therefore, or 2x z 8 Jt 767. + 3x + 5 = (n) Solution. = +7 (.v 2) + 4 (x x Expand the function f(x)=e term containing (x-f-1) where 1 1 l 1 (x) = e* in 2 2) + 3 2) (A- powers of . x + to l the 8 . for all n, p)( 1)=JL. Hence, +6(*+ 1); 0 768. Expand the function /(x) the term with (x 1)*. = lnjt in powers of x 1 up to _ 78 _ [Ch. 2 Differentiation of Functions = sin x in powers of x up to the term containing x e* in powers of x up to the term contain770. Expand f(x) n~ l ing x 771. Show that sin(a /i) differs from 769. Expand 9 / (x) and to the term containing x*. = . + + h cos a sin a 2 by not more than l/2/i 772. Determine the origin . of the VT+x&l+x y* yi+i&l+x~x*, 2 a) b) , and evaluate their errors. 773. Evaluate the error Due 774. to line a catenary in of the shape 775*. \x\ \x\<\ is formula in the weight, a heavy y = a cosh. thread Show own its approximate formulas: Show approximately suspended thread that the by the parabola expressed that for \x\<^a, to within small \x\ for lies , (^-J we have the approximate equality Sec. 9. The L'Hospital-Bernoulli Rule Evaluating Indeterminate Forms for oo 1. Evaluating the indeterminate forms and and u . Let oo the single-valued a (*) are both infinitesimals or both infinites as x functions if is, / (x) the quotient (p (x) ^-4 . \ at x = a, is one of the indeterminate forms -- oo i then 00 . , *-+<* provided that the limit . lim /(*) (p (x) ^lim x-+a f (*) q>' ( x) of the ratio of derivatives exists. or _ V Hospital-Bernoulli Sec. 9] The rule is when a = also applicable fix) If the quotient / ,, _ Rule for Indeterminate Forms 00. yields an again indeterminate at the point form, = a, of one of the two above-mentioned types and /' (x) and the requirements that have been stated for f(x) and q? (x), pass to the ratio of second derivatives, etc. x 79 q>' However, may exist, it should borne be whereas the ratios mind that the in limit of the ratio -^-~ derivatives do not tend to any limit the of (x) satisfy we can then all Example 809). 2. Other indeterminate forms. To evaluate an indeterminate form like 0oo, transform the appropriate product fi(x)*f t (x), wne re lim/, (jt) = and (see K+O. /(*) lim/ 2 (*) = oo, ^^ into thequetient form (the - U * *->a (T^T\ /i (X) ( the f orm -). oo M*) In the case of the indeterminate form oo appropriate difference /,(*) f 2 (x) oo, one should transform the into the product / t (x) first evaluate the indeterminate form 7*7^; r lim if (X) i and l L / 1 7^7^=1, \x x-+a i\ ( x )j then we re- ) duce the expression to the form form (the ). /Tw The indeterminate forms I, 0, 00 are evaluated by first faking loga rithms and then finding the limit of the logarithm of the power [f l (x)]^ (x} (which requires evaluating a form like 0oo). In certain cases it is useful to combine the L'Hospital rule with tht finding of limits by elementary techniques. Example 1. Compute lim JL1 *->o cot (form "). 7 oo x Solution. Applying the L'Hospital rule we have JEfL^llm pL*r x+ocotx jc-o(cot*) lim lim We get the indeterminate form -jp jc-*o however, L'Hospital rule, since Um C-frO We sint X * Hm sin ~~*-H) X * thus finally get " JC->0 COt X we . x do not need to use the [Ch. 2 Differentiation of Functions 80 Example Compute 2. limf-J- x *-M)V sin Reducing xL^ J (form common denominator, we to a lim oo oo). get (4__J_Ulim*.!z^lf x x J x-o x sin x x-+o \ sin 2 2 v (fonn 2 z ' Before applying the L'Hospital rule, we replace the denominator of the 2 2 fraction by an equivalent infinitesimal (Ch. 1, Sec. 4) * sin A;~ x*. obtain ter lat- We o The L'Hospital rule gives lim L) ( Then, in A "~ S = lim m *=lim- a elementary fashion, we find _ ___ 2 x-<> \ sin Example 3. cos 2x 1 x x 2 sin 2 * ,. 1 2 Compute 8 *2 lim (cos 2x) 00 (form I ) X-M) Taking logarithms and applying the L'Hospital lim In (cos X-*0 lim 2*p = X-+Q 31ncos2* X rule, = _ 6 lim we %X x-+Q get = _6 . J^ x Hence, lim (cos 2x) *^e-*. Jf->0 : F ind the indicated limits of functions in the following ples. x '- 77G. lim ,_>, *' Solution. lim " ; X->1 777. lim xcosx v m r ^r 2 "~~ 7/ OJi ~~ 9 * sinx , 779. lir *7on is I-* tj 11IU ' ji x *->il_si n :r 780. lim tan* sin* exam- sec ,. 781. 2tan* 1+COS4* 2 ' x 785. lim- *-cot^ 2 lim^. tan5x 782. S ^Ji 786. lim y s *-" 787. . i . 1. lim [, cos x) cot (1 x x->o . n x lim -r-?=r , ) lim 1 783. 81 n A- x + 1 lim Solution, (1 . ^cos x) cot ^ = Hm , cos A '^- lim X->0 Sill AT ; Forms L' Hospital-Bernoulli Rule for Indeterminate Sec. 9] Sill A linrZi!iJ^( 788. lim(l A*) Ian ~ a 792. . ^ ,V-^l X^X lim arc sin x cot x. 789. -, V lim x" sin 793. ' hnilnxln (x 1). X-+Q ri 790. n>0. lim(jc e?"*), v^o 791. lim x sin 794. lim ^ *i f^ \ A n"^ lr % ] . Solution. A = [- A n 1 A' 1 , = lim lim j ^^MiiA-l A 795. lim 796. lim (A 1) x * 11 A = lim j \nx [ -1 *^ A -7 T- A h~? 2 A' ] L__l limf-^^' * 1 X y 797 ~ >l ^Vc _^/ A) AO J ^} 2cosx/ 2 798. lim A;*. Solution. s lim p We = lim j~ have 0, ** = r/; whence lim//=l, In y=?x that In is, A". ImiA^ lim In l. t/ = limjtln x = 82 (Ch. 2 Differentiation of Functions 799. limx*. 804. li V-H tan a 800. limx4 * "*. \ 805. Hmftan^f) 4 / 1 X-+l\ 1 801. linue s/n 806. lim (cot x) ln *. *->0 *. X-H) cos 802. lim(l-*) 803. lim(l+x ta 807. 2 )*- lta(I) x / x-*o \ ". 808. lim (cot x)* in *. X-+0 809. Prove that the limits of X a) cannot sin* be found by the L'Hospital-Bernoulli rule. Find these limits directly. 810*. Show that the area of a circular segment with minor and central angle a, which has a chord (Fig. 20), is AB=b CD=A approximately with an arbitrarily small relative error when a ->0. Chapter III THE EXTREMA OF A FUNCTION AND THE GEOMETRIC APPLICATIONS OF A DERIVATIVE Sec. 1. The Extrema 1. Increase and of a Function of One Argument of tunctions. Tlu Junction y f(x) is called on some interval if, fo. any points x and x 2 which belong to this interval, from the inequality A',/ (A,) (Fm. 21&)]. I! f(x) is continuous on the interval [a, b] and /' (x)>0 [/' (A') decrease increasing (decreasing) l i ffxj xz i, X (a) Fifi. Fig. 22 21 In the simplest cases, the domain of definition of f (x) may be subdividincrease and decrease of the funca finite number of intervals of tion (intervals of monotonicity). These intervals are bounded by ciitic-' or f' (x) does not exist]. points x [where /'(jc) Example 1. Test the following function for increase and decrease: ed into = Solution. We find the derivative t/' Whence y' = for x=l. On = 2x a 2 = 2(* number scale 1). we get two intervals of monot- oo<x0, and, hence, the function /(*) increases in the interval (1, +00) (Fig. 22). + Extrema and the Geometric Applications 84 Example Determine the intervals 2. of and decrease of increase [Ch. 3 a Derivative of the func- tion Solution. = , i o\\2 oo<*< Example 2 Here, 2 *^~ for Hence, the the of discontinuity and function (/' = decreases in the intervals function y and Test the following function for increase or decrease: 3. /i y Solution a is ~ *v s . 5 *y a ' 3 Here, (2) = 2 find the points x l Q, we Solving the equation x* -x 0, x,= l 1, *2 at which the derivative y' vanishes. Since y' can change sign only when passing through points at which it vanishes or becomes discontinuous (in the given case, y' has no discontinuities), the derivative in each of the intervals 1, 0), (0,1) and (1, (00, 1), ( +00) retains its sign; for this reason, the function under investigation is monotonic in each of these intervals. To determine in which of the indicated intervals the function increases and in which it decreases, one has to determine the sign of the derivative in each of the intervals, To determine what the sign of y' is in the interval 00, inter1), it is sufficient to determine the sign of y' at some point of the val; for example, taking x= 2, we get from (2) f/' 12>0, hence, y'>Q in the interval (00, increases Similar1) and the function in this interval we find that y' = 1 ~ ' v the in Y\ /'A\ I interval (0,1) ** (here, we can interval (1, x=l/2) and y'>0 use in the +00). the function being tested inThus, creases in the interval ( oo, 1), decreases in the interval 1) and again increases in the interval (1, -f oo). 2. Extremum of a function. If there (1, exists a two-sided neighbourhood of a point X Q such that for every point X^X of this Q neighbourhood we have the inequality f(x)>f(x Q ) then the point x is called the J Fig 23 minimum point of while the number mum the / (x ) of the function y neighbourhood of the point function is y f(x), called the mini- f(x). Similarly, if x lf the inequality any point xj^x l of some f(*) The Extrema Sec. 1] The of a Function of One Argument 85 and absence of an are given by the following rules: extremum sufficient conditions for the existence continuous function / (x) + of a 1. If there exists a neighbourhood (X Q 6, * 6) of a critical point * such that /'(x)>0 for X Q d<x<xQ and /'(jt)<0 for xQ <x<xQ d, then * is the maximum point of the function / (*); and if /' (*)<0 for * 6<*<x is the minimum point of the and /' (x)>0 for x <x<xe function 6\ then * + + /(*)- Finally, function 2. if f there for / (x). r If f (XQ) f' ( XQ ) (* if unchanged sign ) =Q = 0, is some 0<|jc positive number 6 such then x X Q |<6, is that /' (x) retains its not an extremal point of the then X Q is the maximum and /"(*<,)<(), )>0, then x is the minimum point; but if f )^0, then the point X Q is not an extremal point.$

and

f"

and /'"

(*

(*

point; (*

)

= 0,

More generally:

let the first of the derivatives (not equal to zero at the function f (x) be of the order k. Then, if k is even, the is an extremal point, namely, the maximum point, if f (k) (* )<0; and it is the minimum point, if / (ft) (x )>0 But if k is odd, then A- O is not n extremal point. Example 4. Find the extrema of the function

point x point X Q

the

of

)

<

i/

==2*

+3

j

Solution. Find the derivative (3)

V

x

Equating the derivative

Whence,

x-

:

x-= function

if

we -/i,

find

then

is

we

get:

x l =-

a sufficiently small

Hence,

/'<0*).

and //max=-l. Equating the denominator r/

to zero,

the critical point

where h

\+h,

y'

v

*,

1.

From formula

(3)

we

have:

if

positive number, then /y'>0; but I is the maximum point of the

f

of

the expression of y' in (3) to zero,

we

get

=

\\e find the second critical point of the function A'2 there 0, where for*/! we have no derivative //' For *== /i, we obviously have //<0; is the minimum point of the function y, and //>0. Consequently, * 2 m jn (Fig. 24). It is also possible to test the behaviour of the function 1 at the point x by means of the second derivative

whence is

=

i/

/=--4^=

=

1 is the maximum I x, and, hence, *, point of the function. 3. Greatest and least values. The least (greatest) value of a continuous function f (x) on a given interval [a, b] is attained either at the critical points of the function or at the end-points of the interval [a, b].

Here, r/"<0 for

p

to determine the sign of the derivative y', one can *) If it is difficult calculate arithmetically by taking for h a sufficiently small positive number.

86

Extrema and the Geometric Applications

Example

of a

Find the greatest and least values

5.

on the interval

Derivative

[Cfi.

3

of the function

P/2

Solution. Since

it

follows that the critical points of the function y are *,

=

1

and

Y

Fig. 24 of the function at these points and the values function at the end-points of the given interval

Comparing the values

we conclude the

(Fig.

x=l (at *=2 /i

point

at the point

that the function attains

25)

J

the

minimum

point),

and

its least

the greatest value

Determine the

812. 813. 814.

y=l

m=l,

at

A4 = ll

o

(at the right-hand end-point of the interval).

intervals

of decrease

and increase

1ions: 811.

value,

of the

4*

jf.

= (* 2) y = (A:+4) = *'(*- 3).

*>-

i

2

.

{/

s

817.

=

818.

= (x

.

{/

of the func-

The Extrema

Sec. 1]

819.

of

a Function

y^ \-V~x.

822.

y = 2e*

823.

= x -f sin x. y = x\nx. = arcsin(l-f-x).

820. y

821.

One Argument

of

'

y

825

-

^T"

g

We

find

the

zero,

we

*

for extrema:

= +

Solution.

-'*.

_. 2~<*.

24

t/

Test the following functions x* 826. y 4*4-6.

z

derivative

of

the

given function, value of the argument x= 2. Since i/'<0 when x< 2 is 2, and y'>Q when *> 2, it follows that *= the minimum point of the function, and #min 2. We get the same result of the second derivative at the critical point y"~< by utilizing the ~ sign

Equating

to

y'

the

get

critical

=

827. y 828. {/ 829. (/

--

= =

Solution,

.

!

We

find the derivative

= 6*

y'

4-

6x

12

=6

+*

2 (jc

2).

2 Equating the derivative y' to zero, we get the critical points x,= and *,= !. To determine the nature of the extremum, we calculate the second derivative ^"^ 6 (2* 4-1). Since /( 2 2)<0, it follows that x,= is the maximum point of the function y, and #max 25. Similarly, we have the function y and t/*(l)>0; therefore, x 2 =l is the minimum point of

=

i=

2.

<-

12 )

2 '

840.

1

(*

I)

y-

2)'.

841.

836. V

= rr4=^.

837. t/=

w=J/(^

OQQ ooy.

it */ ===

Determine the indicated

843. y

=

844.

=

847.

citi ^. v -+ sin ^k*. I

least

intervals

=

/

,_

846. y

1)'.

O sin cin ZA Ov z

842. #

(if

f/

/1ft // M oto.

= x'e-*. = -. X -

ar/* fan ^ idii ^t. /t-~drc

interval

is

^

of the functions on the not given, determine the

and greatest values the

ln(l+*).

je

845>

^_.

838.

=

t/

_

88

greatest

Extrema and the Geometric Applications

and

[Ch. 8

of a Derivative

values of the function throughout the domain

least

of definition). 849. 850. 851.

= rih&. -= y x(lOx). y= sin + cos A;

imum

interval

= 2x* + 3*

A;.

12*

a)

b)

2

on the interval on the interval

that for positive values of

856. Determine

nomial

V = x* on the

854. y

= arc cos x.

Show

-

4

4

852. # 855.

853

!/

*we

+

[

1,3].

1

1,6];

f

[10,12],

have the inequality

coefficients p and q of the quadratic triso that this trinomial should have a minJt= 1. Explain the result in geometrical terms.

the

y*=x*+px + q

=

3 when t/ 857. Prove the inequality e*

> +x 1

when x

4* 0.

Solution. Consider the function

In the usual

way we

find

lhat

this function

has a single

minimum

/(0)

Hence,

/(*)>/ (0) and so e* as

we

set

> +x 1

when x when x

0,

^ 0,

out to prove.

Prove the inequalities: x sin x 858. x

<

^< o

when *>0.

^

when

~
when

859.

cos*>l

860.

A:

JL

861. Separate a given positive number a into two summands such that their product is the greatest possible. 862. Bend a piece of wire of length / into a rectangle so that the area of the latter is greatest. 863. What right triangle of given perimeter 2p has the greatest area? 864. It is required to build a rectangular playground so that should have a wire net on three sides and a long stone wall it on the fourth. What is the optimum (in the sense of area) shape of the playground if / metres of wire netting are available?

Sec.

1]

The Extrema

of

a Function

of

One Argument

89

865. It is required to make an open rectangular box of greatest capacity out of a square sheet of cardboard with side a by cutting squares at each of the angles and bending up the ends of the resulting cross-like figure. 866. An open tank with a square base must have a capacity of v litres. What size will it be if the least amount of tin is used? 867. Which cylinder of a given volume has the least overall surface? 868. In a given sphere inscribe a cylinder with the greatest volume.

869.

In a given sphere inscribe a cylinder having the greatest

lateral surface.

870. In a given sphere inscribe a cone with the greatest volume. 871. Inscribe in a given sphere a right circular cone with the greatest lateral surface. 872. About a given cylinder circumscribe a right cone of least volume (the planes and centres of their circular bases coincide). 873. Which of the cones circumscribed about a given sphere has the least volume? 874. A sheet of tin of width a has to be bent into an open cylindrical channel (Fig. 26). What should the central angle cp be so that the channel will have maximum capacity?

D N

I

M Fig. 27

875. Out of a circular sheet cut a sector such that when made into a funnel it will have the greatest possible capacity. 876. An open vessel consists of a cylinder with a hemisphere at the bottom; the walls are of constant thickness. What will the dimensions of the vessel be if a minimum of material is used for a

given capacity?

=

OB of the door of a ver877. Determine the least height h so that this door can pass a rigid rod tower of length /, the end of which, M, slides along a horizontal straight line AB. The width of the tower is (Fig. 27).

tical

MN

ABCD

d
90

Extrema and the Geometric Applications

M

of

a Derivative

[Ch. 3

878. A point (x # ) lies in the first quadrant of a coordinate plane. Draw a straight line through this point so that the triangle which it forms with the positive semi-axes is of least area. 879. Inscribe in a given ellipse a rectangle of largest area with sides parallel to the axes of the ellipse. 880. Inscribe a rectangle of maximum area in a segment of the parabola y* 2px cut off by the straight line x 2a. ,

=

881.

On

=

the curve y

=

find a point at

,

1

-\-

Xt

which the tangent

forms with the A>axis the greatest (in absolute value) angle. 882. A messenger leaving A on one side of a river has to get to B on the other side. Knowing that the velocity along the bank that on the water, determine the angle at which the is k times messenger has to cross the river so as to reach B in the shortest possible time. The width of the river is h and the distance between A and B along the bank is d. 883. On a straight line AB=a connecting two sources of light A that (of intensity p) and B (of intensity
M

square of the distance from the light source). suspended above the centre of a round table At what distance should the lamp be above the table

portional to the 884. A lamp of radius

r.

is

an object on the edge of the table will get the greatest illumination? (The intensity of illumination is directly proportional to the cosine of the angle of incidence of the light rays and is inversely proportional to the square of the distance from the so that

light source.)

885. It is required to cut a beam of rectangular cross-section ont of a round log of diameter d. What should the width x and the height y be of this cross-section so that the beam will offer maximum resistance a) to compression and b) to

bending?

I i/J

Note. The resistance of a beam to compresof its crossis proportional to the area to the product of the section, to bending width of the cross-section by the square of sion

its

height.

886. A homogeneous rod AB, which can rotate about a point A (Fig. 28), is carrying a load Q kilograms at a distance of a cm from A and is held in equilibrium by a vertical force P applied to the free end B of the rod. A linear centimetre of the rod weighs q kilograms. Determine the length of the rod x so that the force P should be least, and find P mln Fig.

2

.

Sec. 2]

_

The Direction

887*. The centres on a single straight locity v strikes strikes C of mass

fi,

of Concavity.

Points of Inflection

91

of three elastic spheres A, B\ C are situated Sphere A of mass moving with vewhich, having acquired a certain velocity,

M

line.

m. What mass should B have so that C will have the greatest possible velocity? 888. N identical electric cells can be formed into a battery in different ways by combining n cells in series and then combining the allel.

in pargroups (the number of groups is ] The current supplied by this battery is given by the formula

resulting

,

where

NnS

~~

the electromotive force of one cell, r is its internal and R is its external resistance. For what value of n will the battery produce the greatest <

is

resistance,

current? 889. Determine

the

diameter y of a circular opening in the which the discharge of water per second Q will be greatest, if Q = cy Vhtj, where h is the depth of the lowest point of the opening (h and the empirical coefficient c are

body

of a

dam

for

constant). 890. If x lf # 2 ..., x n are the results of measurements of equal precision of a quantity x, then its most probable value will be that for which the sum of the squares of the errors ,

(*-*,) 0=2 1=1 value (the principle of least squares).

is

of least

of

Prove that the most probable value of x the measurements.

is

the arithmetic

mean

Sec. 2. The Direction of Concavity. Points of Inflection

1. The concavity of the graph of a function. We say that the graph of a differentiable function y f(x) is concave down in the interval (a,b) [concave the arc of the curve is below (or up in the interval (a p 6,)] if for for the tangent drawn at any point of the interval (a, b) lt above) or of the interval (a,, &.)] (Fig. 29). A sufficient condition for the concavity downwards (upwards) of a graph y f(x) is that the following inequality befulfilled in the appropriate interval:

a.<x
a<x<6 =

rw
irw>oj.

2. Points of inflection. A point [* f (jc )] at which the direction of concavity of the graph of some function changes is called a point of inflection ,

(Fig. 29).

Extrema and the Geometric Applications

92

of

[Ch. 3

a Derivative

point of inflection x of the graph of a function derivative f (* ) or /" (x ). Points at which Q or f (x) does not exist are called critical points of the second kind. f'(x) The critical point of the second kind x is the abscissa of the point of inflec* an * the intervals x 6 tion if I" (x) retains constant signs in .d

For the abscissa

y

f (x)

there

is

of the

=

no second

<x<

< <

+ 6,

where 6 is some positive number; provided these signs are

x?

Jc

And it is not a point of inflection if the signs of f (x) are the same in the above-indicated intervals.

opposite.

y~f(x)

Example 1. Determine the intervals of concavity and convexity and also the points of inflection of the Gaussian curve

I

Solution.

We

have

i

I

bx

a,

X

b,

and Fig. 29

Equating the second

we

derivative y* to zero,

find the critical points of tHe

second kind

*

=

and

7=r

*o

=

T=-

+

These points divide the number scale OO
Fig. 31 -f(this is obvious if, for example, we take one point in each lively, +, of the intervals and substitute the corresponding values of x into y ) Therefore: ,

1)

the curve

curve

is

is

concave up when

concave down when

oo< -=^

F

2

x

<

<x<

7= and F2 ==

V

2

.

F=

< x <-f oo;

2) the

V 2

The points

(

-=^

\V2

,

r=]

are

VeJ

points of inflection (Fig. 30). It will be noted that due to the symmetry of the Gaussian curve about the #-axis, it would be sufficient to investigate the sign of the concavity of x this curve on the semiaxis +00 alone.

< <

__

Sec. 3]

Asymptotes

Example

Find the points

2.

of

93

inflection of the graph of the function

y=*/7+2.

Solution.

It

is

We

have:

obvious that y" does not vanish anywhere. Equating to zero the denominator of the fraction on the

right of (1), we find that y" does not exist for x 2. Since y" for 2 and f/"<0 for 2, it follows that ( 2,0) is the point of inflection (Fig. 31). The tangent at this point is parallel to the axis of ordinates, since the first derivative y' is infinite at x 2.

x<

rel="nofollow">

*>

Find the

intervals of concavity and the points graphs of the following functions: 6x* 12x + 4. 891. y = x* 896. y cosx.

of

inflection

of the

+

892. 893.

y = (x + l)\ y = -4r o

897.

898.

.

X-\-

894. 895.

ff

=

X

9

i

899.

. ,

12

900.

\2x.

y=i/4x*

= y=x y=x

sin*. 2

In x.

= arc tanx y = (l+x*)e*. //

x.

Sec. 3. Asymptotes

1. Definition. If a point (#,/) is in continuous motion along a curve f(x) in such a way that at least one of its coordinates approaches infinity (and at the same time the distance of the point from some straight line tends to zero), then this straight line is called an asymptote of the curve. 2. Vertical asymptotes. If there is a number a such that y

Jim /(v)--=

then the straight line x 3

a Inclined asymptotes.

is

If

00,

an asymptote (vertical asymptote). there are limits llm X

->>

+ 00

K

and lim [/(*)X-++ 00

=

Ml = *i.

then the straight line y k l x+b l will be an asymptote 0, a right horizontal asymptote). asymptote or, when ^

=

If

there are limits llm

(a

right inclined

Extrema and the Geometric Applications

94

of a Derivative

[Ch. 3

and Urn

=

+

k zx y b^ is an asymptote (a left inclined asymptote horizontal asymptote). The graph of the function y f(x) (we assume the function is single-valued) cannot have more than one right (inclined or horizontal) and more than one left (inclined or horizontal) asymptote. Example 1. Find the asymptotes of the curve

then the straight or,

when

fe

2

= 0,

line

=

a left

Solution. Equating the denominator

to zero,

we

get

two

vertical asyinp-

lotos-

x= We

x=l.

and

1

seek the inclined asymptotes. For x

kl

=

lim

l

lim

x)

(// *

+ oo

X

= lim

we obtain

~l,

lim

*-+o> b =-

>

v

2

}^x z x

y^2

=0,

\ \

S

-/ Fig. 32

hence, the straight line y

=x

the right asymptote. Similarly,

is

when*

oo,

we have fc

fc

a

=

Hm

=

~=

1;

lim AC->~

Thus, the totes

is

Example

2.

y= -x

(Fig. 32). Testing a curve for asympconsideration the symmetry of the curve. Find the asymptotes of the curve

left

simplified

asymptote if

we take

Is

into

__

Sec. 3]

95

Asymptotes

Solution. Since

lim

t/

=

oo,

=

is a vertical the straight line x asymptote (lower). Let us now curve only for the inclined right asymptote (since x>0). We have:

k= b

lim (y *-*+ 00

lim

=

X++OD X x)

=

lim #->+<

test

the

1,

\nx

oo.

Hence, there is no inclined asymptote. If a curve is represented by the parametric equations x cp(0i */ ^(0 then we first test to find out whether there are any values of the parameter / for which one of the functions cp (t) or \|> (/) becomes infinite, while the other remains finite. When (p(/ )=oo and ty(t ) = c, the curve has a horizontal asymptote y c. When \j)(f ) = oo and (p(V ) = c, the curve has a vertical asymptote x = c. If
=

=

lim

then the curve has an inclined asymptote y kx+ b. If the curve is represented by a polar equation r /(cp), then we can find its asymptotes by the preceding rule after transforming the equation of the curve to the parametric form by the formulas x r cos cp /((p) cos q>; r sin

) sin (p. y

=

Find the asymptotes of the following curves: 901.

11

903. y

=

^rr.

-,

=

908. u

=x

909. y

= e-

2

910. i/=

.

911.

905.

y^Y^^l.

912.

906.

y==-

913

907.
.

r

* ~~"

-

914. x

=

/;

j/

=

*

915. Find the asymptote of the hyperbolic

spiral r

=

.

of a Derivative

Extrema and the Geometric Applications

96

[Ch. 3]

Sec. 4. Graphing Functions by Characteristic Points In constructing the graph of a function,

find its

first

domain

of definition

and then determine the behaviour of the function on the boundary of this domain. It is also useful to note any peculiarities of the function (if there are any), such as symmetry, periodicity, constancy of sign, monotonicity, etc. Then find any points of discontinuity, bending points, points of inflection, asymptotes, etc. These elements help to determine the general nature of the graph of the function and to obtain a mathematically correct outline of it. Example 1. Construct the graph of the function

The function exists everywhere except at the points x 1. and therefore the graph is symmetric about the point

Solution, a)

The function 0(0,

This simplifies construction of the graph

0).

The

b)

lim

odd,

is

discontinuities

are

x=

hence, the straight lines

t/=oo;

and

1

jc

and

1;

lim

and

oo

J/=

V-M + O are vertical asymptotes of the

#=1

X->--10 graph. c)

We

seek inclined asymptotes, and find

,=

lim X -> +

bl

oo

lim

-x

= 0,

y

oo,

#->-t-oo

thus, there is no right asymptote. From the symmetry of the curve it follows that there is no left-hand asymptote either. d) We find the critical points of the first and second kinds, that is, points at which the first (or, respectively, the second) derivative of the given function vanishes or does not exist.

We

have:

,

x=l,

The

derivatives y' and \f are nonexistent only at that is, only at points where the function y itself does not exist; and so the critical points are only those at which y' and y" vanish. From (1) and (2) it follows that

when x= when x =

y'=Q r/"

y' retains a

Thus,

(-V3, intervals

l), (

(1,

00,

V$\ and x= 3. in each of the intervals constant_ sign 1), 3), = ( (l, 3, V$)

and (V~3

1),

(1,

t

0),

+00), (0,

1),

and (1,

/ 3)

(

00,

in

each

and

(3,

J/T), of

+00).

the

To determine the signs of y' (or, respectively, y") in each of the indicated intervals, it is sufficient to determine the sign of y' (or y") at some one point of each of these intervals.

Sec

Graphing Functions by Characteristic Points

4]

97

It is convenient to tabulate the results of such an investigation (Table I), calculating also the ordinates of the characteristic points of the graph of the function. It will be noted that due to the oddness of the function r/, it is enough to calculate only for Jc^O; the left-hand half of the graph is constructed by the principle of odd symmetry.

Table I

e)

Usin^ the results

function (Fig

of

the investigation,

33).

-/

Fig. 33

4-1900

we

construct

the

graph

of

the

Extrema and the Geometric Applications

Example

2.

of

a Derivative

[Ch. 3]

Graph the function In x

x Solution, a) The domain of definition of the function is 0<x<-f-oo. b) There are no discontinuities in the domain of definition, but as 0) of the domain of definition we have approach the boundary point (#

= limw = lim

JL? =

oo

X

X-*0

JC->

we

=

Hence, the straight line jc (ordinate axis) is a vertical asymptote. c) We seek the right asymptote (there is no left asymptote, since x cannot tend to oo ):

k=

lim X<-++

=

.

The

00

-^ X

lim # x->+<

=

0;

= 0.

right asymptote is the axis of abscissas: find the critical points; and have d)

j/

= 0.

We

Inx

1

y 3

y'

and

y" exist at all

of the

points

y' = Q

domain

when ln*=l,

that

definition of the function and

of is,

when x =

o

(/'=0

We

when

form a table,, including io the characteristic points

Inx^y,

the it

is

that

is,

when x~e*l*.

characteristic points (Table 11). In addition useful to find the points of intersection of

34

=

=

1 the curve with the coordinate axes. Putting / 0, we find * (the point of intersection of the curve with the axis of abscissas); the curve does not intersect the axis of ordinates the e) Utilizing the results of investigation, we construct the graph of lunction (Fig. 34).

-h

CM

CO

CM

O

'

" >

non

> C a c.u

-~

Extrema and the Geometric Applications

100

its

of

[Ch. 3

a Derivative

Graph the following functions and determine for each function domain of definition, discontinuities, extremal points, inter-

vals of increase and decrease, points of inflection of the direction of concavity, and also the asymptotes. 916. y

=x

9

3x*.

"

9

918.

u = (x

919.

y-

921. 922. 923.

924. 925.

\

= = y= y= = (/

(/

926. y==

928. 929.

930.

__

=,-

,6

3*'+! 932. 933. 934. 935.

936.

938.

_

y = 2x + 2-3'l/(xl-

963

z

l

.

'

#=

its

graph,

Sec.

an Arc. Curvature

Differential of

/>]

964. y

=

.

*

sin

+

-7-

y=

976.

.

4 /

\

978. ,,=

967.

979. ^

= y = *-l-sin*. y = arc sin (1 /F).

970. 971. 972.

and 973.

974.

953.

982 tan

.

983.

984

.

<>arcsin

f/

= lnA:-arc tan*. In cos x. y = cos^ = arc tan(ln = arc sin In (*' 4-1). ,/

,/

985.

je).

y==x

-^-

987. y

= lnsin*.

A good

exercise

ples 826-848. Construct the

is

to

>

j

Af

i/

K

= garcun* = n sin x

,,

A;.

= x arc tan - when when * = 0. y= = 2 arc cot*. = + arc tan*.

975. y

arc cosh

I

965. */= sin*- sin 2*. 966. (/ COS*-COS2*. 968.

101

*

=

graph the functions indicated following functions

the

graphs of

parainelrically. l 2/. 988. x=--t* //----/ 2t, sin/ (a>0). 989. x=--acob*/, te~ 990. jc /e', y fl / 4-g991. x i/=2/ + ea (cosh / a (sinh/ 992. x /), i/

in

Fxam-

represented

+

y^a =

= = =

l

.

1

.

,

=

I)

_

(a>0).

Sec. 5. Differential of an Arc. Curvature

1. Differential of an arc. The differential of an arc s of represented by an equation in Cartesian coordinates x and y the formula

dshere,

the equation of the curve

if

a)

2

J/~(d*)

//

= /(*),

b)* = /,Urt.

then ds

then

= 0,

ds^-

d)

^(*,

f/)

of the

',

form

ds

then

*

2

(dy)

-

= q>(0, y = +(0,

c)

is

+

ds-

V F* + F

then

'

.

2 /;

V F'

a is

plane curve expressed by

Extrema and the Geometric Applications

102

of

a Derivative

[Ch. 3

Denoting by a the angle formed by the tangent (in the direction increasing arc of the curve s) with the positive ^-direction, we get

of

dx

cos

a = -3-

,

ds

dy -r

sina

.

ds

In polar coordinates,

Denoting by p the angle between the radius vector of the point curve and the tangent to the curve at this point, we have a 008 P

sin

= dr

p

of the

'

/

2. Curvature of a curve. The curvature K of a curve at one of its is the limit of the ratio of the angle between the positive direcpoints and N of the curve (angle of contintions of the tangents at the points

M

gence) to the length of

the

M ^MN^\s

arc

K=

iim

As \\hore

a

point

M

is

the angle between

circle

line (/C

the

*

o

AS

when

M

.V

(Fig.

35),

that

is,

= ^,

positive

*

rfs

directions

of

the

tangent jt the

arid the .v-axis.

curvature,

The

Au

= 0)

f

of

curvature

R

is

the reciprocal

of

the

the absolute value of the

e.,

K=

,

where a

is

the

are lines of constant curvature.

of

circle)

and the straight

Sec

an Arc. Curvature

Differential of

5]

103

We have the following formulas for computing the curvature in rectangular coordinates (accurate to within the sign): 1) if the curve is given by an equation explicitly, y f(x), then

2)

if

the curve

given by an equation implicitly, F(x,

is

0,

then

represented by equations in parametric form,

*=

Fxx.

Fx

Fxy '

Flx

y)

F

F'yy

'y

Fv

F'

x

3

/j

3) /

if

(/),

\j)

the curve

is

then f

y y

*,'"

x

^

,

where dx

dy -

^

'

~~dt* In polar coordinates,

when the curve

is

~

given by the

equation

/(q)),

we have r

z

+ 2r'

2

rr"

where ,

r

= dr

d"r

.

and

-

2

dcp

.

dtp

3. Circle of curvature. The circle of curvature (or osculating circle) of a is the curve at the point limiting position of a circle drawn through v and two other points of the curve, P and Q, as P and Q M. The radius of the circle of curvature is equal to the radius of curvature, and the centre of the circle of curvature (the centre of curvature) lies on the in the direction of concavity of normal to the curve drawn at the point

M

>

M

M

M

the curve.

The coordinates X and Y computed from the formulas

X=x-

the centre of

of

L,,

,

-

curvature

-jf-r

of

the

curve

are

.

{j

The evolute

of a

curve

is

the locus

of

the

centres

of

curvature

of

the

curve. If

ture

in the

formulas

for

we regard X and Y

determining the coordinates as the current coordinates

of

of the centre of

curvapoint of the evoof the evolute \vith

a

then these formulas yield parametric equations parameter x or y (or /, if the curve itself is represented by equations parametric form) z Example 1. Find the equation of the evolute of the parabola // x

lute,

.

in

Extrema and the Geometric Applications

104

Solution.

X=

4* 8

1

Y--

,

+ 6*

2

involute

curve

a

of

curve

a

is

the parameter x,

Eliminating

Y

the equation of the evolute in explicit form,

The

[Ch. 3

a Derivative

of

o'

we

+ ^lT")

which the given curve

for

find

an

is

evolute.

MC

The normal

CC

length of the arc

involute P 2

of the

of the evolute

l

is

a

is

the evolute

to

tangent

equal to the corresponding

P,;

the

increment

the radius of curvature CC, in AfC; M,C, that is why the involute P 2 is also called the evolvent of the curve P, obtained by unwinding a taut thread wound onto P, (Fig. 36). To each evolute there corresponds an infinitude of invowhich are related to different initial lutes, lengths of thread. 4. Vertices of a curve. The vertex of a curve of the curve at which the curvature is a point has a maximum or a minimum. To determine the vertices of a curve, we form the expression of the curvature K and find its extremal points. In place of the curvature K we can take the

R

curvature

of

I

points

Since

// J

= sinh a

1

a cosh

, 2

and, hence,

x

J

*= of

is

2

X .

a

0).

=

cosh a

We

follows

it

that

tf

=

a

X

rf/?

have -j- = sinh2 dx a

.

Equating M 6

y

Q,

the point

we

the

A

we

get sinh 2

Computing

(0,

get

we

whence

0,

a

-r-y-

,=

2

A:

cosh2

catenary.

The vertex

=

a a curvature (or of

of

the

catenary

the

sole

putting

into

find

and

the second derivative

a point of the radius of

minimum

the

curvature) of

thus,

and J(/"

= acosh

to zero,

=Q

point *

value x

the

/?

>

(a

I")

ax

it

extremal

a

the derivative -j critical

its

I

the computations are simpler in this case. 2. Find the vertex of the catenary

a cosh

y

.

^

Example

36

Solution.

if

and seek

7-7^

> 0.

Therefore,

the

maximum

f/

= acosh

is,

a).

Find the differential of the arc, and also the cosine and sine of the angle formed, with the positive ^-direction, by the tangent to each of the following curves: 993. *

2

2 */

=a

2

(circle).

+ ^-=l (ellipse). y* = 2px (parabola).

994. ~2

995

+

Sec. 5]

_

996. x 2 / 8

997. 998. 999.

-f

2/

t/

Differential of an Arc. Curvature

=a

2 /'

(astroid).

_

105

y = acosh (catenary). x = a(ts\nt)\ y = a(lcost) (cycloid). x = acos*t, y = asm*t (astroid).

Find the differential of the arc, and also the cosine or sine of the angle formed by the radius vector and the tangent to each of the following curves: 1000. r^atp (spiral of Archimedes). 1001. 1002.

= (hyperbolic spiral). r = asec*-|- (parabola). r

1003. r

= acos*-

(cardioid).

1004. r=za.v (logarithmic spiral). 2 a 1005. r a cos2q) (lemniscate). Compute the curvature of the given

=

points: 1006. y 1007. x*

= x* 4x* ISA' + xy + y* = 3 at 2

+ = =

1008.

=1

at the

coordinate origin.

the point

A

the vertices

at

curves at the indicated

(1,

1).

and 5(0,

0)

(a,

b).

=

*' at the point (1, 1009. * /*, f/ 1). 2 2 2a eos2q> at the vertices cp 1010. r 2 1011. At what point of the parabola t/

=

and

= 8x

is

= n. the curvature

equal to 0.12S? 1

1012. Find the vertex of the curve y^-e* Find the radii of curvature (at any point) of the given x* (cubic parabola). 1013. y .

lines:

=

1014.

5 + S =1

(ellipse).

= -!^. * = acos y = as\n*t (astroid). = a(cosM sin 0; y = a(s\nt

1015. * 1016. 1017. circle).

8

/;

/

A:

/?osO

a

curvature of

the

y* = 2px.

1021. Prove

y = acosh

of

=

1018. r ae k v (logarithmic spiral). 1019. r- a(l -f-coscp) (cardioid). 1020. Find the least value of the radius of

parabola

involute

that

the

is equal to a

segment

curvature

coordinates of the centre of Compute at the indicated points: curves given the

of

the

catenary

of the normal.

curvature of

the

Extrema and the Geometric Applications

106

1022. xy=l at the point (1, 1). x* at the point (a, a). 1023. ay* Write the equations of the circles curves at the indicated points:

of

a Derivative

[Ch. 3

=

= =

x* 1024. y Gjc+10 at the point 1025. y e* at the point (0, 1). Find the evolutes of the curves: 1026. y* 2px (parabola).

1027.

= J + g=l

of

(3,

curvature of the given 1).

(ellipse).

1028. Prove that the evolute of the cycloid

x~a(t is

sin/),

y = a(l

cost)

a displaced cycloid.

1029. Prove that the evolute of the logarithmic spiral r is

also a

1030.

logarithmic spiral with the same pole. that the curve (the involute of a

Show x

is

= a (cos + /

/

sin

the involute of the circle

/),

;c

#=-a(sin

= acob/;

//

/

circle)

/cos

= asm/.

/)

Chapter IV

INDEFINITE INTEGRALS

Sec.

Direct Integration

1.

1. Basic rules of integration. 1) If F' (A-)--- MA), then

where C 2) 3)

4)

an arbitrary constant. f f (x) dx, where A

is

^Af(x)dx=-A [ft

\

U'H:

f f

If

f2

(

v )l

dv

-

( fj

(*)/*

i

and

(x)dx-~F(\-) -f-C

is

f f2

/-cf

a constant quantity.

O)

dx.

(v),

then

In particular,

r

i

ax

a

J

2. Table

II.

of standard

=

\

dA'

f*

III

.

,*r

IV

==

-r-:

\ 2 s J X -f- a dx C \ )

-=2

;5

a*

A'

dx r

V.

integrals.

dx

1

a

\:

arctan

a

I

r

~-,T-

2a

a+v

|-C

=

v:

1

arc cot

+C

(a

-f-C

(a*Q).

yS-

0).

(a^O).

VI.

VII.

a

(>0); fc x d* =

+C '

(a

^ 0). '

__

108

VIII.

IX.

\ f

X

cosx +

sinxdA;=

_

[Ch. 4

Indefinite Integrals

C.

cosxdx=sinx + C. COS 2

XI.

J XII

--= x sin

cotx+C.

2

'

tan

smx

XIII.

XIV.

x

H-C = In

|

cosec x

cot x |-f-C.

cosx f

sinhxdx=coshx-|-C.

XV. dx

XVL XVIL Example f (ax

2

1.

+ bx + c) dx=

f

Applying the basic rules tion,

1,

2,

3 and the formulas of integra-

find the following integrals:

5dVd*.

1031.

1040.

1032-5(6^ + 8^3)^. 1

x

033.

(x

r 1

034.

J (a

1035.

-i

a) (x 4 b) dx.

+ bx

J

)* clx.

yZpxdx.

-

(V

1042 ' U4 ^'

3

_

-- dxv

V x)

j/51

1043.

1036. '~ n

T

1037. J

(nx)

"

1045

^x '

dx.

}

1046

'

1047

'

1038.

1039.

^+ljc-/x-dx.

f

^4+^

'

,,

'

Sec

_

Direct Integration

1]

2

tan *dx;

1048*. a)

109

1049. a)

2

Jtanh *d*.

b)

b)

1050.

Jcoth'jtdx.

$3Vdx. differential. Rule 4 considerably sign expands the table of standard integrals: by virtue of this rule the table of integrals holds true irrespective of whether the variable of integration is an independent variable or a differentiate function. 3. Integration under Example of the the 2. 2 where we put u Exa m p.e We 5* We 2 took advantage of Rule 4 and tabular integral xdx 1 d(x f 1. 2 1 ) 3. 2 implied u Example 2. 4. jc and use was made , ( x 2 e xl} dx-^~ ( e*' d 3 (jc ) of - Rule 4 and tabular i-e^ + C by integral V. virtue of Rule 4 and tabular integral VII. In examples 2, 3, and form before making use of of we reduced the given / ( \ This type 4 a tabular dx\ transformation is Some common transformations ples 2 and a) 3, I (u) the following du t where a = (p (x). called integration under the differential sign. which were used of differentials, b) (a ^ 0); b) xdx = Using the basic rules and formulas ,053. to in Exam- are: dx=^d(ax-\ tegrals: integral integral: of 2 ^d(x ) and so on. integration, find the following in- 112 Indefinite Integrals 1140. f-^f-. sinh x 1143. H41. f-*_. J cosh x 1144. 1142 /f \ . ' Find the indefinite < 1 Ad 46. ,,47. integrals: (x 1/5=1? dx. i v* (cothxdx. J v J sinh x cosh x 1145. [tanhxdx. J J r* [C/i. 1163. A \f I J?=CT1 f-^-. ][email protected] i *J " M -j f^d,. 1165 1148. \xe-**dx. 1151. ,, 67 . Jl_!d*. -- i 1168. i/^c' V e J ,- jcdx JV^L l \ J I 1169. ~ cc^s sinx sin A: -f- 1 Jx + cos* 1 i ftan/JC 1166. U50. djt x ' - - A* cos x * dx. 2 51 f v "-^) v ^ sin SA: 1170. J ^tan^-2 > ' ff 2 -|--^- )-^-. 2x*+l/ 2x*+l ^ ,) 1157. ^a ilax cosxdx. 7^Ti 1161. " J M7.:. !!!_!-. + !' ^ ^ to f fl _t ^ H76. ! K4-tan'x * (< jsin -Jdx. ( .rfy. [ 1174. Jlan'aArdx J 3< 5 1173. dAt> f-i^=. 1160. I179 1172. j -j^* . ii77. f _ si J sinajtcosax' 4 Integration by Substitution Sec. 2] J 113 l/T=3 Sec. 2. Integration by Substitution 1. Change of variable in an indefinite integral. Putting where t is a new variable and cp is a continuously differentiable function, we will have: The attempt is made to choose the function q> in such side of (1) becomes more convenient for integration. Example 1. Find Solution. It is natural to put t = V~x 1, whence A- a way =/ 2 -}- 1 that the right and dx = 2tdt. Hencu, Sometimes substitutions of the form are used. Suppose we succeeded in transforming the integrand f(x)dx to the form f(x)dx=g(u)du where u = q>(jc). t _ 114 If [g(u)du known, that is _ Indefinite Integrals is, [Ch. 4 then we have already made use oi this method in Sec. 1,3. Examples 2, 3, 4 (Sec. 1) may be solved as follows: Actually, Example 2. u = 5# cfw 2; 1 Example 3. u = x*; Example 4. w = jc s ; du du = 5c(x; _ -~ d 2 ' +c_ 1 dx\ 2 , xdx = - 2xdx; = 3x -du. dx 2 x dx . = . - 2. Trigonometric 1) If ~a sin 2) /; If _ substitutions. an integral contains the radical ]fa z x2 the usual thin^ , is to put whence an integral the contains radical V* a2 2 , we put xawct, whence /^x 3) If 2 a2 = a tan an integral contains the radical V'V ^. +a 2 , we put* = atan/; whence It should be noted that trigonometric substitutions do not always turn out to be advantageous. It is sometimes more convenient to make use of hyperbolic substitutions, which are similar to trigonometric substitutions (see Example 1209). For more details about trigonometric Sec. 9. Example 5. Find i ^ and hyperbolic substitutions, dx. see Integration by Substitution Sec. 2] xlant. Solution. Put f J y~x*+l * f V~ian 2 _ 1? dt "~ f J = In sin | / tan 2 J 2 / tan / + / + sec r-r cos 2 cos' t , / i dt . cos ! si " 2 sin J / f sec dt 1 / ~~ f cos dx= Therefore, 1J5 sin 2 J * 4- cos2 2 /-cos/ -J 1 < M-~ * + C = In cos 2 / _ / 1 ^/ ^ J cos tan f / -{- | _ cos/ ~~ , 2 J sin / V\ / -j-tan 2 / 1 tan< find the following 1191. Applying the indicated substitutions, integrals: c) f e) COS A' d* \ J Applying suitable substitutions, 1192. S*(2x+5)"djc. 1 1193. (' J 1194. + *..d*. the following integrals: find >97. n- csinA )" 1198. l+^A- f- J! JxK2t+l ,, 99< ** 1195. r . Applying trigonometric substitutions, find the tegrals: ,201. ('-=*. *' K J ,202. 1203. l -=. 1204*. f J iZEl'dx. ^ following in- __ U6 1205. \Ch. 4 Indefinite Integrals 1206*. (f^+idx x J f ff 2 J x y 4 __ x2 1207. 1208. Evaluate the integral by means r dx J /*(!-*) x=sin of the substitution 2 /. 1209. Find by applying the hyperbolic substitution x = as\n\\t. Solution. We have: ]Aa 2 +x = 2 ]/~a 2 +a suih 2 2 /=a cosh ^ and dx=a cosh /d/. Whence \ ya 2 z -\-x dx= ^ a cosh t-a cosh f ctf = ~~~2~ Since __ x ~ a and we finally get where C 1=^C In a is a new arbitrary constant. 1210. Find C putting x } = a cosh/. Sec. 3. Integration by Parts A formula for integration on tiable functions, then \ by parts. udv = uv H If ( = vdu. and u = i|)(*)are differen- Sec. 3\ _ _ Integration by Parts Example Find 1. \ In*, dv Putting u H7 x In xdx. dx xdx we have da x* 2 v , % Whence ~~9 x* dx Sometimes, to reduce a given integral to tabular form, one has to apply the fcrmula of integration by parts several times. In certain cases, integration by parts yields an equation from which the desired integral is determined. Example 2. Find \ We V e* cos x dx. have = e* cos x dx \ e xd e* sin x (sin x) + \ e \ e x sin dx AT x d (cos x) e x e* sin x sin -\- x e + x cos x \ e* cos x dx. Hence, e* cos x dx \ e x sm .v-j-e^cos x V e x cos x dx, whence cos v dx -- ~ (sin x -f- cos Applying the formula of integration by .v) -f C. parts, find the following integrals: 1211. ^ 1212. \nxdx. Jarclanjcdx. 1213. Jarcsin 1214. A-rfjc. Jjcsiiucr/.v. 1215. Jjccos3A'Jx. 1216. Urfjc. 1219*. 1220*. x sin x cos x d\ a 1222*$

1223.

(jt

+5x+6)cos

x*

\nxdx.

^

1224.

Jln'xd*.

1225.

(^djc.

f^d*. K A

J

Jx-2-*rfx.

1218**.

{

1226.

J

1217.

1221.

JjV'd*.

1227.

1228.

Jjcarctanjcdjc.

Jjcarcsmxdjc.

2

(x ^2A' + 5)^*dA:.

'-

1229.

1230

'

\n (x

+ V T~x*) dx.

118

Indefinite Integrals

1231

1234

-

1232. 1233.

[C/i.

eax *

'

\

JisFr**'235.

mb * dx

4

-

$sin(lnx)dx. Je*sinxdx.$3*cosjtdx.

Applying various methods,

find the following integrals:

2

1236. (x*e~* dx.

1246.

V\-x

J

1237.

v

\e 2

1238. 1239.

(x

*dx.

1247.

-2x + 3)\nxdx.

1248.

2

jttan 2*d;t.

f

2

^x\n~dx.

1249.

1240.

f^dx.

1250**,

1241.

fllL^d*.

1251*.

cos (In x)dx.

J

1242.

1243. 1

244.

*n/ir

1245.

2

f

{ j

arctan3jcdA:.

jc

x

(arc tan

( (arc sin

fare

sin

\

^2

J

5

x

2 A:) 2

jc)

a2)2

2 :

-|

'

1252*.

dx.

1253*.

dx.

\

1//1

^ j

254*.

f

z

-4-

-4^-

x dx. .

,

^.

Sec. 4. Standard Integrals Containing a Quadratic Trinomial

1. Integrals of the form f

mx + n

\

2-7-7

.

7dx.

J

The

principal calculation procedure the form

ax z

is

to reduce

+ bx + c = a(x-}-k)

the

z

(1)

-{-l,

/ are constants. To perform the transformations in (1), it is best to take the perfect square out of the quadratic trinomial. The following substitution may also be used:

where k and

If m=0, then, reducing the quadratic trinomial to the form the tabular integrals III or IV (see Table).

(1),

we

get

Sec

Standard Integrals Containing a Quadratic Trinomial

4]

Example

119-

1.

_

dx

1

-(-*)

H-

g \2 5

-7

2

Qi

.

arc =-o*~7= -o 2

tan

(-TJ+S '*?

4

2

"S?+c. If m&Q, then from the numerator out of the quadratic trinomial

the derivative

-+.

f J

we can take

a.v

2

+^

and thus we arrive at the integral discussed above.

Example

f J

-> ;5=^n

2.

,v d -

i

f"

^J

-

A

'-x-i

IV 2.

-

Integrals of the

5

form

2x-l

are similar to those analyzed above. 0, and VI, if a lar integral V, if a

>

Example

r

J

The

The methods

integral

of calculation

finally reduced

is

< 0.

3.

dx

Example

d*.

I

+

1

f*

dx

1

4.

^+3 ^^2 f *" yV+2x + 2 J ^ ^ + 2x4-2

JC ^^

.

2

4jt

3

to

tabu-

J20

[Ch. 4

Indefinite Integrals

*

3.

of

Integrals

the

form

f J (mx

+ n)

V^ax'

=

+ bx + c

.

By means

of the in-

verse substitution i

mx+n

=

/

these integrals are reduced to integrals of the form

Example

5.

2.

Find dx

We

Solution.

put

whence

4.

Integrals of the form

\

]T ax*-\-bx

+ cdv.

By taking

the perfect square

out of the quadratic trinomial, the given integral is reduced following two basic integrals (sec examples 1252 and 1253): 1)

('

J

V

a*

A'

2

dx

= 4 fa^x* + ^ arc sin ^ 2 (a

2)

J

>

to

+ C; c/

0);

Vl?

Example

6.

sin

Find the following

1256.

integrals:

S' '',c 00

P

J

xdx

^7^+

one

of the

Integration of Rational Functions

Sec. 5]

oKn \o\J.

i

j

---

i

<

-

c r\

<

121

dx

I

\

1270. v2

dy

1

O *7 1

*~

J

1272.

'

2* 2

V~2 {-3* d* r I

J

X

'

f* t

1273.

X 1

1-7/1

J f

/"*l/"o

f~=.

,H4. 1265.

1

1266.

V

[

J

^

>1J -

T

1267.

xdx

V %

J

x

5

-

."~

-,-^ /5, -2,

dx.

1277 197 n 1278

dK.

2

f-1

JYKI

2

,

-v

279

x cJJt

sin

x

dA

(

J

1'-^=.

1268.

g

t'

T^PTTT^TlTInjfdi

i'

J x

^ _ ilnA._ !n 1

m

2

K

Sec. 5. Integration of Rational Functions

t. The method of undetermined coefficients. Integration of a rational function, after taking out the whole part, reduces to integration of the proper rational fraction

where P rator

P

and Q

(x)

is

(x)

(A-) are integral polynomials, and the degree lower than that of the denominator Q (A-).

If

Q(jr)

= (*

a)*.

.

of

the

nume-

.(A'-/)\

/ are real where a, distinct roots of the polynomial Q (x), and a, .... ., K are natural numbers (root multiplicities), then decomposition of (1) into .

.

partial fractions

is

justified:

^ calculate the undetermined coefficients A lt A 2t ..., both sides of the identity (2) are reduced to an integral form, and then the coefficients of like powers of the variable x are equated (llrst method). These coefficients may likewise be determined by putting [in equation (2) or in an equivalent equation] x equal to suitably chosen numbers (second method).

To

122

Indefinite Integrals

Example

4

Find

1.

xdx (*-!)(* +

We

Solution.

[C/t.

"

2

1)

have:

Whence t

a) F/rsf

method of determining the

We

coefficients.

the form x^(A-{- B^ x 2 -{-(2A-{- B 2 )x-\-(A ents of identical powers of x we get:

B

B2

l

)

(x\).

(3)

rewrite identity (3) in Equating the coeffici-

t

Whence

=i

,

,

= _i

Second method of determining the

b) <3),

B

;

we

,,4.

:

1=4-4, Putting x

we

1,

4

i.e.,

=

'/ 4

x=\

in identity

.

get: 1

Further, putting *

Putting

coefficients.

will have:

= 0,

we

=

2

B2 =

i.e.,

-2,

l

/2.

will have:

Hence,

TJ *

\

~~ Av

1

1

T t

4-1

f

\ J Ax

AT

Example

Solution.

2.

1

Find

We

have: "" 2

x(x tind 1

When

=A

2

(*

I)

x

x

I)

+ Bx (x

1)

1(JC

1)*

+ Cx.

(4)

advisable to combine the two methods of determining coefficients. Applying the second method, we put * in identity (4). We get 1=4. Then, putting jc=l, we get 1=C. Further, app2 lying the first method, we equate the coefficients of x in identity (4), and get-

Hence,

solving this example

it

is

= 4 + 0, 4=

1.

fl-=

=

i.e., l,

B= and

1.

C=l.

In t egration

Sec. 5]

Rational Functions

of

123

Consequently,

/=

f dx

dx

\

p

,

\

'

J

the polynomial

If

dx

r- = ln jc17+ J (*l)

f*

\ J X

Q

1

,

2

;

,

,

'

JC

,

,

'

A-

ib of

~

r-f C.

1

'

has complex roots a

(x)

,

,

In

JC '

1

multiplicity k, then

partial fractions of the form (5)

will enter into the

expansion

Here,

(2).

Bk are undetermined coeflicients which are determined and A lt B lt .., A k by the methods given above For k~\, the fraction (5) is integrated direct,

ly;

k>\,

for

use

is

of the reduction

~]

q

Example

and make the substitution 3.

method;

+

z sable to represent the quadratic trinomial x

here,

it

px~{-q in the form

A--J-

=

z.

j2

^

is

(

first

x-\-~

\

-f-

Find

Solution. Since A

then, putting x -\-2---z, r==

r

*\._ dz=z

jrc

wo

2 -|

4x

got 2

r

_j_^

tan ?

Hit i!ini

r

-

-

2. The Ostrogradsky method.

is

(A:)

derivative Q'

X

and

Y

the greatest

-- -- arc

-

Q

If

P(x)

where Q,

5-

i

z=

(A)

has multiple roots, then

"-

p

A'(v)

common

tan

r

(.Y)

A

divisor of the polynomial

(6)

Q

(x)

and

its

(A-);

undetermined coefficients, whose degrees by unity than those of Q, (A-) and Q 2 (x). The undetermined coeflicients of the polynomials X (x) and Y (x) are computed by differentiating the identity (6). Example 4. Find (A-)

(x)

are polynomials with

arc, respectively,

less

C }

dx

U'-

124

[Ch. 4

Indefinite Integrals

Solution.

Ax z + Bx + C

dx

we

Differentiating this identity,

?Dx

.

2

-\-Ex

+ F ax .

get

Dx z +Ex

\=(2Ax-\-B)(x*

3x*(Ax* + Bx

1)

+ C) + (Dx* + Ex + F)(x*

Equating the coefficients of the respective degrees of x,

4 = 0;

D = 0; E

F

D + 3C = 0;

2fl--=0;

we

+ 24 = 0;

B

i).

have:

will

+ F==

1;

whence

C = 0; D = 0; ~; O

5= and, consequently,

C 9

^ ~ _

To compute the r

-^

into partial

8

1)

on the

integral

__ 2

x

1

3^ -l

2

(x

= 0; dx

P

3 J x8

right of (7),

F= -4O 1

we decompose

the

fraction

fractions: 1

x8 lhat

is, 1

Putting

#=1, we

= L (x + + 1) + MX (x 2

A:

1)

+N

(A:

(8)

1).

L=-.

get

<j

Equating the coefficients of identical degrees of x on the ,of

we

(8),

find

right

and

.

or

Therefore,

r

_

dx

\yS J *

i

p

^\V

o J x

1

i

dx

r \_

1

1

Q\

J j

~~ =ll

3

]X

11-1 '

6

and

^^^^^(J^\\ + ^ ln \ x Find the following

1280

'

^ +^^

r

l

integrals: '

1282

'

mian

^^+ C

'

left

Integrating Certain Irrational Functions

Sec. 6]

1284 ' **

C

J

1285

'

5 *' + 2 * - 5* + 4* dx*

12Q1 29J '

f j (Ii_4jt

-

125

dx

+ 3)<J

dx c J *(*

+ !)*

-~

d* C

1296.

-r-g^gdx. ^"* " I

R V2

C*

(

Av

t

C/

Q

rt

^4

.--.

,

*

I

I

J

'

.

1288.

-290.

.Tjpfc

l291 '

J^Sf*.

'292.

J^d*.

1299. .300.

Applying

l306

different procedures, find

integrals:

the integrals:

'

I3 7 '

l309

find the following

l312

rf *-

'

-

Sec. 6. Intagrating Certain Irrational Functions

1.

Integrals of the

f.rm i

/? <s

a rational function and p lt q v p a

,

q z are whole numbers.

12J

[Ch. 4

Indefinite Integrals

Integrals of form (1) are found by the substitution

where n is the least Example 1. Find

common

multiple of the numbers

q2

,

...

* dx a I

Solution.

f

_

The substitution

____dx

2,x

f 2z'dz

J V"5jc=T-- J/2?=l""j

z

2

2

1

-^ =2

4

to

an integral

of the

form

r^i " 2

1

J

=2 = (1+

2

/23T-i)

+

Find the integrals: 13 15.

f-^^djc. /*-! A

r

1316.

('J^Zdjc.

1321.

J .v-f-2

J

dx

c

1322.

7,7==.

J

1U7. .

^.---

-. ---('-. J

/v+l

.

I-

--^7= A (jv

(o

-A)

V

1-

x

1323.

.

v+1

)/(v+l) 1324.

1325.

,320.

2.

Integrals of the form "

r

where P n Put

(x)

is

ai*

+

- dx,

t,

x

(2)

+c

a polynomial of degree n

dx

a polynomial of degree (n with undetermined coeffi1) number. The coefficients of the polynomial Q n -i(x) and the number K are found

where Q n _i(x) cients and A, is

is

a

by differentiating identity

(3).

Integrating Certain Irrational Functions

Sec. 6]

Example

2.

Whence

Multiplying by V^* we obtain

2

+4

and equating the coefficients

of identical

degrees oi

A-,

=

-

;

D-0;

X=_

Hence, -

3. Integrals

of the

form dx

i;

a)

n

V ax

are reduced to integrals of the form (2) by the

They

A

substitution:

a

Find the integrals: 1326

.

1327.

j__^_. (-f^dx.

1329.

1330.

v

-^ fv

-^^rfjc.

1331. FA'

Vl-i

4.

2

X+l

Integrals of the binomial differentials

x

(5)

t

where m, n and p are rational numbers. Chebyshev's conditions. The integral (5) can be expressed in terms of a finite combination of elementary functions only in the following three cases: 1) if p is a whole number; "

2)

=z

s t

is

if

where

m 3)

a

n

if

s

is

whole number. Here, we make the substitution a + bx n =

the denominator of the fraction p;

+p

is

a

whole number. Here, use

is

of the substitution

128

(Ch. 4

Indefinite Integrals

Example

3.

Find

m=-^;n = -r ;p = -^;

x.

,

Solution. Here,

we have here Case 2 The substitution

yields *

= (z

4

3

I)

\Y dz

(z*

y

e

1

Find the

+ J/T

f (2

z

T

1335

e

),fe=J2ef_

dx

'

^-

T

133G 0. f 2

J

* f_-

,334.

Hence,

.

~

1333. r

=2.

integrals: *

1332.

:

Therefore,

= 12 where

2

integrability.

dx=\2z*

;

-1+1 -T

=

.v:

i'

337.{

.

J

(2

^

+ ,

3

.

3

3 JC

)

r*

y

1+

Sec. 7. Integrating Trigonometric Functions

1. Integrals of the form :

where 1)

m If

and n are

an odd positive number, then we put

is '

We

n

\

do the same Example 1.

if

0)

integers.

m^2/j+l ,

'm..

s l(]Zk

n

( sin 10

x cos " xd ( cos

is

^)

\

cos 2 x) k cos" xd (cos

( 1

an odd positive number.

x cos 8 * dx

=

f sin 10 x (1 sin

11

x

sin

2

sin x) 13 .v

d

(sin *)

=

A:).

Sec.

7]

_

__

Integrating Trigonometric Functions

the integrand 2) If m and n are even positive numbers, then formed by means of the formulas sin

*= y (1

2

sin JCGOS

Example

= =

2

If

2

3x

1

3-

4

cos 6* dx

cos 12* n ^

1

\

=

3* dx

--

f /

J \ x

m=

sin

-- -

sin 6x1 p \

^>

3)

f cos

2.

cos

cos 2*),

,

=

= sin .

, 2

x=

2

*= y (1 + cos 2*),

3* sin 3*) 2

sin

2

3x dx

=

'

f \

c

GA:

,

,

Zc

.

2

sin

(sin 6A:

c \ dx cos 6^ ,

2C 6x cos fi o^) ax v

2

.

=

=

/

sin 12AC

and

[i

n=

v are integral negative

numbers

of

parity, then

_

C

dx cos

J

v

In particular, the following integrals reduce to this case:

Example

3.

f

2

-^-= JC sec xd (tan *)= Jf (l+tan*x) = tan x-p--=-o tan '* + C.

J COS X

8

d (tan

x)

jc

_1 r ""-

tan 2

5-1900

(I) is trans-

sin 2*.

f (cos 1

129

-

~

identical

130

form

4) Integrals of the

(or, respectively, cot*

Example tan

1

x

=

f tan

5.

P.

x

= cosec* x

tan*xdx=

\ .

tv

(or

cot w xdx),

V

where

m

is

an

in-

1).

l)dx=

6.

=

tan*xdx =

2

V tan x (sec*x

tan* x

.

(sec*x

\

w xdx

number, are evaluated by the formula

tegral positive

=

[Ch. 4

Indefinite Integrals

-5

1)

\

dx=-^-^ ~

tanx + x+C. ,

,

,

3 J 5) In the general case, integrals 7 m>n of the form (1) are evaluated by means of reduction formulas that are usually derived by integration by parts. 5

Example

\

J cos* x

sin

=

\

cos

J

dx

A;

1

1

C cos x

2^^"j -^

f

. ,

cos x

c

sin

x

Find the integrals: 1338. 1339. 1

340.

1341.

1352.

Jcos'xdx.

J

C \

sin sin

2

sin*

x

8

-jr

*

cos

5

1346.

'T

^

-~-

*

dx.

x

J

sm*xcos*xdx. sin

4

1 A:

cos x dx.

Jcos'Sxdx.

1353.

x

1343.

345.

-2

8

x cos x dx.

-

1342.

1

COS

sin

Jsin'xdx.

J

1344.

{;

J

1354. J

1355.

JsecMxdx.

1356.

Jtan

1357.

!

5jcdA:.

jjcot'*d*.

1358. 1359.

1360.

' COS X

1362. 1363.

1364.

J

sin'xi/cosxdx.

f J

V sin x cos*

dx .

f-==. J l^tan*

*

_

_

Integrating Trigonometric Functions

Sec. 7]

2. V

cos

Integrals of the form

mx cos

1)

sin

2) sin

3)

mx

sin

V

cos

nxdx,

sin

\

131

rnx

sin

nx dx and

nx dx. In these cases the following formulas are used;

mx cos nx = -~ mx

sin

nx

cosm*cos ^^

Example

7.

I

= -^

= sin

-9-

(m + n) x + sin (m

[sin

(m

[cos

[

c

s

9* sin

x

n)

n) x] ;

cos(m

+ n) x]\

n)*-fcos (m-f/i)

(m

xdx=

8* -^ [cos

\

x].

cos

lOjt]

dx=*

Find the integrals: 1365.

^

1366.

J

sin

cosSxdx.

sin 10* sin 15

1367.

f cos

1368.

f sin

3.

3jc

~ cos |-

sin

A:

1370.

d*.

djc,

1371.

~ djc.

1372.

^-

J cos(aA:

sin

$a>/ f sin (co/ -f-cp) cos x cos* 3x dx ^ sin x sin 2* sin J dt. . 3* dx. Integrals of the form f where R 1) 1369. is /? cos x) dx, (sin *, (2) a rational function. By means of substitution x . tan t, whence 2* integrals of new variable Example form (2) are reduced t. 8. Find f J Solution. f* J Putting to integrals __ *E 1 + sin x + cos tan -pr^*' we wil1 x have of rational /. functions by the 132 we have 2) If the identity R( Ihen we can use 4 [Ch. Indefinite Integrals sin*, the cos*) s/? * tan substitution cos*), (sin*, = to reduce the / integral (2) to a rational form. Here, sin jg = COS*= t ,- and *=arc Example tan/, dx <3 > Jnn-'- tan* = f, sin 2 * = t 2 dt dx = ^, :j -jz have will dt dt p * note denominator that r 1 the integral + C = -7=2 arc tan F (3) it is ( /T tan *) + evaluated faster divided by cos 2 *. is useful to apply C. the numerator and if procedures (see, for artificial 1379). Find the 1373 2~) of the fraction are first In individual cases, example, /" (/ 2 K = i = -^=arctan We . 2 Find 9. Solution. Putting we . ' 1374. J 3 integrals: + 5*cos* C^ 1382 *' ' *!. J sin * + cos * cos* . , dx J 3 sin 2 2 J sin * iooyi* 1384 *1 385. -|- P \ . (1 ^^. T I39 ' X r-. dx. 3 cos*) Sin2x <*- P COS jj , ^ - p-j-^-rfx S ' 1389*. cos 2 * 3sin *cos * si n T J. \ ' dx P 1386. 1388 * JiHiJT 5sln*cos* J 100T 1387. 2 ** 1383*. f 10 or mi-. + 5 cos * Is. n *-6 sin* + 5 ^ ^ ' sin*) (3 sin*) o-.j |;;;;;i::;>. ' Integration of Hyperbolic Functions Sec. 8] 133 Sec. 8. Integration of Hyperbolic Functions Integration of hyperbolic functions is completely analogous to the integration of trigonometric functions. The following basic formulas should be remembered: sinh 2 cosh 2 X 1) 2) sinh 2 3) cosh 2 *=l; *= ~ (cosh 2* x= Y (cosh 2x+ 4) sinh x cosh x Example 1. 1); 1); = ~ sinh 2x. Find cosh 2 *djt. V We Solution. have 2 f cosh x djc-=r~(cosh 2x + \)dx Example 2. ( cosh 8 We xcU= f jsinh 2x + ^ Find \ Solution. = cosh 3 ****. have cosh 2 xd (sinh x)= J (l + sinh 2 Jc)d(sinhA:) = = smh x-\ . , , Find the 1391. integrals: 1397. Jsinh'jtd*. 1392. 1398. ^cosh'xdx. 1393. sinh'* cosh* 1394. 1396 Sec. 9. Integrals C 2 J sinh . Using of the ^ A: cosh 2 1402. . A; Trigonometric and Hyperbolio Substitutions for Finding Form (1) R is a rational function. 134 Indefinite Integrals _ [CH. 4 the quadratic trinomial ax* + bx-\-c into a sum or difference Transfor Transforming integral (1) becomes reducible to one of the following types t of squares, the of integrals: 2) R (z, 3) R (z, The 1) 2 2) 2 3) 2 m* + Vz 2 z*) dz\ m z latter integrals = msin/ = mtan/ = msec/ Example 1. or z or 2 or z ) dz. are, taken respectively, by means = m tanh = msinh/, = m cosh/. /, Find dx ==/. (x+\Y Solution. Putting We have *+l = tan2, we then have dx sec 2 dx = f tan \ Example Solution. Putting 2. Find We have - 2 zdz = sec*zdz = x zsec2~"J and ~ dz = f cos 2 . of substitutions: Sec. 11] we finally Using Reduction Formulas 135 have Find the integrals: 1403. S/3-2jt-x'd*. 1409. J W-6*-7d*. " 1404. 1410. 1405. 1406. Sec. 10. Integration of Various Transcendental Functions Find the integrals: J (* 1416.$*

1417. 1418.

1419.

1420.

+ I)

2

1415.

2

]

cos

2

e

2x

1421.

dx.

2

1422.

SJCC/A;.

x s\nxcos2xdx.

^

(*-/= J

1423.

f^lnj^djc. *

J 2

$e *sin x \^e jcdjt. 1424. smxsmSxdx. 1425. x ^ A'e 2 cos x dx. 1426. I 1 J J ? In (x jc + /FT?) djc. arc cos (5jc 5 smxsinhxdx. Sec. 11. Using Reduction Formulas Derive the reduction formulas for the following integrals: 1427. /..Jj-^,,; 1428. / =$

find /.

and

sln"xdje; find / 4 and

/,.

/,.

136

[Ch. 4

Indefinite Integrals

/w

=

1430. 7 n

=

1429.

Sec.

J^h;;

find

(x n e~*dx\

I*

find

and / 10

7 4-

.

12. Miscellaneous Examples on Integration

dx

1448.

xdx f

JO+jt x

5

dx

J *'-2*+2 1433.

P

T

J

)

xdx

1449.

C

..

2^ j

j y^i

1450.

dx.

KI-J

1

^

r

+1

x

dx

t

2 '

dv

dx

-

dx

1452.

dx

1453.

dx

P '

J (x* r

+ 2)

i

f-7^ A

'

t/

dx

y l/^v2 * r

-

1455.

Jx-2A' +

l

dx

C__^L_ 1457.

'

Jf *

VT^X*

,458. '

J 5jt

1442.

f

J 1460.

1443.

-1

1461

dx -

1444.

IHB^T^

1462.

+

J/ Jl a!+i)' (

1446.

ax

'-^-

1463.

r

1464. P

1447.

y2

j^===d.

f^ ^

^

dA;

1465.

sin' A;

\

COS '

cosec'SxdA;.

Miscellaneous Examples on Integration

Sec. 12]

1466.

1

M67.J.-(f,4 '

l469

-

dx,

fsinh^l-A

f

J

J 2sinx

X

+ 3cosx-5 "

J 2-+w-*-

A

f.

-

,)sinh'*" dx

1470.

A:

d)c

f*

!468

sinh ^ cosh

484.

1485

137

]>

1488.

C_^_ ,. 2e*

2* J e

d.v

.471.

1489.

"

j sin

2x

sin

jc

dx

1472.

rfx.

J

i

j1474.

1476. 1477. 1478.

1479. 1480.

JV? + A:

a*

J

sin x dx.

2

Jxe^djc.

J(x

1493.

J/?Tirfx.

J

Y

i

sin*

2>

J

.j-^z

y

\

(sin

C-

>

(

x

-s-

1)

src tdn x

*

dx.

1494.

\

1495.

(Varc

1496.

J cos(lnx)dA:.

1497.

J (x

1498.

Jxarctan(2x+3)dx.

1499.

fare sin

dv

>-

I

/\

\nYT=xdx.

x'

!

10~ *cfx.

1492.

/

^*V'djt.

r*

1483.

sin*

CJiLdx.

2

J

1481.

1491.

aX

<

f

1

1

sinyd.v.

3x) sin 5*

*^i*

cos

"2"^*.

+ cos x)*'

1500.

V^ dx.

rfjc.

Chapter

V

DEFINITE INTEGRALS

Sec. 1. The Definite Integral as the Limit of a

Sum

1. Integral sum. Let a function f (x) be defined on an interval and a=xc <Xt < < x n = b is an arbitrary partition of this interval n subintervals (Fig. 37). A sum of the form .

.

.

into

(1)

where /

is is

= 0,

1,

2,

...

(n

1),

called the integral sum of the function f (x) on [a, b]. the algebraic area of a step-like figure (see Fig. 37).

Geometrically,

S,,

10

Fig. 37

Fig. 38

2. The definite integral. The number of subdivisions n tends to zero,

from

is

x=a

limit of the sum S nt provided that the to infinity, and the largest of them, Ax/, called the definite integral of the function f (x) within the limits to * &; that is,

=

max

(2)

A*j

-> o

^

The Definite Integral as the Limit

Sec. 1]

of

a

Sum

139

it is integrable on [a, b]\ If the function / (x) is continuous on [a, b], i.e., the limit of (2) exists and is independent of the mode of partition of the interval of integration [a, b] into subintervals and is independent of the choice of points / in these subintervals. Geometrically, the definite integral (2) is the algebraic sum of the areas of the figures that make up the curvilinear trapezoid aABb, in which the areas of the parts located above the #-axis are plus, those below the jc-axis, minus (Fig. 37).

The definitions of integral sum and definite integral are b. eralized to the case of an interval [a, b], where a Example 1. Form the integral sum S n for the function

>

by dividing the interval into n equal parts and chooscoincide with the left end-points of the subintervals x i+l ]. What is the lim S n equal to? n -+ CO 9 Ax. and tc/ J Solution. Here, -. Whence n n

on the interval ing [x it

gen-

naturally

points

[1,10]

that

|/

= 101 = Hence

(Fig. 38),

S n -58-L.

lim n

=

.

2

->> oo

Example 2. Find the area bounded by an and the ordinates * = 0, and x = a (a >

jc-axis,

Partition

Solution. parts

=

.

the

arc of the parabola 0).

base a into n equal

Choosing the value

of

the

tion at the beginning of each subinterval,

y

func-

we

will

have

The areas of the rectangles are obtained by multiplying each yk

Summing, we

by the base

A*=

(Fig.

39).

get the area of the step-like figure

Using the formula

sum

for the

of the squares of integers,

2>

n(n+\)(2n+\) 6

Fig. 39

= x*

t

the

140

we

[Ch.

Definite Integrals

5

find

= g'n(n-l)(2n-l) 6H*

S and, passing

'to

the limit,

S-

lim n -*

w

'

we obtain >

Sn =

lim n

-* co

(-D(2-l) = al 6n

>

Evaluate the following definite integrals, regarding them as the limits of appropriate integral sums: b

1501.

1502.

i

1503.

dx.

x*dx.

a

-2

T

10

1504.

J(0 t +g*)
J2*dx. S

and g are constant.

1505*.

8

j

x dx.

i

1506*. Find the area of a the hyperbola

curvilinear

by two ordinates: x = a and x = b 1507*. Find

trapezoid

(0
bounded

by

and the x-axis.

X

=

sintdt.

Sec. 2. Evaluating Definite Integrals by Means of Indefinite Integrals

1. A definite integral with variable upper limit. continuous on an interval [a, b], then the function

If

a is

the aritiderivative of the function F' (x)

= f (x)

2. The Newton-Leibniz formula. o

f (x)\

that

is,

a<*<6. F' (*) = /(*)

for If

th ^n

a

function

f (t)

is

Sec. 2]

_

Evaluating Definite Integrals by Indefinite Integrals

The antiderivative F

Example

1.

(x)

is

computed by finding the

_

indefinite integral

Find the integral

Find

Find the derivatives

of the following functions:

X

\={\ntdt 1510.

(

f(jc)=Td/.

1512.

/

*

1513. Find the points of the

extremum

of the function

X

y

= j!lild/

in the region

Applying the Newton-Leibniz formula,

l^~-

1515.

find the integrals:

X

1

1514.

* rel="nofollow">0.

1516.

X

1

f

Jdt.

J

~*

^-. *

1517.

J*cos/<#. '

-.

Using definite integrals, find the limits of the sums:

1520.

lim

n-

Rl

[Ch.

Definite Integrals

142

Evaluate the integrals: 8

A

.

2*

J(jt*

P

+ 3)d*.

1522.

dx

I Jl

4

1523. cos* a da.

1536. Jl

1524.

2d*.

JK* 2

2

1537.

sin'cpdcp.

* ]'

1 1539. Jl

JF

4

1540.

1

71

4~ l

iL 8

***

1541.

f

dx

JF=3F 1831. 1543. .L

coshA:dA;.

Ins

4

1532.

\ j

2

sec a da.

f*

*'

rfjt

5 J cosh !' In

.

J

sinh*xdjc.

5

Sec. 3]

_ _ Improper Integrals

Sec. 3. Improper

143

Integrals

1. Integrals of unbounded functions. If a function f (x) is not bounded any neighbourhood of a point c of an interval [a, b] and is continuous and c<x<>b then by definition we put for in

a<*
t

C-B

b

(f(x)dx =

J a

lim C

^a

b

f(x)dx

+e

C

lim

f(x)dx.

(1)

-*%+e

are finite, the improper inteIf the limits' on the right side of (1) exist and a or c thj b, gral is called convergent, otherwise it is divergent. When c definition is correspondingly simplified. If there is a continuous function F (x) on [a, b] such that F'(x) f(x)

=

=

=

when x

then

c (generalized antiderivative),

*

F(a).

If

a<*<6

|/(*)|
and

f

(2)

F(*)dx converges, then

the

in-

a

tegral (1) also converges (comparison test).

and

lim

X-+C

when x-+

then

c,

for

1)

f

(x)

\

x m

c

\

oo,

oo,

A

^

0,

i.

e.,

f(x)~

the integral (1)

1

If

converges,

the function / (x)

2) for

*

the

when

continuous

is

^ I

m^>l

then we assume

\f(x)dx= J

lim b ->

oo

(3)

\f(x)dx J

and depending on whether there the respective integral Similarly,

is

is a finite limit or not on the called convergent or divergent.

b

lim

f(x)dx=

a -^

I'

C \

m<

integral (1) diverges. 2. Integrals with infinite limits.

<

^A

I/WK^W

(f(x)dx

f

J

oo

f(x)dx=

)

a

and the

integral

of (3),

b

oo

and

right

[p(x)dx

lim a->

oo

converges,

\f(x)dx. J

then

the

infe-

a gral (3) converges as well. If

/W^rO

oo,

then

1)

gral (3) diverges.

and for

lim

m>

1

f (x)

xm

= Ajt<x>

the integral

(3)

t

A^Q,

converges,

i.e., 2) for

/(x)~-4 when

m
the inte-

144

__ Definite Integrals

Example

&_

f 2 J x

[C/i.

5

1.

lim

M

f 4? + e-*oJ* "f"* *2

e -K> J

-i

=

lim

lim (I-lU e-*oV (l-l e

e->oVe

/

e

and the integral diverges. Example 2. 6

QO

= Example

3.

lim lim (*rT^2= 2 &->< ft-oo J

(arc tan 6

1+^

arc

tan

())

=

.

2

Test the convergence of the probability integral

(4) *r

Solution.

We

put C

00

1

00

e~ x

*

dx=

C

2

dv+

C

e~ x dx. *

1

The first of the two integrals on the right 2 the second one converges, since e~x <e~*

dx=

not an improper integral, while

is

lim

when

(

x^\

e~ b

+ e~

and l

)=e~

hence, the integral (4) converges. Example 4. Test the following integral for convergence:

w r j

^^

y

J?

'

Solution.

When x -++&>, we have

Since the integral

converges, our integral (5) likewise converges. Example 5. Test for convergence the elliptic integral

dx

l

'

t

Sec. 3]

145

Improper Integrals

Solution. The point of discontinuity of the integrand the Lagrange formula we get

where

*<*,
Hence,

for

*-+

I

is

x=l. Applying

we have

V

1

1

/

/T ^

2

\\-xj

1

'

Since the integral 1

i

v dx

ff nr^y converges, the given integral

(6)

converges as well.

Evaluate the improper integrals

(or establish their divergence):

xlnx a

P AY

1552.

-^.

f1

1560.

I a

i

*L a

oo

1553.

J.

1561.

Jcotjcdx

146

_ 00

1562.

kx

\e-

dx

GO

f -arc tan*

_

Definite Integrals

(*>0).

^

d*-

1565.

1566

(Ch.

5

-

1564.

Test the convergence of the following integrals:

-

100

1567.

f

5

j)

.-

**..,_

/*+2 */*+*

1

.

1571.

C

J

./*

..

^/l-jc

1570.

1574*. Prove that the Euler integral of the

fiist

kind

(beta-

function)

B(P, q)=* converges when

p>0

and

q>0.

1575*. Prove that the Euler integral of the second kind (gam-

ma-function)

converges for

p>Q.

Sec. 4. Change of Variable in a Definite Integral

a<x
If a function f (x) is continuous over and *=-q>(0 It a function continuous together with its derivative cp' (t) over a<^<6, where a=*9(a) and &=cp(P), and /[
Sec. 4]

of Variable in

Change

a Definite Integral

147

then

Example

1.

Find x*

We

Solution.

V~a*x*dx

(a

> 0).

put

asin

t\

dt.

Then

P=

?

arc sin

= arc sin l

=y

and, Therefore,

.

we

consequently,

we

shall

can

take

a

= arcsinO = 0,

have

IL a

C x2

2

VV

x2

=a

4

a2

dx=\

sin

2

2 1

a 2 sin 2

J/~a

jl

JL

2

Z

f sin

2

/

cos

2

/

d/ =

1

a cos

t

dt

IL t 2

f sin 2/

d/ =

-^-

C -^-

(1

Jtfl

1576.

Can the substitution

Transform the following indicated substitutions:

A:

= COS/

definite

be

integrals

in the integral

by means

of

the

8

1577.

f(x)dx,

.

x

= arc tan/.

i

1578

.

ly=,

1581. For the integral b

148

__ Definite Integrals

[Ch.

5

indicate an integral linear substitution

which the limits

as a result of

of integration

would be

and

1,

respectively.

Applying the indicated

evaluate

substitutions,

the

following

integrals:

1583. a

1584.

x \Ve -\dx,

n

o

n 2

1586.

Evaluate

the

following

integrals

by

means

of

appropriate

substitutions:

1587.

f^=

A~

i5 8 9.

fi^^d*. *

1590.

x

*

'

2 2

1538.

____

8

J

f J 2*4-

%=.

1^3*+

1

Evaluate the integrals: a

1591.

C J

dx y

xV

x2

1593.

e + 5x4-\

tyax

J

x*dx.

o 271

1594.

fg-f 53 cos x

.

J i

1595. Prove that

o

if

f(x)

is

an even function, then

Integration by Parts

Sec. 5]

But

if

f(x)

1596.

is

Show

an odd function, then

that 00

00

00

~'T=

-00

1597.

Show

149

dx

-

that !L 2

1

arc cos

x

J

1598.

Show

x

o

o

that

T C

f(s\nx)dx =

T f

Sec. 5. Integration by Parts functions If the interval [a, b], then

u

u

(x)

(x) v' (x)

and

dx

Applying the formula

v (x)

are

continuously

= u (x) v(x) for

t;

integration

differentiate on

(*) u' (x) dx.

by

parts,

the

(1)

evaluate the

following integrals: *L 00

2

1599.

^xcosxdx.

1603. oe

e

1600.

Jlnxd*.

1604.

J

1

1

1601.

J*V*dx. o

1602.

Je*sinxdx.

1605. o

e- aJC cos6A:dA:

(fl>0).

__

150

[Ch. 5

Definite Integrals

1606**. Show that for the gamma-function (see the following reduction formula holds true:

From

+ =

T (n n\, \) that for the integral

this derive that

Show

1607.

2

= = [sm /= sin" xdx n

is

1575)

a natural number.

2

n

f

n

if

Example

[

cos

71

the reduction formula n

n ~*

n

holds true.

Find /, if n is a natural Devaluate I 9 and 7 10 1608. Applying repeated

number. Using the formula obtained,

.

integral (see

Example B(p,

<7)

parts,

evaluate the

=

where p and 9 are positive 1609*.

by

integration

1574)

integers.

Express the following integral

in

terms of

B

(beta-

function):

m= if

m

sin

and n are nonnegative

w

x cos

n

integers.

Sec. 6. Mean-Value Theorem

1. Evaluation of

If

integrals.

f(x)^F(x)

^f(x)dx^^F(

66

f(x) and
where

m

is

the interval

the smallest [a,

b]>

for

a<*<6

M

(1)

and,

besides,

(p(*)^0, then

6

j

9

W ^.

(2)

a

a

and

then

X )dx.

mJq>(x)d*<J/(*)9(*)dx<M a

a<*<&,

a

a

If

for

b

b

is

the

largest

value of the function /

(x)

on

Mean-Value Theorem

Sec. 6]

In particular,

if

151

m (ba) <.[f(x)dx*^M

(6

(3>

a).

a

The inequalities

(2)

and

may

(3)

be replaced,

respectively,

by

their equiva-

lent equalities: b

b

a

and b

-a), a

where c and

Example

are certain 1.

numbers lying between a and

b.

Evaluate the integral 5L

M 2

Solution. Since

(Xsin'x^l, we

have

Jl

<

2

2'

that

i/l K 2

'

is,

1.57

2. The mean value

of a function.

1.91.

The number b

is

called the

mean value

of the function / (x)

on the interval

1610*. Determine the signs of the integrals without evaluating

them:

n b)

152

[Ch. b

Definite Integrals

Determine (without evaluating) which of the following

1611. integrals

is

greater: i

i

V\+x*

a)

dx

or

J

dx\

1

1

2

2

x sin x dx

or

e x*dx

or

b) ^

J x sin

2

x dx\

2

c)

\

Find the mean values

of the functions

on the indicated

inter-

vals:

1612. /(x) 1613. f(x) 1614.

=x =a

0<x
2 ,

f(x)=sm

2

x,

1615.

1616. Prove that ( J o

dx lies

r

1/2+^

Find the exact value

0.70.

between

^2

of this integral.

Evaluate the integrals: n 4

4

1617.

+ jfdx.

1620*. JT

+

1618

.

2

1

r

djf

J gqpjj.

2JI

1619.

f

1622. Integrating by parts, prove that 200JI

10071

43

0.67 and -4= V^2

The Areas

Sec. 7]

of

153

Plane Figures

Sec. 7. The Areas of Plane Figures

1. Area in rectangular coordinates. If a continuous curve is defined in f(x) [f(x)^Q], the area of the rectangular coordinates by the equation y curvilinear trapezoid bounded by this curve, by two vertical lines at the

a Fig.

X

b

40

=

a and x 6 and points x is given by the formula

by

a

segment

of

the jc-axis

(Fig. 40),

b (1)

Example straight

1.

lines x

Compute

=

\

and

.v

the 3,

area bounded by the parabola and the x-axis (Fig. 41).

b

y-fi(*) Fig. 42

Solution. The sought-for area

Fig. 43

is

expressed by the integral

y

X = -~ ^

X

the

154

Definite Integrals

Example the

2.

Evaluate the area

(Ch.

bounded by the curve # = 2

y*

*/

5

and

/-axis (Fig. 42).

Solution. Here, the roles of the coordinate sought-for area is expressed by the integral

axes

are changed

and so the

=

where the limits of integration y l 2 and i/ 2 =l are found as the ordinates of the points of intersection of the curve with the t/-axis,

X

a Fig. 44

In the

curves

/

Fig. 45

more general case, if the area S is bounded by two continuous and y = f 2 (x) and by two vertical lines x = a and x = b, where

= M*)

/i(*XM*) wn

n

a<*<&

(Fig. 43),

we

will then have:

(2)

Example

3.

Evaluate the area 5 contained between the curves

y
=

If

set 1

=2 of

x 2 and

l /

equations

=JC 2

(3)

(3)

simultaneously, we find the of formula (2), we obtain

and # 2 =1. By virtue

=

=

the curve is defined by equations in parametric form x q>(/), y y(t\ the area of the curvilinear trapezoid bounded by this curve, by two

then

The Areas

Sec. 7]

a and #=&), and

vertical lines (x integral

Plane Figures

of

by a segment

155

of the

*-axis

is

expressed

by the

f, are determined from the equations and 6 =
where a

and

/,

= (p(/j)

equations

Due

Solution.

we /

2

xc

y l

parametric

= a cos = bsint. c

/,

i

it is sufficient to compute the area of a the result by four. If in the equation jc aco

symmetry,

multiply

and then x

put x

first

= 0.

to the

then

I

\

its

a,

we

get the limits of integration

/,

= = y and

Therefore, ji

2

-i-S

=

(b

sin

a

(

sin /)

dt=ab(

sin

2 /

dt

=^

S nab. 2. The area in polar coordinates.

and, hence,

If a curve is defined in polar coordinates by the equation r~f (
Fig. 47

Fig. 46

which

correspond

to

the

values

cp,

=a

and

cp,=

p,

is

expressed

by

the

integral

Example

5.

Find

(Fig. 47).

the

area

contained

inside

Bernoulli's

lemniscate

156

__

Definite Integrals

Solution.

By

virtue of the

[Ch. 5

curve we

of the

symmetry

quadrant of the sought-for area:

Whence S = a2

_ determine

first

one

.

=

z

1623. Compute the area bounded by the parabola y 4x x and the x-axis. 1624. Compute the area bounded by the curve y \nx, the e. ;t-axis and the straight line x x (x 1625*. Find the area bounded by the curve y 2) 1) (* and the x-axis. 1626. Find the area bounded by the curve y* x, the straight 8. line y=l and the vertical line x 1627. Compute the area bounded by a single half-wave of the and the Jt-axis. sinusoidal curve 1628. Compute the area contained between the curve y ianx,

=

=

=

=

y=smx

=

the x-axis and the straight

line

x=~

.

=

1629. Find the area contained between the hyperbola xy m*, and x 3a (a>0) and the x-axis. the vertical lines 1630. Find the area contained between the witch of Agnesi

=

x^a

u= x- + a

y

2

and the x-axis.

2

1631. Compute the area of the figure bounded by the curve 8 and the y-axis. the straight line y 1632. Find the area bounded by the parabolas y**=2px and

=

y~x*, x

2

= 2py.

1633. Evaluate the area

and the straight line f/ = 1634. Compute the area

bounded by the parabola y = 2x

x*

x.

of a segment cut off by the straight 2 2x from the parabola y = x 1635. Compute the area contained between the parabolas y*=*x*,

line

y=3

.

*/=Y and 1636.

the straight line y

Compute the area contained

and y = 4

y=

1637.

Agnesi

1638.

between

the

parabolas

2

|x

.

Compute the area contained z between the witch

y=

and the

= 2x.

1

and the parabola

x

f^.

Compute the area bounded by the curves y**e*> straight line

x=l.

of

_

The Areas

Sec. 7\

of

Plane Figures

1639. Find the area of the figure

~

= f2

and the straight

l

__

157

bounded by the hyperbola

linex = 2a.

1640*. Find the entire area bounded by the astroid

1641. Find the area between the catenary

y = a cosh the (/-axis and the straight line

,

= ~(e + 2

f/

!)

=

x 1642. Find the area bounded by the curve a*y* 1643. Compute the area contained within the curve

2

(a*

2 ;c ).

~

1644. Find the area between the equilateral hyperbola x* y* the x-axis and the diameter passing through the point (5,4).

= 9,

1645. Find

the

and the ordinate Find

1646*.

between

area

x=l

the curve y

(x>l). bounded

area

the

the

x-axis, 9

the by J

asymptote x = 2a (a>0). 1647*. Find the area between the strophoid

and

~i,

2

cissoid

r/ y

X = 2ax o

its

y*

=x

(

x ~~ a)2

2Jfl^~~

X

and

asymptote (a>0).

its

1648. circle

and

2

jt

the area of the two parts into which divided by the parabola if =2x. 2 2 the area contained between the circle x y

the

Compute

+

2

r/

=--8 is

1649. Compute the parabola x*=l2(y 1). 1650. Find the area contained within the astroid jc

1651. Find the cycloid

the

= acos

8

/;

y= b

sin

+

= 16

8

/.

bounded by the x-axis and one arc

area

f

\

x = a(t # = a(l

of

sin/), cos/).

1652. Find the area bounded by one branch of the trochoid

tefr***/and a tangent

to

it

at its lower points.

158

5

(Ch.

Definite Integrals

1653. Find the area bounded by the cardioid cos2/),

--

sin20.

1654*. Find the area of the loop of the folium of Descartes

*-r+T"

*-

a (1-f coscp). 1655*. Find the entire area of the cardioid r 1656*. Find the area contained between the first and second turns of Archimedes' spiral, r acp

=

(Fig. 48).

1657. Find the area of one of the leaves of the curve r acos2cp. 1658. Find the entire area bound2 2 a sin4cp. ed by the curve r 1659*. Find the area bounded by the curve r asin3cp. 1660. Find the area bounded by Pascal's limagon

=

=

=

Fig. 48

1661.

Find

the

= 2 + cos

r

area

and the two half-lines

cp.

bounded by the parabola

9=4-

and 9

1662. Fin'd the area of the ellipse r

= y. r = -r 1

+ e cos

1663. Find the area bounded by the curve

=

r

r

=a

sec

2

^

(e
cp

= 2acos3cp

a. lying outside the circle r 1664*. Find the area bounded by the curve x*

and

+ y* = x* + y*.

Sec. 8. The Arc Length of a Curve

1. The arc length

in rectangular coordinates.

y=f(x) contained between two

The arc length s of = a and x~b

points with abscissas x

2 8

2 8

a

curve

is

' ' Example I. Find the length of the astroid * +/'*-= a (Fig. 49). Solution. Differentiating the equation of the astroid, we get

The Arc Length

Sec. 8]

For this reason, we have

of a

Curve

159

for th* arc length of a quarter of the astroid:

Whence s = 6a. 2. The arc

length of a curve represented parametrically. If a curve is
=

s

where

t

l

and

t2

=

"

are values of the parameter that correspond to the extremities

of the arc.

49

Fig

Example

2.

Fig. 50

Find the length

one arc

of

=

x a(t [ \ j/=:a(l Solution.

We

have

of the

cos/)-

cosf) and -^

-^=a(l

= asin/.

/=2a The limits of integration of the arc of the cycloid. If a curve is defined the arc length s is

^=

and

*

2

cycloid (Fig. 50)

sin/),

= 2ji

by the equation

r

sln~-d/=

C

correspond

= /(cp)

Therefore,

to

in polar

the

extreme points

coordinates, then

P

where a and p are the values the arc.

of

the

polar

angle

at

the extreme points of

160

[Ch.

Definite Integrals

Example

The

3.

Find the

entire curve

is

the

length of

entire

described by a point as

cp

curve

r

=

asin-|-

ranges from

5

(Fig. 51).

to 3ji.

Fig. 51

Solution.

She curve

We

have

r'

= a sin

2

-^- cos

,

o

o

therefore the entire arc length of

is

8JI

s=

J J/a*

1665.

y*

8JI

= x*

sin

-|

+o

Compute the

cos'

sin*

-f-

-|-

arc

length

of

=a d

sin

j

2

-f-

rfq>

=^

semicubical

from the coordinate origin to the point x = 1666*. Find the length of the catenary y =

A

1667.

x=0

to

Compute

B(b,h). the arc length of

x=l.

the parabola y

curve y

of

the

of

the curve y

*=K8.

*=

1670. Find the to

1671.

between

length of the curve y

arc

to

= e*

= lnx

= 2}/"x

from

lying between

from x

= /3

= arc sin (e~*)

from

= In secy,

lying

jc=l.

Compute the t/

=

and

j/

=

arc length -5-

of the

curve x

.

o

1672. Find the arc length of the curve x

=1

from the

(6,a) to the point

1668. Find the arc length the points (0,1) and (l,e). 1669. Find the arc length to

parabola

4.

acosh-^-

vertex

.

= e.

= ^-y

2

-^\ny

from

Volumes

Sec, 9]

of Solids

161

1673. Find the length of the right branch of the tractrix

a+vV-t/

2

from y = a to

y

t/=&(0<6
1674. Find the length of the closed part of the- curve

= x(x

a

3a)

.

1675. Find the length of the curve to

t/

= ln ( coth ~-J

from

x=b (0
=

to

/

= 7\

1677. Find the length of the evolute of the ellipse

1678. Find the length of the curve

x=- a (2 cos/ y = a(2 sin/ 1679. Find the length of the

cos 2/)

2

-y

which

is

cut

off

I

/

first

1680. Find the entire length of 1681. Find the arc length of r = asec

f

sin 2/).

turn of Archimedes' spiral

the cardioid r that part of

by a vertical

line

= a(l + coscp). the

passing

parabola

through

the pole. 1682. Find the length of the hyperbolic spiral rq rel="nofollow">= 1 from the point (2,'/ 2 ) to the point C/,,2). ae m v, 1683. Find the arc length of the logarithmic spiral r r a. lying inside the circle

=

=

1684. Find

r=l

to

the arc

length

r = 3.

of

the curve

= -g-

(r

+

from j

'

Sec. 9. Volumes of Solids

.

1. The volume of a solid of revolution. The volumes of solids formed by the revolution of a curvilinear trapezoid [bounded by the curve y&f (x) fhe AT- ax is and two a and x vertical lines x b\ about the x- and '(/-axes are J

=

6

-

1900

t

162

Definite Integrals

[Cft.

5

expressed, respectively, by the formulas: b

b

Vx =ji

1)

J a

y*dx\

2)

V Y =2n

J

xr/dx*).

a

Example 1. Compute the volumes of solids formed by the revolution of a bounded by a single lobe of the sinusoidal curve # = sinx and by the segment O<;*<JT of the x-axis about: a) the x-axis and b) the j/-axis. figure

Solution.

a)

V^^-ji

b)

Vy=2n

\

xsinxdx=2jt(

xcosx + sinx)Jc ~2ji t

.

The volume of a solid formed by revolution about the t/-axis of a figure bounded by the curve x=g(y), the (/-axis and by two parallel lines y = c and t/

= d,

may

be determined from the formula

from formula

obtained x and y.

(1),

given

above, by

interchanging

the

coordinates

If the curve is defined in a different form (parametrically, in polar coordinates, etc.), then in the foregoing formulas we must change the variable of integration in appropriate fashion. In the more general case, the volumes of solids formed by the revolution about the x- and (/-axes of a figure bounded by the curves /! /! (x) and y 2 f z (x) a and x b are, respectively, [where f$$x)^f z (x)] and the straight lines x equal to = t = = a and b Example circle x* 2. + (y Find the volume of a torus formed by 2 a 2 (6^a) about the x-axis (Fig. 52). &) = the rotation of the *) The solid is formed by the revolution, about the (/-axis, of a curvilinear a, x trapezoid bounded by the curve y b, f(x) and the straight lines x and 0. For a volume element we take the volume of that part of the solid formed by revolving about the 0-axis a rectangle with sides y and dx at a distance x from the (/-axis, Then the volume element dVy=2nxydx, whence = = b V K =2JI { xydx. = = Volumes Sec. 9] Solution. We of Solids 163' have I/, =6 I^a 2 x 9 and y 2 = t>+ Va*x 2 Therefore, (the latter integral is taken by the substitution x=asi K -a x a Fig 52 The volume of a solid obtained by the rotation, about the polar axis, of a sector formed by an arc of the curve r F((p) and by two radius vectors ipr-=a, = = 3 Vp = ~ 2 C JT \ r 8 sin cpd q>. a This same formula is conveniently used when seeking the volume obtained by the rotation, about the polar axis, of some closed curve defined in polar coordinates. Example r = asin2(p 3. Determine the volume formed by the rotation about the polar axis. Solution. = 2.--n\ rsinq>d(p = 8 yJia C sin 8 .L 2 = ^ Jia 3 8 f J sin 4 cos 8 cp dcp = ^ lOo jia 8 . 2(p sin q> dcp = of the curve 164 __ \Ch. 5 Definite Integrals 2. Computing the volumes of solids from known cross-sections. If S S(x) the cross-sectional area cut off by a plane perpendicular to some straight line (which we take to be the x-axis) at a point with abscissa *, then the volume of the solid is is where *, and x 2 are the abscissas of the extreme cross-sections of the solid. Example 4. Determine the volume of a wedge cut off a circular cylinder by a plane passing through the diameter of the base and inclined to the base at an angle a. The radius of the base is R (Fig. 53). Solution. For the *-axis we take th? diameter of the base along which the cutting plane intersects the base, and for the (/-axis we take the diameter of the base perpendicular to it. The equation of the circumference of the base is *2 + =R 2 j/ The area 2 . of the section ABC at a distance x from the origin is 2 S(x) = area A ABC = -^ ABBC = -^yy tana =^- tana. 1 1 r/ Therefore, the sought- , for volume of the wedge is R R y= 2~ f y2 tanad*=tana (* (R 1 1 *)<& = y tana R . 1685. Find the volume of a solid formed by rotation, about the x-axis, of an area bounded by the x-axis and the parabola 1686. Find the volume of an ellipsoid formed by the rotation of the ellipse ^r +| = 1687. Find the the x-axis', of - 8 l volume about the x-axis. formed by the rotation, about an area bounded by the catenary y = acosh the of a solid xa. , and the straight lines 1688. Find the volume of a solid formed by the rotation, about a the x-axis, of the curve j/=sin x in the interval between x xr-axis, = and x = n. 1689. Find the volume of a solid formed by the rotation, about the x-axis, of an area bounded by the semicubical parabola if = x s the x-axis, and the straight line x== 1. 1690. Find the volume of a solid formed by the rotation of the same area (as in Problem 1689) about the {/-axis. 1691. Find-, the volumes of the solids formed by the rotation of an area bounded by the lines y = e*, x = 0, y = about: a) the x-axis and b) the y-axis. 1692. Find the volume of a solid formed by the rotation, about the t/-axis, of that part of the parabola j/ 2 = 4ax which is cut off by the straight line x = a. , Sec. 9] _ _ Volumes of Solids 165 1693. Find the volume of a solid formed by the rotation, about the straight line x a, of that part of the parabola y*=4ax which is cut oft by this line. 1694. Find the volume of a solid formed by the rotation, about the straight line y p, of a figure bounded by the parabola = = 2 t/ = 2p* and the straight line * = -- . 1695. Find the volume of a solid formed by the rotation, about the x-axis of the area contained between the parabolas y x* L = and y= Y*. 1696. Find the volume of a solid formed by the rotation, about the x-axis, of a loop of the curve (* 4a)tf =ax(x 3j) 1697. Find the of the cyssoid y* volume A8 = ^ _x of a solid about its generated by the rotation asymptote x = 2a. 1698. Find the volume of a paraboloid of revolution whose base has radius R and whose altitude is //. 1699. A right parabolic segment whose base is 2a and altitude h is in rotation about the base. Determine the volume of the resulting solid of revolution (Cavalieri's "lemon"). 1700. Show that the volume of a part cut by the plane jc 2a off a solid formed by the rotation of the equilateral hyperbola x* tf^c? about the *-axis is equal to the volume of a sphere of radius a. 1701. Find the volume of a solid formed by the rotation of a bounded by one arc of the cycloid x=-a (/ sin t), figure y=^ a (\ cos/) and the x-axis, about: a) the x-axis, b) the y-axis, and c) the axis of symmetry ot the figure. 1702. Find the volume of a solid formed by the rotation of acos 8 /, y bsm*t about the //-axis. the astroid * 1703. Find the volume of a solid obtained by rotating the cardioid r a(l -hcostp) about the polar axis. 1704. Find the volume of a solid formed by rotation of the acos 2 = = = = = + 166 of [Ch. 5 Definite Integrals 1708. A circle undergoing deformation the points of its circumference lies on moving so that one the y-axis, the centre is + describes an ellipse ^^-=1, and the plane of the circle is perpendicular to the jq/-plane. Find the volume of the solid generated by the circle. 1709. The plane of a moving triangle remains perpendicular to the stationary diameter of a circle of radius a. The base of the triangle is a chord of the circle, while its vertex slides along a straight line parallel to the stationary diameter at a distance h from the plane of the circle. Find the volume of the solid (called a conoid) formed by the motion of this triangle from one end of the diameter to the other. 1710. Find the volume of the solid bounded by the cylinders z 2 z* a*. a and y 1711. Find the volume of the segment cut off from the ellip"' + = = u 2 2 = * by the plane x = a. paraboloid |21712. Find the volume of the solid bounded by the hyperbo- tic loid of + z ^ -f one sheet 1713. Find the ^-=1 and rj volume the planes 2 X2 of the ellipsoid ^2 = U2 Z and z li. + ^ + ^2"= ! 10. The Area of a Surface of Revolution Sec. The area of formed by the rotation, about a and x f(x) between the points x a surface arc of the curve y the formula = the = b, is an expressed by x-axis, of b b Vl+y'*dx (ds = 2 is (1) the differential ol the arc of the curve). Zfta Fig. 54 If surface of the curve cbtained from formula the equation x is is (!) represented differently, the area of the by an appropriate change of variables. Sec. The Area 10] of a Surface of Revolution Example 1. Find the area of a surface formed by 2 x-axis, of a loop of the curve 9i/ ;t(3 x)* (Fig. 54). Solution. For the upper part of the curve, when 167 rotation, about = / = -- (3 mula x) y~x. Whence the differential of the arc (X*<3, ds= X ~^ r_dx. 2 the area of the surface (1) V the we have Fromfor- x 9 = 2n ( J Example of the 2. \(6 Find the area x cycloid =a of a surface s\nt)\ (t 2 y = a(l V formed by the rotation of one arc cost) about its axis of symmetry (Fig. 55). Solution. The desired surface is formed by rotation of the arc OA about the straight line AB, the equation of which is x na. Taking y as the independent variable and noting that the axis of rotation AB is displaced relative to the #-axis a distance na, we = will Y have Passing to the variable (na /, we obtain da d y n = 4na 2 ( ( nsin /sin y + sinf sin ~- Fig. 56 dt-j The dimensions of a parabolic mirror AOB are indicated required to find the area of its surface. 1715. Find the area of the surface of a spindle obtained by rotation of a lobe of the sinusoidal curve about the 1714. in Fig. 56. It is ys'mx #-axis. 1716. Find the area of the surface formed a part of about the the tangential curve t/ = tan* by the rotation from Jt = to of A:==^-, jc-axis. 1717. Find the area of the surface formed by rotation, about the x-axis, of an arc of the curve y e-* 9 from x to x = = 168 \Ch. 5 Definite Integrals 1718. Find the area of the surface (called a catenoid) formed about the x-axis from by the rotation of a catenary # = acosh x = to x = a. 1719. Find the area of the surface of rotation of the astroid 3 x'/'-i -a /'/ 3 3 ' about the y-axis. 1720. Find the JC = -^/ 2 y area the surface of rotation of the curve of y=\ from In (/about the *-axis to y = e. 1721*. Find the surface of a torus formed by rotation of the 2 x* b) ^=a* about the *-axis a). (y 1722. Find the area of the surface formed by rotation of the + circle ellipse ^+^ = about: 1 1) the *-axis, 2) the y-axis (a>6). 1723. Find the area of the surface formed by rotation of one arc of the cycloid x cos t) about: a) the a(t sin/) and r/ a(l x-axis, b) the y-axis, c) the tangent to the cycloid at its highest = = point. 1724. Find the area of the surface formed by rotation, about the j^-axis, of the cardioid x t/ = a (2 cost = a(2sin/ cos 2/), \ sin2/). / 1725. Determine the area of the surface formed by the rotation 2 2 a cos2 of the lemniscate r = = Sec. 11. Moments. Centres of Gravity. Guldin's Theorems 1. Static moment. The static moment relative to the /-axis of a material A having mass m and at a distance d from tha /-axis is the quantitv point Mi~md. The static moment relative to the /-axis of a system of n material points in the plane of the axis and at distances with masses m,, m 2 ..., m n lying d lt d 2 ..., d n is the sum , , n M i= m A2 = / (1) ! where the distances of points lying on one side of the /-axis have the plus sign, those on the other side have the minus sign. In a similar manner we define the static moment of a system of points relative to a plane. If the masses continuously fill the line or figure of the x#-plane, then the static moments about the x- and /-axes are expressed ^respectivex and not as the sums (1). For the cases of ly) as integrals and geometric figures, Itie density is considered equal to unity. M My Sec. is Moments. Centres of Gravity. Guldin's Theorems II] In particular: the arc length, for 1) the curve * = *(s); we have y=y(s) whare t L = V(dx)* + (dy)* is M Y =. (s) ds\ x (s) 2) for a plane figure bounded by th3 curve y a and y b we obtain = bounded by the Find the static straight lines: Solution. Here, y (2) = y(x), ihz Jt-axis and two t b = b II x\y\dx. (3) the x- and /-axes of a triangle a a 1. ds Fig. 58 b Example s the differential of the arc); Fig. 57 vertical lines x parameter L y (ds the 169 moments about ~-f-^ a b ~ . ) = l, x = 0, = // Applying formula (Fig. 57) (3), we obtain ab* and 2. Moment of inertia. The moment of inertia, about an /-axis, of a imfe2 point of mass m at a distance d from the /-axis, is the number l t -=-tnd . of n material of an a The moment of inertia, about /-axis, system points with masses m lt m 2t ..., n is the sum rial m 170 __ [Ch. 5 Definite Integrals where d lf d 2 ..., d n are the distances of the points from the /-axis. In the case of a continuous mass, we get an appropriate integral in place of a sum. Example 2. Find the moment of inertia of a triangle with base b and altitude h about its base. Solution. For the base of the triangle we take the x-axis, for its altitude, the y-axis (Fig 58). Divide the triangle into infinitely narrow horizontal strips of width dy t which play the role of elementary masses dm. Utilizing the similarity of we triangles, obtain j = bL dm h y . n , dy and Whence 3. Centre of gravity. The figure (arc or area) of mass M coordinates of the centre of gravity of a plane are computed from the formulas moments of the mass. In the case of geometnumerically equal to the corresponding arc or area. For the coordinates of the centre of gravity (x, y) of an arc of the plane curve y connecting the points A[a f (a)] and B [6, f (b)], f (x) (a^x^b), where MX ric figures, and My are the static the mass M is = t we have B B t> b ( -A\yds 2 y 1^1 +(y') dx a b S 2 '} The coordinates dx of the centre of gravity (x, y) of the curvilinear trapezoid x^b, Q^y<^f(x) may be computed from the formulas b ~~ X b -.(y^dx ^xydx a *~~ - O f y a_ ^ ,3 b where S=f y dx is the area of the figure. a a There are similar formulas for the coordinates of the centre of gravitv 6 y of volume. Example 3. Find the centre of gravity of an arc of the semicircle ,,2^n*- i,i^n\ (pig 59) Sec Moments. Centres 11] We Solution. Guldin's Theorems of Gravity 171 have and Whence f, ' yds-- xJ .. ,. yV-.v 2 ^= y a J 2 Hence, a 4. Guldin's theorems. Theorem 1. The area of a surface obtained by the plane curve about some axis lying in the. same plane intersecting it circumference of is to circle equal the rotation of an arc of as the curve and not the product of the length of the curve by the described by the centre of gravity of the arc of the curve. Theorem 2. The volume of a solid obtained by rotation of a plane figure about some axis lying in the plane of the figure and not intersecting it is equal to the product of the area of this figure by the circumference of the circle described by the centre of gravity of the figure. Fig. 59 1727. Find the static moments a segment of the straight line x o lying between the axes. y o about the coordinate axes of __ 172 Definite Integrals 1728. Find the static about its moments of a rectangle, [Ch. 6 with sides a and &, sides. 1729. Find the static moments, about the x- and t/-axes, and the coordinates of the centre of gravity of a triangle bounded by the straight lines x + y Q. a, x Q, and y 1730. Find the static moments, about the x- and (/-axes, and the coordinates of the centre of gravity of an arc of the astroid = = _t_ * = _i^ -If/* > ==aT > lying in the first quadrant. 1731. Find the static moment of the circle r = 2asin about the polar axis. 1732. Find the coordinates of the centre of gravity of an arc of the catenary y = a cosh ~ = x= a to x a. Find the centre of gravity of an arc of a circle of radius a subtending an angle 2a. 1734. Find the coordinates of the centre of gravity of the arc of one arch of the cycloid from 1733. x = a(t sin/); y = a(\ cos/). 1735. Find the coordinates of the centre of gravity of an area bounded by the ellipse 2 -|-=l and the coordinate axes ( 1736. Find the coordinates bounded by the curves of the centre of gravity of 1737. Find the coordinates of the centre of gravity bounded by the first arch of the cycloid x = a(t sin/), r/ = a(l cos/) and the jc-axis. 1738**. Find the centre of gravity of a hemisphere lying above the ;q/-plane with centre at the origin. of an area an area of radius a 1739**. Find the centre of gravity of a homogeneous right circular cone with base radius r and altitude h. 1740**. Find the centre of gravity of a homogeneous hemi- sphere of radius a origin. lying above the jo/-plane with centre at the &?c. 12] Applying Definite Integrals 1741. Find the its moment to Solution of Physical Problems 173 inertia of a circle of radius a about of diarneler. 1742. Find the moments of inertia of a rectangle with sides a and b about its sides. 1743. Find the moment of inertia of a right parabolic segment with base 26 and altitude ft about its axis of symmetry. 1744. Find the moments of inertia of the area of the ellipse 2 JC U2 ^ + ^=1 about its principal axes. 1745**. Find the polar moment of inertia of a circular ring and R t (R that is, the moment of inertia with radii R^ about the axis passing through the centre of the ring and perpen} dicular to its plane. 1746**. Find the moment of inertia of a homogeneous right circular cone with base radius R and altitude H about its axis. 1747**. Find the moment of inertia of a homogeneous sphere about its diameter. of radius a and of mass 1748. Find the surlace and volume of a torus obtained by rotating a circle of radius a about an axis lying in its plane and at a distance b (b>a) from its centre. 1749. a) Determine the position of the centre of gravity of M 2 t 2 T an arc of the astroid x -\-i/ T = a* lying in the first quadrant. b) Find the centre of gravity of an area bounded by the curves 2 = 2px and x* = 2py. t/ 17f>0**. a) Find the centre of gravity of a semicircle using Guldin's theorem. b) Prove by Guldin's theorem that the centre of gravity of a triangle is distant from its base by one third of its altitude Sec. 12. Applying Definite Integrals to the Solution of Physical Problems 1. The path traversed by a point. If a point is in motion along some curve and the absolute value of the velocity o~/(/) is a known function of the time t, then the path traversed by the point in an interval of time is '. * Example 1. The velocity of a point o Find the path ing the during s = 0. 1/ covered by the point ol motion. commencement this interval? 8 in is m/sec. 7=10 the interval of time sec followis the mean velocity cf motion What 174 [Ch. 5 Definite Integrals Solution. We have: 10 p t* .-Jo.irt-o.i T = 250 metres and =-=25 2. The work of the x-axis, of a force. then the work If m/sec. a variable force X=f(x) acts in the direction over an interval [x ly x z ] of this force is A= 2. What work has to be performed to stretch a spring 6 cm, if kgf stretches it by 1 cm? Solution, According to Hook's law the force X kgf stretching the spring kx, where k is a proportionality constant. by x m is equal to 100v. we get and l kgf, 100 and, hence, X 0.01 Putting x Whence the sought-for work is Example a force of 1 = X= m 0.08 0.06 A= = = X= 100 x dx = 50 x = 0. 18 2 j kgm 3. Kinetic energy. The kinetic energy of a material velocity v is defined as point of mass m and mv* The mv m 2% kinetic ..., mn energy of a system of n material points with masses having respective velocities t; lf v 2 ..., v n is equal to , , To compute the kinetic energy of a solid, the latter is appropriately partitioned into elementary particles (which play the part of material points); then by summing the kinetic energies of these particles we get, in the limit, an integral in place of the sum (1). Example 3. Find the kinetic energy of a homogeneous circular cylinder of density 6 with base radius R and altitude h rotating about its axis with angular velocity CD. Solution. For the elementary mass dm we take the mass of a hollow cylinder of altitude h with inner radius r and wall thickness dr (Fig. 60). We have: Since the linear velocity of kinetic energy is the mass dm is equal to t; = /-co, the elementary Sec. 12] Applying Definite Integrals to Solution of Physical Problems 175 Whence 2 r 9 dr^= nco 6/? 4 fc 4. Pressure of a liquid. To compute the force of liquid pressure we use Pascal's law, which states that the force of pressure of a liquid on an area S at a depth of immersion h Is where y is the specific weight of the liquid. Example 4. Find the force of pressure experienced by r submerged vertically in water so that its diameter is radius water surface (Fig a semicircle of with the flush 61). Solution, We partition the area of the semicircle into elements strips parallel to the surface of the water. The area of one such element (ignoring higher-order infinitesimals) located at a distance h from the surface is ds The pressure experienced by where y is ^ 2xdh = 2 this V'* element h2 dli. is the specific weight of the water equal to unity. the entire pressure is Whence r =2 The velocity of a body thrown vertically upwards with velocity V Q (air resistance neglected), is given by the 1751. initial C J ITS Definite Integrals \Cfi. 5 formula = . ff'. where t is the time that elapses and g is the acceleration of gravAt what distance from the initial position will the body ity. be in t seconds from the time it is thrown? 1752. The velocity of a body thrown vertically upwards with initial velocity v (air resistance allowed for) is given by the formula where t is the time, g is the acceleration of gravity, and c is a constant. Find the altitude reached by the body. 1753. point on the x-axis performs harmonic oscillations A about the coordinate origin; its velocity given by the formula is where t is the time and t; co are constants. Find the law of oscillation of a point if when / = it had an abscissa * = 0. What is the mean value of the absolute magnitude of the velocity of the point during one cycle? 1754. The velocity of motion of a point is v = te~"' Qlt m/sec. Find the path covered by the point from the commencement of motion to full stop. , 1755. A rocket rises vertically upwards. Considering that when the rocket thrust is constant, the acceleration due to decreasing weight of the rocket increases by the law find the velocity at any instant of time /, is zero. Find the altitude reached at lima /==^~^ if / = the (a initial ftf >0), velocity / r 1756*. Calculate the work that has to be done to pump the water out of a vertical cylindrical barrel with base radius R and altitude H. 1757. Calculate the work that has to be done in order to pump the water out of a conical vessel with vertex downwards, the radius of the base of which is R and the altitude H. 1758. Calculate the work to be done in order to pump water out of a semispherical boiler of radius R 10 m. 1759. Calculate the work needed to pump oil out of a tank = through an upper opening (the tank has the shape of a cylinder with horizontal axis) if the specific weight of the oil is y, the length of the tank H and the radius of the base R. 1760**. What work has to be done to raise a body of massm from the earth's surface (radius R) to an altitude ft? What is the work if the body is removed to infinity? Sec. 12] Applying Definite Integrals What work *2 =10 cm? 1762**. and Problems 177 = Two 1761**. on the lie to Solution of Physical electric charges * 100 CGSE-and e l==200 CGSE x-axis at points * and *, 1 cm, respectively. will be done if the second charge is moved to point A length p=10kgfcm = cylinder with a movable piston of diameter / = 80cm 2 . is filled with What work must be done steam at to halve the a D = 20 cm pressure volume of steam with temperature kept constant (isoihermic process)? 1763**. Determine the work performed in the adiabatic expan3 of air sion and pressure (having initial volume u =l 2 8 10 m ? p _=l kgf/cm ) to volume u, 1764**. A vertical shaft of weight P and radius a rests on a bearing AB (Fig. 62). The frictional force between a small part a of the base of the shaft and the surface of the support in contact with it is F==fipa, const is the pressure of the shaft where p on the surface of the support referred to unit area of the support, while pi is the coefficient of friction. Find the work done by the frictional force during one revolution of the the = m ft = shaft. 1765**. Calculate the kinetic energy of a and radius R rotating with angular velocity G> about an axis that passes through its centre perpendicular to its plane. 1766. Calculate the kinetic energy of a right circular cone of mass rotating with angular velocity CD about its axis, if the radius of the base of the cone is R and the altitude is H. 1767*. What work has to be don? to stop an iron sphere of 2 me'res rotating with angular velocity w = 1,000 rpm radius R j about its diameter? (Specific weight of iron, y = 7.8 s/cm .) 1768. A vertical triangle with base 6 and altitude h is submerged vertex downwards in water so that its base is on the surface of the water. Find the pressure of the water. 1769. A vertical dam has the shipa of a trapezoid. Calculate the water pressure on the dam if we know that the upper base 20 m. a 70 m, the lower base 6=50 m, and the height h 1770. Find the pressure of a liquid, whose specific weight is y. on a vertical ellipse (with axes 2a and 26) whose centre is submerged in the liquid to a distance h, while the major axis 2a of the ellipse is parallel to the level of the liquid (h^b). 1771. Find the water pressure on a vertical circular cone with radius of base R and altitude H submerged in walei vertex downwards so that its base is on the surface of the water. disk of mass M M = = = 178 Definite Integrals [C/i. 5 Miscellaneous Problems = 1772. Find the mass of a rod of length / 100 cm if the linear density of the rod at a distance x cm from one of its ends is 6 = 2 + 0.001 x 2 g/cm. 1773. According to empirical data the specific thermal capacity of water at a temperature is = 0.9983 What from /C (0^/<100) 5.184xlO- + 6.912xlO5 2 7 / quantity of heat has to be expended to heat to 100 C? / . g 1 0C of water 2 of of 1774. The wind exerts a uniform pressure p g/cm on a door width b cm and height h cm. Find the moment of the pressure the wind striving to turn the door on its hinges. 1775. What is the force of attraction of M a material and mass on a material point of mass a straight line with the rod at a distance a from one length / rod of m lying on of its ends? 1776**. In the case of steady-state laminar (low of a liquid through a pipe of circular cross-section of radius a, the velocity of flow v at a point distant r from the axis of the pipe is given by the formula difference at the ends of the pipe, |i is is the pressure the coefficient of viscosity, and / is the length of the pipe. Determine the discharge of liquid Q (that is, the quantity of liquid flowjng through a cross-section of the pipe in unit time). 1777*. The conditions are the same as in Problem 1776, but the pipe has a rectangular cross-section, and the base a is great compared with the altitude 26. Here the rate of flow u at a point M(x,y) is defined by the formula where p Determine the discharge of liquid Q. 1778**. In studies of the dynamic qualities of an automobile, use is frequently made of special types of diagrams: the velocities v are laid off on the Jtr-axis, and the reciprocals of corresponding accelerations a, on the (/-axis. Show that the area S bounded v 2t and v l and v by an arc of this graph, by two ordinates v by the *-axis is numerically equal to the time needed to increase the velocity of motion of the automobile from v l to v 2 (accelera- = tion time). = Sec. 12] Applying Definite Integrals 1779. a A downward of the horizontal beam vertical load beam, and of to Solution of Physical of length uniformly support Problems 179 / is in- equilibrium due to distributed over the length reactions A and fi(yl==5==-y-j t M directed vertically upwards. Find the bending moment x in a cross-section x, that is, the moment about the point P with abscissa x of all forces acting on the portion of the beam AP. 1780. A horizontal beam of length / is in equilibrium due to support reactions A and B and a load distributed along the kx, where x is the distance length of the beam with intensity q from the left support and k is a constant factor. Find the bend- = ing moment Note. Mx in cross-section x. The intensity of load distribution is the load (force) referred to unit length. 1781*. Find the sinusoidal current during a cycle T quantity of heat released in a conductor by an alternating with resistance /?. Chapter VI FUNCTIONS OF SEVERAL VARIABLES Sec. 1. Faiic Notions 1. The concept of a function of several variables. Functional notation. variable quantity 2 is called a single-valued function of two variables jc, y, if to each set of their values (x, //) in a givm range there corresponds a unique value of z The variables x and y are called arguments or independent variables. The functional relation is denoted by A * Similarly, we Fxample = /(*, y). define functions of three or more arguments. Express the volume of a cone V as a function of 1. eratrix x and of its base radius y Solution. From geometry we know that the where h is the altitude of the cone. But h This is y volume ** y 2 - of a cone its is Hence, the desired functional relation. of the function z^f(x.y) The value point is P that (a.b). is, gen- when x=^-a and y at a bt denoted by / (a,b) or f (P) Generally speakthe geometric representation of a func- ing, tion like z f in (x,y) nate system X, Y. Z Example 2. Find/ is (2, Fig. 63 Solution. Substituting r=2 and t/= 3, we find a rectangular a surface (Fig. 3) and/1, coordi63). if Sec. Basic Notions 1) Putting *=1 ~ and replacing y by , we [SI have will 1 SL\ *' thai is, /(l. 2. Domain )=f(*. 0). of definition of a function. By the domain of definition of a function ? f(x, y) we understand a set of points (*, r/) in an jq/-plane in which the given function is defined (that is to say, in which it takes on definite real values) In the simplest cases, the domain of definition of a function U a finite or infinite part of the jo/-plane bounded by one or several curves (the boundan, of the domain). = Similarly, for a function of three variables u f(x, y, z) the definition of the function is a volume in At/z-space. Example 3. Find the domain of definition of the function domain of 1 > < 2 x2 function has real values if 4 or x* + y 2 4. y inequality is satisfied hy the coordinates of points lying inside a circle of radius 2 with centre at the coordinate origin. The domain of definition oi the function is the interior of the circle (Fig 64). Solution. The The latter Fig. 64 Example 4. Fiy Find the domain z Solution. The first term of definition of the function = arc sin or | function * when + |/ xy of the function The second term has when 65 real defined for is values .The domain if xr/^O, of 1<~ in i.e., definition of or two cases: the entire | is shown in Fig. 65 and includes the boundaries of the domain. Functions 182 of Several Variables [Ch. 6 3. Level lines and level surfaces of a function. The level line of a function 2 f(x, y) is a line / (*, y)-C (in an *r/-plane) at the points of which C (usually labelled in the function takes on one and the same value z = drawings). The iace level surface of a C, at the z) = / (x, y, function of three arguments u~f(x, points of which the function takes value u~C. Example 5. Construct the the function z x*y. Solution. The equation of = has the form x 2 y Putting C = 0, 1784. Find /(1/2, 3), /(I, -1), z __x y i 2, ~ -j .... level lines of the level lines . we get a family 1782. Express the volume V of a regular tetragonal pyramid as a function of its altitude x and lateral edge y. 1783. Express the lateral surface S of a regular hexagonal truncated pyramid as a function of the sides x and y of the bases and the altitude z. Fig. 66 Find 1, or y is a sura constant z) t on lines (Fig. 66). of level 1785 =C y if 1 x, f( f(y,x), y), 2 1 , if f(*,y) 1786. Find the values assumed by the function at points of the parabola y = z , and construct the graph function 1787. Find the value of the function Z 2 at points of the circle x 1788*. Determine f(x), 1789*. Find f(x, y) if rrr - +y*=R if 2 . of the Sec. __ Basic Notions 1] 183 = V7+/(l/x 1). Determine the functions / y= 1. Letz = */(-V Determine the functions f and 1790*. Let * z if z =x and when 1791**. z if when x=\. 1792. Find and sketch lowing functions: a) b) z==Y\ x z=\-\-V 2 domains the of definition of the fol- 2 y \ (x = 2 - x + arc cos z-l/l-^ + z = arc sin X 2 = V 7IT 2 = c) z d) //; 2 e) f) ' ; r g) h) 1793. Find the domains of the following functions of three arguments: a) u=-\x + b) u y + z\ = ln(xyz): c) w=--arc sin jc+arc sin # d) u = V\x 2 2 y + arc sinz; z\ 1794. Construct the level lines of the given functions and determine the character of the surfaces depicted by these functions: a) z^x + y; b) 2 = ^+y c) z = x*-y*; d) z = z = 2 ; e) O^ 1795. Find the level lines of the following functions: a) z b) 2 = ln(*'+f/); = ar d) z = /(y a*); 1796. Find the level surfaces of the functions pendent variables: x y-\-z; a) u * ' ' b) u = c) u = J = + of three inde- _ _ [Ch. 6 Functions of Several Variables 184 Sec. 2. Continuity 1. The limit of a function. = /(*, y) there e > 2 is P A number A is 1 called the limit of a function approaches the point P (o. is a 6 > such that when < Q < 6, where Q = |/\x the distance between P and P', we have the inequality as the point (x, y) I ft), if a)* for + U/ any 6) 1 y)A\<e. f(x, y) = A. /(*, In this case we write lim X-K2 y-*b 2. Continuity and continuous at a point points of discontinuity. P (a, b) A function z=f(x, y) is called if lim f(x x-+a 9 0) = /(a, b). A function that is continuous at all points of a given range is called continuous over this range A function /(AT, y) may cease to be continuous either at separate points (isolated point of discontinuity) or at points that form one or several lines (lines of discontinuity) or (at times) more complex geometric objects. Example 1. Find the discontinuities of the function _*+> Solution. zero. But ** The function -r/ function has for = its or y will be meaningless if the denominator becomes 2 x is the equation of a parabola. Hence, the given discontinuity the parabola y x2 . 1797*. Find the following limits of functions: a) c)li n liin(*?^)sinl; M; e)lin-L; 1798. Test the following function for continuity: ft x /) = / V\x* I if when * 2 -r-// 2 1. 1799. Find points of discontinuity of the a) z = ln; c) e- functions: Sec. 3] Partial Derivatives 1800*. Show that the function SL when when -{ is 185 ** jt + =^=0, = = continuous with respect to each of the variables x and y sepabut is not continuous at the point (0, 0) with respect to rately, these variables together. Sec. 3. Partial Derivatives t. Definition of a partial derivative. example, y constant, we get the derivative If f(x,y), then assuming, for z which is called the partial derivative of the function z with respect to the variable x. In similar fashion we define and denote the partial derivative of the function z with respect to the variable y It is obvious that to find partial derivatives, one can use the ordinary formulas of differentiation. Example 1. Find the partial derivatives of the function y we Solution. Regarding y as constant, dz 1 1 get = 1 t dx , tan x , cos 2 dz we 2. Find the 2. x 2* y partial derivatives of the following function of three a = *y z + 2* 3y + 2 + 5. ? Euler's theorem. A function if for every real factor k degree n 2x' y / 1 arguments: Solution. . I/sin y have will y Example y y y Similarly, holding x constant, x f (x, y) fs called a we have f (kx, ky) --= ttf homogeneous function of the equality (x, //) A [Ch. 6 Functions of Several Variables 186 rational integral function will be and the same degree. The following relationship holds of degree n (Euler's theorem): xfx (x, y) Find the 1801. z = homogeneous for a if all homogeneous its terms are of one differentiable function + yfy (x, y) = nf (x, y). partial derivatives of the following functions: 1808. 2 3axy. = 1809. 2 = 1810. 2 = 1811. 2 = 1806. 1807. z = arc tan 1814. Find /;(2, and fy (2, 1) 1815. Find /;(!, 2, 0), /i(l, 2, 0), /(*, y,z) Verify Euler's theorem ples 1816 to 1819: 1816. f(x,y) 1817. z if 1) xy f(x,y) ft (l, 2, 0) +L . if = ln(xy-\-z). on homogeneous functions = Ax + 2Bxy-Cy 3 t in Exam- 1818. / . = J- 1820. ) , where r = : 2 1821. Calculate 1822. Show that 1823. Show that 1824. Show that 1825. Show J826. Find that and y 57, = 2, if x+ , g+fj+S-0, g+g+g-1, -.(,. y), 2 it if = ln(*' = u = (x-y)(y-z)(z- X if 2 ). if -*+J=J. Sec. 4] _ Total Differential of a Function 1827. Find z--^z(x y) knowing that t (J2 = X 2 -4- 2 = slny J/ ^and z(x, y) _ 187 when *=1. Through the point M(l,2, 6) of a surface z = 2x*+y* drawn planes parallel to the coordinate surfaces XOZ and YOZ. Determine the angles formed with the coordinate axes by the tangent lines (to the resulting cross-sections) drawn at their common point M. 1829. The area of a trapezoid with bases a and b and alti1828. are tude h is equal to S= l /,(fl + &) A. Find g, g, g and, using the drawing, determine their geometrical meaning. 1830*. Show that the function 0, has partial derivatives fx (x, y) and fy (x, y) at the point (0,0), although it is discontinuous at this point. Construct the geometric image of this function near the point (0, 0). Sec. 4. Total Differential of a Function 1. Total z = /(*i y) increment of a The function. total increment of a function the difference is Az-Af (x, + Aj/)-f (*, #)--=/(*+ Ax, y). a function. The total (or exact) differential of a function z f(x, y) is the principal part of the total increment Az, which is linear with respect to the increments in the arguments Ax and A//. The difference between the total increment and the total differential of 2. The total differential of + is an infinitesimal of higher order compared with Q At/*. \^&x* function definitely has a total differential if its partial derivatives are continuous. If a function has a total differential, then it is called differenThe differentials of independent variables coincide with their incret table. ments, that is, dx=kx and dy=ky. The total differential of the function z /(x, y) is computed by the 'formula the function A = dz . =^ dX + , d2= dz . , dy - d-y Similarly, the total differential of a function of three arguments u is computed from the formula . du Example 1. = du -3- dx . dx , -j' du ^- d y dy , du . dz. + -rdz . For the function f(x find the total increment t and the y)=x* + xyy total 2 differential. =/ (x, y, z) [Ch. 6 Functions of Several Variables J88 Solution. f(x+Ax, = 2x y + Ay) = (x -f Ax) Ax + AA 2 +x - = [(2x + y) A* + (x 2 -f (x + A*) (y Ay + y Ax + Ax2 Ay 2y) A^l -f- Ay) 2y Ay + (A* + AA> Ay (y -f Ay a Ay) 2 ; = Ay (2x + y) A* -{-(* 2y) Ay is the total differential of Here, the expression d/ = 2 2 infinitesimal of higher order is an the function, while (A* -f AJC* ) comared with VAx 2 +Ay 2 compared Example 2. Find the total differential of the function AyAy . Solution. 3. Applying For sufficiently Q= y Av 2 -f Ay 2 , the total differential of a function to approximate calculations. small |AA:| and |Ay| and, hence, for sufficiently small we have for a differentiate function z f(x t y) the approxor = Az^dz imate equality dz . 3. Example The Solution, we = altitude of a cone is // 30cm, the radius of the base the volume of the cone change, if we increase by = 10cm. How will 3mm and diminish R fl H by The volume 1 mm? of the cone is V = -~-nR 2 H. The change o in volume replace approximately by the differential AV =^ dV = -j n (2RH dR + R* dH) = = lji( 2.10.30.0.1 3 + 100.0.3) = s 01 Example 4. Compute 1.02 approximately. Solution. We consider the function z^x^. considered the increased value of this function Ay 0.01. The initial value of the function z lOjiss 31. 4 cm*. - = = In x 8 Hence, 1.02 - 01 The desired number may be when jc=l, y = 3, Ajc = 0.02, s l =l, Ay = 3-1.0.02+ 1- In 1-0.01 =0.06. ^ 1+0.06=1.06. = 1831. For the function f(x,y) x*y find the total increment total differential at the point (1, 2); compare them if 0.1 f Ay 0.2. a) Ax=l, A//-2; b) A* 1832. Show that for the functions u and v of several (for and the = = example, two) variables the ordinary rules of differentiation holcb v du udv Sec. 4} _ Find the total Total Differential of a Function _ differentials of the following functions: 1841. z- In tan i. x 1842. Find df(l, 1836. z 1837. K = si = ln(x*+y*). = lnl+i. /*, 1838. z 1844. 1839. 1845. = arc tan if ~*' = u = u= 1846. w -2-+ = arctan^. 1847. Find d/ + arctan-*. 1848. tj) 1), 1843 - " 2=yx?. 1840. 2 189 (3, 4, 5) if = One = side of a rectangle is a 10 cm, the other & 24cm. a diagonal / of the rectangle change if the side a is increased by 4 and b is shortened by 1 How will mm mm? the change and compare it with the exact value. A closed box with outer dimensions 1849. and 6 cm is made of 2-mm-thick plywood. Approximate 10 cm, 8 cm, Approximate the volume of material used in making the box. 1850*. The central angle of a circular sector is 80; it is desired to reduce it by 1. By how much should the radius of the sector be increased so that the area will remain unchanged, if the original leng:h of the radius is 20 cm? 1851. Approximate: 2 a) (1.02)'- c) sin32-cos59 calculating (0.97) sin 60 ; b) . . (when converting degrees into radius and three significant figures; round off the take last digit). Show that the relative error of a product is approximasum of the relative errors of the factors. 1853. Measurements of a triangle yielded the following data: side a=-100m2m. side fc angle 1852. tely equal to the ABC C 601. To what degree side c? 1854. The oscillation from the formula of period = 200m3m, accuracy can T of a we compute the pendulum is computed 190 _ _ [Ch. 6 Functions of Several Variables where is / the length of the of gravity. Find the error, a result of small errors A/ 1855. pendulum and g is the acceleration when determining T, obtained as a = and Ag = (J in measuring / and and P (x, The distance between the points P (* , g. V equal to Q, while the angle formed by the vector P P the x-axis is a. By how much will the angle a change dx, y dy)? point P(P is fixed) moves to P l (x y) with is if the + + Sec. 5. Differentiation of Composite Functions 1. The case of one independent variable* If z f(x, y) is a differentiate function of the arguments x and y, which in turn are differentiate functions of an independent variable /, then the derivative of the composite function z puted from the formula dz ____i dt '"dxdt^dydt = /[ ^(01 ma X be com- ' In particular, if / coincides with one of the arguments, for instance then the "total" derivative of the function z with respect to x will be: x, dzdy Example 1. Find ~, if where * = cos/, y y, Solution. e'*+ 2 From formula ^3(-sinO + Example 2. * 8X+ (1) ^-2^ Find the partial derivative z Solution. t*. we have: = e*y t ~ and the total derivative , if where y jZ^syety. From formula (2) we obtain 2. The case of several independent variables. If z is a composite function of variables, for instance, z where *=q> (u v) f(x,y), y=ty(u t v) (u and v are independent variables), then the partial derivatives z with respect to u and v are expressed as several = independent t dz dy ~ Sec. 5] and _ Differentiation of Composite Functions ~~-~ _ dzdy dz__dzdx 191 * In all the cases considered the following formula holds: . dz dz -j- . dx dx dz . . + dy 5- dy y (the invariance property of a total differential). Example p Find ^- and ^du dv 3. z=:f(x, if , x=uv, y= where y), Solution. Applying formulas (3) and we (4), . get: and Example Show 4. that the function z Solution. ment x* The function + y*=t, (x* +y 2 satisfies ) the equation = * dx = dy depends on x and y via the intermediate cp argu- therefore, dz__dzdt___ 2 and dzdt \z . , . rt , Substituting the partial derivatives into the left-hand side of tion, we get t/ that is, *~^ i)2 the function z satisfies the given equation. 1856. Find ~ if 2 = , where x = e t y 1857. Find u t y = \nt. if = lnsin-=, f 1858. Find where x = 3t\ y = yt'+l. y if j u = xyz, where ^=/ +l, y=ln^, I z the equa- 192 _ _ Functions of Several Variables ^ 1859. Find if where x = r, I860. Find if z 1861. Find = uv g 1862. Find g where u = , ~ and 2 = arc ~ and ~ and j z = f(u, v), ~ ow and = arc tan 1865. Find ~ z 1866. Show = #*. (x). if where a = A: ~ 1 , where x = usmv, where w = x^/ + ^. where x = 3*), cp sin ip, R =R e then = 1868. where Show =a if cos and *_a. if = /(*. // if | - 1867. Find v^e 1 t/ if = /(u), that */ where y = t oy , and and tan if 1863. Find 1864. Find = cos #, sin x, v if = xy 2 [Ch. y> *). that where y= ( if / is a differentiable function, dz dz then cos 9 cos \|>> sin 9, _ Sec. 6] 1869. Show _ that the function w = f(u, where u = x + at, v = y + bt dw dT= a Show satisfies the 1871. , , dw dt that the function =J + equation Show v), satisfy the equation dw 1870. 19S Derivative in a Given Direction that the function equation x-^ + y -^ = j 1872. Show that the function satisfies the satisfies the (A- The side y ) ^- + xy-^ = m increases at the rate of a rectangle x -^20 the other side f/ 30 decreases at 4 m/sec. What the rate of change of the perimeter and the area of the rect- 1873. = of 5 m/sec, is 2 2 equation m angle? 1874. What is The equations of motion of a material point are the rate of recession of this point from the coordinate origin? 1875. Two boats start out from A at one time; one moves northwards, the other in a northeasterly direction. Their velocities are respectively 20 km/hr and 40 km/hr. At what rate does the distance between them increase? Sec. 6. Derivative in a Given Direction 1. The derivative of a function function z 7- 1900 = /(#, and the Gradient in a given direction. y) in a given direction / = PP, is of a Function The derivative of a where f(P) and If the function z where a (Ch. 6 Functions of Several Variables 194 is are values of the function at the points P and differentiate, then the following formula holds: / (P,) is the angle formed by the vector / with the x-axis (Fig. 67). Y e^ P(*,y) Fig. 67 In similar fashion we define the derivative in a given function of three arguments u f(x, y z). In this case = du du = du cos a + du 5- cos p + 3- cos v, dz dx dy , / for a , r ' dl direction t T (2) where a, P, Y are ^ ne angles between the direction / and the corresponding coordinate axes. The directional derivative characterises the rate of change of the function in the given direction. 2 2 at the point 3 = dz -. dx = 4*; . (dz\ =4; A 3\dxjp dz Here, sina Applying formula (1), we = sin 120 = get The minus sign indicates that the function diminishes at the given point and in the given direction. 2. The gradient of a function. The gradient of a function z f(x, ij) is 3 vector whose projections on the coordinate axes are the corresponding par- = Derivative in a Given Direction Sec. 6] tial 195 derivatives of the given function: dz , , dz (3) The derivative of the given function in the direction / the gradient of the function by the following formula: is connected with ~ That is, the derivative in a given direction is equal to the projection of the gradient of the function on the direction of differentiation. The gradient of a function at each point is directed along the normal to the corresponding level line of the function. The direction of the gradient of the function at a given point is the direction of the maximum rate of increase of the function at this point, thlft on its is, when /=grad z the derivative -^ takes greatest value, equal to In similar fashion u -=/(*, y, z): we define the gradient du . , du of function a . , du of three variables, . The gradient of a function of three variables at each point is directed along the normal to the level surface passing through this point. Example 2. Find and construct the gradient of the function z~x*y at the point P(I, 1). X2 J X Fig. 68 Solution. Compute the partial derivatives and their values at the point P. _ i dxjpdz ^T" Hence, grad 7* z = 2t+J (Fig. 68). s \y'* . 9.2 _ _ Functions of Several Variables 196 1876. Find the = x* derivative of the function z xy2y* the direction that produces an angle at the point P(l, 2) in of 60 with the x-axis. = x* 1877. Find the derivative of the function z at the point Af(l, 2) in the direction from point tf (4, [Ch. 6 + 6). _ + 1 xy* 2x*y point to the this = + 1878. Find the derivative of the function z lnYx* y* the point P(l, 1) in the direction of the bisector of the at first quadrantal angle. 1879. Find the derivative of the function u = x* 3yz + 5 at in the direction that forms identical the point Af(l, 2, 1) angles with all the coordinate axes. 1880. Find the derivative of the function u = xy + yz -)- zx at the point M(2, 1, 3) in the direction from this point to the point N(S, 5, 15). eP + e*) \n (e* 1881. Find the derivative of the function u at the origin in the direction which forms with the coordinate axes x, y, z the angles a, p, y, respectively. 1882. The point at which the derivative of a function in any direction is zero is called the stationary point of this function. Find the stationary points of the following functions: = a) + z-=x* = x* + y*= u 2y*-{ z*xyyz b) z c) 1883. Show that the derivative of the function z + = any point of the ellipse 2x* y* C* along the ellipse is equal to zero. 1884. Find grad z at the point (2, 1) if at 1885. Find grad z at the point (5, 3) if 1886. Find grad u at the point 1887. Find the magnitude and (1, 2, 3), point (2, 2, 1) direction taken normal to the u=xyz. if of grad u at the if 1888. Find the angle between the *=ln-j- = at the points A (1/2, 1/4) gradients and 5(1, 1). of the function Sec. 7] _ Higher-Order Derivatives and Differentials 1889. Find the steepest slope of the surface = _ 197 + x' z 4y* at the point (2, 1, 8). 1890. Construct a vector field of the gradient of the following functions: = je-f y: a) z = x* + y c) z z \ Sec. 7. Higher-Order Derivatives and Differentials 1. Higher-order partial derivatives. The second partial derivatives of a function z /(*, y) are the partial derivatives of its first partial derivatives. For second derivatives we use the notations 32z d (dz\ Derivatives of order higher than second are similarly defined and denoted. If the partial derivatives to be evaluated are continuous, then the result of repeated differentiation is independent of the order in which the differentiation is performed. Example 1. Find the second partial derivatives of the function z~ arc tan . y Solution. First lind the dz first == partial _J_ dz^ dy~~ Now m derivatives: j__ y 1_ J , differentiate a second time: ^L = d . dz z f y \- 2 *y . dxdy We note that the so-called so- different "mixed" partial derivative may be found in a el way, namely: d / dxdy~~dydx dx( 2 y* _ 198 f(x t We = d(d*z) d*z and, generally, z function function: similarly define the differentials of a function z of order higher than two, for instance: If [Ch. 6 Higher-order differentials. The second differential of a of this y) is the differential of the differential (first-order) 2. z _ Functions of Several Variables = /(x, differential of dz = d(d n - l z). y), where x and y are independent variables, then the second the function z is computed from the formula (1) Generally, the following symbolic formula holds true: formally expanded by the binomial law. z f (x, (/), where the arguments x and y are functions of one or eral independent variables, then it is If = "- * If x and i/ sev- +2 are independent variables, then d 2 jt becomes identical with formula (!) Example 2. Find the total differentials = 0, d 2 y = Q, and of the first formula and second orders (2) of the function z Solution. First method. We = 2;t 2 3xyy 2 . have *-<*-* I*-*. Therefore, dz = fr Further we have ~ ^-4 djc whence it 2 ' J*!L-_ 3 2 ^l ' djcd^"" dy* follows that Second method. Differentiating we find Differentiating again and remembering that dx and dy are not y, we get x and = (4dx 3dj/) dx (3d* dependent on + 2d(/) dy =4dx*6dx dy2dy z . Higher-Order Derivatives and Differentials Sec. 7] if Find if 1894. Find if 1893 . ^ = arc tan A 1 Find 1895. . . xy if 1896. Find all second partial derivatives of the function Find 1897. ^~- dxdy dz if u 1898. Find , if z 1899. Find f(0, = 0), = sin (xy). = (l 1900. Show /^(O, 0) f xt/ (Q, 0), 4 that if = arc sin 1901. 1902*. Show Show that = y if that for the function if 199 _ _ [Ch. 6 Functions of Several Variables 200 = 0] [provided that f(0, 0) we have rxy(' 0)=-1, 1903 - &"= Find z g where z y(x, z 1906. u if = f(x, = f(u, v), r Show a)* = arc tan + (y = lny , satisfies the b)\ (x, t) =A + sin (akt 2 d^u_ ~~~ a d^u 2 Show dx sin Kx u(x // , ' that the function (x-x i , cp) equation of oscillations of a string dt* (where * conduction Laplace equation that the function u 1909. y), that the function Show satisfies the = <((x, Laplace equation = Y(x 1908. z), that the function / where y, where u u 1907. f(u, v), y). Show satisfies the 0)=+1. = xy. where u = x* + y*, v 1904. Find /;,(0, y ' z > , z > '>- )* + (y-y n )* e a are constants) satisfies ^a'/^-u^U*^ the equation of heat _ Higher-Order Derivatives and Differentials Sec. 7] Show 1910. that the function _ 201 are arbitrary twice differentiable functions, satiscp and \|) the equation of oscillations of a string where fies Show 1911. satisfies that the function the equation x 1912. Show satisfies the 1913. 2 + 2wy dxdy + Jy , 5-,2 dv r- , 2 x-, dy* l = n0. that the function equation Show -7 ' d su that the function z = f[x + y(y)] satisfies the equa- = u(x, which tion d22 dz ~ dx dx dy 1914. Find u-^u(x, y) dzd 2z 2 dy dx ' if dTSy^^ 1915. the form of the function u Determine satisfies the equation 1916. Find d*z if 2 1917. Find d u if 11 1918. Find d*z . if - rn \l^ 1919. = (f} \^/F Find dz and d*z z =u v whpr^ vviitivx / t> -r y2 j\r -1- /y* . | y if where u =~ , v = xy. y), _ _ 202 Functions of Several Variables 1920. Find d'z z = ax, where u v), v = by^ v = ye*. if = f(u, 1922. Find d*z 6 if z~f(u, 1921. Find d*z [C/i. where u = xey v), , if z = e x cos y. 1923. Find the third differential of the function = * cos y + y sin x. z Determine all third partial derivatives. 1924. Find df(l 2) and d*f(l, 2) if 9 = x + xy + y* 2 f(x, y) 4\nx \Q\ny. 2 1925. Find d /(0, 0,0) f(x Sec. 8. + 9 y, =x z) if z Integration of Total Differentials t. The condition for a total differential. For an expression P (x, y)dx-}~ Q(* y)dy> where the functions P (x, y) and Q (x, y) are continuous in a simply connected region (in D) the total ficient that D together with their first partial derivatives, to be some function u (x, y), it is necessary and suf- differential ol aq^ap ~~ dx Example is a total t. Make sure that the expression differential of some function, and Solution. In the given case, = 1, ' dy find that function. P = 2x + y Q t x+2y. Therefore, ,5 = -- and, hence, where u It is is the desired function. given that -= But on the other hand and Finally we have 2jt + #; therefore, = x + y' (y) = x + 2y, whence q>' (y) = 2y, (p(f/) = = Sec. ti\ _ 2. The 203 case of three variables. Similarly, the expression P(x, where P _ Integration of Total Differentials z)dx y, + Q(x, z)dy y, + R(x t z)dz, y, R(x, y, z) are, together with their first partial functions of the variables x, y and 2, is the total differential of some function u (x y, z) if and only if the following conditions (x, y, z), Q(x, y, z), continuous derivatives, t are fulfilled: dQ^W dR^dQ dP^dR ' dx Example 2. Be sure ' dy dy dz dx dz that the expression the total differential of some function, and find that function. f 2 3jc Solution. Here, p 1, 3x, R 2yz+\. We establish 30 the fact thai is = dQ dx + = Q=z + = dP = O . dy dR = dQ -r = dz dy dP c. rr = OR = r V. / dx dz and, hence, ~ where u We is the sought-for function. have hence, u= On 2 (3x + 3y$$dx

= x* + 3xy

+
z).

the other hand,

du -

-

3

~ = dtp

dz

whence

y^

=z

2

and ~P

dz

= 2f/z+l.

two variables q>(#, 2) tion for total differential find q>:

of

whose is

The problem reduces partial derivatives are

fulfilled.

We

that

x

is,

y(y,

= ^2 + 2 + C, 2

e)

And

finally,

to finding the function

known and

the condi-

_ _ Functions of Several Variables

204

Having convinced

are

total

differentials

[C/i.

6

yourself that the expressions given below of certain functions, find these functions.

1926.

ydx + xdy.

1927.

(cosx+3x*y)dx + (x'y

1

)

1928-

1930. 1931.

-dx^dy. y x dx + *

-

*

y

r

*

z

dy.

1932. Determine the constants a and 6 in such a the expression 2 z 2 (ax + 2xy + y ) dx-(x + 2xy + by*) dy

should

be

a

total

differential of

some function

z,

manner

and

find

that

that

function.

Convince yourself that the expressions given below are some functions and find these functions.

total

differentials of

1933. 2

+

+ 3z)dx+(4xy + 2y + 8xy* + 2) dx +

z)dy

+ (3x

l)dy

(x

1934. (3x 2y* 1935. (2xyz3y*z

y

2)dz.

2

y

xdx + ydy + zdz 1938*. Given the projections of a force on the coordinate axes

v _.

v_

y

to

where A, is a constant. What must the coefficient K be for the force to have a potential? 1939. What condition must the function f(x, y) satisfy for the expression f(x,

y)(dx

+ dy)

to be a total differential?

1940. Find the function u

if

du = f(xy) (ydx + xdy).

Sec. 9]

_

_

Differentiation of Implicit Functions

Sec. 9. Differentiation of Implicit Functions

205

1. The case of one independent variable.

If the equation f(x, y) 0, where differentiate function of the variables x and y, defines y as a function of x, then the derivative of this implicitly defined function, provided

/

that

a

is

y)

(*

y) ?= 0,

f' (x, y

may

be found from the formula

dy dx

f' x

=

(**y)

f'y(x,y)'

found

are

derivatives

Higher-order

by successive differentiation

formula

of

a) Example

1.

Find

-

dx

-~

and

Solution. Denoting the

if

dx 2

left-hand

side

of this

equation by

f (x, y),

we

find

the partial derivatives

f'u

(x

t

y)-=3(x

Whence, applying formula

z

+y

(1),

z 2

(

2y

.

)

we

get

find the second derivative, differentiate with respect to x the first derivative \vhich we have found, taking into consideration the fact that y is a functiun of x'

To

x -~ dx

y J

dx

F

(x,

2

dx\

y

y J

x

y J

~

(

)

y J

\

2

y

'

2

if

2. The case of several independent variables. Similarly, 0, where F (x, y, z) is a differentiate function y, z)

x,

y and

z,

Fz

(x

y,

z)

if

of

the equation the variables

defines z as a function of the independent variables x and y and then the partial derivatives of this implicitly represented 0, function can, generally speaking, be found from the formulas t

^

'

dK

F'z

(x,

dlJ

y, z)

F'g

(x, y,

z)

another way of finding the derivatives of the function ating the equation F (x, y, z)=0, we find

Here

is

dF

Whence

it

is

_,

,

dF

J

,

dF

,

rt

possible to determine dz, and, therefore, dz TT-

dx

.

dz

and 3dy

.

z:

different^

_ _ Example

we

[Ch. 6

Functions of Several Variables

206

Find

2.

and

-T-

if

j-

Solution. First method. Denoting the left side of this equation find the partial derivatives

F'x

(x, y,

z)

= 2x,

Applying formulas dz

_

d*~

F'x(x> F'z (x

t

(2),

y. *)

F'

y (x, y, z)

we

40-z+l, F z

(x,

y,

2)

^ ~"

2)

(x,

//,

z),

= 6z-0.

get

_

2x

~~ y,

=

F

by

Jy(*.

dz^ dy~

'

6*

z)

y

^

(x>

!/,

^

1

z

4//

62

/

z)

Second method. Differentiating the given equation, we obtain 2x dx

4f/

Whence we determine dz

t

dy

+ 6zdz

that

is,

zdy + dy = 0.

y dz

the total differential of the implicit func-

tion: \

z}dy

4//

~ dy we Comparing with the formula dz = -Q- dx -\,

2x

dz

3. A system

6z

y

dy

F(x, G(x,

f

and v as functions

u

y,

62

y

system of two equations

of implicit functions. If a

\ defines u

\4yz

dz '

dx

see that

t

i>)

y, u,

o)

= 0, =

of the variables

x and y and the Jacobian

dF_ dF_

D(F, G) D(u, v)'

dudv

dGdG du dv

then the differentials of these functions (and hence their partial derivatives as well) may be found from the following set of equations . dF dF dF dx ^ dv =0, ^+ -z- dyy + -^~ du ^ + ^dx dv

'dF

,,

,

dy

,

Example

3.

3dx

-r~

dy

,

^du dG -^--3y dv du *

l

.

.

.

t

/Q

'

The equations

define u and v as functions of x and ^ w; find

and -,,dx dx dy dy

rr?

.

.

Differentiation of Implicit Functions

Sec. 9]

Solution.

First

207

method. Differentiating both equations with respect to #

we obtain du

dv

.

whence __

?ff

dx~~ Similarly

we

u ~^~y x y

dv __ u-\-x dx~~ x y

'

'

find

_

v

dy~~

x

du

+y

*

dy~~~ x

y

we

Second method. By differentiation differentials of all four variables:

x du

_

dv '

y

two equations that connect the

find

du 4- dv = dx + dy, + u dx + y dv + v dy =4).

Solving this system for the differentials du and dv,

we obtain

xy Whence

_ dx

'

x

y

(}y__-f-x

dx~~xy

f

x

t)f/

_

dv

y

v -}-x '

'

dy~~x

y

4. Parametric bles x and y

is

representation of a function. If a function represented parametrically by the equations z

= z(u,

z

of the

varia-

v)

and

then the differential equations

be found from the following system

may

of this function

of

dx

dx

Knowing dz ^~=pr and 3-

dz

dx

the differential

.

dy

~^. ^

dz^p

du

dx dv

dz

dz

xdu

dv

,

,

.

5~du-\--5- dv,

d*

+ qdy,

.

dv.

we

find the partial derivatives

_ _

208

[Ch. 6

Functions of Several Variables

Example

The function

4.

arguments x and y

of the

z

is

defined by the

equations _.

,

dz

,

dz

Find ^- and 3ox

.

dy

we

Solution. First method. By differentiation connect the differentials of all five variables:

dx

From

the

first

= du + dv

find three equations that

,

two equations we determine du and dv: 20 dx

"^

dy

_

,

2u dx

dy

'

'

2(v

2(u

u)

w)

Substituting into the third equation the values of du and have: d dy

-

rfy

just found,

we

= 6wu (u '

Whence 3-

=

3au,

3-

^jc

dt/

=TT 2

(w-fy). v ;

Second method. From the third given equation we can find

+ 3^; *=3Jf dx ^ dx Differentiate the respect to y:

two equations

first

the

first

system we

dy

.

we

da

dv a'

dy

u v

find

= _l__

dy~~2(u Substituting the expressions

__

dx~ u

~

dv_

v)'

and

~

^ (5) '

with respect to x and then with

dy

v

dx~~ v the second system

first

l

dy

find du___

From

= 3'Jf + 3t,'f!.

f

dx

From

f dy

1

dy~~2(vu)' into formula (5),

we obtain

_

Sec. 9]

_

209

Differentiation of Implicit Functions

1941. Let y be a [unction of x defined by the equation y*

Find ma *

dy dhj di' d?

1942. y

Show

that

and ana is

^

dx"'

a function defined

^=

1943. Find

and explain the

% ^

1944. Find

by the equation

if

y=\+y x

and

g

.

y

if

result obtained.

=

A;

and (g) (g) \ax J K~\ \ax jx=i

1945. Find

z

if

Taking advantage of the results obtained, show approximately the portions of the given curve in the neighbourhood of the point

*=1. 1946.

The function y In

dx

1947.

and

Find

dx

c,. j Flnd

5i

and

2

and

9

s

-1-

2//

^

variables x and y

and

^ dy

4

8

z

3;q/z

2r/

is

defined by

+ 3 = 0.

%-

~ dx

A:

Find

if

dx z

^2

1949. Find

1950.

arc tan

function z of the

x ,

|?T7 =

dx*

1948. The the equation

^2

defined by the equation

is

cosy

The function

and

for the -j-

if

-|-

y cos 2

+zcosx= 1.

defined by the equation

z

is

x*

+y

2

system

z

of

2

xy

= 0.

values *==

1,

//

= 0,

z

= l.

__

210

dz 1QR1 1951. KinA

Fmd,

1952. /(*, y,z) 1953.

equation

Functions of Several Variables

l>

,

= 0.

= (p(x, y), = 0. ty(x, 2

f/)

d* 2

&*

dz

1955. 2

is

if if

,

where y Find ~.

is

a

function of x

defined by the

if

a function of the variables x and # defined

equation 2x*

[Ch. 6

Show

2 Find dz and d z,

1954.

d*Z ,

__

+ 2y*

-4-

8xz

z*

z

+8-

by the

.

Find dz and d z for the values x = 2, f/^=0, 2^=1. 2 1. What are the 1956. Find dz and d 2, if In z=jc + i/ 4-2 first- and second-order derivatives of the function 2? 1957. Let the function 2 be defined by the equation 2

where

1958.

an arbitrary differentiate

is

Show

constants.

Show

is

satisfies the

02,

y

bz)

1960.

y=

xq> (z)

f(y y) ,

Show

+

&, c are

Q,

an arbitrary differentiate function of two arguments, equation

a-+b - = 1959.

a,

that the function 2 defined by the equation

F(x where F

and

function

that

o|>

= 0.

that

(2)

Show

the

that

function

satisfies the

1

xfx +y~ = z. 2

defined

the

by

equation

equation

d*zf(

1961. The functions y and 2 of the independent variable x are 2 2 2 2 2 2 defined by a system of equations * 2 ^0, x 32 4. 2#

+#

+

+

=

_ _

Sec. 10]

Change

2H

of Variables

1962. The functions y and z of the independent variable x are defined by the following system of equations:

Find dy,

dz, d*y, d*z. 1963. The functions u and v of the independent variables x and y are defined implicitly by the system of equations

Calculate du

du

d2u

d*u

dzu 2

~dy

for x=--Q,

y

'

dv dv dx' d~y

62 v '

dz v

d*v

~d\*'

dxdy

j

dy*

1.

1964. The functions u and v of the independent variables and y are defined implicitly by the system of equations

Find du, dv, d*u,

d*v.

The functions u and

v of the variables x and defined implicitly by the system of equations

1965.

x

.v

= (p(w,

v),

y are

//=-i|)(w, v).

(b du du du 51' cty' d~x> fy'

^ and g? x = M cos u, y w sin u, 2 = x = u + v, y = u and ~ v,z = uv. b) Find = c) Find dz, where r and are functions of the variables 1967. e = F(r, 1966. a)

Find

,

y

if

if

\l

jc

,

.

,

cp)

x and y defined by the system of equations

Find

2

ax

and

.

dy

1968. Regarding z as a function of x

x = a cos

Sec.

q)

cos

i|),

and

y, find

y = b sincpcosij), z

~

and ~,

if

=

10. Change of Variables

When

changing

variables in differential

them should be expressed

in

expressions,

the derivatives in

terms of other derivatives by the rules of

entiation of a composite function.

differ-

[Ch. 6

Functions of Several Variables

212

\. Change of variables in expressions containing ordinary Example 1. Transform the equation x2 y

dx putting

*

=

derivatives.

y-

Solution. Express the derivatives of y with respect to x in terms of the derivatives of y with respect to /. We have

dx

dy

dy

dt

dt

dx

_

dt

dt

'

t*

dt

Substituting

the

expressions

equation and replacing x by

the

of -r-

we

,

derivatives

found

just

the given

into

get

or

dt*

Example

2.

ruij -"'

Transform the equation x

g + ^y_g =0

.

taking y for the argument and x for the function. Solution. Express the derivatives of y with respect to x in derivatives of x with respect to y.

terms of the

dx^Jx'' dy

"~

'

dx*

dx{ dx \

^

}

2

dyl dx \dx~~ \ dy J

-

fdx\ (Ty)

dx

-

'

fdx\* (dy)

dy

Substituting these expressions of the derivatives into the given equation, will

have

d*x dj

1 '

dy

\dy

L=o

dx

dy

'

we

Sec.

or,

10]

_ _

213

of Variables

Change

finally,

Example

3.

Transform the equation

~xy'

dx

by passing to the polar coordinates ,v=rcoscp, Solution. Considering

dx = cos

cp

f/

= rsinq>.

as a function of

r

dr

cp,

d// = sin

r sin cp dcp,

(1)

from formula dr

cp

+ r cos

cp

(1)

we have

dcp,

whence dr .

.

.

.

d//_sm cp dr -f cos cp 3J T dr

cos

dcp __ ~~

cp

-

sin cp r

,

r

r sin cp T d(pr

cos

cp

3- -f

r cos cp ^

_rcos

d
-- r sin

dr coscp

or,

-

r

--

dr cp

Putting into the given equation the expressions for sin

,

\

cos (p Y

dcp

cpH-

rcoscp

r sin q)

x,

rsm

//,

and

-^

,

we

will

have

cp

rsincp*

cp

after simplifications,

2

Change of variables in expressions containing Example 4. Take the equation of oscillations of

and change

it

new independent

to the

partial derivatives. a string

variables a and p, where

a = .v

at,

Solution. Let us express the partial derivatives of u with respect to x and t terms of the partial derivatives of u with respect to a and p. Applying the formulas for differentiating a composite function

in

du___du

du__du da

du dp

da,

9

du dp

we get du

du

,

^T-=^( da

d/

^'^

du__du

^

a)

du

+ ,

du -35"

dp

Q

a (du is

\dp

^'

_^_,^f

du\ 3~

I

day

1

_ _

214

Functions of Several Variables

..

[C/i.

6

v

~

same formulas:

Differentiate again using the

dt 2 ~~dt

d2u

d 2 u\,

x

2 (d u

,

d*u

dz u

dx*~dx\dx)

da\dx

_d

2

u

d'u

d*u

/)ft 8

'

rtn

'

'

/)ft

/}R2

we

Substituting into the ^iven equation, 2

2

/d a

d2 u

-=

for

5.

2 Transform the equation x 2 ^- + y

the

x

y

new independent *

d*u\

= 0.

r

Example

d2 u

Jd u

d u\

d u

have

will 2

2

2

=z

taking

,

w

and

variables,

=

2

-g-

for z

x

,v,

the

new

function.

and Solution. Let us express the partial derivatives ypartial

derivatives

^~

and

.

^

To do

On

x

dx v

t

dy

oy_

-^^,

dx

dz ^

.

2

the other hand,

dw

dw =

:;

du

du

dw ~\

dv

dv

Therefore,

dw -3-

du

,

du

dw

dx

dv = + -5dv x ,

,

dz

2

z

= 2

or

dw

,

i^ w_f^x__m dy\^,^x

&*_

.

Whence

dw

1

dw\

,

.

z

2

and, consequently,

dz^_ dx~~

2

dw ~~ 1 dw \ J z x*dv J ~~du \x

/

terms of the

differentiate the given relation-

this,

ships between the old and new variables:

u

~ in

dw

.

Sec.

Change

10]

215

of Variables

and dz __ z^

dw

2 dy~~y dv

'

Substituting these expressions into the given equation,

we

get

or

Transform the equation

1969.

2

x j--

putting x=--e*. 1970. Transform the equation

putting

A:

cos/.

T

rm 1971

69

Transform the following equations, taking y as the ar-

gument:

The tangent of the angle [A formed by the tangent line and the radius vector OM of the point of tangency (Fig. 69)

1972.

MT is

expressed as follows: tan u=^

216

_ _ Functions of Several Variables

Transform

this

by

expression

to

passing

1973. Express, in the polar coordinates x the formula of the curvature of the curve

1974.

[C/i.

polar

= r cos

6

coordinates:

q>,

y=

r sin

cp,

Transform the following equation to new independent

variables u and v: dz

dz

if

u = x

v =x +y 2

t

*-*3&

'

2 .

1975. Transform variables u and v:

the

following dz

if

=n

equation

new independent

~

dz

.

to

= x, 0=-p

11

1976. Transform the Laplace equation
,

dM_ n ~-"

dx*~*'dy

z

to the polar coordinates A:

= rcoscp,

y

= rsm
1977. Transform the equation

X

2

^2_ y ^__ ~ U> dx* 2

2

ar/

u=*xy and

putting

y

=

.

1978. Transform the equation dz

dz

,

yTx- K jy=(y-^ z

>

by introducing new independent variables

and the new function w=\nz (x + y). 1979. Transform the equation dx*

taking u

>=

= x + y,

for the

v

new

=^

for

function.

the

new independent

variables

and

Sec. 11]

_

The Tangent Plane and the Normal

to

a Surface

_

217

1980. Transform the equation

putting u

= x+y,

v

=x

w = xyz,

y,

where

w=w(u,

v).

11. The Tangent Plane and the Normal to a Surface

Sec.

1. The equations of a tangent plane and a normal for the case of explicrepresentation of a surface. The tangent plane to a surface at a point to (point of tangency) is a plane in which lie all the tangents at the point various curves drawn on the surface through this point. The normal to the surface is the perpendicular to the tangent plane at the point of tangency If the equation of a surface, in a rectangular coordinate system, is given in explicit form, z function, then f (x, y), where f (x, y) is a differentiate z ) of the surface is the equation of the tangent plane at the point (x f/ it

M

M

z-*o=/i(* z

0o)(X-*o)

.

,

+ /i(*o,

)0

,

r

-0o).

and X, K, Z are the current coordinates

Here, t/ ) f (x the tangent plane. The equations of the normal are of the form

where

,

Z

F,

.Y,

Example the surface z

1.

(i) of

the point of

are the current coordinates of the point of the normal. Write the equations of the tangent plane and the normal

=

2

at the point

y

M (2,

the

of

M

=v '

dx

to

1,1).

Solution. Let us lind the partial derivatives their values at the point

M

(fo\ '

given

function and

=2

\dxjM

*~ Whence, applying formulas or 2x-|-2f/

10

z

= ^-i- =

-, which

2. Equations it

representation implicitly,

and F

(X Q

,

which

t/

,

z

)

is

(1) is

and

(2),

we

will

the equation

have

of the

z

1

=2(*

2)

+2

tangent plane and

(r/-|- 1)

^

==

the equation of the normal.

of the tangent plane and the normal for the case of implicof a surface. When the equation of a surface is represented

= 0,

the corresponding equations will have the form

_ _ Functions of Several Variables

218

which

[Ch. 6

the equation of the tangent plane, and

is

XXQ F'X

(**,

Yy

_ *

y

ZZQ

_

Q

Z 0) F'z (*0. J/0.

(*0> 00. *0)

Fy

)

which are the equations of the normal. the equations of Example 2. Write s 8

the tangent plane and the normal to a. z a at a point for which x the surface 3;q/z 0, z/ Solution. Find the z-coordinate of the point of tangency, putting x 0, a 8 whence z a. Thus, the z* a into the equation of the surface: .j/

=

=

=

=

point of tangency

is

F

Denoting by

M (0,

and

Applying formulas

(3)

+ a=:0,

a).

the left-hand side of the their values at the point Af:

which

and

we

(4),

~r~

=

n

'

wn

^

equation,

~2 5

the

Qj/

cn are

a

e q ua ^ions of

^ ne

)

/?sina,

of

paraboloid

the normal.

and

revolution

z

= x*+y

2

the

the equaindicated

at

the

point

;

cone

b) to the c) to

the

i

points: a) to '

find

get

1981. Write the equation of the tangent plane tions of the normal to the following surfaces at

0-

we

the equation of the tangent plane,

is

x

or

=

=

,

(x, y, z)

partial derivatives

or ^-(-z

a,

=

the

= at the x*+y* + z = 2Rz

^ + -^

y-

2

sphere

point at

the

(4, 3, 4);

point

(ffcosa,

/?).

1982. At

what point

of the ellipsoid ~2

y2 _4. "" ^ f_4.^ __ f.2

a

2

b

2

c2

1

*

does the normal to it form equal angles with the coordinate axes? 1983. Planes perpendicular to the A:- and #-axes are drawn z* 169. through the point (3, 4, 12) of the sphere x* y* Write the equation of the plane passing through the tangents to the obtained sections at their common point M. 1984. Show that the equation of the tangent plane to the central surface (of order two)

M

+

ax

2

+ by

2

2

-\-cz

=k

+ =

Sec. 11]

_

The Tangent Plane and the Normal

M (x

at the point

Draw

#

,

,

z

to

the

+

Draw

1986.

which cuts

off

_

219

a Surface

has the form

)

surface x x to the 4
to

+ 2tf + 3z = 21

2

2

= 0.

^a+fi + 'T^l

to the ellipsoid

tangent

a

planes

tangent

plane

on the coordinate axes. equal segments 2 2

z* 1987. On the surface je 2*^=0 find points at which the tangent planes are parallel to the coordinate planes. s 1988. Prove that the tangent planes to the surface jq/z a of the tetrahedron of constant volume with the planes form coordinates. r f 1989. Show that the tangent planes to the surface }/ x-\-\/ y

+y

=m

+

-\

is

\/

z^Ya

cut

on the coordinate axes, segments whose sum

off,

constant.

Show

1990.

that the cone

^-i-f!

= -^

and the sphere

are tangent at the points (0, b,c). 1991. The angle between the tangent planes drawn to given surfaces at a point under consideration is called the angle between two surfaces at the point of their intersection. At what angle does the cylinder x*-\-y* R* and the sphere

=

(x-R)

2

2

!-
-#

2

intersect at the point

Affy,

^-^,

OJ?

1992. Surfaces are called orthogonal if they intersect at right angles at each point of the line of their intersection. Show that the surfaces x*+y* z* r t (sphere), y xiany 2 (** (plane), and z y*)ian*-ty (cone), which are the coordinate surfaces of the spherical coordinates r, cp, tj?, are mutually ortho-

=

gonal. 1993.

z^xf(~ * \

Show )

=

+ =

+

that all the planes tangent lo the conical surface

at the point

M

(,v

,

//

z

,

),

where x

+ 0,

pass through

/

the coordinate origin. 1994*. Find the projections of the ellipsoid /'

+z

1

on the coordinate planes. 1995. Prove that the normal revolution z

= /(/^ +

a

2

f/

)

(/'

xy

at

+ 0)

1=0 any point

of

the

intersect the axis

of

surface of rotation.

220

_

Functions of Several Variables

_

[C/i.

6

Sec. 12. Taylor's Formula for a Function of Several Variables

of all orders f (x, y) have continuous partial derivatives (rc+l)th inclusive in the neighbourhood of a point (a, b). Then Taylor's formula will hold in the neighbourhood under consideration:

Let a function

up

to the

-a) + f'y (a, b)(y-b)]

+

where

In other notation, )

+ -Jy

[/tfX

if

...+[^^ or

-j

df

(x, y)

+

2 rf

/ (x, y)

+

.

.

.

The particular case of formula (1), when a b Q, is called Maclaunn's formula. Similar formulas hold for functions of three and a larger number of variables. A3 Example.' Find the increment obtained by the function f (x, y) when passing from the values 1 to the values *,-- -{-//, 1, y

=

*=

1

Solution. The desired increment may be found by applying formula calculate the successive partial derivatives and their values at

First

given point

(1, 2):

= 3- 1+3.2=9, = /;il,2)=-6.4 + 3.1 -21 fxx (\, 2) = 6-l=6 /;

x

f

(1,2)

f

^(1,2)= -12.2= -24,

f

(2).

the

Taylor's Formula for a Function of Several Variables

Sec. 12]

are

All subsequent derivatives into formula (2), we obtain:

~

2

[/i

-64-2/z k- 3-f

1996.

powers

Expand h and

of

2 fc

f (x i fe

+ ^y

24))

(

ft,

y

zero.

identically

8

-

[/i

+ k)

G

+ 3/z

2

+

fc.

in a series

of

2

221

these

Putting

3/ife

_

-0-f *'(

positive

results

12)]

integral

if

=

2

+

+

* 1997. Expand the function f (x, y) 6x 2xy 3y* 4 by Taylor's formula in the neighbourhood of the point 2/y

(-2,

1).

1998. Find the increment received by the function ^-x*y when passing from the values x=l, {/=! to

4*

tegral

the

yz

neighbourhood

of

(1, 1, 1).

Expand

powers

of

f (x /?,

t

k,

/ (x, y, z)

2001.

f(x, y, z)

z-l-4 by Taylor's formula in

3#

the point 2000.

= Jc'-fy + 2* +2xy 1

Expand the function

1999.

f(x,y)~

=--=

ft,

x

*

/)

in a series

of

positive in-

if

/,

2

z-|

fr,

//-J

and

// 4 z

2

2A:e

-

2jr//

Expand the following function

2yz.

a

Maclaurin's series

in a

Maclaurin's series

in

up to terms of the third order inclusive: /(.Y,

2002.

//)

= ?* sin//.

Expand the following function

up to terms

of order four inclusive: / (x, //)

= cos x cos y.

2003. Expand the following function in a Taylor's series in the neighbourhood of the point (1, 1) up to terms of order two inclusive: /(*.

{/)

= {/*

2004. Expand the following function in a Taylor's series in the neighbourhood of the point (1, 1) up to terms of order three inclusive:

_ _

222

[Ch. 6

Functions of Several Variables

2005. Derive approximate formulas (accurate to second -order terms in a and P) for the expressions

if

and |p| are small compared with unity. 2006*. Using Taylor's formulas up to second-order

|a|

terms,

approximate

^O98;

1/T03;

a)

2 01 -

b) (0.95)

.

an implicit function of x and y defined by the + y = 0, which takes on the value z= 1 for x= 1 equation and y=l. Write several terms of the expansion of the function z in increasing powers of the differences and y 1. 2007. z z

is

9

2xz

x\

Sec.

13. The

Extremum

of

a Function

of Several

Variables

1. Definition of an extremum of a function. We say that a function f(x,y) has a maximum (minimum) f (a, b) at the point P (a, b), if for all points P' (x, y) different from P in a sufficiently small neighbourhood of P the inequality /(a, b) > f(x, y) [or, accordingly, /(a, b) < f (x y)] is fulfilled. The generic term for maximum and minimum of a function is extremum. In similar fashion we define the extremum of a function of three or more t

variables.

2. Necessary conditions for an extremum. The points at which a differentiate function f (x, y) may attain an extremum (so-called stationary points) are found by solving the following system of equations:

0)-0,

t' x (x.

f't/

(x

t

y)-Q

(1)

(necessary conditions for an

extremum). System (I) is equivalent to a single 0. In the general case, at the point of the extremum equation, df(x, #) P (a, b), the function f (x, y), or df (a, ft) = 0, or df (a, b) does not exist. 3. Sufficient conditions for an extremum. Let P (a, b) be a stationary point of the function f(x, y), that is, df (a, &)- 0. Then: a) if d*f (a b) < for dx z dy*>Q then /(a, b) is the maximum of the function f(x, //); b) if d z f(a, ft)>0 for d* 2 -}- di/ 2 then /(a, b) is the minimum of the function 0, 2 /(* 0); c ) if d /(a, ft) changes sign, then f (a, b) is not an extremum of /(v, //). t

+

t

>

The foregoing conditions are equivalent

= f'y (a,

ft)

-0

Then: I) if P(a, ft), namely (or

C>0);

2)

A=f xx

and

if

then the question (which is to say,

>

A

0,

(a,

then

maximum,

a

A

<

0,

then

ft),

B~fxy

the if

an extremum

(or is

to the

ft),

C

<

following: let

C = /^(ci,

has

function

A<

there

(a,

an

0),

f[ (a,

We

ft).

form

extremum at a minimum,

and

no extremum

at

P

(a

t

b)----

the

the point if

A

>

A==0. remains open

ft);

3)

if

of the function at P (a, ft) requires further investigation). 4. The case of a function of many variables. For a function of three or more variables, the necessary conditions for the existence of an extremum of it

Sec.

13]

_

The Extremum of a Function

of Several Variables

_

223

are similar to conditions (1), while the sufficient conditions are analogous to the conditions a), b), and c) 3. Example 1. Test the following function for an extremum:

Solution. Find the partial derivatives and form a system of equations

(1):

or r

**

\ xy

we

Solving the system

Let us find

(

(2,

P 4 (_2,-1).

P,(-l,-2);

1);

second derivatives

tiie

d2 z adx 2

=c

6.v,

d2z 3 T-

dxdy

For the pomt

= 6, A^=4C

2

P

t

:

= 36

<

144

0.

6r/,

for

c = 6x

dy

each stationary point.

=6. B =

A = (g} 2

\dx Jp

d*z T-22

= ry

B2

and form the discriminant A=^/4C 1)

/

get four stationary points:

Pt

P,(l,2);

+ *_5-0, 2 = 0.

=12, C=(g) = (fL\ \dy J p, \dxdyjp, 2

l

Thus, there

2) For the point P 2 4 --12, B^6, C-12; the function has a minimum. This minimum function for A -2, y~\' :

no extremum at the point P,.

is

A

= 144

is

36

equal

> 0,

>

/I

0.

the value

to

3012^28. 6; A = 36 ^-6, 12, C^ ^- 12, B = 6, C= 12; A = 144

At P 2 of

the

)

i

3) For the point P 9 no extremum. 4) For the point P 4

:

:

At the point

P4

fi---

the function has a

maximum

equal to

2 ma x

144

36

< 0.

>

0,

^

There

A

< 0.

6-f-30-{-

4- 12 ---28 5*.

of

Conditional extremum. In the simplest case, the conditional extremum

a function /(A,

attained <|)(jr,

on

//)

is

a

maximum

condition

that

its

or

//),

minimum

this function which is related by the equation the conditional extremum of a funcwe form the so-called Lagra
arguments

(coupling equation). To find given the relationship q> (A-,

w)

tion /(A-,

the

of

are

=

function F(A-,

where X

is

y)-=f(

an undetermined

multiplier, and

we seek

the ordinary

of this auxiliary function. The necessary conditions for the to a system of three equations:

with three unknowns to

x,

t/,

X,

determine these unknowns.

from which

it

is,

generally

extremum

extremum reduce

speaking,

possible

_ _

[Ch. 6

Functions of Several Variables

224

The question of the existence and character of a conditional extremum is solved on the basis of a study of the sign of the second differential of the Lagrange function: -

yi for the given

system

dx 2

y

dxdy

of values of x, y,

dy

2

h obtained from

(2)

or the condition

that dx and dy are related by the equation

<

Q, and a Namely, the function / (x y) has a conditional maximum, if d*F conditional minimum, if d 2 F 0. As a particular case, if the discriminant A of the function F (x, y) at a stationary point is positive, then at this point there is a conditional maximum of the function / (x, y), if A (or C 0), and a conditional minimum, if A 0) (or C In similar fashion we find the conditional extremum of a function of three or more variables provided there is one or several coupling equations (the number of which, however, must be less than the number of the variables) Here, we have to introduce into the Lagrange function as many undetermined multipliers factors as there are coupling equations. Example 2. Find the extremum of the function t

:

>

<

>

>

z=:6

4*

<

3y

provided the variables x and y satisfy the equation

x*-\-y*=\ Solution. Geometrically, the problem reduces to finding the greatest and 6 4.v plane z 3y for points of its intersection with the cylinder ji 2 -f// 2 =l We form the Lagrange function least values of the e-coordinate of the

F(x, y)--=6

We

have

T=~

following system

4

= ***-

+ 2>jr,

-

4x-3f/-l-M* 3

+ 2X#.

The

of equations:

: i

Solving this system

we

find i

-

5

^-"2"'

X

_

4

'-~5~'

and 2

____^_ ~" 2

1

___

^~""5"

Since

dx 2 it

follows that

-*"'

dxdy

=0,

f

2

2

-|-{/

1).

necessary conditions yield the

Sec.

The Extremutn

13]

54

~__

x

and

-=-

o

2

3 = --, o

Function of Several Variables

thend 2 /7 >0,

and,

225

the function

4^3

consequently, 5

=

=

-- and f/ -=O D and, consequently, the function at this point has a conditional

has a conditional z

F
then d

/

of a

minimum

at this point.

If

K-

-^ Z

x

,

,

Thus,

= 6 + i5 + -=ll,

z max

A

6. Greatest and smallest values of a function. rentiable in a limited closed region attains its greatest (smallest) value either at a sta-

function

that

is

diffe-

tionary point or at a point of the boundary of the region.

Example

3.

Determine the greatest and

smallest values of the

function

in the region

A'<0, [/<0, x + y^z

3

Solution. The indicated region is a angle (Fig. 70). 1) Let us find the stationary points: I

z' K

|

7 {J

~ 2xy 1=0, ^ 2y x -}-1^0; \-

whence x-= 1, //-= At A1 the value to test for

tri-

70

M

and we get the point 1 the function ZM

1;

of

=

It

(

1,

is

not absolutely necessary

an cxtrcmum

1)

Let us investigate the function on the boundaries of the region. we have 2 [/ 2 -f-f/, and the problem reduces to seeking the When A greatest and smallest values of this function of one argument on the interval 2)

=

3^//^0.

Investigating,

=

(2 sm ) v _

(3,0);

When

find

get

(Zsm) y =*

x-[-y

=

z

that

(2g r ) x=0

T" at thc 3

A-

P oint

we

find

that

(z &m ) x

+

,,^^^~r

we

(~ V 2

will

3

we

the

at

point

3);

(0,

l^)

x z -\-x. Similarly,

3 or //-=

=6

l

at the point (0,

When //~0 we point

we

at the point

'

(

(2g r ) v=0

=6

at

the

)

have /

that

find

z

= 3A

3 -^

2

-f-9A'-j-6.

3 \ ~

,

J

;

Similarly

(2 gr )

+ =

3 metres coincides with (z gr ) x =o anc (^r).jf=o- ^ n * ne straight line jc ^ we could test the Function for a conditional extremum without reducing to a function of one argument. 3) Correlating all the values obtained of the function z, we conclude at the stationary 1 that z gr 6 at the points (0, 3) and (3, 0); z sm ^

=

point

8-1900

M.

=

226

_ _

[Ch. 6

Functions of Several Variables

maximum and minimum

Test for

the following

functions of

two variables:

2013.

= (x \)*+2y*. = (x I) 2y*. 2x y. z = x* + xy + tf = z *y (6 xy)(x>0 2 = x +y 2x*+4xy 2y z =x

2014.

z=l

2008. z

2

2009. z 2010. 2011. 2012.

9

= 2 =

2015. z 20,6.

4

4

.

(

Find the extrema 2 2017. a = x

of the 2

+ +z

=

2018.

z

xy

^++

Find the extrema

functions

following

2

f/

+

+x

of

three variables:

2z.

-(^>0, y>0, z>0).

of the following implicitly represented func-

tions:

2019*. x* 2020. x

9

+ y* + z* 2x+4y 62 11=0. y*3x + 4y + z* + z 8 = 0.

Determine the conditional extrema 2021. z

= xy

for

+ 2y z = x* + if

2022. z^=x

for

2023.

for

2027.

= cos + cos W= 2y + 2z ^=^ + +2 u = xtfz*

2028.

u^xyz

2

2024. z 2025.

2026.

A:

2

2

-=

+

a

A:

8

f/

f/

for// for

for

of the following functions:

*

A: 2

A:

+ -

= ~-.

r

+

for .v-f-y

-

+ z=

12(jc>0,

provided x-(~j/+e=5,

xy+yz+zx=8.

2029. Prove the inequality

if

x^zQ Hint: Seek the

maximum

of the

function u

= xyz

provided

Finding the Greatest and Smallest Values of Functions

Sec. 14]

in

227

z=l+x + 2y

2030. Determine the greatest value of the function the regions: a) x>0, y^*0, x y^l\ b)

+

2031. Determine the greatest and smallest values of the func2 2 #* tions a) z y in the region x?+y *^l. x*y and b) z 2032. Determine the greatest and smallest values of the func-

=

=

tion

z

= sinx-h sin y-f- sin (x + y)

in

the

O^jt^-2.-

region

2033. Determine the greatest and smallest values of the func1 x' tion z 3xy in the region y*

=

+

^

0^x^2,

14. Finding the Greatest and Smallest Values of Functions

Sec.

Example t. It is required to break up a positive number a into three nonnegative numbers so that their product should be the greatest possible. Solution. Let the desired numbers be x, y, a x We seek the maxi//.

mum

the function / (x, y) xy(a to the problem, the closed triangle x^O, y^zQ, x of

According

+ y^a

x y). function

/ (x,

is

y)

considered inside a

(Fig. 71).

Fig. 71

Solving the system of equations

f'(x,

we

will

have the unique stationary point

(-TTI

triangle. Let us test the sufficiency conditions.

>

(/)

=

20,

T We

(x, y)=a2x2y, f IXIf

^

or

* ne

j

have f" yy'(x,

/)

=

of

the

_ _

228

[Ch. 6

Functions of Several Variables

Consequently,

And

so

at

f-g-,

~)

the function reaches a

the contour of the triangle, this

maximum

to say that the product will be greatest, a3 greatest value is equal to is

maximum.

Since f(x, */)=0 on

will be the greatest value, if

x

=y=a

x

//

which

= --, and

the

.

-^=-

Note

The ploblem can

also be solved

extremum, by seeking the maximum that x

+ y + z = a.

2034.

From among

all

of the

by the methods of a conditional function u xyz on the condition

rectangular

=

parallelepipeds

with

a

given volume V, find the one whose total surface is the least. 2035. For what dimensions does an open rectangular bathtub given capacity V have the smallest surface? 2036. Of all triangles of a given perimeter 2p, find the one that has the greatest area. 2037. Find a rectangular parallelepiped of a given surface S

of a

with greatest volume. 2038. Represent a positive number a in the form of a product of four positive factors which have the least possible sum. on an x^- plane, the sum of 2039. Find a point (x, y), the squares of the distances of which from three straight lines = Q, f/ = 0, x y+l=0) is the least possible. (x 2040. Find a triangle of a given perimeter 2p, which, upon being revolved about one of its sides, generates a solid of greatest volume. 2041. Given in a plane are three material points P, (x lf # ), P*( x z' #2)* ^i(*> #3) w ith masses m lf m 2 m 3 For what position of the point P(x, y) will the quadratic moment (the moment of inertia) of the given system of points relative to the point P 2 P2 P2 P 3 P 2 ) be the least? (i.e., the sum m.P.P 2 3 2042. Draw a plane through the point c) to form (a, b a tetrahedron of least volume with the planes of the coordinates. 2043. Inscribe in an ellipsoid a rectangular parallelepiped of greatest volume. 2044. Determine the outer dimensions of an open box with a given wall thickness 8 and capacity (internal) V so that the smallest quantity of material is used to make it.

M

t

,

+m

+m

.

M

t

Sec.

14]

Finding the Greatest and Smallest Values

of

229

Functions

2045. At what point of the ellipse w2

r2

-Ta2

does the tangent line to angle of smallest area? 2046*. Find the axes

2047.

Inscribe in surface.

a

z

'

form with the coordinate axes a

it

of

5x

+ TT-1 b tri-

the ellipse

2 -f-

8xy + 5tf

given

= 9.

sphere

a

cylinder

having

the

greatest total

2048. The beds of two rivers (in a certain region) approxi2 = 0. y mately represent a parabola y = x* and a straight line x It is required to connect these rivers by a straight canal of least length. Through what points will it pass? 2049. Find the shortest distance from to the straight line x __ ~~ 1

_

y

~~ 3

the

point

M(l,

2,

3)

2 '

2

2050*. The points A and B are situated in different optical media separated by a straight line (Fig. 72). The velocity of

*1

Fii<.

72

Fig. 73

light in the first medium is v l9 in the second, v 2 Applying the Fermat principle, according to which a light ray is propagated along a line AMD which requires the least time to cover, derive the law of refraction of light rays. 2051. Using the Fermat principle, derive the law of reflection of a light ray from a plane in a homogeneous medium (Fig. 73). 2052*. If a current / Hows in an electric circuit containing a .

resistance

/?,

proportional

then the quantity of 2

to /

/?.

heat

released in unit time is / into

Determine how to divide the current

230

_ _

currents / t

#i

[Ch. 6

Functions of Several Variables

,

72 ,

so

#a

R*> possible?

/,

by means

that

the

of three wires,

generation of

whose resistances are

heat would be the

least

Sec. 15. Singular Points of Plane Curves

1. Definition of a singular point.

=

is

f(x, #) at once:

2. Basic

called a singular point

if

A

At

point Af (x c y Q ) of a plane curve coordinates satisfy three equations ,

its

M (*

let

the

a node (double point) (Fig. 75); is either a cusp of the first kind (Fig. 76) or of c) if second kind (Fig. 77), or an isolated point, or a tacnode (Fig. 78).

the

types of singular points.

a singular point

,

#

),

second derivatives

be not

all

equal to zero and

then: a)

b)

if if

A>0, A<0,

A = 0,

then then then

M M M

is

A = /1C-B 2

,

an isolated point (Fig.

74);

is

Fig. 74

Fig. 75

When solving the problems of this section it is always necessary to draw the curves. z x* has a node if a ax* 0; an Example 1. Show that the curve y isolated point if a 0. 0; a cusp of the first kind if a 2 z Solution. Here, f (x, y)z=zax Let us find the partial derivatives and equate them to zero:

=

<

+ x'y

fx

(x,

t/)

=

.

+

=

>

Singular Points of Plane Curves

Sec. 15]

system has two solutions:

This

coordinates of the point

Hence, there

is

a

N

do not

0(0, 0) and the

satisfy

unique singular point

(0,

N

231

(

equation

--~a

Oj,'btit

the

0).

Fig. 78

Fig. 77

Fig. 76

t

of the given curve.

Let us find the second derivatives and their values at the point 0: ,

4=20, 0=0,

a =0 fa>0

Pig.

Fig. 80

79

Fig. 81

Hence, if

a>0,

if

a

if

2 {/

=x

< 0,

a--^0, 8

or

#=

then then then

A<0 A

>

Aj^O. y which

Y^\

about the x-axis, first kind (Fig. 81).

is a node (Fig. 79); and the point is an isolated point (Fig. 80); and The equation of the curve in this case

exists only is a tangent.

when Jc^O;

the curve

Hence, the point

M

is

is

will

be

symmetric

a cusp of the

232

Functions of Several Variables

Determine the character

the

of

[Ch. 6

singular points of the follo-

wing curves: 2053. y* 2054. (y

=

x* -\-x\

2055.

=* ay=aV-x'.

2056.

jtyjt

2057. x*

x

+y

2

8

5

.

)

2

9

f/^O.

3axy = Q (folium

2061.

of Descartes).

= (cissoid). = a*(x y (lemniscate). + = (a x)x (strophoid). + = ftV (a>0, 6>0) (x* + y*)(x a)* 3

2058. y*(a x) 2 2059. (x* y*) 2060. (a x)y*

jc

2

z

)

2

(conchoid).

Consider three cases: 1)

a>6,

a

2)

= 6,

a<6.

3)

2062. Determine the change in character of the singular point 2 curve y (x b) (x a)(x c) depending on the values of

=

of the a,

6,

c(a
are real).

Sec. 16. Envelope

t. Definition of an envelope.

The envelope of a family of plane curves curve (or a set of several curves) which is tangent to all lines of the given family, and at each point is tangent to some line of the given family. 2. Equations of an envelope. If a family of curves is

a

f(x

dependent on

a)=0

y,

t

a single variable parameter of the latter are found

metric equations

a has an envelope, then the parafrom the system of equations

= 0, a) = o.

t

f(x. y, a)

\

&(*.

y.

Eliminating the parameter a from the system the form

D(x,

0)

(1)

(1),

we

= 0.

get an equation of (2)

It should be pointed out that the formally obtained curve (2) (the *>ocalled "discriminant curve") may contain, in addition to an envelope (if there is one), a locus of singular points of the given f amily, which locus ts not part of the envelope of this family. When solving the problems of this section it is advisable to make

drawings.

Example. Find the envelope

*cosa+f/sina

of

the family of curves

p

= 0(p = const,

p>0).

Sec. 16]

233

Envelope

Solution. The given family of curves depends on the parameter a. the system of equations (1):

J \

Solving the envelope

system

for

x

*cosa + y sin a x

sin

x and y

a+y t

cos

p = 0,

a

= 0.

we obtain parametric equations

= pcosa,

r/

of the

=

Squaring both equations and adding, we eliminate the parameter

frig.

Form

a:

82

Thus, the envelope of this family of straight lines is a circle of radius p with centre at the origin. This particular family of straight lines is a family of tangent lines to this circle (Fig. 82).

2063. Find the envelope of the family of circles

2064. Find the envelope of the family of straight lines

(k is a variable parameter).

2065. Find the envelope of a family of circles of the same radius R whose centres lie on the jc-axis. 2066. Find a curve which forms an envelope of a section of length / when its end-points slide along the coordinate axes. 2067. Find the envelope of a family of straight lines that form with the coordinate axes a triangle of constant area S. 2068. Find the envelope of ellipses of constant area S whose axes of symmetry coincide.

Functions of Several Variables

234

2069. Investigate the character

the "discriminant curves" a constant parameter):

of

of families of the following lines (C

=

=

;

=

+ x) (y

d) strophoids (a

is

8

cubic parabolas y (x C) 2 b) semicubical parabolas t/ (x 2 C) (x c) Neile parabolas y* a)

[Ch. 6

C)*;

;

=*

8

C)

2

x).

(a

Fig. 83

2070.

The equation

trajectory of a shell fired from a v at an angle a to the horizon

of the

with initial velocity point resistance (air disregarded) is

Taking the angle a as the parameter, find the envelope of all trajectories bf the shell located in one and the same vertical plane ("safety parabola' ) (Fig. 83). 1

17. Arc Length of a Space Curve

Sec.

The dinates

where

differential of equal to

an arc

of a

space curve in rectangular Cartesian coor-

is

current coordinates of a point of the curve.

z are the

x, y,

If

are parametric equations of the space curve, then the arc length of a section of

it

from

t

=

t

l

to

t

=

t2

/a

=

f j

is

M

1/7 <M + (dy\* + ( dz y (-dr 2

,

,

(-df)

The Vector Function

Sec. 18]

of a Scalar

235

Argument

In Problems 2071-2076 find the arc length of the curve:

=

2071. x

t,

y=/

2

2-~-'

,

from

to

t

= 2.

= 2 cos y = 2 sin z = -|- from == to = = *' cos y = e* sin z = e from = to arbitrary t from JC==0 to x==6 y = 4~' 2 = 4~ * = 3(/, 2jcy = 92 from the point (0, 0, 0) to M (3, 3, 2). = aarcsin~, z = -|-ln ^j from the point 0(0,0,0)

2072. x

/,

t

/

1

t,

it.

t

2073.

A:

2074.

/,

f

/,

.

-

f

2075. 2076.

f/

M(*

to the point

,

j/

z

,

).

2077. The position of a point for any time by the equations

Find the mean velocity Sec.

=

/

of

(f>0)

f

motion between times

f

=

l

is

defined

and ^=10.

18. The Vector Function of a Scalar Argument

1. The derivative of the vector function of a scalar argument. The vector a (0 may be defined by specifying three scalar functions ax (t) a (t) and a z (t) which are its projections on the coordinate axes: function a

9

t

y

The derivative of the vector function a-=a(t) with respect to the scalar argument t is a new vector function defined by the equality

da The modulus

a(t

+ M)-a(t)_da x (t)

.

day (0

.

da f

(t)

of the derivative of the vector function is

da dt

The end-point

of the variable of the radius vector

r=r(/) describes

in

space

the curve

r=x(t)l+y(t)J+*(t)*. which

is

called the hodograph of the vector r.

The derivative

-~

is

a vector, tangent to the

hodograph at the corre-

sponding point; here,

dr [

\

where

s is

dt

[_ \

ds

dt

'

the arc length of the hodograph reckoned from

For example,

Up

some

initial point.

236

__ If

the

parameter

Functions of Several Variables

the time, then -jT

is

t

=

tf

is

_

[Ch. 6

the velocity vector of the

=w

=

is * ne acceleration vector of the and JTS "^T of vector the r. extremity 2. Basic rules for differentiating the vector function of a scalar argument.

extremity

of the vector r,

= m-~-,

2)

-77-

(ma)

3)

-77-

(cpa)==-~-a

4,

7)

where

+ (p-^-

,

m

is

a constant scalar;

where q>(0

a scalar function of

is

<*,_..+..:

a-~=0,

if |

a

|-= const.

Example 1. The radius vector time defined by the equation

r=i Determine the trajectory Solution.

From

Eliminating the

(1)

time

of

a

of

moving point

of

motion, the velocity and acceleration.

t,

we

find that the trajectory of 1

~~

u

z '

differentiating,

we

find the velocity

and the acceleration

The magnitude

any instant

(1)

4"" 3 (1),

at

we have:

x

From equation

is

4t*j+3t*k.

line:

We

/;

of the velocity

note that the acceleration

\*L

\dt*

is

is

constant and

= ]/(_ 8) +6 2

2

is

=iO.

motion

is

a straight

Sec.

_

18]

Show

2078.

where r

The Vector Function

l

the

that

of

a Scalar Argument

vector

and r 2 are radius vectors

_

= (/* 2 equation two given points,

rr,

of

237

r,) is

/,

the

of a straight line. 2079. Determine which lines are hodographs of the following vector functions:

equation

a)

= at -f c\ = a/ + r r

c)

2

b)

ft/;

d)

= a cos -f b sin r = a cosh -f 6 sinh

r

t

t\

/

/,

a, ft, and c are constant vectors; the vectors a and b are perpendicular to each other. 2080. Find the derivative vector-function of the function a(t)a (/), where a(t) is a scalar function, while a(/) a(t) is a unit vector, for cases when the vector a(t) varies: 1) in length only, 2) in direction only, 3) in length and in direction (general case). Interpret geometrically the results obtained. 2081. Using the rules of differentiating a vector functisn with respect to a scalar argument, derive a formula for differentiating a mixed product of three vector functions a, 6, and c. 2082. Find the derivative, with respect to the parameter t, of the volume of a parallelepiped constructed on three vectors:

where

=

a = i + tj+t z k,

2083.

The equation

of

r

motion

is

= 3/cos/~j-4/sinf,

is the time. Determine the trajectory of motion, the the acceleration. Construct the trajectory of motion and velocity and the vectors of velocity and acceleration for times, / = 0,

where

/

.

2084. The equation of motion

is

Determine the trajectory of motion, the velocity and the accelWhat are the magnitudes of velocity and acceleration and / = and what directions have they for time = -y? 2085. The equation of motion is

eration.

r= i cos a cos at -\-jsmacostot + k sincof, where a and o are constants and / is the time. Determine the trajectory of motion and the magnitudes and directions o! the velocity and the acceleration.

-

_

Functions of Several Variables

238

2086. The equation of motion

of

shell

a

\Ch. 6

air re-

(neglecting

is

sistance)

where V Q {V OX v oy v oz } is the initial velocity. Find the velocity and the acceleration at any instant of time. 2087. Prove that if a point is in motion along the parabola ,

,

2

z/=

,

z

manner

in such a

on the x-axis remains constant

that

the

of

projection

[-= const],

velocity

then the accelera-

tion remains constant as well.

2088. A point lying on the thread into a beam describes the spiral

of

a

screw being screwed

where 9 is the turning angle of the screw, a is the radius of the screw, and h is the height of rise in a rotation of one radian. Determine the velocity of the point. 2089. Find the velocity of a point on the circumference of a wheel of radius a rotating with constant angular velocity co so that its centre moves in a straight line with constant velocity V Q .

Sec. 19. The Natural Trihedron of a Space Curve

At any nonsingular point

M

(;c,

//,

sible to construct a natural trihedron

z) of a

space curve r

consisting

of

three

r(t)

is

it

pos-

mutually perpen-

dicular planes (Fig. 84):

M A4jA4

containing the vectors

1)

osculating plane

2)

normal plane

3)

rectifying plane MMjAIj, which

MM*M

S,

2

,

which

is

perpendicular to is

At the intersection we obtain three straight lines; 1) the tangent 2) the principal normal MAf 2 of which are defined by the appropriate vectors: 1)

2) 3)

T=-rr

(the vector of the tangent l'me)\

^-gfx^p

(

NBXT (the

the vect

r

f tne

binormal);

vector of the principal normal)]

The corresponding unit vectors

T

o

and

-r-^

the vector

JV

;

3)

;

~

and

at

perpendicular to the first

MM^

all

TT

two planes.

the binomial

MM

9t

The Natural Trihedron

Sec.

19]

may

be computed from the formulas

dr

t= dF

v

;

of

a Space Curve

239

ds

=

dt ds

If X K, Z are the current coordinates of the point of the tangent, then the equations of the tangent have the form f

X-x = Y

Z

y==

'

3C

*

z

(I)

'2

if

Normal Rectifying

plane

plane

Osculating plane

Fig. 84

where of

Tx

--

;

Tv=

~

the line and the plane

T 2 = -~

,

we

;

from the condition

get an equation

of the

Z

2)

of

perpendicularity

normal plane:

= 0.

T x Ty 7\ by B x

(2)

Bv

B z and A^.

(1) and (2), we replace get the equations of the binomial and the principal normal and, respectively, the osculating plane and the rectifying plane. Example t. Find the basic unit vectors T, v and p of the curve If

in

Wy,

equations

^V^,

,

at the point

i

at this point.

Solution.

and

,

,

= 1.

Write the equations mal

,

we

We

have

of

the tangent, the principal normal and the binor-

[Ch. 6

Functions of Several Variables

240

Whence, when

1,

we

get dt

=

dt

X

2 3 ^= 026 1

dt i

J

>'

1

2

3

662 Consequently, T

Since for

= <+2/+3A

P=

3/-ay+*

V

?=1 we have *=1, y=l, 2=1,

~

'

-1U-8/+9*

follows that

it

"~

""

2

1

3

are the equations of the tangent,

x\_y ~~

l_z ~~ 3

3

1

1

are the equations of the binomial and

*-l

z-1

y-1

9

8

11

are the equations of the principal normal. If a space curve is represented as an intersection of F(JC, y, z)

then in place

of

the vectors

-^-

= 0, and

G(x, TT-Z

y, 2)

two surfaces

= 0,

we can take

the vectors dr{dx, dy, dz}

and d 2 r {d*x, d*y, d z z}; and one of the variables x, y, z may be considered independent and we can put its second differential equal to zero. Example 2. Write the equation of the osculating plane of the circle *2 at its point

M(l,

1,

+ + z = 6, 2

2

J/

x

+ y + z^Q

(3)

2).

Solution. Differentiating the system (3) and considering x an independent we will have

variable,

x dx

+ y dy + z dz

--=

0,

and dx*

+ dy + y d*y + dz + z d*z .= 0, 1

2

d 2 (/ Putting

x=l, y=\>

z~2,

we

get

The Natural Trihedron of a Space Curve

Sec. 19]

Hence, the osculating plane

dx

{dx,

t

241

defined by the vectors

is

-

and

0}

|o,

2

yd*

,

jdx*\

or 1,

{1,

and

0}

1,

{0,

1}.

Whence the normal vector of the osculating plane J

B=

=

1

1

-1 and, therefore,

its

equation

is

lj

k

1

is

-l(x-l that

as

is,

it

should be, since our curve

is

located in this plane.

2090. Find the basic unit vectors

x^l at the point

f

= -g-

T, v,

y=sin/,

cosf,

p of the curve z

=

t

*

2091. Find the unit vectors of the tangent and the of the conic spiral

principal

normal

at an arbitrary point. Determine the angles that these lines with the z-axis. 2092. Find the basic unit vectors r, v, p of the curve

y

=

x*,

z

make

= 2x

2. at the point x 2093. For the screw line

y

= asmt,

z

= bt

write the equations of the straight lines that form a natural trihedron at an arbitrary point of the line. Determine the direction cosines of the tangent line and the principal normal. 2094. Write the equations of the planes that form the natural trihedron of the curve x*

one

2

-1- 1/

+ * = 6, 2

x

2

if

-1-

z

2

-4

points M(l, 1, 2). the equations ot the tangent line, the normal z t* the and t*, /, osculating plane of the curve * plane y at the point (2, 4, 8).

at

of

2095.

its

Form

M

=

=

=

_ _

[Ch. 6

Functions of Several Variables

242

2096.

Form

and binormal

the equations of the tangent, principal normal, an arbitrary point of the curve

at

Find the points at which the tangent to to the plane x 10 = 0. 3y-\- 2z

this curve is parallel

+

2097. Form equations of the tangent, the osculating the principal normal and the binormal of the curve

at the point / at this point.

= 2.

Compute the

direction cosines of the binormal

2098. Write the equations of the tangent and plane to the following curves: a)

x = R cos 2

b)

z=x*+y*, x = y *

C)

2

/,

y = R sin /cos/,

+ y + z = 25, 2

2

z

= Rsmt

for

/

x+z=5

=

=

2/3,

at the point (2,

normal Find the equation y = x at the coordinate origin. 2100. Find the equation of the osculating plane

*

normal

the

?-;

at the point (1,1, 2);

2099

z=x

plane,

of

3).

plane to the curve

the

z

if,

*,

(/

= <>-',

2101. Find 2

* 2 b) * 2

a)

c)

JC

to the

curve

=

= ty2

0. at the point / the equations of the osculating plane to the curves:

2

+y + 2 = 9, x y = 3 at the point (2, 1, 2); = 4y, x' = 24z at the point (6, 9, 9); + z = a y fz = 6 at any point of the curve (x 2

2

2

2

2

2

2

2

2

,

QJ

y ot z

).

Form

the equations of the osculating plane, the principal normal and the binormal to the curve

2102.

z

y 2103.

Form

= x,

*

2

=z

at the point (1,

1,

1).

the equations of the osculating plane, the princi-

normal and the binormal to the conical screw-line A; = /COS/, j/=/sin/, z~bt at the origin. Find the unit vectors of the tangent, the principal normal, and the binormal at the origin. pal

Sec. 20. Curvature and Torsion of a Space Curve

1. Curvature.

By

the curvature of a curve at a point

number *

/(

R

=

lim AS-+O

JL, As

f

M

we mean the

Curvature and Torsion

Sec. 20]

of

a Space Curve

243

the angle of turn of the tangent line (angle of continence) on a curve MN, As is the arc length of this segment of the curve. segment R is called the radius of curvature. If a curve is defined by the equation r=r(s), where s is the arc length, then

where

(p

is

of the

For the case

we have

parametric representation of the curve

of a general

(1)

2. Torsion. By mean the number

torsion (second curvature) of a curve

at

a

point

M

we

-1 r-l-ihn As-*o As Q

the angle of turn of the binormal (angle of contingence of the where second kind) on the segment of the curve MN. The quantity Q is called the radius of torsion or the radius of second curvature. If r=r(s), then is

drd 2 rd sr ds ds 2 ds 3

dp_

ds

where the minus sign

is

direction, and the plus If /=/(/), where t

taken when the vectors

when

sign,

and v have

the

an arbitrary parameter, then

is

1

=

dr d 2r dV d/d?" d7 -

(2)

Q

~d,

dt

Example

1.

Find the curvature and the torsion of the screw-line r

Solution.

We

same

not the same.

= i a cos

t

-\-j

a sin

/

+ k bt

t

+ja cos t + kb

(a

> 0).

have

~= _/ a

_

sin

t

2

d r

==

/

a cos

/

= _/ a sin/

/a

sin

/,

cos/.

Ja

Whence k

J a sin a cos

t

/

a cost b -a sin

/

=

i

ab sin

/

jab cos

t

+ a*k

Hence, on the basis

formulas

of

1

JDL

R ~(

we

(2),

+ 6i)

a

"fl'

/.

get

a

VaT+b*^ i

fl

and

(1)

+a

1

and 1

~

Q

a*b

a 2 (a 2

b

=

+ b*)~~ a + b 2

2

'

Thus, for a screw-line, the curvature and torsion are constants. 3 Frenet formulas:

dr

_

v

^v '

7s~"~R

2104. Prove

___ ~~

5s

T

iP

^P_

v

'

ds~~~~o"'

~~~~R~T~~Q

the curvature at all points of a line is a straight line. 2105. Prove that if the torsion at all points of a curve is zero, then the curve is a plane curve. 2106. Prove that the curve that

zero, then the line

if

is

x=\+3t + 2t is

a

2 ,

y

= 22t + 5t\ z=lt*

plane curve; find the plane in which it lies. Compute the curvature of the following curves:

2107. a)

x = cost, y = s'mt, 2 2 2 z = l, y -| //

b) x*

2108. curves: a)

b)

z

= cosh

= e'cos/,

=

at the point / 0; at the point (1, 1,

Compute the curvature and

y = e sint, z x^acosht, y asiuht.

je

/

-2x + z = Q

1).

torsion at any point of the

i

e*\

z

= at

(hyperbolic screw-line).

2109. Find the radii of curvature and torsion at an arbitrary point (x, y, z) of the curves: 2

x 9 b) x

a)

= =

2110. Prove that the tangential and normal are expressed by the formulas acceleration v2 dv VOif T <MJ V V,

w

WT-^T,

Wv-

components

of

R

where v is the velocity, R is the radius of curvature of the T and v are unit vectors of the tangent and principal trajectory, the to curve. normal

Sec. 20]

2111.

=

_

_

la cost

tion w. 2112.

Curvature and Torsion of a Space Curve

A

+ja

sin

in

is

point t

+ btk

The equation

Determine,

at

trajectory and acceleration.

times 2)

the

/

of

24S

uniform motion along a screw-line r =* with velocity v. Compute its accelera-

motion

is

and /=!: 1) the curvature of the tangential and normal components of the

Chapter VII

MULTIPLE AND LINE INTEGRALS

Sec. 1. The Double Integral

in

Rectangular Coordinates

1. Direct computation of double integrals. The double integral of a continuous function f (x, y) over a bounded closed region S is the limit of the corresponding two-dimensional integral sum

f (x,

y)dx dy =

lim max A*i max Ar// -

(1)

c

&y k = yk+l yk and the sum is extended over those xg, and k for which the points (*/, y k ) belong to S. 2. Setting up the limits of integration in a double integral. We distinguish two basic types of region of integration.

where A*

-

t

values of

x,

= Xf +l

i

x

o Fig. 85

1)

The region

of integration

Fig. 86

5

(Fig. 85)

is

bounded on the

left

and

right

while the vaFig. 85). In the region S, the variable x varies from x l to x riable y (for x constant) varies from 9, (x) to j/ 2 q> 2 (x). The integral (1) ma>

^=

=

The Double Integral

Sec. 1]

247

in Rectangular Coordinates

be computed by reducing to an iterated integral by the formula *a

<Pa (X)

f(x, y)dy,

J (S)

<PaU>

where x

is

held constant

when

\

calculating

/(x, y) dy.

2) The region of integration S is bounded from below and from above by the straight lines y y and y yt(y z >yi), and from the left and the (y)], right by the continuous curves x = ^> (y) (AB) and x = 2 (y) (CD) [t|? 2 (y) each of which intersects the parallel y = Y (y* Y y t) at only one point l

(Fig. 86). As before,

^^

i|?

l

^ <

we have $2 Vl JJ/(*. y)dxdy=\dy (S) here, in the integral j/i \ f(x, y) dx we (U) f (x. J i|?, y)dx t ((/) consider y constant. If the region of integration does not belong to any of the above-discussed types, then an attempt is made to break it up into parts, each of which does belong to one of these two types. Example 1. Evaluate the integral / D Solution. Example 2. Determine the limits of integration of the integral (x, (S) y)dxdy 248 _ _ Multiple and Line Integrals 2 [Ch. 7 2 = * 1 the region of integration 5 (Fig. 87) is bounded by the hyperbola y 2 and x 2 (we have in view the region constraight lines * taining the coordinate origin). Solution. The region of integration ABCD (Fig. 87) is bounded by the 2 and x 2 and by two branches of the hyperbola straight lines if = = and by two = x- y=yT+* that is, it belongs to the first and type. We = (/--1 have: dx f(x, y)dy. J J Evaluate the following iterated integrals: 21 2113. 2114 35 \dy$$jf + 2y)dx. 2117. ^pTv?31 2118. Jdy J rrfr. dcp j a o sin Jt_ 2115 2 X 2dU 8 COS (p ' 00 Write the equations of curves bounding regions over which the following dduble integrals are extended, and draw these regions: 2 2121. 2-t/ Jrf/ f(x, y)dx. X+9 3 2122. J JdxJ IX 4 2125. f(x, y)dy. 10-y dy o JdxJ/(*. y)dy. K25-JC- 3 3 2123. 2X 3 2124. 00 ^dx f(x, ^)dx. 2126. f(x, y)dy. X+2 2 y 1 d* -ix* f(x, y)dy. Set up the limits of integration in one order and then in the other in the double integral JJ/(*. y)dxdy (S) for the indicated regions S. Sec. The Double Integral 1] 2127. C(0, S is a rectangle S S is a triangle is a in Rectangular Coordinates with vertices 0(0, 0), 249* 4(2,0), 5(2, 1), 1). 2128. 2129. C(0, with vertices 0(0, 4(1, 0), trapezoid with vertices 0(0, 0), A 0), (2, 0), 5(1, 5(1, 1). 1), 1). 2130. C(2, 7), 2131. 0), (0, S parallelogram with a is D(l, 5). S a is whose circular sector arc end-points are vertices 045 A 4(1, 2), 5(2, 4), with centre at the point and 5 f - 1, 1) (Fig. 88). (1, 1) Fig 89 right parabolic segment 405 bounded by the a segment of the straight line 54 connecting and parabola the points 5(~-l, 2) and 4(1, 2) (Fig. 89). 2133. S is a circular ring bounded by circles with radii r=l and /?-=2 and with common centre 0(0, 0). 2134. S is bounded by the hyperbola if x? \ and the circle 2132. S is a 504 x 2 2 | ^9 (the region containing the origin is meant). 2135. Set up the limits of integration in the double // (x, integral y)dxdy (S) if the region S b) * c) x 2 2 h// defined by the inequalities is 2 e) ; |- //* *; Change the order //<x 2 = 2a; :a. of integration in the following double integrals: \2X 2136. f(x, y)dy. 2137. JdJ/(x. y)dy. Multiple and Line Integrals 250 2138. \dx \ f(x, y)dy. 2141. \ d* dy \ f(x t y)dx. - VTZx^tf a 2139. - [Ch. 7 * f(x, y)dy. J 2142. \dy V*-* /(*, yi ~2 2 20 2140. Jdx f(x J 9 y)dy. RVT A 2143. J -V d*J/(x, o sin 2144. \dx t/)dy+ f(x. y)dy. o x /)d/. J/(jc, Evaluate the following double integrals: ( where S is a triangle with vertices 0(0, [ xdxdy 2145. t 0), (S) A(\, 1), and B(0, 1). A(2,0)X Fig. 90 2146. ^xdxdy, where the region Fig. 91 of integration Sis bounded (S) by the straight line passing through the points A (2, 0), fi(0, 2) of a circle with centre at the point C(0, 1), and and by the arc radius 1 (Fig. 90). The Double Integral Sec. 1] * 2147. Vr a , \ \ JJ 2 in Rectangular Coordinates ===, where S x*y* is 251 a r part of a circle of radius (S) a with centre at 0(0, 2148. lying in the first where S a y dx dy V** 0) 2 t is quadrant. triangle with vertices (S) A 0(0,0), 2149. -1), and (I, fl(l, 1). y*dxdy, where S \j !/"*# triangle with vertices a is (S) 0(0, 4(10, 0), rr 2150. ed J J (S) 1), and fl(l, 1). - dxdy, where S ey by the parabola y* is =x a curvilinear triangle and OAB bound- x = Q, (/=! the straight lines (Fig. 91). 2151. ff^Ti, - the parabola 2152. extend: y=7f and Compute the is a parabolic segment the straight line y integrals \dx [ j j = x. _rt 2 tfsmxdy; c) 2 bounded by and draw the regions over which they i-f-cosjc at a) where S / x* sin' , 1 " b) COS* When solving Problems 2153 to 2157 the drawings first. 2153. Evaluate the double integral it is abvisable to make (5) if S line is a region x = p. bounded by the parabola y* = 2px and the straight 2154*. Evaluate the double integral ^xydxdy, extended over the region S, which 2 2 and an upper semicircle (x 2) # is + = bounded 1. by the #-axis _ _ Multiple and Line Integrals 252 (Ch. 7 2155. Evaluate the double integral dxdy /25=5 f (S) where S the area of a circle of radius a, to the coordinate axes and lies in the gent 2156*. Evaluate the double integral is which first circle is tan- quadrant. ^ydxdy, (S) where the region S is bounded by the axis of abscissas and an of the cycloid = R(t y = R(\ x sin/), cos/). 2157. Evaluate the double integral (S) in which the region of integration nate axes and an arc of the astroid 2158. Find the region mean value S is bounded by the coordi- of the function f(x, y)=--xy 2 in the SJO^Jt^l, 0 Hint. The mean value of a function f(x, y) the region in 5 is the number 2159. Find the mean value of the square of the distance of 2 2 from the coordia point (x, y) of the circle (x nate origin. M Sec. 2. Change ^R af+y of Variables in a Double Integral 1. Double integral in polar coordinates. In a double integral, when passing from rectangular coordinates (x, y) to polar coordinates (r, cp), which are connected with rectangular coordinates by the relations # we have = /'Cos(p, y r sin ip, the formula ^{ (S) f (* rel="nofollow"> y)dxdy=(( (S) (r cos q>, r sin cp) r dr e/cp, (1) Sec. 2] _ Change of Variables in a Double Integral _ 253 a integration (S) is bounded by the half-lines r the curves r and r r 2 ((p), where r l (q>) r,(cp) r z f the region r=^p(a of = and and and a^rp^p, P F C f r) r (q), dr = dcp /-J a (S) < F f \ dcp r) r rfr, (cp, r, (cp) r 2 (cp) where F (cp, In evaluating the r sin (p). r)~/(rcos(p, \ integral F ((p, r)rrfr 'i ( we hold the quantity (p constant. the region of integration does not belong to one of the kinds that has been examined, it is broken up into parts, each of which is a region of a If given type. 2. Double if integral in curvilinear coordinates. the more general case, In double integral in the (x, y)dxdy from the variables x, y to the variables u, v, which it is required to pass are connected with x, y by the continuous and differentiate relationships A-^cp(w, i), between the points region S f of the of t|?(K, I/ v) both directions, continuous) correspondence that establish a one-to-one (and, in and the points region S of the .v//-plane if the Jacobian the of some UV- plane, and w *^ D(u retains a constant f dx dy du da y) , V' dx dy dv dv t sign in the region 5, then the formula (^ y) dx dy - I [cp (u, v), (w, v) ] | / | du du (6) holds true The limits of the new integral are determined from general rules on the basis of the type of region S' Example 1. In passing to polar coordinates, evaluate where the region 5 is origin (Fig 92). Solution. Putting x J/'l x 2 radius a circle of rcoscp, if^ Y"\ // (r /? rsincp, cos 2 cp) = ! we with centre at the coordinate obtain: (r sin cp) 2 = ^1 r8 Multiple and Line Integrals 254 Since the coordinate r in the region to 2jt, it follows that varies from S varies 2Jt ff Pass to to 1 for any q>, and q> 1 y\x* coordinates r and cp and set up the limits of to the new variables in the following polar with integration from (Ch. 7 respect integrals: 2160. 2162. K, 2161. y)dy. Jd*/(]/J JJ/(x, y)dxdy, \-'/ where S a triangle is bounded by the straight lines (/ = #, y~ x, 2163. * -i 2164. ^f(x, y)dxdy, where S is bounded by the lemniscate (S) Fig. 92 2165. Passing to polar coordinates, calculate the double inte gral (S) where S C(f,0) is a semicircle (Fig. 93). of diameter a with centre at the poin _ Sec. 2] Change of Variables in a Double Integral _ 255 2166. Passing to polar coordinates, evaluate the double integral (S) extended over a region bounded by the circle jc* y* = 2ax. 2167. Passing to polar coordinates, evaluate the double + in- tegral (S) integration S is a semicircle of radius a with centre at the coordinate origin and lying above the #-axis. r 2168. Evaluate the double integral of a function f(r, where the region of = = = V a*-x* a Jdx Vx' + J tfdy. 2170. Passing to polar coordinates, evaluate (S) where the region S is a loop of the lemniscate 2171*. Evaluate the double integral ll T/^R; (S) extended over the region S bounded by the ellipse x x* -j + ^^ u passing to generalized polar coordinates: X = ^wo r Y , y 2172**. Transform c p* \dx\f (x y)dy t o (0 ax and c: rel="nofollow">0) by introducing new variables u 256 _ __ Multiple and Line Integrals 2173*. = x + y, Change the variables u v = xy [Ch. 7 in the integral i i \dx\f (x,y)dy. 2174**. Evaluate the double integral (S) where S is bounded by the curve a region -_ k b2 Hint. Make ~~fi 2 ' 2 the substitution br sin cp. The area of y >, Sec. 3. Computing Areas 1. Area in rectangular coordinates. a plane region S is (S) If the region S defined is by the a^x^b, inequalities q> (x) ^y ^ \|) (x) , then b S = op \dx X) J a is ( cp dy. (x) 2. Area in polar coordinates. If a region S in polar coordinates r and defined by the inequalities a^cp^p, / (cp)^/' (q>), then F P S = ffrdcpdr= C a (S) 6/9 q> P) C /-dr. /(q 2175. Construct regions whose areas are expressed by the in- _ tegrals x+2 2 a)) dx J j d; dy; b) j dy d*. J Evaluate these areas and change the order of integration. 2176. Construct regions whose areas are expressed dy the tegrals ji arc tan a) ] n_ T Compute 2 * d(f sec J 2 rdr\ o b) J n "T" these areas. a(i+coscp) ^9 \ a in- Sec _ _ Computing Areas 3] 2177. Compute the area bounded by the straight lines x 257 y, 2178. Compute the area lying above the x-axis and bounded 3a. 4ax, and the straight line x-[-y by this axis, the parabola y* = = 2179*. Compute the area bounded by the ellipse 2180. Find the area bounded by the parabolas f z y =10x4- 25 and y = 6x + 9. 2181. Passing to polar coordinates, find the area bounded by the lines 2 2 z Q. x x -[-y* 4x, 2x, x, y y -y = = = = 2182. Find the area bounded by the straight line r cosq)=l circle r~2. (The area is not to contain a pole.) 2183. Find the area bounded by the curves and the r = a(14coscp) and r acosq>(a>Q). bounded by the 2184. Find the area ( * , ~*~ V 4 line (/!V__ " ~ **_//? "9 J ' 9 4" 2185*. Find the area bounded by the ellipse (x2y 2 !-3) 4 (3* -\-4y- 2 1) - 100. 2186. Find the area of a curvilinear quadrangle bounded by 2 z the arcs of the parabolas x --=ay x* a,x, y by y* x(Q<. = y = = y new variables u and x2 = uy, if u, and put vx. 2187. Find the area of a curvilinear quadrangle bounded by the arcs of the curves if=ax, if a, xy bx, xy (Q<.a<.b, = = 0 new variables u and u, 9 -1900 v, y*~vx. and put = (Ch. 7 Multiple and Line Integrals 258 Sec. 4. Ccmputing Volumes The volume V a cyltndroid bounded above by a continuous surface of low by the ph-ne 2 0, and on the sides by a right cylindrical surface, which cuts out of the ju/-plane a region S (Fig. 94), is equal to * = /(*, y), be 2188. Use a double integral to express the volume of a pyravertices 0(0, 0)", A(\, 0, 0), fl(l, 1,0) and C(0, 0, 1) ol the limits Set integration. 95). up (Fig. mid wiih C(0,0,1) Fig. 94 Fig. 95 In Problems 2189 to 2192 sketch the solid expressed by the given double integral: 2189. f J dx x f (1 J Z-X 2190. y)dy. _ 2193. Sketch the solid V a* a tegral f dx - x* ( YC? 2191. value is x)dy. 2192. whose volume tfy* dy\ whose volume is expressed by the in- reason geometrically to find the of this integral. 2184. Find the volume of a solid bounded by the elliptical 2 x + y=\, and the coordi2x* -f f/ 1, the plane paraboloid z nate planes. 215. A solid is bounded by a hyperbolic paraboloid z x* tf and the planes (/ 0, e 0, x=l. Compute its volume. = + = = _ Sec. 5] 2196. planes */ A solid = 0, is = 0, 2 th Computing _ Area* of Surfaces 2 bounded by the cylinder x y = x. Compute its =a 2 +-z 2 259 and the volume. Find the volumes bounded by the following surfaces: = y\ x* -\-y* --='*, 2 = 0. = = 2J/T, * + 2 = 6, 2 = 0. y"x, y = x* +y\ y = x z //=!, 2 = 0. x -}-*H-2 = a, 3*4-*/ = a, ~x4-tj=^a, + -1, y = *' = 0, 2 = 0. = 2ax, 2 = a*, 2 = p* (a rel="nofollow"> p). x if 2197. az 2198. 2199. 2200. 2201. 2202. f/ 2 , */ = 0, 2 = 0. , 2 -h In Problems 2203 to 2211 use polar and generalized polar coordinates. 2203. Find the entire volume enclosed between the cylinder 2 2 2 2 a x 2 -\-y a and the hyperboloid x -f- if 2* 2204. Find the entire volume contained between the cone 2 2 2 z* = a 2 and the hyperboloid x 2 -f- if 2(* +if) 2 x + /* f 2205. Find the volume bounded by the surfaces 2a2 = . = . = ** 2 4-// 2 2 =a 2 , 2 = 0. volume 2206. Determine the of the ellipsoid 2207. Find the volume of a solid bounded by the paraboloid 2 2 2 -= x 2 H- // 2 and the 3a -f- 2 (The volume lying sphere * 4inside the paraboloid is meant.) 2208. Compute the volume of a solid bounded by the jq/-plane t 2 2 2 2 the cyl inder x -*- y 2a^, and the cone x -f if = 2 2209. Compute the volume of a solid bounded by the jq/-plane, 2 2 2 the surface 2 =-ae~ <*'J4 " >, and the cylinder x + y = /? 2210. Compute the volume of a solid bounded by the *f/-plane, ^ 202 = . = . J . the paraboloid z =^+ 2 5 , = and the cylinder ^-h| 2 2211. In what ratio does the hyperboloid 2 2 2 divide the volume of the sphere x 4- t^ 2212*. Find the volume of a solid bounded 4- 2~. 2 jc ^3a 2 2 -h// 2 2 = a* ? by the surfaces Sec. 5. Computing the Areas of Surfaces The area o of a smooth single-valued surface z on the jci/-plane is the region S, is equal to -JJ = f(x t y), whose projection Multiple and Line Integrals 260 [Ch. 7 %+j+= 2213. Find the area of that part of the plane = 1 lies between the coordinate planes. 2214. Find the area of that part of the surface of the cylin2 2 der x 4 y R 2 (z^O) which lies between the planes z mxand which z = = = nx(m>n>Q). 2215*. 2 Compute 2 2 =z the area of that part which is situated y bounded by the plane y -\-z-a. cone x 2216. Compute 2 2 y cylinder 2 2 2 2 x y +z =-a 2217. Compute + the area =ax + x of that which part of out of cut is of the in the first the it surface of the octant and is surface of the by the sphere . sphere x 2 2 y -\- + z 2 the area =a 2 of that part of the y2 cut out by the surface -^ surface of the ..2 + ^=1. 2218. Compute the area of that part of the surface of the 2 2 ax paraboloid y + z = 2ax which lies between the cylinder tf and the plane x= a. 2219. Compute the area of that part of the surface of the 2 2 2ax which lies between the xy-plane and the cylinder x + y = cone x 2 2 -\-y =z 2 . 2220*. Compute the area of that part of the surface ol the 2 2 = 2ax. cone x 2 y 2 which lies inside the cylinder x 2 z f/ 2221*. Prove that the areas of the parts of the surfaces of the 2 2 z 2 2az arid x 2az cut out by the cylinparaboloids x + y y 2 2 der x -\-tf are size. of R equivalent 2222*. A sphere of radius a is cut by two circular cylinders whose base diameters are equal to the radius of the sphere and which are tangent to each other along one of the diameters of the sphere. Find the volume and the area of the surface of the remaining part of the sphere. 2223* An opening with square base whose side is equal to a is cut out of a sphere of radius a. The axis of the opening coincides with the diameter of the sphere. Find the area of the surface of the sphere cut out by the opening. of 2224*. Compute the area that part of the helicoid = + = = = e = carctan ders x 2 X 2 -i-y which =a 2 lies in the first octant between the cylin- and Sec. 6. Applications of the Double Integral in Mechanics 1. The mass and static moments ot a lamina. If S is a region in an jq/-plane occupied by a lamina, and Q (x, y) is the surface density of the lamina at the point (x, y), then the mass of the lamina and its static M Sec. 6] _ _ Applications of the Double Integral in Mechanics M moments x and double integrals A4 MY relative to the --= H Q (x, y) dx and t/-axes are x- Mx= J dy, jj (S) yQ J expressed 261 by the dx dy, (x, y) (S) M Y =^ J*e(x, y)dxdy. (1) (5) If the lamina homogeneous, then Q is 2. The coordinates centre of gravity of a lamina, then - My _i_ y where M is tive to the formulas (1) const. (x, y) of the centre of gravity of a lamina. M -M ~_ // M M x My coordinate axes (see 1). we can put Q=l. the 3. The moments lamina relative of to the x- /X= S S and , t/-axes are, y'Q (x, y) dx dy, of its is moments The moments static rela- homogeneous, then in of inertia 01 a respectively, equal to /r= *2 Q J J (*. y) *x dy. (2) (S) (S) The moment are lamina a lamina. of inertia the ' the mass of the lamina and If is (x, y) x d . J ' C If inertia of a lamina relative to the origin is (3) Putting Q(X, //)-=! in formulas inertia of a plane iigure. 2225. Find the mass (2) and (3), of a circular we get the geometric lamina of radius proportional to the distance of a point density and is equal to 6 at the edge of the lamina. is 2226. OB = a A lamina has the OA = b, and its shape of a right moments R if of the from the centre triangle with legs and density at any point is equal to the distance of the point from the leg 0/4. Find the static moments of the lamina relative to the legs 0/4 and OB. 2227. Compute the coordinates of the centre of gravity of the sin* area OmAnO (Fig. 96), which is bounded by the curve // and the straight line OA that passes through the coordinate origin and the vertex A (-^ , Ij of a sine curve. 2228. Find the coordinates of the centre of gravity of an area bounded by the cardioid r = a(\ + cosij)). 2229. Find the coordinates of the centre of gravity of a circular sector of radius a with angle at the vertex 2a (Fig. 97). 2230. Compute the coordinates of the centre of gravity of an area bounded by the parabolas // 4.x f 4 and if 2x4-4. 2231. Compute the moment of inertia of a triangle bounded 2 relative to the #-axis. 2, y 2, # by the straight lines x y = + = = = [Ch. 7 Multiple and Line Integrals 262 2232. Find the ters d and moment D(d a) of inertia of an annulus with diamerelative to its centre, and b) relative to diameter. 2233. Compute the moment of inertia of a relative to the axis passing through its vertex the plane of the square. 2234*. C)mpute the moment of inertia of the parabola if = ax by the straight line x its square with side a perpendicularly to segment cut a =a straight line // = relative to oil the a. Fig. 96 2235*. Compute the moment of inertia of an area bounded by 4 and the straight line x-\-y the hyperbola xy 5 relative to the straight line x y. 2236*. In a square lamina with side a, the density is proportional to the distance from one of its vertices. C:>mpute the moment of inertia of the lamina relative to the side that passes through this vertex. 2237. Find the moment of in?rtia of the cardioid r a(l cos = = = = + = Sec. 7. Triple Integrals 1. Triple integrals in rectangular coordinates. The triple integral of the /(*, y, <) extended over the region V is the limit of the corresponding threefold iterated sum: function 62 = lim max max max \x\ A'/j - o ~> o Az/c -> o 2 2 2^ f i fa (*t> &/> d z ' Sec. 7] 2o3 Triple Integrals Evaluation of a triple integral reduces to th? successive computation ordinary (onefold iterated) integrals or to the computation double and one single integral. three Example Z d*d/dz, JJ*y V /=. where the region V We is defined by the inequalities have X 1 2 ' Example one Compute 1. Solution. of the of 7T dy Evaluate 2. 2 2 AT extended over the volume of the ellipsoid t Z // -- - -|- 2 = -j- 1 . Solution. a ^ J x2 dxdy dz - a x2 dA- J J J -a (V) S yz is V We ^ + -^^1 -- the area of the ellipse S vvzz -=Jib I/ x*S ^ yz dx t -a (5^.) M^ \\here dydz= ^^ j^2 2 JC = cons ^ an(^ =*n \-V r a l-~-c 2 a- therefore finally get -a (V) 2. Change of variables in a triple integral. . 0. If in tha triple integral 2)dxdydz from the variables AT, //, z to the variables it is required to pass u, ay, which are connected with *, //, z bv the relations x ^(u v, w), y = ty(u,v,w), z = X(". u ^)i where the functions cp, i|), x are: 1) continuous together with their paitial first derivatives; , t 2) in one-to-one (ind, tween the points some region V in both directions, continuju?) correspondence beV oi xf/z-space and the points of of the region of integration of l/l/l^-space; [Ch. 7 Multiple and Line Integrals 264 3) the functional of these functions determinant (Jacobian) dx dx dx ~dv dw dii D (u, v, dy dy dy da dv dw w) dz dz dz dw da dv we can make region V, then retains a constant sign in the use the for- of mula = J \f(x,y,2)dxdydz (V) \ IT ( w \ f \ L? w )> > ty( u > y * w )< In particular, 1) for cylindrical coordinates X get / cp, rcosrp, spherical coordinates the radius vector) (Fig. 99), x we have r, // h (Fig. 98), where z^-//, rsinrp, r\ 2) for r du dv dw. Fig. 99 Fig. 98 we \ / Example r 2 3. = r cos cos i|) cp, ap, r ( is the longitude, q), z \\> the latitude, where cos 9, f/ = /-cosi|3 sin / sin v|\ i[). Passing to spherical coordinates, compute JSJ (V) where V is a sphere of radius R. Solution. For a sphere, the ranges of the spherical tude), \|) (latitude), and r (radius vector) will be coordinates fp (longi- Sec We 265 Triple Integrals 7\ therefore have f f f Vx 2 z -\-y 3. Applications sional A'//z-space -\-z*dxdydz=\ of dcp dty f r A'COS \|?dr f The volume triple integrals. = Ji# 4 of a region of . three-dimen- is The mass of a solid occupying the region V M ----- C f f Y z) v ( is d (V) where \(x,y,z) is the density of the body at the point (*,//,*). The slal/c moments of the body relative to the coordinate planes are M yy = "(V)' Myz = (* (* I ' )} 0')" MZX ~- ' \ \ \ Y (A f/ 2) f/ dx dy dz. (V) The coordinates gravity are of the centre of M f _ J ~~ _ M ' ' Al then we can put Y (*, y> z) = If the solid is homogeneous, mulas for the coordinates of the centre of gravity. The moments of inertia relative to the coordinate axes are T- J J (y* + **) Y (*, U, 2) dx dy I in the for- dz; (V) (V) l =J J J (* (V) Putting of inertia of Y(*0 ^i^21 in * nese formulas, we get the the body. A. Evaluating triple integrals Set up the limits of integration in the triple integral J J (V) for the indicated regions V. ^f(x,y, z)dxdydz geometric moments 266 Multifile V 2240. is a tetrahedron Jr X V 2241. is V 2242*. 2243. V is + ^fl 2 2 J/ = 0, 2 = 0. 1 2 , = 0, = /f. 2 volume bounded by the surfaces a 000 2245. = 0, cone bounded by the surfaces a Compute the following 2244. X bounded by the surfaces a cylinder is bounded by the planes yJrZ =\, JC' [Ch. 7 and Lire Integrals integrals: + +-2+1 * V djt dy J J a 2246 . fd* 00 I ] (' J 1-X 1 2247. rfy .1 .) dx 1-JC-t/ dy J xyzdz. J o o p 2248. Evaluate d* dy dz ' 1)3 where V is the region ol integration planes and the plane x-\~y-[z\. 2249. Evaluate bounded by the coordinate r r r (V} where V (the region 2 paiaboloid 2cu^x of 2 -\-y integration) is the and the sphere X? common of the part 2 + y* + 2 ^3a 2 . 2250. Evaluate (V) where spheres V (region x 2 + y* \-z* of integration) ^R' and x 2 is the + \f + z 2 common ^ 2Rz part of the gee. __ Triple Integrals 7] 267 2251. Evaluate ^zdxdydz, (V) where V is a volume bounded by the plane j. + half of the ellipsoid -+ z = and the upper -J.==l. 2252. Evaluate (V) V where Xz is 2 IJ + "^r the interior of the ellipsoid ~^r 2253. Evaluate V where 2 2 = (jc region (the 2 2 hi/ ) of is integration) and the plane bounded the by cone. = h. z 2254. Passing to cylindrical coordinates, evaluate where V 2 jc the surfaces x* is a region bounded by a -=z and containing the point (0,0, R). -| // 2255. Evaluate 2 2 J 21 - jc a j transforming it to cylindrical coordinates. Evaluate first 225(5. - - di/ dz, J transforming it to cylindrical coordinates. 2257. Evaluate first VR*-X* R Vfla-jir*-0a J \dx -/? first transforming \ -/f^TJa it dy J (A: + o to spherical coordinates. +y* -\- z* = 2Rz 9 [Ch. 7 Multiple and Line Integrals 268 2258. Passing to spherical evaluate the integral coordinates, (V) where V B. 2 is the interior of the sphere x -\-y Computing volumes by means 2259. Use a triple integral bounded by the surfaces to 2 +z 2 triple integrals of compute the volume of a solid 2260**. Compute the volume of that part of the cylinder 2 2 x 2 -f tf = 2ax which is contained between the paraboloid* + y = 2az and the xy-plane. 2261*. Compute the volume of a solid bounded by the sphere 2 2 2 2 2 2 x +y +z =a and the cone z ---x + /y" (external to the cone). 2262*. Compute the volume of a solid bounded by the sphere 2 2 z x 2 +y +z = 4 and the paraboloid x +if=-3z (internal to the paraboloid). 2263. Compute the volume of a solid bounded by the xy-plane, z 2 2 2 z 2 a (internal ax and the sphere x the cylinder x -f- z to the cylinder). 2264. Compute the volume of a solid bounded by the paraboloid +y = = +y + -~ = 2 -i - and the plane C. Applications to of triple integrals mechanics and physics M 2265. Find the mass 0 <*/ Fig. and the plane (l(x, y, z) point the den- if y, (x, = x-\-y-\-z. z) is 2266. Out of an octant of the 2 2 2 2 c z x .> 0, x y sphere cut a solid OABC */^0, bounded by the coordinate planes + + z^O 100 -+- = sity the of a rec- Q^x^a, tangular parallelepiped at x--=a. 1 (a c, & (Fig. 100). < , Find the mass of this body if the density at each point (x, y, z) is equal to the 0-coordinate of the point. 2267*. In a solid which has the shape of a hemisphere 22*0, the density varies in proportion to the Improper Integrals Dependent on a Parameter Sec. 8] 260 distance of the point from the centre. Find the centre of gravity of the solid. 2268. Find the centre of gravity of a solid bounded by the 2 2 4x and the plane x=2. paraboloid // +2z 2269*. Find the moment of inertia of a circular cylinder, whose altitude is h and the radius of the base is a, relative to the axis which serves as the diameter of the base of the cylinder. 2270*. Find the moment of inertia of a circular con^ base, a, and (altitude, /i, radius of density Q) relative to the diameter of the base. 2271**. Find the force of attraction exerted by a homogeneous cone of altitude h and vertex angle a (in axial cross-section) on a material point containing unit mas^ and located at its vertex. 2272**. Show that the force of attraction exerted by a homogeneous sphere on an external material point does not change if the entire mass of the sphere is concentrated at its centre. = on a Parameter. Sec. 8. Improper Integrals Dependent Improper Multiple Integrals 1. Differentiation with respect to a parameter. In the case of certain imposed on the functions / (.v, a), f'a (x, a) and on the corresponding improper integrals \vc have the Leibniz rule restrictions (.v, a) dx = a Example fa (A-, a) dx. differentiating with respect to a parameter, evaluate By 1. \ 'i ~~ > dx \ (a > 0, p > 0). Solution. Let Then ~ da Whence F equation. (a, We p) = have Whence C(p) 2a 1 - Ina + C(p). To 0= = -^-lnp. ~ In P Hence, find + C(P). "2a* C(p), we put a = in the latter Multiple and Line Integrals 270 2. Improper double and a) An infinite region. If region 5, then we put triple integrals. a function f (x, y) is continuous in y) \{f(x, U [Ch. 7 dx dy = lim "S\ ( f (x, y) an unbounded dx dy, (1) W where a is a finite region lying entirely within S, where a -+ S signifies that we expand the region o by an arbitrary law so that any point of 5 should enter it and remain in it. If there is a limit on the right and if it does not depend on the choice of the region o then the corresponding improper inte, gral is called convergent otherwise it is divergent. is nonnegative [f (x, //) , If the inU'grand / (,v, vergence of an inirioper integral is it y)^Q], then on the right of (1) lo exist at least for the region 5. b) A discontinuous function. If a function / (x, y) uous in a bounded closed region S, except the point (*, for the con- and sufficient for the limit one system of regions o that exhaust necej-sary y)dxdy=\\m is everywhere contin- P (a, b), then we put f(x,y)dxdy. (2) where S 6 is a region obtained from S by eliminating a small region of dia meter e that contains the roint P. If (2) has a limit that does not depend on the tyre of small regions eliminated from 5, the improper integral under is called convergent, otherwise it is divergent. /(A, //)^>0, then the limit on the ripht of (2) is not dependent on the type of regions eliminated from S; for instance, such regions may be circles consideration If with centre at P. of radius The concept of of improper double integrals is readily extended to the case triple integrals. Example 2. Test for convergence dxdy , 3) is} where S is the entire j/-plane. Solution. Let a be a circle of radius origin. Passing to polar coordinates for If p< then then 1, lim o - a lim 7 (a) O -* S 7(o)= r P * = lim Q /(a) =00 and with Q p^ 1, centre at the coordinate we have the integral diverges. But if p> 1, ~f QC and the integral converges. For p=l we have Sec. 8] _ Q 2JT / Improper Integrals Dependent on a Parameter (a)= Cdcpf -f^l-==jiln(l+Q r * J J o o 2 lim/(a) ); ~r Q -* oo for p> = oo, that _ is, 271 the integral diverges. Thus, 'the integral 2273. Find (3) /' (*), converges 1. if 2274. Prove that the function + u = 00 r *f' 2 \ J ^ + w j ,2 , (l/ dz 2) QC satisfies the Laplace equation *u dx* is ' ~c)y* F 2275. The Laplace transformation defined by the formula Find d) d ~U + ^ *-Q F(p), if: = cos p/. /(/) a) /(O-l; b) (p) /(/)=e; 2276. Taking advantage of the (ormula compute the integral \ xn 2277*. Using the formula evaluate the integral ~ l for the \nxdx. c) function /(/) /(/) = sinp/; 272 _ Multiple and Line Integrals Applying differentiation with respect __ \Ch. 7 to a parameter, evaluate the following integrals: GO 2278. " 00 e 2279 . rno^ 228 - "-""* f ~**~ e ~'X ( ax arc tan dx > 0, > 0). smrnxdx (a>0, p>0). . dx e-<"d* 2282. (a - (o^O). Evaluate the following improper integrals: GO QC 2283. x_ 2284. 2285. \dy\ev dx. y \\ c) c/ *, ^ 4 ~r 2 , where S is a region defined by the inequali- / (5) ties #^5 2287. 1, The Euler-Poisson integral defined by the formula 00 may also be written in the uate / by multiplying coordinates. 2288. Evaluate CO form I=\c-^dy. Eval- these formulas and then passing to polar 00 GO dz 273 Line Integrals See. 9] Test for convergence the improper double integrals: 2 2289**. where S 55 \nVx*-\-y dxdy, a is circle * 2 + 2 f/ (S) U 2290. where S TTiriva a is defined by region the ine- (6') quality x 2 2291 *. 2292. tf^\ -| I I , ("exterior" of the circle). . J_ 2 d xd CCr JJJ . 2 t (x T~ y where S , * ^\ \ , z a square is where V , a is < A; | | 1 , \y\^l. defined region by the ) (V) inequality x 2 -\- if z -\- 2 ^ ("exterior" of a sphere). 1 Sec. 9. Line Integrals 1. Line integrals of the (Irst type. Let / (x, y) be a continuous function = (p (A-) [a<: \ <;&] be the equation of some smooth curve C. */ Let us construct a system of points M, (A-,-, //,) (/~0, 1, 2, .... n) that break up the curve C into elementary arcs Al _,M = As and let us form the and I integral max 2^ sum S n As/ -^ 0, is AV ^ s r ^ ne #/) ' - - t t im it of this sum, when n -* oo and called a line integral of the first type n lim "->>ac (cfs is 2= / /(xf, r/,-) As/= \ / (x, i/) ds c i the arc differential) and is evaluated from the formula b ' \ J f (A !/) rfs C In the case - ' " of rt ], \ J a f(x parametric we have t cp (,v)) y~\ + (q>' representation of 2 (A-)) the c/.vr. curve C: .Y = q>(/), Also considered are line integrals of the first type of functions of three variables f (x, y, z) taken along a space curve. These integrals are evaluated in like fashion A line integral of the lirsl type does not depend on the direction of the path of integration; if the integrand / is interpreted as a linear density of Hie curve of integration C, then this integral represents the mass of the curve C. 274 Multiple and Line Integrals Example 1. Evaluate the [Ch. 7 line integral where C is the contour of the triangle ABO with vertices A (1, and 0(0, 0) (Fig 101). Solution. Here, the equation AB is (/=l x, the equation and the equation OA is */ = 0. We therefore have AB 2. Line uous BO 0), OB B is (0, 1), x = 0, OA integrals of the second type. If P (x, y) and Q (x, y) are continand f/ = functions B A Fig. x varies, then the corresponding X 101 line integral of the second type is expressed as follows: P (x, y)dx + Q (x, y)dy = C [P (x, cp (*)) + (x) Q (x, q> (x))\ dx. a In the more general case when the curve C is represented parametrically: y ty(t), where / vanes from a to 0, we have = = y) Similar formulas hold space curve. dy + for a ' [P ( Une (0, integral of (9(0, the second type (OJ taken over a A line integral of the second type changes sign when the direction of the path of integration is reversed. This integral may he interpreted mechanically P (x, y), Q(x, y) } along the as the work of an appropriate variable force { curve of integration C Example 2. Evaluate the line integral Cyi ____ ____ C is the upper half clockwise. Solution. We have where 2 y dx 275 Line Inlegtah Sec. 9] + x* dy = 2 [b sin 2 the of t-(a sin *--=a cost, ellipse /) +a 2 / -b cos /] dt ab* f sin 8 1 dt = + a*b jt 3. The case of a total differential. traversed sin* oo cos 2 n = =b y / dt = -t a&. ?t the If C cos 8 a of integrand line integral second type is a total differential of some single-valued function U^--U(x, */), that is, P (x, y)dx-\-Q(x, y)dy-=dU(x, y], then this line integral is not dependent on the path of integration and we have the Newton-Leibniz formula (x 2 yj of the , P(x, y)dx where y ) is In particular, if (x lt } y)dy = U(x 2 + Q(x, the initial and (* 2 #,) is the the contour of integration C , , yJ U terminal is (x l9 point then closed, (1) the of path (2) of integration C is contained entirely within some S and 2) the functions P(r, (/) and Q (x, y) together with tlvir partial dcnviitives of the first order are continuous in 5, then a is the necessary and sufficient coiditioi foi th? existence of the function identical" fulJilment (in S) of the equality If 1) the contour simnlv-connectH reaio.i U 2_^ dX (3) (d) ~dy If conditions one and two are not ful(see integration of total differentials) does not guarantee the existence of a jillcd, the presence of condition (3) single-valued lunction U, and formulas (1) and (2) may prove wrong (see Problem 23:2) We give a method of finding a function U (x, u) from its total diflerential based on the use of line integrals (which is yet another method of integrating a total differential). For the contour of integration C let us take a broken line P P (Fig 102), where P (.v y ) is a fixed [oint and 0, y and dy (x, y) is a variable point. Then along P P, we have y We get: we have dx and along P t M U (x t y)-U II M 1 (J , ~ M (x , y )= P (x, y) dx + Q (x, y) dy = y (x. Similarly, integrating with respect to P C P 2 M, we have x u U(x. y)-U(*v */ )-Q(* y)dy+^P(x, , Vo x* y)dx. y)dy. [Ch. 7 Multiple and Line Integrals 276 + 2y) dx+ (2x 6y) dy = dU. = O, y = Q. Then Find U. Example 3. (4x Solution. Let A-O U(x, y) = or + 2y) dx + C= U where C (0, 0) is 2 -i-c, 3// an arbitrary constant. Y Uo --4/I X XQ 102 Fig. 4. Green's formula for a plane. If C is the boundary of a region S and the functions P (x, y) and Q (x, y) are continuous together with their firstorder partial derivatives in the closed region S-j-C, then Green's formula holds: (S) here t'^e remain C circulation about the contour chosen so that the region S should is to the left. 5. Applications of line integrals. = 1 ) (p An y area dx= bounded by the closed contour C (p is xdy (the direction of circulation of the contour is chosen counterclockwise). The fo lowing formula for area is more convenient for application: 2) Z The work (x, y t X= X(x, y, of a force, having projections accordingly, the work of a force field), z) (or, 2), Y=Y (x, y, along a path C z), is _ Sec. 9] __ 277 Line Integrals expressed by the integral A= ^Xdx + Ydy + Zdz. c If the force has a potential, i.e., if there exists a function potential function or a force function) such that (a dV dx =A dU dU j TT , L -7 , A *j) //2, = where (v l5 (*J, '/J. (*i, l/ lt Xdx + Ydy + Zdz^ J (*|. {/,, , dU the initial and (x 2 Zj) is , = U(x is tli2 2293. xyds, where J C 2t y tt z 2 )-t/ (x lt y, z } ), terminal point of the path. A. Line Integrals of the First Evaluate the following equal to 2-J z2) is Z.) J Zi) f/ 1? (x, y, z) dz dy then the work, irrespective of the shape of the path C, (*. U=U Type line integrals: the contour of the square is | x\ + \ y\ =a (fl>0). s 2294. \ r c K --^= ,. 2 A: if |- , connecting the points 0(0, ^xyds, where 2295. C where a segment of the straight line is \ \- C and A 0) (I, a quarter is 2). of the c lying the in 2296. first + = l fi> quadrant. where C ifds, ellipse ^i is the first arc of the cycloid x =a (t sin /)> c a cos (1 /). }x 2297. 2 +y 2 ds, where c circle x ---- a (cos / 2 2298. ^ (x (-/sin/), y 2 2 -\- y d, ) C is an arc = a(smt where C is of the involute of the tcost) I0^/=^2ji]. an arc of the logarithmic spi- c ral r^ae m v(m>Q) 2299. niscate r 2300. + y) JU c = a cos2 2 where C is (0, a) to the point 0( oo, 0). the right-hand loop of the lem- 2 J (A: \-y)ds, c q/2 A from the point > where C is an arc of the curve JK =^ Multiple and Line Integrals 278 2301. \ %2 (* . ~T~ y i i # = acos/ y = as\nt, f 2 2302. J t/"2# is the where C is turn of the screw-line first = bt. z 2 -I- C where *2 . [Ch. 7 z ds y the x circle 2 + y +z = a 2 2 2 , c x=y. 2303*. Find the area of the cylinder y = surface lateral o x 2 bounded by the planes z the parabolic of = 0, x~0, 2 = x, = 6. // ae cos?, 2304. Find the arc length of the conic screw-line C x sin /, z ae from the point 0(0, 0, 0) to the point A (a, 0, a). 2305. Determine the mass of the contour of the ellipse 1 y = ae* 2 2- + = yi -T2- = l, if l the linear density ot it at each point M (x, y) is equal to \y\. 2306. Find the mass of the first turn of the screw-line A; = a cos/, y = asmt, zbt, if the density at each point is equal to the radius vector of this point. 2307. Determine the coordinates of the centre of gravity of a half-arc of the cycloid x = a(t sin /), y = a(\ cost) [0 2308. Find the moment of inertia, about the z-axis, of the turn of the screw-line x = a cos/, !/ bt. asin/, z 2309. With what force will a mass distributed with uni2 2 form density over the circle x -f y 2 a z 0, act on a mass m located at the point A (0, 0, &)? = = first M = , = B. Line Integrals of the Second Type Evaluate the following line integrals: 2310. 2 J (x 2xy)dx-\- (2xy+tf)dy, where AB 2 parabola y = x from the point 2311. ^ (2a y)dx \ -xdy, ^4(1, where C to 1) is an AB is the point arc an arc of of the B(2, the 4). first c arch of the cycloid x a(t a(l sin/), # cos/) which arc runs in the direction of increasing parameter = = 2 /. taken along different paths emanating J 2xydxx dy OA from the coordinate origin 0(0, 0) and terminating at the point 2312. A (2, ia) 1) (Fig. .103): the straight line Om A\ Lim Sec. 9] parabola OnA, b) the 279 Integrals the axis oi symmetry of which is the of which is the (/-axis; c) the parabola the axis of OpA, symmetry x-axis; broken line d) the e) OBA\ OCA. the broken line 2 2313. 2xydx ^x dy J as in Problem 2312. OA (x+u)dx (x x* + y* i/)dy counterclockwise. Fig. ^tfdx + x*dy, where C 2315. 103 the upper half of the ellipse is c x^acost, y = 2316. bs\r\t traced clockwise. cosydxsmxdy \ taken along the segment bisector of the second quadrantal point A is 2 and the ordinate of d x!f(l' 2317. dll} ^* (f = , where C angle, B is is if the AB of the abscissa oi the 2. the right-hand loop ol the a lemmscate co2(p traced counterclockwise. 2318. Evaluate the line integrals with respect to expressionswhich are total differentials: r* ( - (a. 4) 8) (2. a) rt xdy -\-ydx, ^ (2, f (1. J c) i) (0, 0) 1) 1) ydx d) b) (O r 2) 1 (i. xdx + ydy, 2) xdy , (along a path that does not intersect the _ Multiple and Line Integrals 280 (x, y) dx + dy x ~ry f J e) f) p a th ong a ^at [C/t. not (joes intersect 7 the = 0), straight line A;-f-y (* 2 (a i ___ J/ 2 ) , q> J (*;, 2319. Find the antiderivative functions evaluate the integrals: integrands and 0) (3, a) of the (x< J + 4xy') dx 4 (6x V 4 5# ) dy , (-2, -1) o) (i, * J ~~~ y b) ! x \ j -1) (0, straight line y (, C) 1) x ^ y) \ (the integration path does not intersect the = x), (x+Wx + ydy (the integration j a. i) the straight line y A\ f I , J (0, \V X ! x* 0) 2320. = + y* path does not intersect x), yA J fix I y / \ I \}^xZ + y z fat J Compute /=. f xdx + yfy taken clockwise along the quarter of the X* ellipse -5 the first quadrant. 2321. Show that if f(u) is a continuous closed piecewise-smooth contour, then W 2 + ^=! that lies in function and c 2322. Find the antiderivative function a) b) c) du = (2x du = (3* 2 du = d)' dtt U if: + 3y)dx + (3x4y)dy\ = 2xy + y 2 ) dx 2 (x 2xy + 3y*) dy\ C is a Sec 281 Line Integrals 9] Evaluate the curves: 2323. line taken along the following space integrals z)dx+(zx)dy + (x ^ (y C where tj)dz, is a turn c of the screw- line i / = asin/, corresponding to the variation 2324. of the parameter ydx + zdy + xdz, where (p c I \ ;t /y ^ C is / from to the circle = /?cosacos/, = /?cosa sin = ^sina (a = const), /, traced in the direction of increasing parameter. 2325. xydx + yzdy ( + zxdz, where OA is an arc of the OA circle situated on the side of the A'Z-plane where //>0. 2326. Evaluate the line integrals of the total (, a) 4. zdz, xdx-\-ydy ] (1, 0, -3) (i. i. i) (3, 4. 5) b) x differentials: 8) // z dx f 2 x dy - xd\ r \- xy dz , \-\idij-\-zdz j rA'.|.y" J - +2 ' (0, 0, 0) -M ^v d) yte + mv + 'y** f J (i. xyz , (the integration path is situated D in the first octant). C. Green's Formula 2327. Using Green's formula, transform the line integral / = 2 \f* -\ if dx + y [xy + In (jc 4 c where the contour C bounds the region S. K?T?)] dy, _ _ Multiple and Line Integrals 282 [C/t. 7 2328. Applying Green's formula, evaluate / = 2 (x? if) -t- dx + (x + y) 2 dy, where C is the contour of a triangle (traced in the positive direcverlices at the points A (I, 1), fl(2, 2) and C(l, 3). with tion) Verify the result obiained by computing the integral directly. 2329. Applying Green's formula, evaluate the integral x*y dx + xif dy, c where C 2330. chord is Find y the circle x* is + if = R* traced counterclockwise. A parabola AmB, whose axis is the #-axis and whose AnB, is drawn through the points A (1,0) and 8(2,3). (x + y)dx(x y)dy directly and by applying Green's AmBnA formula. 2331. Find e* y [y* dx (1 -f xtj)dy\, \- if the points A and B AmB on the #-axis, while the area, bounded by the integration path AmB and the segment AB, is equal to S. 2332*. Evaluate Consider two cases: lie ^ifc^f. when when the origin is outside the contour C, the contour encircles the origin n times. b) 2333**. Show that if C is a closed curve, then a) where s is the arc length and n is the outer normal. 2334. Applying Green's formula, find the value of the integral I = (j)[xcos(X, n)+ysm(X, n)]ds, c where ds is the differential of the arc the contour C. 2335*. Evaluate the integral taken along the contour of a A (1, is traced counterclockwise. 0). fl(0 f 1), C(-l, 0) and n is the outer normal to square with vertices at the points 1), provided the contour and>(0, 2S& Line Integrals Sec. 9] D. Applications of the Line Integral Evaluate the areas of figures bounded by the following curves: a cos/, y bs\nt. 2336. The ellipse x acos 3 /, #-=asin 8 /. 2337. The astroid jt x a (2 cos/ cos 2/), 2338. The cardioid y a (2 sin/ = = = = = sin 20- A 2339*. of loop the folium (a>0). Descartes x* +if 3zxy =Q = axy. 2340. The curve (x + y)* 2341*. A circle of radius fixed circle of radius integer, find the area by some point of the R of r of (cardioid). A circle of 2342*. r R and is rolling without outside sliding along a n is an Assuming that it. bounded by the curve (epicycloid) described moving circle. Analyze the particular case radius a fixed circle of radius r R and is rolling inside without sliding along D Assuming that it. is an integer, find the area bounded by the curve (hypocycloid) described by some point of the moving circle. Analyze the particular r> when case r =j (astroid). 2343. A field is generated by a force of constant magnitude F Find the work that the field does in the positive jt-direelion when a material point traces clockwise a quarter of the circle 2 2 x -^-y ^=R lying in the first quadrant. 2344. Find the work done by the force of gravity when a material point of mass m is moved ironi position A (JC P // l? zjz 2 ) (the z-axis is directed vertically upto position B (x 2 // 2 , , wards). 2345. Find the work done by an elastic force directed towards the coordinate origin if the magnitude of the force is proportional to the distance of the point fiom the origin and if the point of application of the force traces counterclockwise a quarter of the ellipse 2346. ^s4-^i=l Find the lying in the potential first and determine the work done by the = 0, K:=0. Z-=rng (force a) X rial is point B(* b) Uv moved from quadrant. of a force R {X, Y, Z\ force over a given path if: of gravity) and the mate- function position A y l9 zj where jx (x lt to position *i)'- x= ?, K=-^. Z=-* f = const and Yx* 4 if f (Newton attractive force) and the material point moves from position A (a, b, c) to infinity; r -\- _ _ Multiple and Line Integrals 284 X= c) force), x* 4- Sec. 2 # = = and k*x, Y k*y, Z the initial point of the + z = /? 2 2 , [Ch. 7 where k k*z, = const (elastic path is located on the sphere while the terminal point is located on the sphere 10. Surface Integrals 1. Surface integral of the first type. Let function and z=-cp(*, y) a smooth surface S. The surface integral of the first type '(x, //, z)dS= n is f (x, //, 2) be a continuous the limit of the integral sum lim -> 00 fas I .3 is the area of the /th element of the surface S, the point (*/, y lt belongs to this element, and the maximum diameter of elements of partition tends to zero. The value of this integral is not dependent on the choice of side of the surface S over which the integration is performed. If a projection a of the surface S on the jo/-plane is single-valued, that to the z-axis intersects the surface S at only is, every straight line parallel one point, then the appropriate surface integral of the first type may be calculated from the formula where AS/ z/) (*' y) S dx dlJ- (a) Example where S is 1. Compute the surface integral the surf ace of the cube 0<Jt Let us compute the sum of the surface integrals over the the cube (z=l) and over the lower edge of the cube (z 0): 00 2. Surface integral is of 00 00 The desired surface upper edge obviously three times greater = and equal to = If P P(x, //, z), Q Q (*, y, z), continuous functions and S + is a side of the smooth surface S characterized by the direction of the normal n {cos a, cos p, cos Y}. t'hen ihe corresponding surface integral of the second type is expressed as follows: R = R(x, integral of the second type. y, z) are P dy dz + Q dz dx+ R dx dy= f f (P cos a -f Q cos p + R cos Y) dS. Sec. 10] 285 Surface Integrals When we pass to the other side, S~, of the surface, this integral reverses sign. then the direcIf the surface 5 is represented implicitly, F (x, y, z) 0, tion cosines of the normal of this surface are determined from the formulas = 1 OF COSBQ^ COSa==-pr^-, D dx dF 1 -FT^ D dy , COS VT = OF 1 -rr- -T D , dz where and the choice should of sign before the radical of the surface S. with the side be brought into agreement = = P (.v, //, z), Q Q (x, //, z), If the functions P continuously differentiable and C is a closed contour bounding a two-sided surface S, we then have the Stokes' formula 3. formula. Stokes' R = R(x, y, z) are (j) C dQ\ = rr\fdR 3-- 5- 1 dz \ \ JJ l\dy 5 c s a J Id? dR\ Tdx + -3 ^\dt fi cos a l + ^ , j dP\ fdQ T- 3-2- \dx 1 dy J 1 cos v y _ dS, | where cos a, cos p, cosy are the direction cosines of the normal to the surface S, and the direction of the normal is defined so that on the side of the normal the contour S is traced counterclockwise (in a rigiit-handed coordinate system). Evaluate the following surface integrals of the 2347. 8 (* 4 tf)dS, where S z first the sphere x +// is type: 2 2 -{-z = a*. 6 2348. 5 Vx s. 2 -\-tfdS where 5 is the surface lateral the of s + g_?6 i= =0 [O^z^bl cone Evaluate the following surface 2349. \ \ integrals of the second type: yz dydz -\-xzdz dx-\- xydxdy, where 5 is the external s side of the surface of a tetrahedron y =Q t 2 2350. = 0, x+y + z = a. Nzdxdy, where S t is bounded by the planes the external side of the ellipsoid x dydz-\-y*dzdx + z*dxdy, 2351. o where S is +? = 2 the 2 a (z side of the surface of the hemisphere +// 2352. Find the mass ot the surface of the cube Os^z 1, if the surface density at the point Is equal to xyz. O^y^l, < 0, A: external ^0). O^x^l, M (x, y, z) _ _ Multiple and Line Integrals 286 [C/i. 7 2353. Determine the coordinates of the centre of gravity of a 2 2 homogeneous parabolic envelope az^je + # Find 2354. the moment (0 inertia of a part of =V the lateral of < < x 2 -f y 2 [0 z about the z-axis. surface of the cone z ft] 2355. Applying Stokes* formula, transform the integrals: 2 (x a) - yz) dx + (y 2 zx) dy + 2 (z xy) dz\ c b) (j) ydx-\-zdy + xdz. c Applying Stokes' formula, find the given integrals and verify the results by direct calculations: 2356. (f (y + z)dx + (z + x)dy + (x + y)dz, where C the circle is c 2357. x)dy + z)dx^(z (y 2 2 JC -|-// (x y)dz, where C is the ellipse X + 2=l. =1, + (x-{-y)dy + (x + y + z)dz, where C is the curve = acos/, z = a (sin + cos/) [0 2358. ()xdx c / // 2 2 z A /IflC with vertices 2360. In what case / (a, 0, 0), is B C (0, (0, a, 0), 0, a). the line integral = Pdx + Qdy + Rdz c over any closed contour equal to zero? 11. The Ostrogradsky-Gauss Formula Sec. If Q C =Q thMr 5 is a closed R= (A% y, z), first partial smooth surface bounding the volume V, and P P (x, y, z), are functions that are continuous together with derivatives in the closed region V then we have the Ostro/? (v, (/, e) t gradsky- Gauss formula where crsa, cos p, surface S cosy are the direction cosines of the outer normal to the Applying the Ostrogradsky-Gauss formula, transform the following surface integrals over the closed surfaces S bounding the rel="nofollow">ec. 11] _ The Ostrogradsky-Gauss Formula /olume V(t*osa, cosp, lormal to the surface 2361. JJ _ 287 cosy are direction cosines of the outer S). xydxdy+yzdydz -\-zxdzdx. s 2362. J J 6 x 2 dy dz + y* dz dx +z 2 d* dy. 2363. Using the Ostrogradsky-Gauss formula, compute the following surface integrals: 2o65. x*dydz + y*dzdx + z*dxdy, where S J the external is s side of the surface of the 2366. of a \ \ V cube O^x^a, O^r/^n, O^z^a. xdydz + ydzdx + zdxdy, where 5 the external side is pyramid bounded by the surfaces x + y-{-z = a, x = Q,y = Q, z=-0. 2367. x* dydz-\-if dzdx = z* dxdy, where 5 the is external ^ side of the sphere x 2368 2 ~a cos p + 2 2 2 f 2 ^(jc cosa-t y // 2 2 -|-z 2 . cos y) d5, where S is the exter- o nal total surface of the cone 2369. Prove that direction, then if S is a closed surface and / is any fixed where n is the outer normal to the surface S. 2370. Prove that the volume of the solid surface S is equal to = -3 V bounded by the M where cose, cosp, cosy are the direction cosines normal to the surface S. of the outer Multiple and Line Integrals 288 [Ch. 7 Sec. 12. Fundamentals of Field Theory 1. Scalar and vector of the point surfaces f t/, (x, A fields. = /(/>) = /(*, = C, where y, z) z), C= defined by the scalar function is a point of space. The const, are called level surfaces of the scalar scalar field P where is (x, y, z) field. = A vector field is defined by the vector function of the point a a(P)~ a(r), where P is a point of space and r=xi-\-yj+zk is the radius vector of the point P. In coordinate form, a a v j-\-a z k, where a x ~a x (x, y, z), ax i a and a z a z (x, //, z) are projections of the vector a on the y ay(x, y, z), coordinate axes. The vector lines (force lines, flow lines) of a vector field are found from the following system of differential equations + = ~~~' dx__dy _dz A scalar or vector 'ry; - that field where " V = ^3-+y^- + ^y is not depend on the time is called nonstationary. does t time, depends on the t. The Gradient. Th vector it if stationary; ~ it Hamiltonian operator the t or (del, called is nabla), is = called the gradient of the field U f (P) at the given point P (ci. Ch. VI, Sec. 6). The gradient is in the direction of the normal n to the level surface at the point P and in the direction of increasing function U, and has length equal to dn~~ If the direction is \dx given by the unit vector = (the derivative of the function 3. Divergence and rotation. + ay j+a z k the scalar is The rotation U in - cos / - cos p + a+ the direction The divergence diva-^ + (curl) of a vector field {cos a, cos p, cos Y}, /). of a vector field \ \ S is i the integral an dS = a (P) ~ a^i \~ ^ + ^^Va. a (P) = ax + ay j + az k 4. Flux of a vector. The flux of a vector field a(P) in a direction defined by the unit vector of the normal S cosy is the vector da y da z to the surface - then \ \ s an dS \ \ (ax cos a -|- a y cos p through a surfaces ujcosa, cos p, COSY} + az cos Y) dS. S If S is a closed surface bounding a volume V, and n is a unit vector of the outer normal to the surface S, then the Ostrogradsky -Gauss formula holds, which 289 Fundamentals of Field Theory Sec. 12] in vector form is ff (\ r\ r* div 5. Circulation of a vector, the work a along the curve C is defined by vector f a dr = C a s ds \ V C a dx dy dz. The of a Held. integral of the line the formula a x dx -f- a dy -f a z dz y (0 C and represents the work done by the field a along projection of the vector a on the tangent to C). curve the C (a s is the If C is closed, then the line integral (1) is called the circulation of the vector field a around the contour C. If the closed curve C bounds a two-sided surfaces, then Stokes' formula holds, which in vector form has the form adr= f f rotadS, /i where n is the vector of the normal to the surface S; the direction of the vector should be chosen so that for an observer looking in the direction of n the circulation of the contour C should be counterclockwise in a right-handed coordinate system. 6. tial Potential and solenoidal fields. The vector iield a(r) is called poten- if U, where Uf(r) is a scalar function (the potential of the field). For the potentiality of a field a, given in a simply-connected domain, it be non rotational, that is, rota = 0. In it is necessary and sufficient that that case there exists a potential U defined by the equation dU ~a x dx-}- a v dy -f- a 2 dz. If the potential U is a single-valued function, then a dr \ U (B) U (A); AB in particular, the circulation of the vector A a = 0; If a is equal to zero: m adr=Q. vector field a (r) is called solenoidal if at each point of the field div in this case the flux of the vector through any closed surface is zero. the field is at the same time potential and solenoidal, then div (grad U)=.Q and the potential function U "Er++5-<>. is or harmonic; AU=0 ' that where is, it satisfies the Laplace A= *'=;+>+> isthe Laplacian operator 2371. Determine the level surfaces of the scalar field where r U = F(Q), 10-1900 \fx*+y*-\-z*. where What will the level surfaces be of a field 290 _ _ Multiple and Line Integrals [Ch. 7 2372. Determine the level surfaces ot the scalar field U = arc sin - 2373. Show that straight lines parallel to a vector c are the vector lines of a vector field a(P) c where c is a constant vector. 2374. Find the vector lines of the field a CD*// 4 CDJC/, where CD is a constant. 2375. Derive the formulas: = t = a) + C K) = C gTad(UV) = Ugrad grad (t/ = 26/ grad grad(C l i/ 2 1 gradf/ +C 2 gradV, where C, and C2 are constants; b) 2 c) j\ ) d) V grad(/ U AfU\ =-* grad e) grad (^ J = cp' (t7) grad U. 2376. Find the magnitude and the direction of the gradient 3 x* z of the field U if 3xyz at the point A (2, 1, 1). Deterthe mine at what points gradient of the field is perpendicular to the z-axis and at what points it is equal to zero. 2377. Evaluate grad f/, if U is equal, respectively, to: a) r, = b) r\ c) j , + + d) /(r)(r = /?+^qr?). = 2378. Find the gradient of the scalar field U cr, where c is What will the level surfaces be of this field, and what will their position be relative to the vector c? a constant vector. 2379. Find the derivative of the function given point P(x, y, this point. 'In z) x U = ^ + y^ + ~ala in the direction of the radius vector r of will this derivative be equal to the what case magnitude of the gradient? 2380. Find the derivative of the function U= in the di- rection of /{cosa, cosp, cosy}. In what case will this derivative be equal to zero? 2381. Derive the formulas: a) div^aj + C^^Cjdivaj + Cjjdivajj, where C and C l constants; b) div (i/c) c) = grad /, where c is div((/a) = grad U-a+ (/diva. 2382. Evaluate di 2383. Find div a where r = for the central 2 are a constant vector; vector field a(P) = /(r)~ , Sec. Fundamentals of Field Theory 12] 291 2384. Derive the formulas: + C a = C rota + C rota where C and rot(/c) = grad U-c, where c is a constant vector; rot (Ua) = grad U a + U rot a. rot(C a l a) 1 2 2) l 1 f 2 , C t 2 are constants; b) c) 2385. Evaluate the divergence and the rotation of the vector respectively, equal to: a) r\ b) re and c) f(r)c, where c a if is a constant vector. a is, 2386. Find the divergence and rotation of the field of linear velocities of the points of a solid rotating counterclockwise with constant angular velocity o> about the z-axis. 2387.' Evaluate the rotation of a field of linear velocities co r of the points of a body rotating with constant angular = velocity = = + , = + a) the lateral surface of the cone 0<e j force F= ^ of a mass w, located point of at the coordinate through an arbitrary closed surface surrounding this point. 2394. Evaluate the line integral of a vector r around one turn of the screw-line * = /?cos/; y = Rsmt\ z = lif from / = origin, to = 2n. 2395. Using Stokes' theorem, evaluate the circulation of the vector a zk along the circumference x* + if x*tfi R 2 z=-0, 2 2 jt taking the hemisphere z J/"/? if for the surface. 2396. Show that if a force F is central, that is, it is directed / = +j+ = \ = towards a fixed point and depends only on the distance r from where f(r) is a single-valued continuous point: function, then the field is a potential field. Find the potential U F = f(r)r, this of the field. 2397. Find the potential by a material point of mass nates: a= equation 10* ~r. Show U m of a gravitational field generated located at the origin of coordi- that the potential U satisfies the Laplace _ _ Multiple and Line Integrals 292 (Ch. 7 2398. Find out whether the given vector field has a potential find if the potential exists: U and a) b) c) a a a= 2399. Prove that the central space field lenoidal only is U, when = ~, f(r) 2400. Will the vector field a constant vector)? where k is a = r(cxr) = /(r)rwill be so- constant. be solenoidal (where c VIII Chapter SERIES Sec. 1. Number Series 1. Fundamental concepts. A number series 00 a,+a t +...+a +...= tt is called convergent if its has a finite limit as n > (1) tt sum The quantity oo. l S= lim S n n -+ oo is then called the sum number of the series, while the is partial 2a n-\ called the remainder of the series. If Sn the limit lim n -* does not exist (or is QO infinite), the series is then called divergent. If a series converges, then lim a n Q (necessary condition for convergence). n-*oo The converse not true. For convergence of the series (1) it is necessary and sufficient that such that for n any positive number e it be possible to choose an and for any positive p the following inequality is fulfilled: is N (Cauchifs test). The convergence or divergence subtract a finite number of 2. Tests of convergence a) Comparison test *! I. of series a is not violated if for >N we add or terms. its and divergence If + *,+ .. of positive series. after a certain n n = +*,!+. ..^ , and the series (2) converges, then the series (1) also converges. If the series ( J) diverges, then (2) diverges as well. It is convenient, for purposes of comparing series, to take a geometric progression: 00 2 aq n=o n (a * 0), 294 [Ch. 8 Series which converges which is a |^| for \q\^\, and diverges for the harmonic series divergent series. Example The 1. series ++ + + converges, since here a = 1 J_ "~n-2 n 2*' while the geometric progression 1 n=i whose ratio Example is = <7 , The 2. converges. series diverges, since of b) its In ln_3 ln_2 general term is /i greater than the term corresponding the harmonic series (which diverges). Comparison test II. there If exists a finite and nonzero limit lim ? y. b n n if a n -^b n ), then the (in particular, the same time. Example 3. The series (1) and series diverges, since 1" 1 whereas a series with general Example 4. The term n n J 2 diverges. series _J_ 1 _J converges, since while a series with general term ^ converges. (2) converge or diverge at Sec. 1] c) __ Number D'Alembert's Let test. rel="nofollow"> an 295 Series certain a (after n) and there be let a limit -l = lim n Then the it not is < series (1) converges if q the series known whether Example 5. an -> GO q. and 1, diverges if q > If 1. = <7 1, then convergent or not. is Test the convergence of the series 1+1+1+ 2 2 ' 22 s Solution. Here, and ?-H = lim 2 Hence, the given series converges. d) Cauchy's test. Let of^^O (after a certain n) and lim n Then (1) converges if q<\, let there -> OD n V and diverges if q>\. When q=l t remains open. e) Cauchy's integral test. If a n f(n), where the function f (x) rnonotomcally decreasing and continuous for jc^a^l, the series of the convergence of the series be a limif n/~ = = the question is (1) positive, and the integral 00 " / (x) dx converge or diverge at the same time. By means converges of series if of the integral test p> may 1, be and diverges tested by it may be proved that the The convergence of a large number if p<^\. comparing with the corresponding Dirichlet series (3) Example 6. Test the following series for convergence --^" -U--L + -L+ Solution. We * 1 i - r "' i have 1 ~ Dirichlet series __1_1 J^ 4/i 2 ' __ 296 [Ch. 8 Series = Since the Dirichlet series converges for p 2, it follows that on the basis of test II we can say that the given series likewise converges. 3. Tests for convergence of alternating series. If a series comparison + |fll+..., l+.-. composed values of the of the absolute (4) terms of the series converges, But if (1) con- (1), then (1) also converges and is called absolutely convergent. verges and (4) diverges, then the series (1) is called conditionally (not absolutely) convergent. For investigating the absolute convergence of the series (1), we can make use [for the series (4)] of the familiar convergence tests of positive series. For instance, (1) converges absolutely if < lim n -> oo \ n In the general case, the divergence of gence of But (4). n (4) ?2_J lim if - GO I /KI< lim or 0-n n I > 1 (1) or f not does 1. from follow lim /\a n n -> oo K \> 1, the diver- then not only does diverge but the series (1) does also. Leibniz test If for the alternating series *!-* conditions the following are (*^0) +*3- **+ fulfilled: 1) (5) ^ b ^b bl 2 s ^. . . ; (5) bn lim 2) n then - = oc converges. In this case, for the remainder of Rn the series the evaluation holds. Example 7. Test for convergence the series Solution. Let us form a series of the absolute values o! the terms of this series: Since . lim lim I 1 n - oo , the series converges absolutely. Example 8. The series converges, since the conditions of the Leibniz converges conditionally, since the series 14- diverges (harmonic series). -* 4- 4-4 test are fulfilled. This series __ Sec Number /\ Series 297 Note. For the convergence of an alternating series it is not sufficient that general term should tend to zero. The Leibniz test only states that an alternating series converges if the absolute value of its general term tends to zero monotonically. Thus, for example, the series its diverges despite the fact that its general term tends to zero (here, of course, the monotonic variation of the absolute value of the general term has been S 2k = S'k + S"k violated). Indeed, here, and S k = cc(S k lim k - Sk lim limit fe exists is a partial and , sum finite (S k is where of the a is harmonic partial sum series), whereas the o f the convergent geo- -* metric progression), hence, S 2fe =oo. lim k -> 00 On the other hand, the Leibniz test is not necessary for the convergence series: an alternating series may converge if the absolute value of its general term tends to zero in nonmonotonic fashion Thus, the series of an alternating .22 ~3 a _ 42 ~~'""(2n 1)' converges (and it converges absolutely), although the Leibniz test is not fulfilled: though the absolute value of the general term of the series tends to zero, it does not do so monotonically. 4. Series with complex terms A series with the general term c n a n = + 00 z -]-ib n (i 1) converges and only if, if, the series with real terms 2 an n=i 00 and 2&,, conver g e n a* same time; the SC B n=i =S= n The series (6) definitely converges series whose terms are the moduli 5. Operations on series. a) that A is, convergent series and of the may 8 i be +'2Xn = (6) i is called absolutely convergent, terms of the series multiplied +a 1 (6), +...+fl n + ...=S t if the converges. termwise by if a, then in this case \ any number fc; __ 298 Series b) sum the By two convergent (difference) of fli _ [Ch. m series + flt+-..+fl n +...=5 + *,+ .. .+*+...=, (7) If (8) *! we mean a series + (a bl) (a, c) The product of l b z) If verges d) terms verges cn + the series = a,bn + a A-I + . . + (a n . (7) and bn ) . . . =S, S2 . the series is (8) + + c>+...+cn +... +0M = 2 cl where 8 *> (9) t ) the series (7) and (8) converge absolutely, then the series (9) also conabsolutely and has a sum equal to S S 2 If a series converges absolutely, its sum remains unchanged when the of the series are rearranged. This property is absent if the series conconditionally. . { Write the simplest formula of the the indicated terms: term /ith of the series using 2401. 1+1+J + I+... 2404. i+4 + j+^+... 2402. 1+1+1+1+"- 2405. 2403. 1+1 + 4 + 4+.. + + + ~4 2406. 4+*.+ ^+... f+{ + + + 2407. . - 2409. 1 2410. 1+ 1 H- 1 - 14 ~ 1 1 . . . +... -I- In Problems 2411-2415 it is required to write the first 4 or 5 terms of the series on the basis of the known general term a n . 2411. a n 2412. 2413 = |q^. 2414. a n = 2415 \* din*. a = t^. flrt = Test the following series for convergence by applying the comparison tests (or the necessary condition): 2416. 1 1 +1 ! + ...+( 1)-' + ... Number 2418. + + 0419 J___1_ 444^ ^ 4 ^ 6 . 2 2422. ++ + 2421. ' ' 4^ -I- ' 2i ' ' ' ' + + ... -= + -=-f -= + ^3-4 . ... 4' 1/2-3 02 2423. 2 299 Series ]fn (n +$$ O" OJ + + y+... i- +=;-+.. . -' 4 J/3 3 V'2 Using d'Alembert's (n test lest, the gence: 2427. 98 -*' -l= 2 1 ! 1 1-5 + ''-} ^ 1-5-9 - " following 2 "~' + 4 4-^- +...-' K 1-1) ' i ' j ' - 1-5- 9. ..(4/i-3) Test for convergence, using Cauchy's test: Test for convergence the positive series: 2431. 1+1 + 1+...+1+... 1 + 1 + 1+ ..+ _ 1 2432. . ((H 2433 - r4+4T7 + 7no+ [ )2 ) r/lo/ ^TtjO +... +(3n-2M3n+l 24.i+f -^+... +sSFI+ 2435. 1 ... ^ + 1- + ^+...+^+... ^ 5 .'^ 7 i i I I J_ i i _X. 9 ^ * j_ i * __________ l bones for conver- 300 Series 2439. -+ + +... 2440. 1+ + +... [Ch. ++... 2441> 2442. 1++.+ 1, 4 1-3.5 -3, 1-jT - ... 4^M2~*" 1.3.5.. .(2/1-1) ' "" 4-8-12...4n 2445. "" 1000. 1002. 1004... (998 1-4-7. ..(3n-2) 2.5-8...(6 n !. 5.9... 2447 -7)(6g-4) +2n) " (8_H)(8n -7) 2 ... I, '- 4 1 ' 4'9 ^--^ 1.3.5.7.9 ....^ '"" 2449 2450. 2451. 1-4-9. .. ""* I | ' 1-3-5-7- 9... (4n arcsin. 2455. sin. 2456. 3) !, i=J 2452 152. Ulnfl+l). ^ \ n J 2457. ,. ~ . . ^- ri'lnn-ln In n 00 2453. yin^i. 2458. /J=2 2454 - E -' 2459 - STTOTJ . 8 Number Sec. 1] 2464. Series . cos-2-) ^ 2469. Prove that the series 1) converges for arbitrary q t p>l, p if and for ?>1, if p=l; p=l. and for (_n-i i 2470. l- 2471. 1 2472. l_ + _... + 2473. l_ 2474 + -....ui--f... -- - + /2 K3 l + .. + 3_... + (^S + " ' ... + . ^- -. +(-!)- 2475. __ 2476. -- -- -- ---^--h...+ + . + _. ..+(_!) ^ 7=4 2|^2 1 1 .^+... I 9 a 7=4 3^3 4^41 1 , n+\ n / {) (n 947Q "' l_k!-J-lL_ 7 7.9~t"7.9.11 -i / "TV ii-' U + \) 1-4. 7... (3/1-2) 7.9-11. ..(2/1 5)' + 302 [Ch. 8 Series na sin 2481. 2482. (_l)"l^, ( ly-'tan n^i n ^ Vn 2483. Convince yourself that the d'Alembert test for converdecide ide the quesl question of the convergence of the gence does not series 2a where > whereas by means of the Cauchy test it is possible to establish that this series converges. 2484*. Convince yourself that the Leibniz test cannot be applied to the alternating series a) to d). Find out which of these series diverge, which converge conditionally and which con- verge absolutely: a)' ___ _J l _1_ 1 1 3 v ' ^\ d) i_ +y *~ + 21 j .j; r . 3F _ . 3 " | 3 "L 32 h 5 . 3?+ __u " ' ' ' 3* _ _ T _l + T T + TT T +... ! J ! ] i . i ! . Test the following series with complex terms for convergence: CO 00 ^2+#. 2485. 2488. 00 2486, 00 (2 ' X" 7 1=1 V. 3 1) ''. 2489. =i __ Sec. Number 1] Series V - {/ = X-? r(2-Q + 1" l [ ' [n(3-20-3ij fa 2493. Between the curves and y=-^ X and 303 the right to intersection are constructed segments parallel an equal distance from each other. Will the sum the lengths of these segments be finite? 2494. Will the sum of the lengths of the segments mentioned of their point of to the t/-axis at of Problem 2493 be in curve y = X curve \)^-\ x finite if the is by the replaced ? 00 Form 2495. Does sum the ^-^ and the series of sum converge? this 00 2496. Form the anc* * es * and difference 2497. ]T 2 r Does the formed series by the subtracting series 00 TT = divergent series i QC n the for convergence. '* rt n of sef i es -f 21 n i converge? i sum converges while 2498. Choose two series such that their their difference diverges. CC 2499. Does Form the product of the QC V ~nVn series and V.OTTM2 tt n this product converge? 2500. Form the series (l +1 + 1+ .. . + J~-f- . . . V. Does this series converge? 2501. Given the series 1+1 1+ ...+ ( - : ^+... Estimate the error committed when replacing the sum of this series with the sum of the first four terms, the sum of the first five terms. What can you say about the signs of these errors? 2502*. Estimate the error due to replacing the sum of the series y+ by the sum 2! (2") of its first + ~3$$'2J + " + 5H2") n terms. +" _ __ 304 [Ch. 8 Series 2503. Estimate the due to replacing the sum error the of series 1 i by the sum JL -L-L-L-l. 4- ^2M3M of its first 4- "+/i!^"' n terms. In particular, estimate the accu- racy of such an approximation for n=10. 2504**. Estimate the error due to replacing the sum of the series 1 sum + 22 + 32+ +^i+ ' n terms. In particular, estimate the accuan approximation for n= 1,000. 2505**. Estimate the error due to replacing the sum of the by the of its first racy of such series 1 by the sum + 2(i)'43()V. of its first . n terms. 00 Zl _ I)"" - - 1 does one have n=i take to compute decimals? to 2507. sum its How many two decimal to terms of the series (2/1 have to take to compute its sum to to places? -MIS* three does one two decimal places? to three? to four? 2508*. Find the 2509. Find the sum sum the of series -L + gL + jL + .. . + of the series Sec. 2. Functional Series 1. Region of convergence. The set of values of the argument x for which the functional series /.<*) converges is + /.(*)+.. .+M*)+... called the region of convergence of this series. S(*) = n where S n (x) vergence, is of the series. fl (x) +f called the z lim Sn (1) The function (x), -* QO (x)+ ...+f n (x), and x belongs to sum of the series; R n (x) = S(x)S n the (x) is region of conthe remainder Sec. 2] In _ _ 305 Functional Series the cases, simplest it (1), to convergence of a series holding x constant. is sufficient, when determining the region of apply to this series certain convergence tests, n Diverges ,%f? . -3-101 Determine the region 1. X 104 Fit*. Example Diverges of of the series convergence x+\ " (x+\y ~ (X+\Y " ~ (*+iy TT" 2-2* "~T^r t""'~ n- 2" + 1 t t ' we Solution. Denoting by u n the general term of the series, lim M -*| I \U n Using d'Alembert's we get the we have ' that l, is, harmonic the *'* * lim I'+H Jx+l\ 2 n^*2 n + (n+$$ |jc| series that series 1 1 + -~ (c n and a are gence) | x a | + c, (A-a) + c real < 3 1< x < or To +..., converges (conditionally). Thus, the scries converges when 2. Power series. For any power c 3<x if is, + TT+-Q-+.. o 2 2, test) have ' assert that the series converges (and converges oo<^< if will 1 l we can test, if absolutely), = \ 11 2 (* -t oo the series diverges, 104). (Fig. When x=l which diverges, and when which (in if x= 3 with the Leibniz accord 3^*<1. series fl)'+ . . . +c n (x B fl) + . . . (3) numbers) there exists an interval (the interval of convercentre at the point x a, with in which the series (3) R with > a R the series diverges. In special cases, the converges absolutely; for \x and oo. At the end-points of radius of convergence R may also be equal to a the interval of convergence x R, the power series may either converge or diverge. The interval of convergence is ordinarily determined with the help of the d'Alembert or Cauchy tests, by applying them to a series, the terms of which are the absolute values of the terms of the given series (3). Applying to the series of absolute values \ = the convergence tests of d'Alembert and Cauchy, we get, respectively, for the radius of convergence of the power series (3), the formulas = lim n - rt* and # = Hm - "/|c,,| too ' However, one must be very careful in using them because the limits on the right frequently do not exist. For example, if an infinitude of coefficients cn 306 __ Series \Ch. 8 vanishes [as a particular instance, this occurs if the series contains terms with only even or only odd powers of (x a)], one cannot use these formulas. It is then advisable, when determining the interval of convergence, to apply the d'Alembert or Cauchy tests directly, as was done when we investigated the series (2), without resorting to general formulas for the radius of convergence. z If = x + ty is a complex variable, then z-Zo) (cn = a n + ib n = for the series power n Z +...+C n (Z-Z Q +... (4) ) there exists a certain circle (circle of convergence) z z inside which the series converges z |>fl the series diverges. At points lying on the cirabsolutely; for Iz cumference of tne circle of convergence, the series (4) may both converge and diverge. It is customary to determine the circle of convergence by means of the d'Alembert or Cauchy tests applied to the scries -z |z | , z Jt -f/f/ ) = with centre at the point , whose terms are absolute values of the terms of the given series. Thus, for example, by means of the d'Alembert test it is easy to see that the circle of convergence of the series "" "" 2-2 2 1-2 ~~"'~~ 3.2 s ~""' n .2* is determined by the inequality |z-f 1 |<2 [it is sufficient to repeat the calculations carried out on page 305 which served to determine the interval of convergence of the series (2), only here x is replaced by z]. The centre of the circle of convergence lies at the point z 1, while the radius R of this circle (the radius of convergence) is equal to 2. 3. Uniform convergence. The functional series (1) converges uniformly on some interval if, no matter what e 0, it is possible to find an N such that does not depend on x and that when for all x of the given interval we have the inequality R n (x) where R n (x) is the remainder of the e, = > \ | given series. If \fn( x )\*f* c n 2 cn converges, (rt=l, then the 2, < ...) n>N a^x^b when functional series (I) and number the converges series on the interval n-\ V absolutely and uniformly (Weierstrass' test). series (3) converges absolutely and uniformly on any interval lying within its interval of convergence. The power series (3) may be termwise differentiated and integrated within its interval of convergence (for fl| The power *-) + +c n (xa)+ 2 then for . . l -a)+. ..+ncn (xa)*- +. cQ dx+ . . . any x of the interval of convergence of the series d (xa) dx+ c2 2 (xa) dx+ . . . + cn . . =/(*), (3), (5) we have =f (x), (xa) n dx+...= (6) _ Sec. 2] _ Functional Series 307 number # also belongs to the interval of convergence of. the series (3)J. Here, the series (6) and (7) have the same interval of convergence as the [the series (3). Find the region of convergence of the series: 2510. (-D-'. 2511. n- n- +I 2512. -jbn l)" ( M= 25.9. 1 252 - 2-(jzrk- n-i I 00 O(T | O 1 3D M x l^A- v~^ sin (9n ^'oMI^fl v^ 2/1-4^'*' > O^vQI n=i ri 1 * I =o nr 2514. X 25.5**. " 2 sin - = X^i. /l 2523. = 2516. 2517. 2522 -J- (-l)" fI i V^ /I=l *- nMn *. . X 2524*. 2525. 2-p- Find the interval of convergence of the power series and test the convergence at the end-points of the interval of convergence: 2531. 2532. 30 ^ 2528. yin I X ^rr. n-1 2533. 2 2529. . n=i 2530. 2534. ' -_^". 2535. 308 Series [Ch. CO OCQA 25<JO. V>, ^ 00 n f V"" s ;-= \2rt-f-ly V 1 r* . 0**1 > 2551. ^- ( x l 7^ n . 2n4 n=i CO + 5)* n ~ ^-rr 4 CO 2537. CO 2553. (-!) n=i 2 "' V X (3 ,n 2539. t ^-' 2554. 2541. ^1 * a=i 2556. 2542**. 2^ nl xnl . n=i 2557. +l 21 /I = l)" ( > l (AT-2)" /j" 2544*. 2V=1 2558. 2545. l( 2(-l)"" -^|r. n T 2559*. n=o X(A; 2549. 2 ^S^ 2562. Y. (3ft n= 2550. V n" (* + 3)". 2563. 2)". Y( - 2) 3 >" .^-. ! (n+l) 2"+ l 1)" 8 309 Functional Series Sec. 2] Determine the circle 2564. of convergence: V W. V <=gC 2566. Aarf ^arf fl. n=i n=o . J 00 2565. 2568. 2567 (1-M0 2 ". ' ' + 2i) + (1 + 20 (3 -h 20 2 + (1 . . + . + (l+ 2569. + r=-, + (l-lHI-20 l + (1-00-20. ..(l-/ii) 9*70 257 2571. Proceeding from the definition of uniform prove that the series convergence, ..+*"+... . does not converge uniformly in the interval ( 1, 1), but converges uniformly on any subinterval within this interval. sum Solution. Using the formula for the get, for |jc|< of a geometric progression, we 1, Within the interval ( 1, 1) let us take a submterval [ 1+ct, 1 a], where a is an arbitrarily small positive number. In this subinterval |jc|s^l a, |1 xl^a and, consequently, I *,.(*) < I _ at"* /I ( 1 . ;> To prove the uniform convergence of the given series over the subintervai it is possible to choose a, 1 a], it must be shown that for any e > an N dependent only on e such that for any n > N we will have the ine- [1+ quality I Rn (x) <e I Taking any e>0, is, j Thus> puttin g a) n>N n t \R n (x)\ is indeed < - + > 1 [since ln(l ea ln ln(ea)_ when let a) (n+l)ln(l In (1 - x of the subinterval under consideration. (1 a)"* 1 w+l <ea, us require that e; whence (i a) for all Ns=s f ln(l less than \ a) e 1, we are a)<0] and convinced that x of the subinterval the given series on any sub- for all 1+a, 1 aj and the uniform convergence of interval within the interval ( 1, 1) is thus proved. As for the entire interval ( 1, 1), it contains points that are arbitrarily (- close to Jt=i, and since Hm R n (x) = \im X-+1 X-+1 -- =00, J/I+ . 1 1 X no mattei how large n is, 310 _ __ Series [C/i. 8 points x will be found for which R n (x) is greater than any arbitrarily large N we number Hence, it is impossible to choose an N such that for n e at all points of the interval ( would have the inequality R n (x) 1, 1), and this means that the convergence of the series in the interval ( 1, 1) is not uniform. > < \ | 2572. Using the definition of uniform convergence, prove that: a) the series converges uniformly in any finite interval; b) the series x* converges (-1. throughout uniformly the of interval convergence i); c) the series 1+F + F+- -+;?+... in converges uniformly positive number; the interval (1-1-8, where 8 co) is any d) the series 2 + (x*- *) + 4 jc (x ) 6 8 jc ) -f- (jc x 2n+2 ) + (x " 2 . . . -h . . . 1, converges not only within the interval ( 1), but at the extremities of this interval, however the convergence of the series in (1, 1) is nonuniform. Prove the uniform convergence of the functional series in the indicated intervals: 00 2573 on the inlerval - I" 1 ' !] on 2574. over *he en tire number scale. 2^ ~~2~ 00 I 2575. Applying sums I ( )"~ T= termwise on the interval differentiation [0, and integration, of the series: Y2 V3 y y y.3 y5 \n + ^ + ^+. ..+?-+... x x -*- + --...+(-l)'-*L + ., , + + +... + _ + ... 2576. x 2577. 2578 . v yLfl l 1]. t find the __ Taylor's Series Sec. 3] v8 2579. s + -' .- +(-!)"+2x + 3x* + ... 4 1 l-3* 2 -[5x 2582. 1.242-3A; 4 + +--- ... 2 ...+(/+ 3-4A: 4- Find the sums of the 2583 r 2 "" 1 *- 2580. 2581. 311 !)*""' 4 series: +++ ++ - ^ + l+!+...+^L i+... 2586. /r Sec. 3. Taylor's Series 1. Expanding a function in a power series. If a function f (x) can be of the point a, in a series of expanded, in some neighbourhood a, then this series (called Taylor's series) is of the form powers of A: \xa\ (I) When holds a the Taylor scries is also called a Maclaurin's series. Equation (I) the remainder term (or simply remainder) of the \xa\ when if Taylor series as n * oo. To evaluate fin (x) the remainder, one can make = <*- f(n * l) l fl + (Lagrant*e's form). Example f if We X ) =, smh n even, and ( is x, = (H) / 0<0<1 where (2) = find f" (x) )l the function f (x) cosh x in a series of powers of x. the derivatives of the given function f (x) cosh x, sinh x, ...; generally, f (n) (x) cosh x, f" (.v) cosh x, sinn A, if n is odd. Putting a 0, we get /(0) (A-) 1, Expand 1. Solution. fl use of the formula = = = /'(0)=0, T(0)-l, /'"(0) = 0, ...; generally, / >(0) ")=-0 if n is odd. Whence, from (1), we have: (II = = 1, = if n is even, and (3) To determine d'Alembert the test. interval We of convergence of the series have lini n-> oo (2/1 (3) we apply the 312 [Ch. 8 Series for any*. Hence, the series converges remainder term, in accord with formula .cosh 9*, R n (x)= >9> 1 , it sinh9*, ~ =^e'*', X I"-*'** and therefore \R n (x)\^ n if is odd, and if n is even. * i e1 = . . A series with the general term in with accord the necessary Y\ n ^ \ 1 made immediately evident with any x (this is therefore, |sinh9*| 1 I test); QO<JC follows that - converges for d'Alembert's convergence, interval has the form: n 1 A + R n (x)= Since the in (2), the help of condition for n+l lim and consequently lim #(*) = () for any x. This signifies that the sum of the /2->00 series (3) for any x is indeed equal to cosh*. 2. Techniques employed for expanding in power Making use of the principal expansions I. * = !++*!+. series. (_oo<* ..+fj+... y II. III. IV. and also the formula for the sum of a geometric progression, it is possible, in many cases, simply to obtain the expansion of a given function in a power series, without having to investigate the remainder term. It is sometimes advisable to make use of termwise differentiation or integration when expanding a function in a series. When expanding rational functions in power series it is advisable to decompose these functions into partial fractions. On x= the boundaries of the interval of convergence (i.e., when 1 it converges absoas follows: for m 1 it and lutely on both boundaries; for diverges when for 1 it l; diverges on both bounconditionally converges when x *) and x=l) the expansion IV behaves = daries. > > m^ m^O x--\ 313 Taylor's Series Sec. 3] Example Expand 2. in powers of x *) the function 3 Solution. Decomposing the function into fractions, partial we will have Since (4) and 2 V^ ^^ ... _ i ^ v ^^ \\j 1 n A> t (5) n =o it follows that we finally get (6) /i n=Q =o The geometric progressions and of |*| hence, 3. Taylor's series two variables /(x, point (a, b) (4) n n-Qv and formula (5) converge, (6) holds when respectively, for |x|<-j |x| i.e., = = 0, is 1 when for a function of two variables. Expanding a function into a Taylor's series in the neighbourhood of a //) has the form ... fc If a notation < the Taylor series is then called a Maclaunn's series. (7) Here the as follows: , y) (X x=a !l=t> ~ d*f(x,y) f a) dx* +2 *) dy* + ~W 2 and so Here and henceforward we mean "in positive integral powers". forth. 314 (Ch. 8 Series The expansion occurs (7) if the remainder term of the series fe=l as n * oo. The remainder term may be represented in the form n+i where 0<0< 1. Expand the indicated functions in positive integral powers of *, find the intervals of convergence of the resulting series and investigate the behaviour of their remainders: 2587. a x 2589. cos(* 2 2590. sin *. (a>0). 9*a sin cin^r-i-M * -r--r Jooo. ,-,>* . + a). /^ 1 of the principal expansions I-V and a geometric the expansion, in powers of *, of the following write progression, functions, and indicate the intervals of convergence of the scries: Making use 2592. 3 2593. X2 *7 ~4X 2594. xe~* 2598. cos - 7^rJT2 5 J *. 2599. sin 3* . ,. o . 2 x + * cos 3*. 2600. . x 2595. e \ 2601. 2596. sinh*. 2602. In * . 1 2597. cos2*. 2603. In (1 ,v -f* 2* 2 ). the Applying expand following functions in powers of *, and indicate the intervals in which these expansions differentiation, occur: 2604. (l+*)ln(l+*). 2606. arc sin*. + _ +* 2 2607. In (x V\ ). various the functions in techniques, expand given Applying powers of * and indicate the intervals in which these expansions occur: 2605. arc tan* 2 2 2608. sin *cos *. 2609. 2610. 2611. +O~ 8 (1 2613 ' 2614 - - cosh8 *' Sec 315 Taylor's Series 3} 2615. ln(x 2 + 3x + 2). in/i_L.^/*v + *)** rMl 2618. J. 2616. f'Jl dx. \ o X 2619. 2617. ^L_ [ "-* z ^" * V ^ 1 *4 Write the first three nonzero terms of the expansion of the following functions in powers of x: 2620. tan*. 2623. sec*. 2621. tanh*. 2624. In cos*. 2622. e^ x 2625. where e is 2627. A the eccentricity and 2a is the major axis of the ellipse. tenary J line heavy string hangs, under its own weight, in a ca= acosh and H is the horizontal where a = // J , a q tension of the string, while q that for small *, to the order string hangs in a 2628. powers -f * 2 ^ 8 . 2* 2 5* 2 in a series of of *-| 4. 2629. powers function the , V y^a parabola Expand the weight of unit length. Show 4 of * it may be taken that the is of s 4* f(*)-5* 2 3* +2. in a series of Expand f(x+h) h in a powers of * series of powers of * in a series of 2630. Expand In* 2631. Expand 2632. Expand Xz in a series of powers of 1. 1. *+l. j 2633. 2634. 2635. Expand X2 ~Y~ oX in a series of ~\~ JL Expand ^ Expand e* in a , , 4jc 2636. Expand ]/* in 2637. Expand cos* 2638. 2 Expand cos * 2639*. Expand In* of powers *-f4. powers of x series of powers of * + 2. a series of powers of * 4. 7 in a series of in a series of in a series of powers of * powers of * j in a series of powers _^ of . ^^. T-J |-2. . 316 [Ch. 8 Series 2640. -* Expand in a series of of powers . 2641. What is the the of magnitude error we put appro- if ximately ^2 + 1 + 1 + 1? To what degree of accuracy if we make use of the series 2642. ber j , we will - by taking the sum of its five first 2643*. Calculate the number ing the function arc sin A: in a ple 2606). 2644. How many terms do ^ . -* O u x=l? to three decimals we have num- .., terms when series of cosAr=l calculate the powers of x by expand(see Exam- to take of the series |j+..., in order to calculate cos 18 2645. How many to three decimal places? terms do we have to take of the series * +..., to calculate sin 15 to four decimal places? 2646. How many terms of the series have to be taken to 2647. How many find the number e to four decimal places? terms of the series In do we have to take to calculate cimals? In 2 to two decimals? to 3 de- __ 2648. Calculate \/7 to two decimals by expanding the funcx in a series of powers of x. tion l/S 2649. Find out the origin of the approximate formula + r \/ (f-{-x&a a = 5, + ~- (a>0), evaluate it by means and estimate the_ error. 2650. Calculate J/I9 to three decimals. of Y^3 t putting 317 Taylor's Series Sec. 3] 2651. For what values of x does the approximate formula 2 JC , yield an error not exceeding 0.01? 0.001? 0.0001? 2652. For what values of x does the approximate formula sin X&X yield an error that does not exceed 0.01? 0.001? '/* 2653. Evaluate . to four decimals. y^-dx 1 2 ^e~* dx 2654. Evaluate to four decimals. 1 2655. Evaluate ^ l/^ccosxdx to three decimals. 1 2656. Evaluate dx to three decimals. J 2657. Evaluate _ j^l+^'dx decimals. to four 1/9 2658. Evaluate ^yxe*dx to three decimals. 2659. Expand the function cos(x y) in a series of powers x and y, find the region of convergence of the resulting series and investigate the remainder. Wiite the expansions, in powers of x and y, of the following functions and indicate the regions of convergence of the series: of 2663*. 2660. sin x> sin y. 2 2661. sin(* +y*). , 664 * \n(lxy+xy). arctan j/_ l 2662*. \=*!!. \+xy 2665. f(x, y) wers of ft 2666. function *:= 1 4- ft, 2667. 2668. ~*y and /(*, = ax*-\-2bxy + cy*. Expand f(x+h,y-\k)in po- k. y) = x* 2y* when passing from + 3xy. Find the values the #=1, increment f/ =2 to the of this values y=2 + k. Expand Expand 2. the function e x+y in powers of x -2 and y the function sin(x+y) in powers of x and + __ Series 318 [C/i. 8 Write the first three or four terms of a power-series expansion x and y of the functions: x 2669. e cosy. in 1 2670. *'- (H-*) Sec. 4. Fourier Series let 1. Dirichlet's theorem. conditions in an interval 1) is We uniformly bounded; is say that a function satisfies the the function f (x) b) if, in this interval, that is \f(x)\^M when (a, a <x< b, Dirich- where M constant; 2) of all has no more than them the function are a the of number finite kind first has a finite limit on f (x) of points of [i.e., at each the 0)= left f (g discontinuity and g discontinuity Urn f (I e) and a ~" finite limit on the right /(-{-0)= lim /( + e) (e>0)J; e ->o has no more than a finite number of points of strict extrenium. Dirichlet's theorem asserts that a function /(*), which in the interval of this interval at ji, Ji) satisfies the Dirichlet conditions at any point x ( which /(x) is continuous, may be expanded in a trigonometric Fourier series: 3) f(x)=?+ a, cos x + b v sin +a x 2 cos 2x-\-b 2 sin 2*+ . . . +a n cos nx + + b n sinnx+..., where the Fourier a n and b n are calculated from the formulas coefficients ji = JT J\ (1) ji f(x)cosnxdx(n = Q, 1, ...);&= 2, f J JI -n -jt If x is a point' of discontinuity, belonging to the interval ( of a jt, n), function f (,v), then the sum of the Fourier series S (x) is equal to the arithmetical mean of the left and right limits of the function: SM = ~ At the end-points of the interval 2. Incomplete Fourier s=/(jc)], then in formula series. * = If n and X = K, a function / (*) is even (1) 6 rt -0 (w = l f 2, ...) and ji a 2 r /"=^ J / = 0, 1,2, ...). [i. e., /(- x) = a function If 319 Fourier Series Sec. 4] odd is / (x) /( [i.e., and x) = an then / (*)], =Q t/i = 0, 2 ...) i, IT bn =~ nx dx / (x) sin J\ JT (n = 1 2, , . . .). A function specified in an interval (0, n) may, at our discretion, be contior an odd function; hence, in the interval ( Jt, 0) either as an even it may be expanded in the interval (0, Ji) in an incomplete Fourier series of sines or of cosines of multiple arcs. 3. Fourier series of a period 21. If a function f (x) satisfies the Dirichlet the discontinuities conditions in some interval ( /, /) of length 2/, then at of the function belonging to this interval the following expansion holds: nued r, =a + y x ju , /(*) fli cos -J- . , 2nx juc + &! stay +a , , 2 cos j- 2nx + b sm+.., . . . . 2 nnx . , . nnx , where -M e / -- dx flllX v f (x) cos . . = 0, /-v (/i t r 1, 2, v ...), (2) 6n = -L \f(x)sln^-dx(n=\, ' 2, -i At the points of discontinuity of the function f (x) and at the end-points of the interval, the sum of the Fourier series is defined in a manner similar to that which we have in the expansion in the interval ( Jt, n). In the case of an expansion of the function / (x) in a Fourier series in x-l an arbitrary interval formulas (2) (a, a-f-2/) of length the 2/, limits of integration in should be replaced respectively by a and a-|-2/ functions in a Fourier determine the sum of the series the following in the the points of discontinuity and at the end-points of the interval (x ic, X function itself and of the sum JT), construct the graph of the of the corresponding series [outside the interval ( JT, ji) as well]: Q x when ~~ n Expand interval ji, ( ji), series at = = < ^ Consider the special case when 9fi79 2672. = xr Consider c) fl = 0, 6 the = 1; 2673. f(x) 2674. f(x) 2675. f(x) 2679. = =x = eax = slnax. (0, =~ bx when special cases: 0. d) a=l, 6 2 . . Expand the interval ct > a) = b = l\ 2676. / (x) 2677. f(x) 2678. f (x) the function f(x) 2ji). a = 1, b) a = 1, 6=1; = cos ax. = smhax. = cosh ax. -jr- in a Fourier series in 320 _ Series 2680. in the interval number g the function f(x) Expand =~ in sines of (Ch. 8 multiple arcs Use the expansion obtained n). (0, __ to sum the series: Take the functions indicated below and expand them, in the interval (0, jt), into incomplete Fourier series: a) of sines of multiple arcs, b) of cosines of multiple arcs. Sketch graphs of the functions and graphs of the sums of the corresponding seiies in their domains of definition. x. Find the sum of the following series by means 2681. f(x) of the expansion obtained: I+P+P+... Find the sums of the following number series 2682. f(x) = x by means of the expansion obtained: 2 . 1) 2683. f(x) * +7)2+ 32+ = eax . 1 2684. 1 2^" 32 42 "!" , when /(*)= when -^ x when x when ji Expand the following sines of multiple arcs: 2687. f(x) 2688. 2) i (x = 0<;c^~ * functions, in the when 0<#^^-, when ~n < x < n. x(n interval (0, K), in x). = sin-|. /(A;) Expand the following functions, in the interval sines of multiple arcs: u when (0, Ji), in co- Sec. 4\ __ _ 321 Fourier Series 2690. when 2691. / (x) = x sin x. cos x when cos x when < x^ ~ , Y 2693. Using the expansions oi the functions x and x* in the in cosines of multiple arcs (see Problems 2681 and it) 2682), prove the equality interval (0, 2694**. Prove that + ?L Ji, ( and in = x} the function /(x) if x\ f(~ then , its n) represents an expansion if the function f(x) the interval Ji, ( odd is is it Ji) even and we have is Fourier series in in cosines of arid /fy-f *) =/ expanded in the interval odd multiple (y~*) sines of arcs, then odd mul- tiple arcs. Expand the following functions in cated intervals: 2695. 2696. 2697. 2698. f(x) = \x\ f(x)=*2x f(x) Fourier series in the indi- !<* ( (0<JC< 1). = e* (/<* f(jc)=10 Jt (5<jc<15). Expand the following functions, in the indicated intervals, in incomplete Fourier series: a) in sines of multiple arcs, and b) in cosines of multiple arcs: (0<*< 2699. /(*)=! * 2700. /(*) 2701. = 1). (Q<x 2702. /M-{ 2 _-; ^n 2703. Expand the following function T (3 ' J \ in cosines \ : ) when -|<jr<2, 3_j^ when 2<jc<3. 1 , 11-1900 ^ of multiple Chapter IX DIFFERENTIAL EQUATIONS Sec. 1. Verifying Solutions. Forming Curves. Initial Conditions 1. Basic concepts. An equation F(x, y, where y Differential Equations of Families of of the type (n) y' t/) = 0, (1) y(x) is the sought-for function, is called a differential equation of order n. The function y y(x), which converts equation (1) into an identity, is called the solution of the equation, while the graph of this function is called an integral curve. If the solution is represented implicitly, O(A, f/)~0, then it is usually called an integral Example 1. Check that the function t/ sinjt is a solution of the equation = = Solution. We have: and, consequently, /" +y sin x -f sin x The integral */, Cp .... Cw ) ^ 0. = (2) of the differential equation (1), which contains n independent arbitrary constants C t ..., C n and is equivalent (in the given region) to equation (1), is called the general integral of this equation (in the respective region). By assigning definite values to the constants C,, ..., C n in (2), we get particular , integrals. if we have a family of curves C n from the system of equations Conversely, eters Cj = 0, dx -0. .... (2) and eliminate the param- *-0, dx n we, generally speaking, get a differential equation of type (1) whose general integral in the corresponding region is the relation (2). Example 2. Find the differential equation of the family of parabolas y^C^x-CJ. Solution. Differentiating equation /' = 2C l (* Eliminating the parameters C, and the desired differential equation C 2) (3) (3) twice, and C 2 from (/" we get: = 2C,. equations (4) (3) and (4), we obtain Sec It 1 is Verifying Solutions 1 function verify that tha to easy 323 converts (3) this equation identity. 2. Initial conditions. differential equation y desired particular solution y for the If = f(x, (n) y y, y' into = y(x) (n ~ l) an of a (5) ) the initial conditions and we know the general solution are given y = arbitrary constants from the system of Cn (5) are determined (if this which y(0) i Find the curve = = y'(0) have: l. We Solution. of the family = in formulas (6) 2. i/ Putting # possible) C lt .... C B ), c ..... c ). . *, for is -cp(*o, 3. C, ..... ..., equation C n ), equations Example C lt of '= and (6) 1=0, we obtain (7), 2-C + ^, (?) 2C 2 t , whence (:,=(), and, hence, y C,= l = e~. Determine whether the indicated functions are solutions of the given differential equations: 2704. *//'== 2#, 2705. y 2706. 2707. 2708. 2709. 2710. t2 (jc = 5x*. t/ = x* h //) . djc -h JK dy = 0, = ^=^ . + = 0, = 3sinjc 4 cos*. ~? + = 0, x = C, cos (/ ^/ 2 a) jc -(X. y^* Show 2 f X/ 2 2711. (x 8 t / -f e. that for the given relations are integrals: 11* // // differential equations the indicated _ _ 324 Differential Equations 2712. (x 2713. (xy-x)y" Form differential C lt = t + xy' + yy' equations = Q, 2y' y \n(xy). of the given families (C, C,, C, are arbitrary constants): Cx. 2714. y = = 272l 9718 ii&. 971Q 2719. n *.= 9799 *722. , fo' ? 3re P arameters >- yr 2724. ^ 2720. & M-- = 2 + Ce"^. 2726. Form thej +ay 0,- y .)'-2p* -* y *-. curves of a parameter). ( #//-plane. i 4 ' + = <:'. 2717. *' i y 2715. y Cx*. 2716. *>=-.2C*. [Ch. 9 2725. differential = = equation of all straight lines in the ; 2727. Form the differential vertical axis in the ^y-plane. 2728. Form the differential equation of equation of all with parabolas circles all in the xy-plane. For the given families of curves find the given initial conditions: 2729. x*y* 5. C, 0(0) 2730. 2731. y C sin 2732. y^C.e-x = y=(C^ = the lines that satisfy = l = 0, Sec. 2. First-Order Differential Equations P. Types of first-order differential equations. A differential equation the first order in an unknown function yt solved for the derivative y' is the form t y' where f(x, y) consider the = f(*. of of (i) the given function. In certain cases it is convenient to variable x as the sought-for function, and to write (1) in the is form x'=e(*,y)>$--.

(i')

j

where

gfr

Taking into account that (1)

and

(!')

may

0'

=

^

and

*'

=

j^

the

differential

equations

be written in the symmetric form

P(x, y)dx+Q(x, t/)
where

P

and Q

(2)

y) are knowri functions. By solutions to (2) we mean functions of the form t/ cp(jc) ihat satisfy this equation. The general integral of equations (1) (x, y)

(x,

=

or

and

x=ty(y) (I'),

or

equation

where

325

First-Order Differential Equations

Sec. 21

C

is

(2),

is

form

of the

an arbitrary constant.

2. Direction

The

field.

set of directions

tana = /(x,

y)

called a direction field of the differential equation (1) and is ordinarily depicted by means of short lines or arrows inclined at an angle a. Curves f(x, y) k, at the points of which the inclination of the field has a constant value, equal to k, are called isoclines. By constructing the isoclines and direction field, it is possible, in the simplest cases, to give a

is

Fig

rough sketch of the

field of

intagral

105

curves, regarding

the latter

which at each point have the given direction of the field. Example 1. Using the method of isoclines, construct the

as

field of

curves integral

curves of the equation y'=*x. Solution. rection field,

The family

By constructing the isoclines x~k (straight lines) and the diwe obtain approximately the field of integral curves (Fig. 105).

of parabolas

the general solution. Using the method of isoclines, make approximate constructions of fields of integral curves for the indicated differential equations: is

2733. y'

2734.

= =

x.

-f-

2735. y'=l-ftf

2736.

y'=

2737. y'

=

8 .

326

(Ch. 9

Differential Equations

3. Cauchy's theorem. b < y < B}

If

U\a<x
a function / (#, y) is continuous in some region in this region has a bounded derivative

and

(** y) rel="nofollow"> tnen through each point (* y Q ) that belongs to U there passes one and only one integral curve y y(x) of the equation (1) [cp (* ) #ol4. Euler's broken-line method. For an approximate construction of the

f'y

,

M

(* integral curve of equation (1) passing through a given point replace the curve by a broken line with vertices M,-(x/, #/), where

f

=

2.

(i'0,

Using Euler's method

find (/(I),

if

y(0)=l

(/i

1.

for the

2

We

t/

),

we

(one step of the process),

/i

/)

Example

,

2,

...).

equation

'

-0.1).

construct the table:

Thus,

/(!)=

1.248.

For

the

sake

of

comparison,

the

exact

value

is

i

T ss 1.284

e

Using Euler's method, find the particular solutions to the given differential equations for the indicated values of x: 2738. y' = y, y(0)=l; find y(\) (A-0.1). 2739.

y'-x + y, /(!)-

2740. ^' 2741.

= -X_,

t/(0)

1;

find y(2),

= 2;

-, y(Q)=l;

find
= 0.1). = 0.1). (A = 0.2).

(A

(/t

Differential Equations with Variables Separable

Sec. 3]

Sec.

First-Order

3.

Differential

Equations

with

327

Variables

Separable.

Orthogonal Trajectories

1. First-order equations with variables separable. is a first-order equation of the type

An equation with

variables

separable

X (x) Y

(y)

Dividing both sides of equation -

= f(x)dx

y'

= f(x)g(y}

dx

+ X, (x) Y, (y) dy = Q

(1)

(i)

(!')

by g(y) and multiplying

Whence, by integrating, we

get

the general

by

dx,

we

get

integral of equa-

tion (1) in the form

Similarly, dividing both sides of equation (!') by X, (x) we get the general integral of (!') in the form

Y

(y)

and integrating,

=

=

then the function y is ti Q (r/ )=0, yQ we have directly evident) a solution of equation (1) Similarly, the straight a and will be the integral curves of equation (!'), if a and b are, respectively, the roots of the equations X, (*)() and Y (*/) 0, by the ieft sides of which we had to divide the initial equation. Example 1. Solve the equation If

for

also (as lines x

some value y

is

y-b

=

'

3>

In particular, find the solution that satisfies the initial conditions

Solution. Equation

(3)

be written in the torm

may

dx~~

x

Whence, separating variables, we have

and, consequently, In |

where the arbitrary constant

y

|

=

In

|

x\

taken antilogarithms we get the general solution In C,

is

+ ln C in

|t

logarithmic form. After taking

f where

C=

When

C,.

dividi dividing by y ila contained in the formula

we could' lose = 0.(4) for C '

the solution

=0. but

the

latter

is

328

(Ch. 9

Differential Equations

Utilizing the given initial sired particular solution is

conditions,

we

get

C = 2;

hence, the de-

and,

J2

y ~~

'

x

Certain differential equations that reduce to 2 separable. Differential equations of the form

with

equations

variables

=

reduce to equations of the form (1) by means of the substitution u is the new sough t-for function 3 Orthogonal trajectories are curves that intersect the lines of the given a right angle. If F (x, y, #') (x, y> ort=0 ia is a parameter) at family is the difierential equation of the family, then

where u

=

O

is

the differential equation of the orthogonal trajectories. Example 2. Find the orthogonal trajectories of the family of ellipses

Solution Differentiating the equation (5), tion of the family

Whence, replacing

if

by

^7,

we

find

the

duerential

106

get

the

differential

equation

orthogonal trajectories ~~

~~

integrating,

we have

i

*x*

(family

'

x

it'

of

equa-

0.

(/'

Fig.

we

parabolas) (Fig. 106).

of

the

329

Differential Equations with Variables Separable

Sec. 3\

4. Forming differential equations. When forming differential equations in geometrical problems, we can frequently make use of the geometrical meaning of the derivative as the tangent of an angle formed by the tangent line to the curve in the pos'tive x-direction. In many cases this makes it possible straightway to establish a relationship between the ordinate y of the desired curve, its abscissa x, and the tangent of the angle of the tangent line (/', that is to say, to obtain the difleiential equation. In other instances (see Problems 2783, 2890, 2895), use is made of the geometrical significance of the definite integral as the area of a curvilinear trapezoid or the length of an arc. In this case, by hypothesis we have a simple integral equation (since the desired function is under the sign of the integral); however, we can readily pass to a differential equation by differentiating both sides. Example 3. Find a curve passing through the point (3,2) for which the segment of any tangent line contained between the coordinate axes is divided in half at the point of tangency. Solution. Let (x,y) be the mid-point of the tangent line AB. which by hypothesis is the point of tangency (the points A and B are points of intersection of the tangent line with the y- and *-axes). It is given that OA 2y and OB Zx. The slope of the tangent to the curve at (x, y) is

M

=

M

OA

dy_ dx~ This

is

y x

OB~

'

the differential equation of the sought-for curve. Transforming, d\

we

get

dy _ ~

~x

~T~~y~

and, consequently, In

\nx-\-\ny Utilizing the initial condition, curve is the hyperbola xy 6.

Cor xy

C.

we determine C = 3-2

Solve the differential equations: 2 2 2742. tan A: sin y d* + cos x cot ydy

6.

Hence,

the

desired

= Q.

=

2743. xy'~ // {/'. 2744. xyy' =-. = a(l 2745. // jq/' x fan ydx 2746. 3c 2747. y' tan * //.

\x\ =

+*V). x

+ (l

e

)

sec*

ydy = Q.

Find the particular solutions of equations that indicated initial conditions: x 0. 2748. (1 +e ) y y' = e*\ //= 1 when jt ! when x 2749. (xy* + x) dx-\-(x* yy)dy=* 0; // 2750. r/'sin x

= y\ny\ y~l

when

= = *= -|.

satisfy

= Q.

Solve the differential equations by changing the variables: 2751. y' = (x+y)**

=

2752. i/ (8*42//+l)'. 2753. (2x + 3{/ l)dx-{ (4x 2754. (2x y)dx (4x 2y

+

+ fo/ -

5) dij

= 0.

the

_ _

330

Differential Equations

In

Examples 2755 and 2756, pass

[C/i.

9

to polar coordinates:

2755.

2756. 2757*. Find a curve whose segment of the tangent is equal to the distance of the point of tangency from the origin. 2758. Find the curve whose segment of the normal at any point of a curve lying between the coordinate axes is divided in

two

at this point.

2759. Find a curve whose subtangent is of constant length a. 2760. Find a curve which has a subtangent twice the abscissa of the point of tangency. 2761*. Find a curve whose abscissa of the centre of gravity of an area bounded by the coordinate axes, by this curve and the ordinate of any of its points is equal to 3/4 the abscissa of this point.

2762. Find the equation of a curve that passes through the point (3,1), for which the segment of the tangent between the point of tangency and the *-axis is divided in half at the point of intersection with the y-axis. 2763. Find the equation of a curve which passes through the point (2,0), if the segment of the tangent to the curve between the point of tangency and the t/-axis is of constant length 2. Find the orthogonal trajectories of the given families of curves (a is a parameter), construct the families and their orthogonal trajectories. 2

2764. x 2 2765. t/

+ y =a = ffx. 2

2

2766. xy 2767. (x

.

Sec. 4. First-Order Homogeneous Differential

1. Homogeneous equations.

A

= a. 2

2

a)

t-f/

=a*.

Equations

differential equation

P(x y)dx+Q(x,y)dy = t

(1)

is called homogeneous, if P (AT, y) and Q (x, y) are homogeneous functions of the same degree. Equation (1) may be reduced to the form

and by means it

is

of the substitution y xu, where u is a new unknown function, transformed to an equation with variables separable. We can also apply

the substitution

Example

1.

x-yu. Find the general solution

to the equation

Sec. 4]

_

_

First-Order Homogeneous Differential Equations

Solution. Put y

we

Integrating,

=

get w

ux',

then

-f xu'

Q

=

In In

,

that reduce to

or

whence x In In

y

2. Equations

= eu + u

331

.

x

homogeneous equations.

If

'

and 6 I

!

#2^2

Ue

0,

then, putting into equation (2) x

w

+ a,

j/

= t;-fp,

where

I

the constants

a and P

are found from the following system of equations,

+ c, = 0,

a 2a

+ b$+ c = 0, t we get a homogeneous differential equation in the variables u and v. If u, we get an equation with variables 0, then, putting in (2) a,x 4- b^y separable. 6 Integrate the differential equations: 2768. 0' 2769. =1 277 1. y^-^. - (x-y)ydx-x*dy = Q. = 2 find the family 2771. For the equation (x +y*) dx 2xydy of integral curves, and also indicate the curves that pass through the points (4,0) andj_l,l), respectively. 2772. = + + 2773. xdy -\-ifdx. ydx Vx* 2 2 0. 2774. (4x* + 3xy jf) dy f/ ) dx + (4y 3jvy 1 2775. Find the particular solution of the equation (x = x 2. 2xydy Q, provided that r/=l when Solve the equations: 2776. (2x + = + = 9777 f/./ 2/77. - 3y*)dx+ 1 2779. Find the equation of a curve that passes through the point (1,0) and has the property that the segment cut off b\ the tangent line on the r/-axis is equal to the radius vector of the point of tangency. 2780**. What shape should the reflector of a search light have so that the rays from a point source of light are reflected as a parallel beam? 532 _ _ [Ch. 9 Differential Equations 2781. Find the equation of a curve whose subtangent is equal mean of the coordinates of the point of tangency. 2782. Find the equation of a curve for which the segment cut off on the y-axls by the normal at any point of the curve is equal to the distance of this point from the origin. 2783*. Find the equation of a curve for which the area contained between the #-axis, the curve and two ordinates, one of which is a constant and the other a variable, is equal to the ratio of the cube of the variable ordinate to the appropriate to the arithmetic abscissa. off 2784. Find a curve for which the segment on the y-axis cut by any tangent line is equal to the abscissa of the point of tangency. Sec. 5. First-Order Linear Differential Equations. Bernoulli's Equation 1. Linear equations. A differential equation of the )-y^Q form (1) (x) f of degree one in y and y is called linear. If a function Q(jt)=~0, then equation ).y takes the form (1) =Q (2) and is called a homogeneous linear differential equation. In this case, the variables may be separated, and we get the general solution of (2) in the form y = C-e - P P(X) dx J (3) . To solve the inhomogeneous linear equation (1), we apply a method that called variation of parameters, which consists in first finding the general solution of the respective homogeneous linear equation, that is, relationship (3). Then, assuming here that C is a function of x, we seek the solution of the inhomogeneous equation (1) in the form of (3). To do this, we put into (1) y and y' which are found from (3), and then from the differential equation thus obtained we determine the function C(x). We thus get the general solution of the inhomogeneous equation (1) in the form is ^/x -f J = C(x).e Example I. Solve the equation y' Solution. Solving it we tan **/-}- cos x. The corresponding homogeneous equation get: C ^- r *^I 1 (4) is 333 Bernoulli's Equation Sec. 5] C Considering and differentiating, we as a function of x, dC Putting y and y' into dC 1 cos* . djc we (4), ~ A dx cos x I x sin fi'nd; ~ ' /,x>o2 2 x cos get: sin* C cos 2 * cos* , dC -r-= or ' , dx whence = Ccos 2 *d*==i-* + Hence, the general solution of equation j (4) has the form COS* In solving the linear equation we can (1) make also use of the substitu- tion uv y where u and v are functions of x. f we require that ' (1) will have the form + v'u^Q(x). \-P(x)u]v [u If (5) t Then equation (6) + P(jc)M = 0, (7) then from (7) we find M, and from (6) we find u; hence, from (5) 2\ Bernoulli's equation. A first order equation of the form y' + P (<) y ^ Q (x) we find y. y\ where a 7=0 and a 7= 1, is called Bernoulli's equation ""*. near equation by means of the substitution z the or substitution the y = uv apply directly It 1 r/ It t reduced to is a li- also possible to method of variais tion of parameters. Example 2. Solve the equation y' Solution. This is = y+* VH- Bernoulli's equation. Putting y=^u>v, we ijet u'v + v'u To determine uv +x the function u y"uv we or u' ~w-0 x we have u Putting this expression into (8), f : u j -f v'u require that the relation u' be fulfilled, whence v we get x*. =x V^taT. (8) 334 _ _ Differential Equations whence we [Cfi. 9 find v: and, consequently, the general solution is obtained in the form Find the general integrals of the equations: 2785. ax -=*. x + = x*. 2786. 2787*. (\ 2788. y*dx(2xy Find the particular solutions that satisfy the indicated con- ditions: +y 2789. X y' e* = Q\ y = b when x 1-- * 2790. y' j-2-7 2791. y' yianx = = 0; cos x ; */ y= = = a. when x-0. when jt = 0. Find the general solutions of the equations: 2792. *l dx + JLx = ' Xyy* m 2793. 2xy x 2794. 0dx + ( +x 2795. 3xdy--=y(l sin A: 3y* smx)dx. 2796. Given three particular solutions y, equation. Prove that the expression What ^^ y lt y 2 of a linear remains unchanged for the geometrical significance of this result? 2797. Find the curves for which the area of a triangle formed by the *-axis, a tangent line and the radius vector of the point of tangency is constant. 2798. Find the equation of a curve, a segment of which, cul off on the x-axis by a tangent line, is equal to the square of the ordinate of the point of tangency. 2799. Find the equation of a curve, a segment of which, cut off on the y-axis by a tangent line, is equal to the subnormal. 2800. Find the equation of a curve, a segment of which, cut off on the y-axis by a tangent line, is proportional to the square of the ordinate of the point of tangency. any x. is Exact Differential Equations. Integrating Factor Sec. 6] 2801. Find the equation of the curve for which of the tangent is equal to the distance of the point tion of this tangent with the x-axis from the point 335 the segment of intersec- M (0,a). Sec. 6. Exact Differential Equations. Integrating Factor 1. Exact differential equations. If for the differential equation P(x.y) the equality dU is :p =-^- fulfilled, dx+Q(x y)dy = Q t then equation (1) may (1) be written in the = and is then called an exact differential equation. The gen= C. The function U (x, y) is detereral integral of equation (1) is U (x, y) mined by the technique given in Ch. VI, Sec. 8, or from the formula form (see (x, t/) Ch, VII, Sec. Example 1. 9). Find the general integral 2 (3x Solution. This + 6Af/ 2 ) an exact is 5= i2 X y and, dx of the differential equation + (6x*y + 4y') dy = 0. differential since equation, hence, the equation is of the form - -J K/ = 0. Here, = and whence U= 2 (3^ + 6xy*) dx + q> (y) = x> + 3*V + q> (y). = x y + 4y* (by y -f (y) = and hypothesis); from this we get y* + C*. We finally get (y) f/(r, t/)-= -f-3xV + +C consequently, x -f3^V + / = C is the sought-for Differentiating U with respect to y, q> jt we 6jc l cp' -y = 4(/* '(//) 4 s find f q> s .V . C( , general integral of the equation. 2. Integrating factor. If the left side of equation (l)is not a total (exact) differential and the conditions of the Cauchy theorem are fulfilled, then there exists a function U, U.(A', y) (integrating factor) such that = \i(Pdx+ Qdy) = dU. Whence it is found that the function The integrating factor u, is u, satisfies the (2) equation readily found in two cases: 336 Differential Equations Example Solution. " . and Solve 2. equation Here P = 2*y + rV-f-^ f , _ d# or dx + x*y + ~ 2xy ! ' hence> r dx cty J . , ^^^ + r 9 Q=x*+y* " Since it the [C/t. ' ft= Q x jB, dx follows that 1 fdP ^ JS ----dQ\ and dx=dx Q.\dy dxj -r . rffi Multiplying the equation by which is . ^. \\ \i = e* an exact differential equation. t . In 11 * = ^, - 11 = ^. r we obtain Integrating it, we get the general integral Find the general integrals of the equations: 2802 (x + 2z/)dy = 0. t/)d^+(^ 1 2803. (V + 1/ 2x) dx + 2xydy = Q. 3 8 2804. (jc S^ 8 + 2) dx (3x'y y ) d(/ = 0. + 2805. + ^-^==' 2806. . y y 2807. Find the particular integral of the equation = which satisfies the initial condition #(0) 2. Solve the equations that admit of an integrating factor of the form fi = M*) or M' H'(y) = : + y*)dx2xydy = 0. 4-xy)dxxdy = Q. 2808. (x 2809. y(l 2810. 281 1. (jc cos t/ y sin r/) rfi/ + (x sin */ -( // cos y) dx = 0. First-Order Differential Equation? not Solved for Derivative Sec. 7] Sec. for 7. 337 Equations not Solved First-Order Differential the Derivative 1. First-order differential equations of higher powers. F(x, which for example two equations: is y, two of degree !/') in y', y'=ti(x,y)> If an equation = 0, (I) the.i by solving (1) for y' <*. we get (2) M Thus, generally speaking, through each point (xQt (/ ) of some region of a plane there pass two integral curves. The general integral of equation (1) then, generally speaking, has the form 0(AM/, where and C^OM*, C) 0> 2 {/, C) (x. y, = 0, (3) the general integrals of equations (2). 2 are Besides, there may be a singular integral for equation (1). Geometrically, a singular integral is the envelope of a family of curves (3) and may be obtained by eliminating C from the system of equations tl> 1 OJx, or by eliminating p = t/' 0, C) = 0, from the system F(x t y, (*, y, 0>c F' p)-0, p of C) = (4) equations (x,y,p)=0. (5) We note that the curves defined by the equations (4) or (5) are not always solutions of equation (1); therefore, in each case, a check is necessary. Example 1. Find the general and singular integrals of the equation 2 A'//' Solution. Solving for y' 2A'j/' -f y Q. we have two homogeneous equations: do fined in the region x the general integrals of which are or Multiplying, we get the general integral of the given equation (2* +y C) 2 4 2 (A- + xy) - or (a we (It family of parabolas). Differentiating the general find the singular integral may integral with respect be verified that y-(-jc=0 to C and eliminating C, + Jt-O. is the solution of this equation.) Differential _ Equations (Ch. 9 by differentiating possible to find the singular integral with respect to p and eliminating p. 2. Solving a differential equation by introducing a parameter. If a firstorder differential equation is of the form It is + 2xp also #= then the variables y and x 1 from the be determined may system of equations dq> where p = t/' plays the part of a parameter. Similarly, if y = ty( x #') tnen x and y are determined > from the system of equations Example Find the general and singular integrals 2. of the equation 2 y=y' -xy'+^. Solution. we Making the substitution t/'=p, rewrite the equation in the form Diffeientiating with respect to x and considering ft dp dp p a function of x, we have p-*p-p-*fx +* or -p(2p x)**(2p x), or-j- into the original equation, = l. we have Integrating , we get p = x + C. Substituting the general solution or =y Differentiating the general solution with respect to C and eliminating C, we x^ x* obtain the singular solution: */ -r-. (It may be verified that */ -r- is the = = solution of the given equation.) If we equate to zero the factor 2p x which was cancelled out, we get x x2 into the which is the and, putting p given equation, we get y=-j t pay , same singular solution. Find the general (In Problems curves.) 2812. 2812 and singular integrals and 2813 construct the (/''-^'-M^O. 2 2813. 4y' 9JC-0. of the field equations: of integral _ Sec 8] _ The Lagrange and Clairaut Equations = Q. 2814. yy'*(xy+l)y' + x 2 2815. yij' 2xy'+y Q. 2816. Find the integral curves 339 = M that pass through the point ( y 0, the of equation y'* +y* = 1 . ) Introducing the parameter y' = p, solve the equations: 2820. 4y = x*+y'\ 2817. x=smy'+lny'. 2818. y 2819. = y'*e>". = y'* /2 2 ,* 9 91 ' Sec. 8. The Lagrange and Clairaut Equations 1. Lagrange's equation. An equation of the form = *(P), where p = tf (1) called Lagrange's equation Equation (1) is reduced to a linear equation in x by differentiation and taking into consideration that dy^pdx: is f (p)] dp. If p^q>(p), then from (1) and (2) we (2) get the general solution in parametric form: parameter and f(p)^ g(p) are certain known functions. Besides, a singular solution that is found in the usual way. 2. C'lairaut's equation. If in equation (l)p^<Mp), then we get where p there is may a be raut' s equation + ^(C) (a family of .straight lines). general solution is of the form y-Cx is also a particular solution (envelope) that results by eliminating the parameter p from the system of equations Its There I x~ \ lJ:= Example. Solve the equation Solution. Putting ing dy by pdx, we y'^p we + i. have //^2pv + (3) ; get p dx = 2p dx + 2.v or Solving this lineai equation, we will have *=l dp ~ different! a Ting and replac- __ 340 __ Differential Equations ^_ [C/t. 9 Hence, the general integral will be =l(l To find the singular integral, we form f/-2px + in the usual the system l, = 2*- way. Whence *= 1 2 and, consequently, Putting y into (3) we are convinced that the function obtained Is not a solution and, therefore, equation (3) does not have a singular integral. Solve the Lagrange equations: 2822. y-' = 2824. y = (1 +y' + r=FT 2823. , ,' Find the general and singular integrals of the tions and construct the field of integral curves: 2826. y xy' +y'*. 2827. y xy'+y'. r 2828. 2829. by _ = = = Xy' + V \-\-(y') y = xy' + j,. ij Clairaut equa- 2 . 2830. Firid the curve for which the area of a triangle formed a tangent at any point and by the coordinate axes is con- stant. 2831. Find the curve it the distance of a given point to any tangent to this curve is constant. 2832. Find the curve for which the segment of any of its tangents lying between the coordinate axes has constant length /, Sec. 9. Miscellaneous Exercises on First-Order Differential Equations 2833. Determine the types of differential equations cate methods for their solution: = = 2ju/ + *'; y' b) (*-*/)//' c) d) y' (f/- 8) U = and indi- Miscellaneous Exercises on First-Order Differential Equations Sec. 9] = (* 4+ y sin (x* xy) y' = #'; i) j) x cos f/' k) + 2xy')dx + + (y* + 3*y dy = 0; m) (x' 3xy) dx + (* + 3)dy = 0; 1) 1 ; {/' 341 (x' ) 2 n) Solve the equations: 2834. a) (x ycos^ b) jcln 2835. dy xrfx=(^ y'^dy. 2836. (2xy*y) dx-\-xdy=-Q. 2837. xy' -\- y -^ xtf \nx. 2838. y xy' 2839. f/ or/' jc#' 2840. ^( 2841. (1 = = 2842. + + y'-y^^l. 2843. y^-((/'-l 2xe>)y'. 2844. //'-|-//cosAr=sin.tcosx. 2848. (x'y 2849. x* -\-y-l) dx \ 2845 ' 2846 ' (l-^ ^'-T-^ = - 2847. y' (xcosy-'r a sin 2 = 1. (xy + 2x3yQ)dy = Q. = y' 2850. ./" d.v = (x*y f 2) + y dy J-| dJf = 0. 286 =* X 2862. *. = cosA:. f/y'-hf/ 2852. 2dx 2853.. . . i ,' J , 2854. 2863. 2855. xdy-\ ydx^ifdx. !. 2856. //' (j: |- sin //) = 2857. H? = -P + P / ff = 2^'H-l/l / //'== j (1 +lny 2864. (2e d*- (x 1 2859. x */" -h Zxyy' 2860. ^^H-\t ln^). ) dy x <je y') ^- + 2/' x +y 0. = 0. 2865. 2868. dx = ^ dydx = /' =2 f ^ (;(/' = . t . 4 f x S J 2858. ^' 1 (-y' v 4- 2 \ * y-1 J -1- 1 ) 2867. a(xt}' 2868. xdy- y dx = y* dx. 342 [Ch. 9 Differential Equations 2869. (x* 1 )/ At, dy + (x* + 3xy V^-\) dx = 0. 2870. ^ 5 + 2871. J/oM1 dy+ (x y 2872. xyy'* (x* +y*)y' = xy' + (3* + 2;q/ = 3p 2873. y 2874. 2875. Ya* +x 2 ) dx = Q. . 2 y'Jdx + f* 1 2xy 3y*)dy=--Q. 2 2 -t-4*/ . 2 for the indicated initial con- ditions: 2876. #= y'=^\ y 2881. jc=l. = \\ y=\ for *=1. = 0. y = 2; y = 2 for = = = for 0. ^ + !) !; = cos = for * = 0. #' + for x = 0. x*\ y = 2y = y' = = for * = 0. y'+y 2x\ y 2877. e*- y' 2878. cot ;q/' 2879. e^(^ 2880. for */ jt -f A;; {/ 2882. for x = 0. 2883. xy'=y\ a) //==! for jc=l; b) y for x-0. 2884. 2xy' y\ a) y=l for jc=l; b) y z 2 for x-0; b)y=l forjc=-0; 2885. 2xyy'-\-x 0; a) # = Q for je= 1. c) y 2886. Find the curve passing through the point (0, 1), for which the subtangent is equal to the sum of the cooidinates of the point' of tangency. 2887. Find a curve if we know that the sum of the segments* cut off on the coordinate axes by a tangent to it is constant and \ = = = = y-0 equal to 2a. 2888. The sum of the lengths of the normal and subnormal equal to unity. Find the equation of the curve if it is known that the curve passes through the coordinate origin. 2889*. Find a curve whose angle formed by a tangent and the radius vector of the point of tangency is constant. 2890. Find a curve knowing that the area contained between the coordinate axes, this curve and the ordinate of any point on it is equal to the cube of the ordinate. 2891. Find a curve knowing that the area of a sector bounded by the polar axis, by this curve and by the radius vector of any point of it is proportional to the cube of this radius is vector. 2892. Find a curve, the segment of which, cut off by the tangent on the x-axis, is equal to the length of the tangent. Miscellaneous Exercises on First-Order Differential Equations Sec. 9] 2893. Find the curve, between the contained which the segment of axes coordinate of the divided is 343 tangent into half by the parabola if =--2x. 2894. Find the curve whose normal at any point of it is equal to the distance of this point from the origin. 2895*. The area bounded by a curve, the coordinate axes, and the ordinate of some point of the curve is equal to the length of the corresponding arc of the curve. Find the equation of this curve if it is known that the latter passes through the point (0, 1). 2896. Find the curve for which the area of a triangle formed vector of the point of by the x-axis, a tangent, and the radius 2 is to a constant and tangency equal 2897. Find the curve if we know that the mid-point of the segment cut off on the x-axis by a tangent and a normal to the curve is a constant point (a, 0). . When forming first-order differential equations, particularly in phvsical frequently advisable to apply the so-called method of differenwhich consists in the fact that approximate relationships between tials, infinitesimal h.crements of the desired quantities (these relationships are accurate to infinitesimals of higher order) are replaced by the corresponding relationships between their differentials. This does not affect the result. Problem. A tank contains 100 litres of an aqueous solution containing 10 kg of salt. Water is entering the tank at the rate of 3 litres per minute, and the mixture is flowing out at 2 litres per minute. The concentration is maintained uniform by stirring. How much salt will the tank contain at the end of one hour? Solution. The concentration c of a substance is the quantity of it in unit volume. If the concentration is uniform, then the quantity of substance in volume V is cV. Let the quantity of salt in the tank at the end of t minutes be x kg. The quantity of solution in the tank at that instant will be 100 / litres, problems, is it + and, consequently, the concentration During time c= kg per QQ litre. solution flows out of the tank (the solution contains 2cdt kg of salt). Therefore, a change of dx in the quantity of salt in the tank is given by the relationship This ing, is 2dt litres dt, the of the sought -for differential equation. Separating variables we obtain ln* = 21n(100+0 + lnC C or * (100-M) C= x The constant C 100,000. 100,000 - At is the . .. found from the expiration of ft f =^ 3.9 kilograms of and integrat- u salt. 1 fact one ' that = xvh^n f 0, the tank hour, \ 10, will that is, contain _ _ 344 [Ch. 9 Differential Equations 2898*. Prove that for a heavy liquid rotating about a vertical axis the free surface has the form of a paraboloid of revolution. 2899*. Find the relationship between the air pressure and the 2 altitude if it is known that the pressure is 1 kgf on 1 cm at 2 at an altitude of 500 metres. sea level and 0.92 kgf on 1 cm 2900*. According to Hooke's law an elastic band of length / increases in length klF(k const) due to a tensile force F. By how much will the band increase in length due to its weight if the band is suspended at one end? (The initial length of the band is /.) 2901. Solve the same problem for a weight P suspended from the end of the band. When solving Problems 2902 and 2903, make use of Newton's law, by which the rate of cooling of a body is proportional to the difference of temperatures of the body and the ambient medium. 2902. Find the relationship between the temperature T and the time f if a body, heated to T degrees, is brought into a room = W at constant temperature (a degrees). 2903. During what time will a body heated to 100 cool off and during the first to 30 if the temperature of the room is 20 20 minutes the body cooled to 60? 2904. The retarding action of friction on a disk rotating in a liquid is proportional to the angular velocity of rotation. Find the relationship between the angular velocity and time if it is known that the disk began rotating at 100 rpm and after one minute was rotating at 60 rpm. 2905*. The rate of disintegration of radium is proportional to the quantity of radium present. Radium disintegrates by one half in 1600 years. Find the percentage of radium that has disintegrated after 100 years. 2906*. The rate of outflow of water from an aperture at a vertical distance h from the free surface is defined by the formula where c0.6 and g is the acceleration of gravity. During what period of time will the water filling spherical boiler of diameter 2 metres flow out of cular opening of radius 0.1 in the bottom. m it a hemithrough a cir- 2907*. The quantity of light absorbed in passing through a thin layer of water is proportional to the quantity of incident light and to the thickness of the layer. If one half of the original quantity of light is absorbed in passing through a three-metrethick layer of water, what part of this quantity will reach a depth of 30 metres? Sec. 345 Higher-Order Differential Equations 10] 2908*. The air resistance to a body falling with a parachute to the square of the rate of fall. Find the limit- is proportional ing velocity of 2909*. The is covered with descent. bottom of a tank with a capacity of 300 litres a mixture of salt and some insoluble substance. Assuming that the rate at which the salt dissolves is proportional to Ihe difference between the concentration at the given time and the concentration of a saturated solution (1 kg of salt per 3 litres of water) and that the given quantity of pure water dissolves 1/3 kg of salt in 1 minute, find the quantity of salt in solution at the expiration of one hour. 2910*. The electromotive force e in a circuit with current i, resistance /? and self-induction L is made up of the voltage drop Rl and the electromotive force of self-induction the current = and i Sec. 10. / time at / = 0. when if e^Esmat (E and L^. Determine are constants) o> Higher-Order Differentia) Equations 1. The case of direct integration. If then n i Miles 2. Cases of reduction of order. contain y explicitly, for instance, then, assuming y' we p, I. If differential a equation does not get an equation ot an order one unit lower; F(x, p Example I) p')-0. t Find the particular solution of the equation that satisfies the conditions ^ Solution. Putting Solving we get the = 0, f/' = #'=p, we have latter equation as when x / = p', a linear = 0. whence equation in the function p, 346 Differential Equations From the fact that y'=p = [Ch when x = 0, we have =^ 0, i.e., 9 Cj==0. Hence, or ___ ____ 2 dx~~ ' whence, integrating once again, w? obtain Putting y~ solution is when x 0, we find C 2 = 0. Hence, *2 y^ 2) a If y'=p, we p-?-> y" get desired particular - differential equation does not contain then, putting the x explicitly, an equation of an for instance, order one unit lower: Example Find the particular solution 2. provided that /=!, Solution. Put y' We #' = =p t when then # From it, we the fact that f/'=p equation = 0. tf^p-- and have obtained an equation the argument). Solving of the our equation becomes of the Bernoulli type in p (y is considered find = when r/=l, we have C 1 = 1. Hence, P=y Vy^ or Integrating, we have arc cos Putting y=l x and *=0, we obtain C 2 = 0, =C 2, whence = cosx or t/==secx. Sec. _ __ 347 Higher-Order Differential Equations 10] Solve the following equations: 291 1 . 2912. 2913. /=1 2920. yy" . 2921. y"=\-y'\ 2914. xy" i/' \ yif-y' 2922. = 0. = jfy' + y". /"= 2923. (x -f- + y') = 0. (I -p . !) (x-[- 2) y' + 2915. // = {/". *f/" 2925 y'+T^" =^"- ' ' + t/")=a/". 2918. t/'(l 2919. xy + x0' = t ! + = 0. = r/'ln^. * 2924. = 0. 2916. + x+ 2 + 1+x 2926. xy" /"= 2927. y""-f y"*=l. solutions for the indicated initial con- l. Find the particular ditions: 2928. (1 2*/' -M')z/" l 2929. = 0; = l+t/' 2j/t/"; r/=l, 2930. ytf + i/"=*y"\ y=*l, 2931. xy" 0, t/ t/' y'; = = = Find the general integrals 2932. 2933. 2934. 2935. for for \ JC x= 0. = 0. following equations: f/ = + y" = l; i/V=l; l, xif=V\ (/' = ! = i. = 0. for forjt 0=1, y'=\ for x=l; y=l y= Jt -\-y' ; for A: =e s . + -^'y' = 2 + \nx;y = -^, y' = tor = -i, y'-l for jc= 1. = 2 for x = 0. y'y'' + y'(y\)=*Q; = 2, = = for ^ = 0. 3/Y {/-(-i/" -F 1; y= 2, y' = = foi x = 0. 0; (/=!, -t-y"-2//'' = and = for * = ^j/' + /'* yy" = 0; j/=l for 2939. y"(\-\-\nx) 2941. = 0. satisfy the indicated conditions: 2 2940. * forx=l. 1 for of the =3 y"-yy"=ify'. J = 0. f/i/"-hf/' -y'lnj/ 2937. yy" 2938. j/' yy^Vy^ify'-y'if. yy' = y" + Find solutions that 2936. = 0, = y' = y' j/ \ /-n-ln; (/ < 2942. 2943. 2944. l 1 /' J/ -|- jc / 1, 348 (Ch. 9 Differential Equations 2949. = Q\ y = 0, y' = 2 for = 2. = 0; y=l, #' = 2 for * = 0. 2jr-3/ = 40 t/= 1, -0 for = 0. = 0; #=1, y' = for * = 0. 2yy"-\-y* I/"-*/' -*/; y=-l, y' = l for x=l. 2950. / + e>y-2^' = 0; z + 2945. 2y' (y' 2946. y'y*+yif 2 */' i 2947. A: Gx)-y" I 2948. jc */' ; l */'* 2 i - i //-I, y' =e for ^-- = 0; = 0, 0' = for*=l. = (H-00')/ (l+0'")0'; 0=1, 0'=1 for x = (*+l)0" + x0 = 0'; 0=-2, 0'=4 forx=l. 2951. H-//y"-f 2952. 1 y' ff 2953. Solve the equations: 2954. y' 2955. 0' = = /2 2956. y" j = 4y // . + 2957. yy' y" --= y'* y"* Choose the integral curve passing through x Q. the point (0, 0) and tangent, at it, to the straight line y 2958. Find the curves of constant radius of curvature. 2959. Find a curve whose radius of curvature is proportional to the cube of the normal. 2960. Find a curve whose radius of curvature is equal to the . + = normal. 2961. Find a curve whose radius of curvature is double the normal. 2962. Find the curves whose projection of the radius of curvature on the //-axis is a constant. 2963. Find the equation of the cable of a suspension bridge on the assumption that the load is distributed uniformly along the projection of the cable on a horizontal straight line. The weight of the cable is neglected. 2964*. Find the position of equilibrium of a flexible nontensile thread, the ends of which are attached at two points and which has a constant load q (including the weight of the thread) per unit length. 2965*. A heavy body with no initial velocity is sliding along an inclined plane. Find the law of motion if the angle of inclination is a, and the coefficient of friction is p,. (Hint. plane.) The 2966*. is frictional force is We may consider to the that proportional square if the initial velocity motion where ji/V, ^V is the force of reaction of the the of the is zero.. air resistance velocity. in free fall Find the law of Sec. 11] _ _ Linear Differential Equations 349 2967*. A motor-boat weighing 300 kgf is in rectilinear motion with initial velocity 66 m/sec. The resistance of the water is proportional to the velocity and is 10 kgf at 1 metre/sec. How long will it be before the velocity becomes 8 m/sec? Sec. 11. Linear Differential Equations 1. Homogeneous equations. = The functions 0i q>i(x), f/ 2 if there are constants C, ==(P/iW are called linearly dependent not alt equal to zero, such that !/ = f q> 2 (A:) C lf t ..., ... Cn otherwise, these functions are called linearly independent. The general solution of a homogeneous linear differential equation + P, (x) e/<"- + = coefficients P,-(x) '> //<> w ; th continuous !, (/ where are linearly ...,/ (/,, y tt (fundamental system of solutions). 2. Inhomogcneous . . . + P n (x) y = .... 2, independent The general equations. is n) (1) of the solutions form of equation (1) solution of an inhomogeneous linear differential equation (2) with continuous coefficients where and Y P,- and the right side (x) has the form f (x) the general solution of the corresponding homogeneous equation (1) particular solution of the given inhomogeneous equation (2). If the fundamental system of solutions (/,, y % ..... y n of the homogeneous equation (1) is known, then the general solution of the corresponding inhomogeneous equation (2) may be found from the formula f/ is is a y =C l (x) y, where the functions Cj(x) ing system of equations: '[ (*) + C' t (*' +C = (*)y* :;w !, t 2, (x) y2 + . . . +C n (x) ya .... n) are + + C'n (x) , determined from the follow- /, = 0, +...+c;(o;=o. 0) (the method of variation of parameters). Example. Solve the equation (4) _ _ 350 Differential Equations [C/i. 9 Solution. Solving the homogeneous equation *ir+if'=o we f get (5) Hence, it be taken that may yl and the solution of equation Forming the system the equation is (4) (4) and (3) = \nx may taking t/"+~ = and # 2 = 1 be sought in the form into account that the reduced form of we obtain jt, Whence + and >l and, consequently, y = where A and B are arbitrary constants. 2968. Test the following systems of functions for linear rela- tionships: x a) x, b) ing x + 2 2 x\ 1, d) x, x+1, x +2; 2969. Form a linear a) b) c) d) g) h) x'\ 2 9 e *, e**\ sin *, cos A:, 1; 2 2 sin x, cos *, 1. homogeneous differential equation, know- fundamental system of equations: = sin x, y 2 = cos x\ ==xe*\ y 2 y^e*. 2 y^x* # 2 = * y e* f) ; c) 0, its x\ e) *, 1; 2x , l < f/, = ^x y* = ^x sin ^, f/ 8 = ^ cos A:. 2970. Knowing the fundamental system homogeneous differential equation find its particular solution y that of solutions of a linear satisfies the initial conditions _ Linear Differential Equations with Constant Coefficients Sec. 12] 351 2971*. Solve the equation its knowing particular solution y = -^. l 2972. Solve the equation x*(\nx-l)y"-xy' 4-y = 0, particular solution y =x. knowing its By the method of variation of parameters, solve the following inhomogeneous linear equations. } = a 2973. **(/" 3;t xy' 2 2974*. x*y" + xy' y=x 2975. y'" -f (/'-sec A:. Sec. . 12. Linear Differential . Equations of Second Order with Constant Coefficients t. Homogeneous equations. A second-order linear equation with constant and q without the right side is of the form coeflicients p U) k If l and fc 2 are roots of the characteristic equation .Q then the general solution three ways: 1) 2) 3) of equation (/-CVV + C.e*-* 6, and = *,; y-eV(C, + CV) *, -^(^cnspx-HC^mpA) if kz (1) is (2) t \vritten in one are real and /?, ^ of the following k.\ if (/ 2. Inhomogeneous equations. if *,=a + p and * a -=a The general solution of a pi (p 76 0). linear inhomoge- neous differential equation y" + py' + w=fW (3) be written in the form of a sum: may where y the general solution of the corresponding equation (I) without determined from formulas (1) to (3), and Y is a particular solution of the given equation (3). The function Y may be found by the method of undetermined coefficients in the following simple cases: a e *P n (x), where P n (x) is a polynomial of degree n. 1. f (x) If a is not a root of the characteristic equation (2), that is, (p (a) 96 0, then we put Y e**Q n (x) where Q n (x) is a polynomial of degree n with undetermined coeflicients. If a is a root of the characteristic equation (2), that is, is and side = Y=x 2. = e Qn / (*) \\here r is the multiplicity of the root * [P n (*) cos bx+ Q m (x) sin bx\. (x) t = a(r=l or r = 2). 352 _ __ [Ch. 9 Differential Equations If i cp(a ^ 0, bi) then we put Y = e ax [S N (x) cos bx + T N (x) sin bx], where S^(x) and Tu(x) But if cp(a &/) = 0, then are polynomials of degree K = x re ax where r=l). is fc 2 r the is [Stf (x) cos 6* + T N (x) sin to] the roots a of multiplicity N-max bi {n, m}, , second-order (for equations, In the general case, the method of variation of parameters (see Sec. 11) used to solve equation (3). 2 Example 1. Find the general solution of the equation 2y" y' y 4xe *. 2 & l=u has roots fc,~l and Solution. The characteristic equation 2& = The general solution . (first type) Y homogeneous equation e* + C 2e The right side of the given equation is/ (x) = Y e zx (Ax + B), since n=l and /=0. Difleren(x). Hence, twice and putting the derivatives into the given equation, we is ==C f/ =4xe zx =t ax P n tiating obtain: the corresponding of %,** (4 A X 2 1 = . + 45 + 4^) __ e i* (2Ax + 25 + A) 2 e * (Ax H- B) 4xe zx . zx and equating the coefficients of identical powers of x arid Cancelling out e the absolute terms on the left and right of the equality, we have bA=4 and 4 28 744-5fl = 0, whence 4 = and 5 -=- o 2A Thus, K^ f oH -g-* anc^ * ) = -. Jo ne gei] eral solution of the given equation f Find the general + = is solution of the equation y* xe* y 2y 2 characteristic equation k has a double root 2/f-f- 1 x ft=l The ri^ht side of the equation is o! the form f(x)xe here, 0=1 and n=-l. The particular solution is Y =x*e* (Ax B), since a coincides \nHth 2. the double root k=-\ and, consequently, r Diilerentiating Y twice, substituting into the equation, and equating the Example 2. . Solution. The \ + = coefficients, we obtain /l = , fl equation will be written in the = 0, Hence, the general solution of the given form * Find the general solution of the equation */*-f y= 2 The characteristic equation and -j-l=r() has roots /?, The general solution ot the corresponding homogeneous equation 3, where a~0 and P = l| be* Example fc a = 3, Solution. i. will |see l The right side fe is of the form /' Sec. Linear Differential Equations with Constant Coefficients 12] = 0, 6=1, P n (jc)=0, particular solution Y, where a #=1, a Q OT = *. (*) To _ 353 this side there corresponds the = 0, fc=l, r=l). twice and substituting into the equation, we equate the coefficients of both sides in cos*, xcosx, sin*, and xsmx. We then get four 44 = 1, from which we deter0, 0, 4C = 0, -25 + 20 equations 2A + 2D 1 1 X2 X mine A cos * -f- -j- sin *. 0, C 0, D Therefore, (here, Differentiating = , = = = = = 3. The is = C, cos + C y jc x2 2 sin x is principle of superposition of the sum of several functions and K/(/ = l, then the sum 3, . y'+py'+w^-fiW y is the solution of equation A: x + sin *. -7- solutions. If the right side of equa- corresponding solutions of the equations are the .., n) cos -j- tion (3) 2, 44 K= . 4 4 The general solution (< = i. 2 ..... n). = Y + Y +...+Y n n l (3). Find the general solutions of the equations: 2982. y" + 2y' 2976. tf 5y'6y Q. 0. 2983. 2977. if 4y' 9y 2984. y" + ky 2978. 0. 2985. 2979. iT y 0. 2980. ^_2i/ +2j/ 2981. + 40' +130 0. = = / yy'^Q. = + y= = = f / Find the particular solutions that satisfy the indicated conditions: 2987. y"5tj'-\-4y 2988. y"+ 3tf' +20 2989. 0" 2990. 0^ 2991. = Q\ = 0; = 0, y=5 = 8 for * = for jc0. y=5 0' = = 2 for x = 0. t y' 1 1, + 40 = 0; + 20' = 0; 0=1, 0'=0 for ^ = = a, 0' = for x = 0. f /=; 2992. 0" 2993. 0" + 30'=0; = = 0; = + f ji for x for jc = = and and = = x for x=l. type of particulai solutions inhomogeneous equations: 2994. a) Indicate 0"-40 = A:V b) 0" 12-1900 + 90 the x ; = cos 2x\ = 3. for for the given _ _ 354 Differential Equations [Ch. 9 = sin 2x + e**\ x d) y" + 2y' + 2y = e sir\x\ - 5 c) y" 4y' -f 4y f) Find the general solutions of the equations: = / ^ +# = + ^ + + = ^. + / + t 2995. y" jc 4i/' 4(/ 2996. *' 6. 2997. 2^' ^ 2998. 8^ 7^=14. 2999. y"y^e*. 3000. 3001. 3002. 3003. f 3004. t/' 3005. y" 2y'-{ 5y 3006. Find the solution of the equation y" 0. satisfies the conditions y=l, */'= 1 for x / / . + . = = + 4y = sin x that Solve the equations: W2y 3007. 2) ^-2 2 + x=/l o) Consider sinp/. the cases: 1) p-co. 3008. 3009. 3010. 3011. 3012. 3013. 3014. 3015. 3016. 3017. / t/" y" y" 2y' y" 2(/' y"-\-y' = 8y = e = 5x + 2e x x ^_y'^2;c-~l 8cos2*. . 3e x . + 2r/ + y = ^ + ^ x x 2y + lQy=s\r\3xi-e /_4f/' + 4f/ = 2^ + if 3y' = / . {/" f f/" . . 3018. 3019. Find the solution to the equation that satisfies the conditions /=-, f/ /==l for ^ / Solve the equations: 3020. (/" 3021. j/" 3022. 3023. tf 3024. 3025. y = 2xsmx. 2x 4(/==g sin2A:. / H- 9y = 2x sin 2y'=e = 0, tx p + o>; Sec. _ Linear Differential Equations with Constant Coefficients 12] = + = = 355 x 3026. tf2y' 3y x(l+e' ). 3x 2xe*. 3027. y" 2y' x 3028. */" 4y'+4y xe* 2xe-'* 3029. y" 3y 2y' (x^ l)e*. 3030*. y" y 2x cos x cos 2x. 3031. t/' 2y 2xe*(cosx sinx). = + + = = . + Applying the method variation of following equations: 3032. y" + y ianx. 3033. 3034. 3035. = y" + y = cot x. = -. y" 2y' y" + 2y' + y = 3036. of parameters, solve < the . 3037. 3038. a) -(-*/ b) y" . V y = tanh A:. x 2y = 4x*e \ 3039. Two identical loads are suspended from the end of a spring. Find the equation of motion that will be performed by one of these loads if the other falls. Solution. Let the increase in the length of the spring under the action in a state of rest be a and the mass of the load, m. Denote by x the coordinate of the load reckoned vertically from the position of equilibrium in the case of a single load. Then of one load where, obviously, k tion is x=C and d* -77 Tt = : l cos when = I/ / = 0; and, consequently, - 1 +C 2 I/ sin whence C t =a t. 7*7* The ^ ~" ^ * initial The general conditions yield solu- Jt =u and C a = 0; and so 3040*. The force stretching a spring is proportional to the increase in its length and is equal to 1 kgf when the length increases by 1 cm. A load weighing 2 kgf is suspended from the spring. Find the period of oscillatory motion of the load if it is pulled downwards slightly and then released. 3041*. A load weighing P 4 kgf is suspended from a spring and increases the length of the spring by 1 cm. Find the law of motion of the load if the upper end of the spring performs a vertical harmonic oscillation (/ 2sin30/ cm and if at the initial instant the load was at rest (resistance of the medium is = = neglected). 12* . L 356 _ _ (Ch. 9 Differential Equations m is attracted by each of two 3042. A material point of mass centres with a force proportional to the distance (the constant of proportionality is k). Find the law of motion of the point knowing that the distance between the centres is 26, at the initial instant the point was located on the line connecting the centres (at a distance c from its midpoint) and had a velocity of zero. 3043. A chain of length 6 metres is sliding from a support without friction. If the motion begins when 1 m of the chain is hanging from the support, how long will it take for the entire chain to slide down? 3044*. A long narrow tube is revolving with constant angular velocity o> about a vertical axis perpendicular to it. A ball inside the tube is sliding along it without friction. Find the law of motion of the ball relative to the tube, considering that a) at the initial instant the ball was at a distance a from the axis of rotation; the initial velocity of the ball was zero; b) at the initial instant the ball was located on the axis of rotation and had an initial velocity v 9 . Sec. 13. Linear Differential Equations of Order Higher than Two with Constant Coefficients 1. Homogeneous equations. The fundamental system of solutions y lt Un f a homogeneous linear equation with constant coefficients n y< + a iy n + + an _,y' +a n y = (1) #t > < . is . . constructed on the basis of the character of the roots of the characteristic equation Q. (2) 1) if k is a real root of the equation (2) of multiplicity m, then to this root there correspond linearly independent solutions of equation (1): Namely, 2) then if to m a i the p/ is a pair of complex roots of equation (2) of multiplicity latter there correspond 2m linearly independent solutions equation (1): *x y l e cos PX, #, = = e*x sin px, yt = xe* x cos PX, 2. Inhotnogeneous equations. equation A y4 = xe* x sin PX, m, of ... particular solution of the inhomogeneous (3) is sought on the basis of rules 2 and 3 of Sec. 12. _ Sec. Euler's Equations 14] __ 357 Find the general solutions of the equations: 3058. y / 13 3045. y'" 3048. y' v 2J/" 3t" 3049. y'" 3050. y'v + 4y = 0. , n ' w'v' + ^O. ' *054. w' 1 sss- ^T ' - 306 >- y' v 3062 y'"iJ ai/ = b. ^ - 3060. 3053. n(n-l) ,. - = x' MS,---.. SB: 3067. Find the particular solution of the equation y'"+2y"-{-2y'+y = x that satisfies the initial conditions y (0) 14. Sec. A Euler's = y' (0) = y" (0)=-=0. Equations linear equation of the form n y^f(x) t (I) a, b, A ..... <4,,_,, A n are constants, is called Enter's equation. Let us introduce a new independent variable /, putting where l ax + b-^e 1 . Then and Euler's is equation transformed coefficients. z Fxample 1. Solve the equation x y" Solution. Putting x^e*. we get into a linear equation + xy' +{/ = 1. z ==e Tr ~Sx dx~z==e (~di*~~~di)' Consequently, the given equatioi takes on the form whence = C, cos + C, sin + y = C, cos (In *) + C a sin (In x) + y or t t 1 1. with constant 358 _ _ [Ch. 9 Differential Equations For the homogeneous Euler equation Q the solution (2) be sought in the form may = **. (3) found from (3), we get a characteristic equaexponent k. If k is a real root of the characteristic equation of multiplicity m then to it correspond m linearly independent solutions m- 1 k 2 Putting into (2) tion from which y' (/, , ..., we can y (n} find the t 0i If a = **. a is p* correspond 2m 0, = **-ln*. */, = ** (In*) , ..., y m y 4 =*Mnx.sin(plnx) ..... 2. We into the given characteristic equation Solving find y^C Solve the equations: 3069. 3070. *'g + 3*| + = 0. *V xy' 3(/ = 0. xY-|-x(/'4-4(/ = 0. <, 3071. 3072. 3073. tT = 3074. 3075. 3076. there 1 cos (p I +4l/=0. equation, after Hence, the general solution will be 3068. it = *a In x cos (P In *), ^m-i^*" (In*)*" 3Xy' 2 we to put Substituting it . Solve the equation X *y'' Solution. (\nx) m, then pair of complex roots of multiplicity linearly independent solutions (/ 8 Example =x ^y_ . 46-1-4 = 0. cancelling out x k , we get the Sec. 15] * _ Systems of Differential _ 359 Equations 3077. Find the particular solution of the equation + y = 2x conditions y = 0, *V that satisfies the initial Sec. xy' 15. Systems of Differential Method = */' Equations To find the solution, for first-order differential equations, that of elimination. system of two form them with respect the desired functions, have, for example, we first equation of the system (1) to x. Determining z from the instance, of is, We solved for the derivatives of when *=1. ! of a normal a system differentiate of the one of and substituting the value found, A.. \ / (3) into equation (2), we get a tion u. Solving it, we find equation with one unknown func- second-order (4) where C, and C 2 are arbitrary constants. Substituting function (4) into formula (3), we determine the function z without new integrations. The set of formulas (3) and (4), where y is replaced by \|>, yields the general solution of the system (1). Example. Solve the system z 3 , f +'-'-T* Solution. We differentiate the first equation with respect to x: ^ + 2^dx^+ 4^-4. dx*^ From the first from the second equation we will dx we determine have -^ ax = -5- & ** ^ = / \ -T- +*+ ( -; 4 l+4x -75- & y ~ \ dy TT"T ax 2y and then j Put ting z and j- into the equation obtained after differentiation, we arrive at a secondorder equation in one unknown y: 360 [Ch. 9 Differential Equations Solving it we find: and then We do likewise can in the case of a system with a larger number of equations. Solve the systems: dy 3078. I =z JJ 3085. dx r/ = 0, z = when je = 0. when / = 0. 3079. 3086. 3080. dz : = 0, y=\ dT dx 3081. 4/_// 3087. dy dz 3088*. dx _ xy~ c) 3082. 2 da __ dy __dz_ x-\-y~~~ z dy __ yz zx t ' dz xif isolate the integral curve passing through the point (1, 1, 2). dy 3089. 3083. 3084. dz 3090. .Sec. Integration of Differential Equations by Power Series 16} 361 3091**. A shell leaves a gun with initial velocity u at an angle a to the horizon. Find the equation of motion if we take the air resistance as proportional to the velocity. 3092*. A material point is attracted by a centre with a force proportional to the distance. The motion begins from point A at a distance a from the centre with initial velocity perpendicular to OA. Find the trajectory. by Means Sec. 16. Integration of Differential Equations of Power Series not possible to integrate a differential equation with the help of elementary functions, then in some cases its solution may be sought in the If is it form of a power series: 00 '(* y=2 n= n *o) 0) - o The undetermined series (1) into the of the binomial x = are found by putting the coefficients c n (n \, 2, ...) equation and equating the coefficients of identical powers x on the left-hand and right-hand sides of the resulting equation. We in the can also seek the solution of the equation form of the Taylor's series y(*) ^ ^^ (*-*)" = y (3) and the subsequent derivatives y (n) (x ) where y(x Q) = y Q t/ ) y' (x ) = f (* are found successively by differentiating equation (2) and by (n- 2, 3, ...) , , putting X Q in place of x Example Solution. 1. Find the solution We put y of the equation = whence, differentiating, we get y" = 2.\Ct + 3.2c x+...+n(n-\)cn x n -* + (n+l s + In + 2)(rt-t-l) <:+,*"+... Substituting y and y" into the given equation, Collecting powers together, on of x the left of this and equating to zero the we arrive at the identity equation, the terms with identical of these powers, we will coefiicients 362 Differential Equations [Ch. 9 have cs = 2 = ^~ and so forth. Generally, 3.4-6. 7-... (k =1,2, 3, - ...). Consequently, * (X* c =f/ and c = y'Q Applying d'Alembert's where for < x < + oo Example 2. 3.4-6.?.. . 1 \ .-3* (3fc+ 1) + '" )* (4) . l oo X^ X1 + 3T4 + 3.4.6.7 + '" + test, it is readily seen that series (4) converges . Find the solution of the equation y' Solution. We We put = + = Differentiating equation y' = x + y, we succes/=! + ^=1 + = 2, y'"=y\ /o" = 2, etc. Consequently, have y =\, i^ sively find l. l 1 (/', For the example at hand, this solution *-l The procedure may x) or be written in final form as = 2e* 1 x. similar for differential equations of higher orders. Testing the resulting series for convergence is, generally speaking, complicated and is not obligatory when solving the problems of this section. is With the help of power series, find the solutions of the equations for the indicated initial conditions. In Examples 3097, 3098, 3101, test the solutions 3099, obtained for convergence. 2 3093. y' y x y 3094. y' x 1; y 2y = + = 2 for * = 0. = + = y for x=l. = y for x==a 0' = / + *; = for * = 0. y' = x* y*\ # for (1 x)y' = l+x y\ y = \ 3095. 3096. 3097. f/ Problems on Fourier's Method Sec. 17] 363 = for x = 0. = for * = 0. 0\ 0=1, #' + = 0; y=l, */'=0 for = + = 0; 0=1, 0'=0 for x = = 0; * = a; ~ = for = 0. = 0; 3098*. 3099. 3100*. /' 3101*. y = 0, 1 y' Jt */ */ 3102. * Sec. 17. Problems on Fourier's Method To find the solutions of a linear partial differential equation homogeneous by Fourier's method, first seek the particular solutions of this special-type equation, each of which represents the product of functions that are dependent on one argument only. In the simplest case, there is an infinite set of such solutions M (tt=l, 2,...), which are linearly independent among themselves in any finite number and which satisfy the given boundary conditions. The desired solution u is represented in the form of a series arranged according rt to these particular solutions: u= Cn u n (1) . C n which remain undetermined The coefficients conditions. = Problem. A transversal displacement u u(x with abscissa x satisfies, at time *, the equation from the are found t) t initial of the points of a string _ ~~ a where a string). = 2 -? (T Q the is Find the form tensile of the and Q force string (2) dx* dt* at time t the is if its linear ends x= density of the and * = / are U 2 Fig. fixed u and =~* at the initial (/ Solution. x) (Fig. It that satisfies the is f instant, 107) and required its = 0, 107 the string had the form of a parabola points had zero velocity. to find the solution u = u(x, t) of equation (2) boundary conditions a(0, 0-0, !!(/, 0=0 (3) _ _ 364 \Ch. 9 Differential Equations and the initial conditions = (*,0) We ^ *(/-*), ;<*, 0) = 0. (4) seek the nonzero solutions of equation (2) of the special form =X u Putting this expression into (x)T(t). equation and separating the variables, we get (2) ZL(0--*1W a 2 7tO~~ * (5)' l ' (x) Since the variables x and t are independent, equation (5) is possible only the general quantity of relation (5) is constant Denoting this constant X 2 we find two ordinary differential equations: when by , =Q )*-T(t) and X" (x) + K*X (x) = 0. Solving these equations, we get = A cos a\t + B sin aKt, (*) = C cos Kx + D sin X*. T (t) X where A, B, C, From = D are X(0) = D cannot be equal to this reason, XA = -p we do To Let constants. arbitrary we have (3) (since sinX/ For condition where k is and zero at an integer. determine the constants. and X(/) = 0; hence, C = the same time as C is zero). us It will readily be seen that not lose generality by taking for k only positive values every value h k there corresponds a particular solution kan kan Ak (, cos-j- I whose sum obviously . . equation A k and B k choose the constants satisfy the initial conditions knx . kant\ knx -7- + ^sm-y-J sm-j- cos satisfies We and the so that the (2) . boundary conditions sum of the series Since (4). CD du y & =2* it kan ( ~\~ A follows that, by putting / . * = 0, sm kant we obtain V kant\ -p + ^cos-y-J and 0) 2, 3,...). (3). kant ^ \ 9 tj sin-y- Ihat satisfies the boundary conditions We construct the series f . (k=\ kan , sin knx -^ . (3). should Sec. _ 17] Problems on Fourier's Method Hence, to determine the a Fourier series, . du .. function A k and B^ coefficients is it 365 necessary to expand in = ^- sines only, the function u(x, 0) in _ x (lx) and the ^ ^n (x, 0)' 0. at From familiar formulas (Ch. VIII, Sec. 4,3) we have / ft if & is Ak odd, and Q if & is knx . 32h . even; / kast _ The sought-for solution = * will 2 jix fA Osin , . d* T- J A = 0, be cos / . sin 3103*. its ends, A u sin At the initial instant x= and * = /, had y and the points , form of the string 3104*. At the string 0<jt at A = 0. fl* time initial the / a form string, attached at of the sine curve had zero velocity. Find the it t. time = 0, / of V (2/i-f \)jtx = 0, receive a velocity -~ the = 1. points of a straight Find the form of the = = l are and x string at time t if the ends of the string # fixed (see Problem 3103). 3105*. A string of length /=100 cm and attached at its ends, and *=-/, is pulled out to a distance A 2 cm at point x = = *=;50 cm then released without any impulse. Determine the shape of the string at any time /. 3106*. In longitudinal vibrations of a thin homogeneous and rectilinear rod, whose axis coincides with the jr-axis, the of a cross-section of the rod with t) displacement u = u(x abscissa x satisfies, at time /, the equation at the initial time, and is 9 c^u_ ~~ a 2 d/ 2 f where a = * r (E is Young's modulus and Q is the density of the rod). Determine the longitudinal vibrations of an elastic horiand pulled zontal rod of length /= 100 cm fixed at the end * back at the end *=100 by A/ l cm, and then released without = = impulse. 366 _ _ ' Differential Equations [C/i. 9 3107*. For a rectilinear homogeneous rod whose axis coincides with the Jt-axis, the temperature u = u(x, t) in a cross-section with abscissa x at time /, in the absence of sources of heat, satisfies the equation of heat conduction di where a is a constant. for any time t in a rod Determine the temperature distribution of length 100 cm if temperature distribution u(x, 0) = 0.01 A; (100 x). we know the initial X Chapter APPROXIMATE CALCULATIONS Sec. 1. Operations on Approximate Numbers 1. Absolute The absolute error of an approximate number a which number A is the absolute value of the difference between them. The number A, which satisfies the inequality error. replaces the exact is called the limiting the limits a absolute A^/l^a + A error. or, more The exact number A A briefly, A=a is located within 2. Relative error. By the relative error of an approximate number a 0) we understand the ratio of the absolute replacing an exact number A (A error of the number a to the exact number A. The number 6, which satisfies > the inequality \*-l ^* -, (2) A is called the of the limiting relative error actual practice A^a, we number often take the number approximate 6= for a. Since in the limiting relative error. 3. Number of correct decimals. We say that a positive approximate number a written in the form of a decimal expansion has n correct decimal places in a narrow sense if the absolute error of this number does not exceed half unit of the nth decimal place. In this case, take, for the limiting relative error, the number one / 1 where k is the first significant / is known that 6^ I , . i , . places definitely has narrow 77: we can number a. And conversely, it if then the number a has n correct decimal , ) \ 1) n correct the absolute error of the last decimal If unit the . \10/ meaning 2(k-\in > 1 of the digit \-i when n of decimals the in word. the In particular, narrow meaning if number a the ^ 1 / 1 \ w "if I To) approximate number a does not exceed a for example, are numbers resulting place (such, of an from measurements made to a definite accuracy), then it is said that all decimal places of this approximate number are correct in a broad sense. If there is a larger number of significant digits in the approximate number, the latter (if it is the final result of calculations) is ordinarily rounded off so that all the remaining digits are correct in the narrow or broad sense. [Ch. 10 Approximate Calculations 368 Henceforth, we shall not otherwise (if assume that all digits in the initial data are the narrow sense. The results of intermediate calculations may contain one or two reserve digits. We note that the examples of this section are, as a rule, the results of final calculations, and for this reason the answers to them are given as approximate numbers with only correct decimals. 4. Addition and subtraction of approximate numbers. The limiting absolute error ot an algebraic sum of several numbers is equal to the sum of the limiting absolute errors of these numbers. Therefore, in order to have, in the sum of a small number of approximate numbers (all decimal places of which are correct), only correct digits (at least in the broad sense), all summands should be put into the form of that summand which has the smallest number of decimal places, and in each summand a reserve digit should be retained. Then add the resulting numbers as exact numbers, and round off the sum by one decimal place If we have to add approximate numbers that have not been rounded off, they should be rounded off and one or two reserve digits should be retained. Then be guided by the foregoing rule of addition while retaining the appropriate extra digits in the sum up to the end of the calculations. Example 1. 215.21 -f- 14.182 -f 21 .4-215.2(1) 14.1(8)4-21 4-= 250.8. The relative error of a sum of positive terms lies between the least and greatest relative errors of these terms. The relative error of a difference is not amenable to simple counting. Particularly unfavourable in this sense is the difference of two close numbers. Example 2. In subtracting the approximate numbers 6 135 and 6.131 to four correct decimal places, we get the difference 004. The limiting relative correct in stated) + 0.001 error 6= is .J. +1 0/XM =-4 = 0.25. j .. ,t . 0.004 Hence, not one of the decimals it difference is correct. Therefore, is always advisable to avoid subtracting close approximate numbers and to transform the given expression, if need be, so that this undesirable operation is omitted. 5. Multiplication and division of approximate numbers. The limiting relative error of a product and a quotient of approximate numbers is equal lo the sum of the limiting relative errors of these numbers Proceeding from Ihis and applying the rule for the number of correct decimals (3), we retain in the answer only a definite number of decimals Example 3. The product of the approximate numbers 25.3-4.12=104.236. Assuming that all decimals of the factors are correct, we find that the limiting relative error of the product is of the 6 Whence result, if number the it is correctly, 25 final, mth power of ^0.01 +^0.01=^0.003. decimals of the product is three and the be written as follows: 25.3-4 12=104, or more correct should 3-4.12= 104 2 6. Powers and of the of = 0.3. roots of approximate numbers. an approximate number a relative error of this number The limiting relative error of Is the th the mth root is of The limiting relative error equal to the m-fold limiting an approximate number a part of the limiting relative error of the number a. 7. Calculating the error of the result of various operations on approximate numbers. If Aa lf ... Aa,, are the limiting absolute errors of the appro, Sec. Operations on Approximate Numbers 1] xitnate numbers a t , . . . an , then the limiting absolute error , S = /(a t, may 369 AS of the resuU ..., a n ) be evaluated approximately from the formula The limiting relative error S is then equal to ,**- *Ji/r ".++ = Example 4. Evaluate S ln (10.3+ V^4. 4 ); the approximate numbers 10.3 and 4.4 are correct to one decimal place. Solution. Let us first compute the limiting absolute error AS in the 1 r A& \ ln (a+ V ~b), AS Aa We have general form: S = = Aa=A6=i=2Q; 1^4.4 the approximate is then equal to = 2.0976...; number y is ' ^SOTT^JQ ^TT; we leave equal to we can ^ be sure 2.1, =- f since the relative -r= ^ ; error of the absolute error tne first decimal place. Hence, U UUi> 0005 - 20 ' "2" Thus, two decimal places will be correct. Now let us do the calculations with one reserve decimal: 2.517. log (10. 3+ |/4 4) =5= log 12 4-1.093, In (10 3+ V 4.4)=^ 1.093-2.303 And we pet the answer: 2 52 8. Establishing admissible errors of approximate numbers for a given error in the result of operations on them. Applying the formulas of 7 for the quantities AS or 6S given us and considering all particular differentials = U-M quantities \-~Ac** or the M equal, we calculate the \dak \da k \f\ absolute errors Aa lt ... Aa^, ... o the approximate numbers a t admissible \ , , . .. ,a n , ... that enter into the operations (the principle of equal effects). It should be pointed out that sometimes when calculating the admissible errors of the arguments of a function it is not advantageous to use the principle of equal effects, since the latter may make demands that are practically unfulfilable In these cases it is advisable to make a reasonable redistribution of errors (if this is possible) so that the overall total error does not exceed a specified quantity. Thus, strictly speaking, the problem thus posed is indeterminate. Example 5. The volume off a circular cylinder (equal V to =~2 R* 2R) tan a. at an by a angle To what of a segment", that is, *a solid cut passing through the diameter of the base to the base, is computed from the formula "cylindrical plane a degree of accuracy should we measure the radius _ _ [Ch. 10 Approximate Calculations 370 #=s:60 cm and the angle of inclination a so that the volume of the cylindrical segment is found to an accuracy up to 1%? Solution. If AV, A/? and Aa are the limiting absolute errors of the quantities V, R and a, then the limiting relative error of the volume V that we are calculating is * 3A/? 2Act R "Nn^a ... We assume 3A -^ < m 1 . and _<2Aa 1 . 1 100" Whence cm -=1 60 sin f mm; 2a Thus, we ensure the desired accuracy in the answer to and the angle of inclination a to 9'. Ihe radius to 1 mm 1% if we measure 3108. Measurements yielded the following approximate numbers that are correct in the broad meaning to the number of decimal places indicated: 1207'14"; b) 38.5 cm; c) 62.215 kg. their absolute and relative errors. 3109. Compute the absolute and relative errors of the following approximate numbers which are correct in the narrow sense to the decimal places indicated: a) Compute a) 241.7; b) 0.035; c) 3.14. 3110. Determine the number of correct (in the narrow sense) decimals and write the approximate numbers: a) 48.361 for an accuracy of 1%; b) 14.9360 for. an accuracy of l%\ c) 592.8 for an accuracy of 2%. 3111. Add the approximate numbers, which are correct to the indicated decimals: 3.10 0.5; a) 25.386 + 0.49 2 41.72 0.09; b) 1.2-10 3.124. c) 38.1+2.0 3112. Subtract the approximate numbers, which are correct to the indicated decimals: + + + + + a) 148.1-63.871; b) 29.7211.25; c) 34.22-34.21. 3113*. Find the difference of the areas of two squares whose measured sides are 15.28 cm and 15.22 cm (accurate to 0.05 mm). 3114. Find the product of the approximate numbers, which are correct to the indicated decimals: a) 3.49-8.6; b) 25.1-1.743; c) 0.02-16.5. Indicate the possible limits of the results. _ Operations on Approximate Numbers Sec. /] _ 371 m 3115. The sides of a rectangle are 4.02 and 4.96 (accurate cm). Compute the area of the rectangle. 3116. Find the quotient of the approximate numbers, which are correct to the indicated decimals: to 1 a) 5.684 : 5.032; b) 0.144 1.2; c) : 216:4. 3117. The legs of a right triangle are 12.10 cm and 25.21 cm (accurate to 0.01 cm). Compute the tangent of the angle opposite the first leg. 3118. Compute the indicated powers of the approximate numbers (the bases are correct to the indicated decimals): a) 0.4158'; b) 65.2'; c) 3119. The side a of 1.5 2 . square is 45.3 cm (accurate to 1 mm). Find the area. 3120. Compute the values of the roots (the radicands are correct to the indicated decimals): a) 1/27715; b) j/65^2 ; c) KsTT. 3121. The radii of the bases and the generatrix of a truncated 0.01 cm; r= 17.31 cm 23.64 cm 0.01 cm; / cone are # 3.14. Use these data to compute the 0.01 cm; ji 10.21 cm total surface of the truncated cone. Evaluate the absolute and = = = = relative errors of the result. 3122. The hypotenuse of a right triangle is 15.4 cm 0.1 cm; 0.1 cm. To what degree of accuracy one of the legs is 6.8 cm can we determine the second leg and the adjacent acute angle? Find their values. 3123. Calculate the specific weight of aluminium if an aluminium cylinder of diameter 2 cm and altitude 11 cm weighs 93.4 gm. The relative error in measuring the lengths is 0.01, while the relative error in weighing is 0.001. 3124. to 221 current if the electromotive force is equal and the resistance is 809 ohms 1 ohm. of oscillation of a pendulum of length / is Compute the volts 1 volt 3125. The period equal to where g is the acceleration of gravity. To what degree of accuracy do we have to measure the length of the pendulum, whose period is close to 2 sec, in order to obtain its oscillation period with a relative error of 0.5%? How accurate must the numbers n and g be taken? 3126. It is required to measure, to within 1%, the lateral surface of a truncated cone whose base radii are 2 m and 1 m, and the generatrix is 5 m (approximately). To what degree of 372 _ _ Approximate Calculations [Ch. 1Q accuracy do we have to measure the radii and the generatrix and how many decimal places do we have to take the number n? 3127. To determine Young's modulus for the bending of a rod of rectangular cross-section we use the formula to *P ~~ 1 4 ' flL ' d*bs where / is the rod length, b and d are the basis and altitude ol the cross-section of the rod, s is the sag, and P the load. To of accuracy do we have to measure the length / and the sag s so that the error E should not exceed 5.5%, provided that the load P is known to 0.1%, and the quantities d and 6 cm? are known to an accuracy of 1%, cm, what degree /50 s2.5 Sec. 2. Interpolation of Functions 1. Newton's interpolation formula. Let * x lt ... xn be the tabular values of an argument, the difference of which h 0,l Ax; (Ax/ *,-+, x/; i ., y n are the correspond..., n 1) is constant (table interval) and # y lt ing values of the function y Then the value of the function y for an intermediate value of the argument x is approximately given by Newton's interpolation formula , . = = t , where o= and A# "7 = #, # , A2 t/ = A#, = A# , ... are successive finite the polynomial diilerences of the function y. \\hen x 0, 1, ..., n), x/ U 0, 1, ., n). As (1) takes on, accordingly, the tabular values y { (/ particular cases of Newton's formula we obtain: for n=l, linear inter potation; the of To use Newton's for /i 2, quadratic interpolation. simplify formula, it is advisab'ie first to set up a table of finite differences. If y f (x) is a polynomial of degree n, then = . = & n y. = const and A" +1 f//=^0 and, hence, formula (1) is exact In the general case, if / (x) has a continuous derivative f ln + l} (x\ on the x lt ..., x n and x, then the error interval [a, b], which includes the points * , of formula (1) is | is some intermediate value between *;(/ practical use, the following approximate formula where = is I, ..., n) and more convenient: (), x. For Interpolation of Functions Sec, 2] 373 If the number n may be any number, then it is best to choose it so that within the limits of the given accuracy; in other the difference A" + {/ w words, the differences A # should be constant to within the given places of 1 ^0 decimals Example sin 1. Find sin 2615' the using tabular data sin 26 = 0,43837, 27 -0.45399, sin 28 -0.46947. Solution. We set up the table Here, /i = 60', 26 15' q-- Applying formula have sin that 2615' = 0.43837 26 _!_ ' 60' 4 (1) and using the + + 0.01562 Let us evaluate the error R 2 if */ sun, then |*/ (/l '|*^l, first horizontal line of the table, (0.00014) = 0.44229. 2 Using formula we will we (2) and taking into account have: Thus, all the decimals of sin 2615' are correct. Using Newton's formula, it is alsc possible, from a given intermediate value of the function //, to find the corresponding value of the argument x (inverse interpolation). To do this, first determine the corresponding value q by the method of successive approximation, putting and ?C'H n\ 2! (i-O, 1, 2, ...). Here, for q we take the common value (to the given accuracy!) m + >. Whence x jt cessive approximations q (m} <;< -{- = l approximate the root of the equation smhx of two sue* 374 Approximate Calculations Solution. Taking y = a 10 we have 4.457, ,_5-4.457 ~ 1.009 We (Ch 0.543 5 ^i.oog 0.565-0.435 0.220 2 1.009 =0.538; = 0.538 + 0.027 = 0.565; :0, 538 + 0. 027 = 0. 565. can thus take * = 2. 2 + 0. 565- 0.2 = 2. 2 + 0. 113 = 2. 313. 2. Lagrange's interpolation formula. In the general case, a polynomial of 0, 1, .... n), is given degree n, which for *=*/ takes on given values yf (/ by the Lagrange interpolation formula = *!> (x (x __ x2) . . . (x *o) (* xn ) *,). XQ) (x (x . * 2) . . . (x xn ) Xk ^) (X .(X ' ' ' ' " ^ ' 3128. Given a table of the values of x and y: Set up a table of the finite differences of the function y. 3129. Set lip a table of differences of the function y x* f 1 for the values *=1, 3, 5, 7, 9, 11. Make sure that JC 5jc all the finite differences of order 3 are equal. 3130*. Utilizing the constancy of fourth-order differences, set 10*' +2** 3jt up a table of differences of the function y x* for integral values of x lying in the range l^jt^lO. 3131. Given the table = + = log + 1=0.000, log 2 -0.301, 0.477, log 3 = = log 4 0.602, log 5 = 0.699. Use linear interpolation to and log 4. 6. log 3.1, compute the numbers: log 1.7, Iog2.5, Interpolation of Functions Sec. 2] 375 3132. Given the table sin 10 sin 11 sin 12 = 0.1736, = 0.1908, = 0.2079, sin 13 sin 14 sin 15 = 0.2250, = 0.2419, = 0.2588. = in the table by computing (with Newton's formula, for 2) the values of the sine every half degree. 3133. Form Newton's interpolation polynomial for a function represented by the table Fill 3134*. Form Newton's interpolation polynomial for a function represented by the table Find y for 3135. A x = 5. 5. function Form Lagrange's y for JC = 0. For what x will y = is given by the table interpolation polynomial and find the value of 3136. Experiment has yielded the contraction of a spring (x as a function of the load (P kg) carried by the spring: mm) Find the load that yields a contraction of the spring by mm. 14 376 (Ch. 10 Approximate Calculations 3137. Given a table of the quantities x and y = 0.5 and for x 2: a) the values of y for x of linear interpolation; b) by Lagrange's formula. Compute by means Sec. 3. Computing the Real Roots of Equations 1. Establishing initial approximations of roots. roots of a given equation The approximation /(*)=0 of the (i) consists of two stages: 1) separating the roots, that is, establishing the intervals (as small as possible) within which lies one and only one root of equation (1); 2) computing the roots to a given degree of accuracy on an interval [a, b] and If a function /(A) is defined and continuous of equation (1). /(a)./(6)<0, then on [a, b] there is at least one root or /' (x) This root will definitely be the only one if /' (x) when < > a<x it is advisable to use millimetre In approximating the root paper and construct a graph of the function y = f(x). The abscissas of the points of intersection of the graph with the x-axis are the rools of the equation /(x) 0. It is sometimes convenient to replace the given equation with an equivalent as the absequation (p (#) if (A). Then the roots of the equation are found cissas of points of intersection of the graohs y q)(x) and y ty (x). 2. The rule of proportionate parts (chord method). If on an interval [a, b] of the equation f (A) there is a unique root O, where the function f (x) is continuous on b], then by replacing the curve y f(x) by a chord [a, passing through the points [a, f (a)] and [b, f (b)], we obtain the first approximation of the root = = = 4 t /~\ (2) To obtain we apply formula (2) to that one of ends of which the function f (x) has values of opposite sign. The succeeding approximations are constructed in the same manner. The sequence of numbers cn (n=l, 2, converges to the .) a the intervals second approximation c 2 [a, c,] or [c,, b] , at the . root , that . is, Generally speaking, we should continue to calculate the approximations c,, until the decimals retained in the answer cease to change (in accord ., with the specified degree of accuracy!); for intermediate calculations, take one or two reserve decimals This is a general remark. If the function / (x) has a nonzero continuous derivative /' (x) on the interval [a, b], then to evaluate the absolute error of the approximate root c2 , . . _ _ crt , we can make where 377 Computing the Real Roots of Equations Sec. 3] u= use of the formula min <x< a (/' (x) |. b and f" (x) ^ for 3. Newton's method (method of tangents). If /' (x) a^x^b, where f(a)f(b)<0, f(a)f'(a)>Q, then the successive approximaof an equation f(x)=Q are computed tions * (rt = 0, 1, 2, ...) to the root rt from the formulas (3) Under the given assumptions, = lim * n -> oo To evaluate where u,= min a we can the errors < 1, 2, ...) is mono- . use the formula \f'(x)\. xQ a= = x For practical purposes where sequence x n (n the tonic and ., , = a, is it more convenient = *_, * which yield the a/^.,) i/i to use the simpler = same accuracy l, 2, formulas ...), (3') as formulas (3). = &. f(b)f"(b)>Q, then in formulas (3) and (3') we should put x 4. Iterative method. Let the yiven equation be reduced to the form If * = (4) where \y' (x)\*^r< 1 (r is constant) for a ^x^b. Proceeding from the iniwinch belongs 4o the interval [a, b], we build a sequence of tial value * numbers x lt x 2 ... according to the following law: , , If a<jc rt <6 (n = l, 2, ...), then the limit is the on/r/ roo/ of equation (4) on the interval [a, 6]; that is, xn are successive approximations to the root |. The evaluation of the absolute error of the nth approximation to x n is given by the formula it IS _ xn y \ I ^^s * xn + l j Therefore, if xn and x n + l coincide error for x n will be -r to _r xn within E, I then the absolute limiting . In order to transform equation f(x) Q to (4), we replace the latter with an equivalent equation where the number A, ^ is chosen so that the function -7- [x Kf (x)] = f 1 Kf (jc) .378 _ __ Approximate Calculations should be small in absolute value in the neighbourhood of the point example, we can put 1 X/'(x )=0]. Example 1. Reduce the equation 2x Inx 4=0 to the form (4) initial approximation to the root * = 2.5. Solution. Here, /(x) x lent equation values of A,; |l_ A,( 2\ | The =x this ^ X ] initial = 2x In * \nx k(2x number U = 2.5 =0, equation /'(jc) and 4) close is that I 4; is, 10 [Cft. = 2-~. We * (for for the write the equiva- take 0.5 as one of the suitable of the to root the equation r^O.6. .6 close to 1 reduced to the form is x =x Inx 0.5 (2x 4) or Example equation that of (5) 2. Compute, to two decimal between 2 and 3. places, the root \ of the preceeding lies Computing the root by th Example 1, putting x ? = 2.5. iterative method. We make use of the result We carry out the calculations using formulas with one reserve decimal. *f =2+ *, = 2 + 4- And so 1^:2 45 (we can Become fixed) Let us npw evaluate the Considering that all ^2. 448, In 2.450 2 *4 = 2 (p(*) y In2.458ss2.450, + ~ln2. 448=^2. 448. since stop here error. Here, = 2 + ~lnx approximations to and xn the third decimal place has lie in the interval [2.4, 2.5], we get Hence, the limiting absolute error remark made above, in the approximation to x 9 is, by virtue of the A== ^-^-=0.0012 ^=0.001. Thus, the exact root g of the equation 2 447 we can lies within the limits < g < 2.449; take g^2.45, and all the decimals be correct in the narrow sense. of this approximate number will _ Sec. 3] Computing the Real Roots of Calculating the root by Newton's method. <x<3 Wee We Here, > On the interval 2 we have: /(3)f (3)>0. Hence, the conditions Equations /' (x) of 3 for > 0; /(2)f(3)<0; = 3 fare(x)fulfilled. and * take carry out the calculations using formulas jtj^S =2 4) In 3 0.6(2-3 0.6(2-2 4592 0.6(2-2.4481 0.6(2-2 4477 = 2. 4592 = 2.4481 *, * 4 = 2. 4477 *, (3') with two reserve decimals: 4592; In 2 4592 In 2. 4481 In2. 4477 = 2 4481; = 2. 4477; 4) = 2 4475. 4) 4) At this stage we stop the calculations, since the third decimal place 2. 45. We omit the does not change any more. The answer is: the root, evaluation o! the error; 5. The case of a system of two equations. Let it be required to calculate the real roots of a system of two equations in two unknowns (to a given degree of accuracy): = f(*. 0=0, , 6 . there be an initial approximation to one of the solutions (|, r\) of system Jt x ot y = y Q This initial approximation may be obtained, for example, graphically, by plotting (in the same Cartesian coordinate system) the curves f(x, #) and by determining the coordinates of the points of interand tp (x, #) and let = this . = section of these curves. a) Newton's method. Let us suppose that the functional determinant dtf.jp) d(x,y) xx = Then by Newdoes not vanish near the initial approximation y Q1 y ton's method the first approximate solution to the system (6) has the form where a P O are the solution of the system of two. a x f/ x, = + , j/, = +P . , linear equations The second approximation is obtained in the very same way: = *!+<*!, *, where a p$ v are the solution

Similarly

we obtain

of the

^1

= ^1+ Pi,

system of linear equations

the third and succeeding approximations.

380

_ _ Approximate Calculations

We

b) Iterative method.

the system of equations

(6),

[Ch

10

can also apply the iterative method to solving by transforming this system to an equivalent one * F(x.y).

=

= \ y (D(*, (

and assuming thai

|F>.y)| + |0;<x 0)|
(f)

y)

\F'y (x, y)\

t

+

\<S> v (x.

#|
1

(8)

in some two-dimensional neighbourhood U of the initial approximation (* y Q ) which neighbourhood also contains the exact solution (, rj) of the system. The sequence of approximations (x n y n ) ( = 1, 2, ...), which converges to the solution ol the system (7) or, what is the same thing, to the solution ,

t

,

x>f

(6),

is

constructed according to the following law:

= F(x If

all

(*, y n ) belong

to

U

yz )

t,

then

%

*/,-c

t

=

*

lim n

lim

,

n

-+ oo

-

y n =i\.

oo

The following technique is advised for transforming the system of equa* tions (6) to (7) with condition (8) observed. We consider the system of equations a/

(

which

is

equivalent to

(x.

provided

(6)

a.

that

Rewrite

0.

it

in the

form

Y.

y)^F(x

,

Choose the parameters functions F(v, y)

-jnd

a,

p,

O (x,

y, & such y} will be

that

the

t

y),

derivatives of the zero in the initial approximate solutions

partial

equal or close to

in other words, we find approximation; system of equations

a,

Y.

fi,

o as

of the

,v

,

t/

)

= 0,

Condition (8) will be observed in such a choice of parameters a, P, Y on the assumption that the partial derivatives of the functions / (x, y) and
given the initial approximation to the root *

= 0.8,

= 0.55.

Sec. 3]

_

Solution.

fv (x*

the Real Roots of Equations

Computing

y

)

Here,

= 1.1;

y)-=x*

f (x t

(*<>.

= 1-92, 0o)

+ y*\

t

t

= 1.6,

y ) equivalent to the initial one) ,

is

a(* + // -l) + p(*'-i/)-0, 2 2 \ Yl* + -0-i-a(*'-
J in the

i/ c )

1.

Write down the system (that 2

381

y)=x'-y- fx (x

q> (x,

=

__

/la, pi

\

VI Y. *l

/

form

For suitable numerical values of system of equations 1

Y anc* ^ choose the solution

a, p,

of

the

+ 1.60+1.920 = 0, l.la

p-0,

+ 1.925 = 0, + 1. lY 6 = 0; 1.6 Y

1

i.

y^

we put a =^0.3, p ^r 0.3, Then the system of equations

e.,

i

\

which

=x y=y x

0.3(^

2 2

+

0.5,

b^QA.

1)

0.3(A-

2 r/

0.5 (* -{-#

2

1)

y),

+ 04 (x*y)

t

equivalent to the initial system, has the form (7); and in a sufficiently small neighbourhood of the point (x Qt y Q ) condition (8) will be fulfilled. is

Isolate the real roots of the equations by trial and error, of the rule of proportional parts compute them to

by means

and two

decimal places. s x 1-0. 3138. X 4 * 05* 1.55 = 0. 3139. + 8 4* -1--0. 3140. x Proceeding from the graphically found initial approximations, use Newton's method to compute the real roots of the equations to two decimal places:

-

3141.

JC

3142. 2x

-\

_2jc In*

5-0. 4

= 0.

3143. 2 3144.

x

= 4x.

logjc=y.

Utilizing the graphically found initial approximations, use the iterative method to compute the real roots of the equations to

two decimal places: 8 3147. jc x 3145. x'-5.*M 0.1=0. 2 = 0. = cos*. 3146. 4* Find graphically the initial approximations and compute the real roots of the equations and systems to two decimals: 3151. x- In* 14 = 0. 3* -(-1=0. 3148. A:' 8 2 8 = x 3152. 2* + 3* 5 0. 3149. * +3* 0.5 = 0. 2 4 3153. 4* 7sin* = 0. 3150. * +* 2* 2 = 0.

382

[Ch. 10

Approximate Calculations

3154. x* 3155. e* '

+ 2x + e- *

= 0. = 0.

6

8

"

~~

" Lff

x*

(

157

4

+y

'

\

4

= 0,

\Qgx-l =0.

y

~~

3156. 3158.

Compute

the equation tan

3159.

jc

to three decimals the smallest positive root of

=

jc.

Compute the

roots of the equation x-tanh

x=l

to four

the

integral

decimal places. Sec. 4. Numerical Integration of Functions

1. Trapezoidal formula. For the approximate evaluation

of

b

If (x) [a,

is

a function

continuous on

into n equal parts

b]

= x + ih Xi

a, let

xn

with an absolute error

of

we divide the

[a, b]]

and choose the interval

= b,

of

-

interval of integration calculations

h==

.

be the abscissas of the partition points, and // /(*/) be the corresponding values of the integrand f(x). Then the trapezoidal formula yields y

Let

Q

(x

/

0,

2,

1,

..., n)

a

where M 2 = max f (x) when To attain* the specified accuracy |

\

is

That

h must be of the order value so that

is,

e

when evaluating

the integral,

the

in-

found from the inequality

terval h

of

V^e

.

The value

of

h obtained

is

rounded

off to the smaller

b

a== ~~ n h

be an integer; this is what gives us the number of partitions n. Having established h and n from (1), we compute the integral by taking the values of the integrand with one or two reserve decimal places. 2. Simpson's formula (parabolic formula). If n is an even number, then

should

in the notation of

1

Simpson's formula

b

h Q-

K#o + #n)

+ 4 (y\ + #3+.. +yn -\) + +0n-2)l

(3)

Numerical Integration

Sec. 4]

of Functions

383

holds with an absolute error of (4)

M 4 = max

where

when

! |

(x)

/

\

To ensure

the specified accuracy e when evaluating the interval of calculations h is determined from the inequality

integral,

the

(5)

That

the interval h

is,

to the smaller value so

is

of the order

that n

J/JF, is

The number h

is

rounded

off

an even integer.

Remark. Since, generally speaking, it is difficult to determine the interand the number n associated with it from the inequalities (2) and (5), practical work h is determined in the form of a rough estimate. Then,

val h in

after the result is obtained, the number n is doubled; that is, h is halved. one to the number of decimal If the new result coincides with the earlier places that we retain, then the calculations are stopped, otherwise the pro-

cedure

is

repeated, etc.

For an approximate calculation of the absolute error R of Simpson's quadrature formula (3), use can also be made of the Range principle, according to

which

where 2 and S are the results h and // = 2/i, respectively.

of calculations

from formula

(3)

with interval

3160. Under the action of a variable force F directed along the x-axis, a material point is made to move along the x-axis 4. Approximate the work A of a force F if a to x from x table is given of the values of its modulus F:

=

=

by the trapezoidal

Carry out

the calculations the Simpson formula.

formula and by

i

3161. Approximate

2

J (3*

4x)dx by the trapezoidal

formula

putting rt=10. Evaluate this integral exactly and find the absolute and relative errors of the result. Establish the upper limit A of absolute error in calculating for n=10, utilizing the error formula given in the text.

384

\Ch. 10

Approximate Calculations i

3162. Using the

* *

calculate

Simpson formula,

V

to four

.

J *

*

r

o

decimal places, taking n== 10. Establish the upper limit A of absolute error, using the error Formula given in the text. Calculate the following definite integrals to two decimals: 3,63.

3,68.

Ife.

(-!<,.

1

Jl

-~.

3164. C

3169.

317

-

-

>

2

3166.

[x\ogxdx.

^

8187.

Q | 7i

dr.

fcos* dX. FT \+X Jcos* ,

3172>

3173. Evaluate to two decimal a PPlyi n S the substitution

the

places

^

-7--

improper

Verify the calculations

1

ft

by applying Simpson's formula to the integral

\

+

is

chosen

,so

QO

that

7^722

<

integral

Jy

rr ~r x

J

2

where b

,

*

10

A

plane figure bounded by a half-wave of the sine curve *-axis is in rotation about the x-axis. Using the Simpson formula, calculate the volume ot the solid of rotation to two decimal places. 3175*. Using Simpson's formula, calculate to two decimal 3174,

f/=sin^ and the

places the length ot an arc of the ellipse in the first

y+

2

=

1

situated

Sec. 5. Numerical Integration of Ordinary Differential

1.

^

A method

of successive approximation be given a first-order differential equation

subject to the initial condition

(Picard's

y'^f(x.y) = # when x = *

/

Equations method).

Let

there

<1) .

Numerical Integration

Sec. 5]

Ordinary Differential Equations

of

_

385

The solution y (x) of (1), which satisfies the given initial condition, can, generally speaking, be represented in the form

y(x)=

lira yi (x) -* 00

(2)

t

where the successive approximations y

*//(*) are

from

determined

the

formulas

,

X

(*)=#<>+

If

the right side f(x, y)

R{ and

in this

satisfies,

is

|x

yt-i (*)) dx

defined and continuous

x

in the

neighbourhood

|

Vi)-f(x. i/JKJ-ltfi-tfil

(L is constant), then the process converges in the interval

of

I* ft

(x.

neighbourhood, the Lipschitz condition \f(x,

where

/

= min(a, ^ M/ \

]

and

successive

definitely

(2)

*

Af = max|/U,

/?

approximation

y)\.

And

the error here

is

/?

)

- y n W \<ML* ~

1

The method of successive approximation (PicarcTs method) is also applicable, with slight modifications, to normal systems of differential equations. Differential equations of higher orders may be written in the form of systems of

differential equations.

2. The Runge-Kutta method. Let ,

To do

it be required, on a given interval to find the solution y (x) of (1) to a specified degree of accuracy e.

we choose

this,

the interval of calculations /i=

the interval [* X] into n equal parts so that h* Xf are determined from the formula ,

X|

=x +M

(i=0,

1,

2,

< e.

..., n).

By the Runge-Kutta method, the corresponding values function are successively computed from the formulas

1

3

1900

by dividing

The partition points

/;

=

(/

(x/) of

the desired

[Ch. 10

Approximate Calculations

386

where f

=0,

1,

2,

...,

n and

(3)

*i"

=/(*/ + A,,

correct choice of the interval h

To check the

it

is

to

verify

the quantity

e= The fraction 6 should amount

a

to

few

hundredths, otherwise h has to be

reduced.

The Runge-Kutta method is accurate to the order of h 1 A rough estimate Runge-Kutta method on the given interval [x X] may be obtained by proceeding from the Runge principle: .

of the error of the

,

Rn

y*m

I

Urn

I

'

15

where /i = 2m, y 2m and y m are the results of calculations using the scheme (3) with interval h and interval 2/i. The Runge-Kutta method is also applicable for solving systems of differential equations y'

with given

= f(x,

y> z).

*'

=
y,

z)

(4)

= z = 2 when x x by the Milne method, =(1) X we in some way find

conditions / 3. Milne's method. To solve initial conditions y=^y when X values initial

yi=y(*i),

t/

.

,

Q,

y=0(*i).

/

subject to the the successive

=*/(*)

of the desired function y (x) [for instance, one can expand the solution y (x) in a series (Ch. IX, Sec. 17) or find these values by the method of successive approximation^or by using the Runge-Kutta method, and so forth]. The ap-

proximations y t and y] for the following values of successively found from the formulas

*=

\vhere

fi

= f(x

it

=

-

r/ t

(i=4,

5,

..., n)

are

+~(7"

y^ and7i = /(*i,

Hi)-

To check we

calculate the quantity (6)

_

Numerical Integration of Ordinary Differential Equations

Sec. 5]

387

does not exceed the unit of the last decimal \(y m retained in the y (x), then for f/ we take I// and calculate the next value y/ + 1 10~ w then one has to start from the berepeating the process. But if e/ ginning and reduce the interval of calculations. The magnitude of the initial interval is determined approximately from the inequality h 4 10~ m For the case of a solution of the system (4), the Milne formulas are written separately for the functions y (x) and z (x). The order of calculations remains the same. Example 1. Given a differential equation */'=*/ x with the initial condition y(0)=1.5. Calculate to two decimal places the value of the solution of this equation when the argument is x 1.5. Carry out the calculations If

e/

for

-

t

,

>

,

<

combined Runge-Kutta and Milne method.

a

by

.

We

Solution.

choose the initial interval h from the condition /r*<0.01. let us take h 0.25. Then the entire interval of x to jc=1.5 is divided into six equal parts of length

To avoid involved writing,

integration from 0.25 by means of points x f (i 0, 1, 2, 3, 4, 5, 6); we denote by y f and y^ the corresponding values of the solution y and the derivative y' We calculate the first three values of y (not counting the initial one) by the Runge-Kutta method [from formulas (3)]; the remaining three values we calculate by the Milne method [from formulas (5)] 1/4. */s y& The value of // will obviously be the answer to the problem. We carry out the calculations with two reserve decimals according to a definite scheme consisting of two sequential Tables 1 and 2. At the end of Table 2 we obtain the answer. # #, x =^0, // =1.5 Calculating the value y r Here, / (x, */)

=

.

fl

=

/i

= 0.25.

A^

=i

+

(*<

= 4-D (0.3750 + 2-0. 3906 + 2.0.3926 + 0.4106) ==0. 3920; (

0)

,

*(>

= / (*o. =

/ /

(

0o) h

+~

)=s

(

0+

1

.5000)0.25-0.3730;

(0)

\ + -y- j = (- 0.125+1.5000 + 0.1875) 0.25= 0.3906; fc

/

Xo

/i

,

& (0)

h

/

*i

=

=n *o+y.

0o

\ + -|-) ^ = (0 125+1.5000 + 0.1953)0.25 = 0.3926; ^= 0.25+1.5000 + 0.3926)0.25 = 0.4106; (

= 1.5000 + 0.3920 =1.8920

(the

first

three

decimals

in

tins

approximate number are guaranteed). Let us check:

6

_

=

20 |0.3906Q.3926|_ .1 = 0. 13.

10.37500.39061""

By in

13*

we chose was rather rough. values y t and y 9 The results are tabulated

this criterion, the interval h that

Similarly

Table

1.

we

calculate

the

156

.

388

Approximate Calculations Table

1.

Calculating^,, f(x, y)

Calculating the value of j/

=1. 5000,

^=1.5000, Applying formulas

jr 2 ,

y9

We

have: f(x, y)

=

jf4 .

= 1.8920, = 1.6420, ^'

(5),

we

by the Runge-Kutta Method.

=

y t = 2.3243,

r/,

[Ch. 10

^=1.8243,

/z

y,

= 0.25,

,== 2.8084;

= 2.0584.

find

-^

= 1.5000 + 4

A OK

(2-

1.6420-1.8243 + 2. 2.0584)

= 3.3588;

1+3.3588 = 2.3588; =

02

+ ^(01

h y' t)

. hence, there

is

'

= 2.3243 +

^

3.3588-3.3590

no need

(2.3588

+ 4- 2.0584 +

1

.8243)

1

to reconsider the interval of calculations.

= 3.3590;

Numerical Integration of Ordinary Differential Equations

Sec. 5]

389

= =

We obtain i/ 4 y4 3.3590 (in this approximate number the first three decimals are guaranteed). Similarly we calculate the values of y^ and y^. The results are given in Table 2. Thus, we finally have y

(1.5)

= 4.74.

4. Adams' method. To solve (1) by the Adams method on the basis of the initial data */(x ) we in some way find the following three values /o g of the desired function y (x):

=

[these three values may be obtained, for instance, by expanding y (x) in a power series (Ch IX, Sec. 16), or they may be found by the method of successive approximation (1), or by applying the Runge-Kutta method (2) and so forth]. With the help of the numbers X Q x it * 2 x t and */ y lt y zt ys we calcu,

late

,

q lt q 2

,

2

We

.

,

?, where

= h y' = hf (x 2

2,

// 2 ),

qs

= hy' = hf (x,, 3

then form a diagonal table of the finite differences of

V'=f(x,

I/O

f/2

i/5

y)

q:

_ _ Approximate Calculations

390

Thus,

numbers

the

utilizing

we

= + --

A#,

it,

<7,

e

+ A*/

A? 2

A<7 2 ,

,

calculate, by

i

in

table of differen-

consists in continuing the diagonal

ces with the aid of the

Hhe difference table,

[Ch. 10\

A 2 ?,, A 8

means

of

situated

formula

o

+ IZ A <7,+ 2

-r^

A*<7

O

After

.

And when we know * 4 and #4

(7)

finding

diagonally in

and putting n-- 3

A//,,

we

calculate

we calculate q^~hf(x^ f/ 4 ), difference table and then fill into it the 2 A 3<7,, which are situated (together with 4 ) along finite differences A
#4

*/3

introduce

8-

*/ 4

A# 8 and

,

,

into the

4

,

=

,

.

,

W

m

].

In this

sense

the

Milne (5) and Runge-Kutta Evaluation of the error

is

(7)

equivalent

to the

formulas of

(3).

for the Adams method is complicated and for practical purposes is useless, since in the general case it yields results with considerable excess. In actual practice, we follow the course of the third finite differences, choosing the interval h so small that the adjacent differences A 8 <7f and A*
Example

2.

Using the combined Runge-Kutta and Adams method, calcu(when #1.5) the value of the solution of the x with the initial condition f/(0) 1.5 (see

late to two decimal places differential equation

y'y

Example

1).

We

use the values y lt y z //, that we obtained in the solution Their calculation is given in Table 1. We calculate the subsequent values //4 f/5 #6 by the Adams method (see Tables 3 and 4).

Solution.

of

Example

,

1.

,

,

The answer to the problem is #4 = 4.74. For solving system (4), the Adams formula (7) and the calculation scheme shown in Table 3 are applied separately for both functions y(x) and z(x).

Find three successive approximations to the solutions and systems indicated below.

differential equations

= =

3176. y'=jf+y*\ y(Q) Q. 3177. y' x y 2, z' y-z\ f/(0)=l, z(0) 3178. */" -#; y(0) 0, y'(0)=l.

= + + =

=

= -2.

of

the

Table

3.

Basic Table for Calculating /(*,

y4 y

= -* +

,

*/;

g

fc

by the

=y^0.25

,

(Italicised figures are input data)

Table

-1

153

Auxiliary Table for Calculating by the

_

Approximating Fourier

[Sec. 6]

Coefficients

_

393

/i = 0.2, use the Runge-Kutta method to approximately the solutions of the given differential equations and systems for the indicated intervals: 3179. 0' = x; 0(0) = 1.5 (

Putting the interval

calculate

3180. 3181.

=|

0'

0(1)

0';

y'=*z+

=

z'=0

1.

1

x,

(Kx<2). 0(0) = 1, z(0) =

(0<x
l

combined Runge-Kutta and Milne method or Applying Runge-Kutta and Adams method, calculate to two decimal places a

the solutions to the differential equations and systems indicated below for the indicated values of the argument; 0.5. 3182. 0'=x+0; 0=1 when x=--0. Compute y when x 0. Compute y when x=l. x* 3183. 0' 0; 0=1 when x 0. Compute when x 0.5. 3184. 0' 3; 0=1 when x 20 x 3185. J0' 20+z,

=

= =

I

= =

+

= + z' = x + 20 + 3z;

Compute y and 3186.

f0'

=

3187. 0*

30

=2

3188.

y=

z

when

0V +1=0; and

when x = 0.

2

when x = 0.

1

when x = 0.

1

0'

=

when jc=l.

=1.5.

2

A:(JI)

=

A:

0=1, A:

^ + |-cos2/ =

3189.

= = 0.5. = 0' z

2,

= 2,

0;

Compute y when

Find

z,

z;

when x=l.

Compute

z

when x = 0.5.

z

)2r'=y and Compute

= 2,

=

0;

x = Q,

=

Jc'

when

l

f=-0.

x' (n).

Sec. 6. Approximating Fourier Coefficients

Twelve-ordinate scheme. Let y n of

the function y

and y Q

-f/ 12

We

~f(x n

)

(n

points f(x) at equidistant equidist set

= 0,

1,

xn = -

..., of

up the tables: y* y\

yu

Sums

yz

y* y*

y\9 y> y*

(2j)

Differences (A) Q u l uz Iu U* U 6 M 4

Sums

s

s l s 2 s,

Diilerences

t

tv

t

Sums Differences

12)

be the values

the interval lU,2ji],

Approximate Calculations

394

The Fourier

may

\Ch

=

coefficients a n b n (n 1, 2, 3) of the function 0, y be determined approximately from the formulas: ,

6t

a2

=s

&, s,

+ 0.5(s,

&8

s 2 ),

= 0.50! + 0.866a + a = 0.866 (1, + *,), = a, a 2

10

= f(x)

8,

8,

(1)

where 0.866

.

=

We

have

+

f(x) zz

Other

'

30

10

schemes

are

also

used.

(an cos

nx +

Calculations

b n sin nx).

are

simplified

patterns.

Example. Find the Fourier polynomial for the function y represented by the table

From formulas

(1)

by the use

of

= f(x)

we have

= 9.7; a, = 24.9; = 13.9; 6 = 8.4; 6, 2

2

=i0.3; a a = 3.8; 6,

= 0.8.

Consequently, / (x)

^ 4.8 + (24.9 cos x + 13.9

sin x)

+ (10.3 cos2x

8.4 sin 2x)

+ Using the for

(3. 8

+ cos 3*

+ 0.8 sin 3x).

12-ordinate scheme, find the Fourier polynomials the following functions defined in the interval (0,2:rc) by the

Approximating Fourier Coefficients

Sec. 6]

tables of their values that correspond to the

395

values

equidistant

of the .

argument. 3190.
= 4300 = 5200 y = = 9.72 y, = 8.97 = 8.18 y, = 1.273 y = 0.788 y = 0.495 #, 4

t

= = 6.68 y, = 9.68 - 2.714 = 3.042 y, = 2.134

3191. y

t/,

3192.

J/

/2

f/ 4

t/ s

4 s

= 7400 - 2250 = y, /,

= 3850 = 7.42 y, = 6.81 y, = 6.22

yg

= 0.370 = 0.540 y, = 0.191 y,

y,

= 7600 = 4500 ^, = 250 0,

=5.60 / 10 =4.88 / n =3.67

y.

y,

=-0.357

= 0.437 n = 0.767

#, t/

3193. Using the 12-ordinate scheme, evaluate the Fourier coefficients for the following functions: a)

f(x)

=

(x>

first

several

Chapter 1.

then a b) + b + \b\. Whence (a = & a (^ 6 \<\ab\ & ^ a Hence, |a |a \ab = |a/ + |6 3. a) -2<x<4; \a-b = |a + b) |<|a| + -6| < x < 0; d) x> 0. 4. 24; 6; 0; 0; 0; 6. 5. 3, x > c) .; ~; 0. (*)= --x +

a=(a

Solution. Since

>|a|

and

\b\

Besides,

t

\

I

|

|

|.

|

(

|

<

b) x

I

|

|

|

|

1

1;

|.

|.

|

1;

7.

b)~oo

<*< + oo.

/2<*< +

be 2

+*

* but

16. 19.

oo

2<0, this

22. q>(*)

= 2^

1

i.

5* 2

10. i|>(x)

=

a)

T = JI,

T=

Periodic,

e)

y=

*-|-KO,

or

e.,

should either

x
15.

18.

1

a)

Even, b)odd,

= i[f (x) + / (x)

/

(-x)}

A

7 =ji,

c) periodic,

0<x
if

even, d)odd,

+ ![/ (x)-f (-x)}.

2n

r = -r-,

c)

2,...).

1,

2

3^ + 6x23.

if

i.

Arji
21.

x,

<*<

2^0,

x

K*<2. \<x<

2<x<2.

2 n, b) periodic, -^ u

nonperiodic. 27.

is,

oo

13. a)

l<;t<2. Solution. It 2)<0. Whence (A:+ 1) (*

Thus,

impossible.

0<*<1. 17. 20 K*<100.

e) odd.24. Hint. Utilize the identity

26.

that

-

-

14.

l<x<2-

e.,

+00).

2), (~-2, 2), (2,

2<0;

^

is 1,

-4-< <

00,

(

*=0, |jc|^ K2. 2

or ^

<^< Jt

12.

b)

x'^0,

A'+l^O, Jt^*2,

oo;

.

d) periodic

C<

m q^ when = m +q qil +q /^A;^/,-!-^; 28.

= ^i^i + ^2(^ when + /2<^
/1

37.

/

-^-;0;

/

2

7-. 38.

a)

/.

y

=2 x=

29. (p[x|)(jc)]

=Q

when

2l 2

l

2Jf

;^[(p(x)]

y

1,

rel="nofollow">

s

= 2^

(x

l

l

l

2)

30. x. 31. (x

when #

>

1,

when

+ 2) y
.

= when x = when jc< and * = 2, / > when 1 <x< 2, 1; b) < when oo < x < 1 and 2 < x < -f oo; c) z/>0 when oo<x< -|- oo; when x = 0, x = ]/~3~and A;= /" "3, r/>0 when V%< x<0 and d) j/ = 1

i/

i/

1

oo);

b)

and

l<x<-f

x=V y +

\

y<0 whenO<^
oo,

and

x

=

39. a)

*

= -- (y

3)

=

*

c)

=

\

if

~

y lanyf

= arc tan 43. a) c)

*/

#=

y=u = arc l

a)

^=--;

d)

VlogT;

when

2

x

d)

= 2-10>

oo<#< +

f

(

oo);

x

e)

=

* = # when oo<0<0; x=/l/ when = 2x 5; b) # = 2a = cos.r, c) = u u= x 42. a) # = sin x; b) sinw, a = 3
,

2

2

.

,

y = 2(x*

!/""<

,

,

t/

c)

= -cosjc

oo);

40.

41.

.

oo<#< +

(

397

JC |

|

oo<x<0 and

1)

=

and

|x|
if

b) y = log (10 = y x when 0<x< + oo.

< 1/2S;

10*), 46.

|*|>1.

if

-oo<x
Hint. See Appen-

\j

dix VI, Fig. 1. 51. Hint. Completing the square in the quadratic trinomial 2 we will have y y and yQ (4ac b 2 )j4a. a(x * ) where * b',2a Whence the desired graph is a parabola y ax* displaced along the *-axisby See .Appendix VI, 53. Hint. XQ and the t/-axis along yQ by

= +

=

=

=

.

The graph

is

See

Hint.

58.

2.

Fig.

a hyperbola

Appendix

y=

,

shifted

VI,

along

Fig.

the

3.

*-axis by

Hint.

61.

X Q and along 2

the t/-axis by (

x+

yQ

Taking the integral

Hint.

62.

.

\ (Cf. 61*). 65. Hint. See

Appendix VI, Fig.

part,

4.

we have y=-~ o

--13 9

/

/ '

67.Hint. See Appendix VI,

Appendix VI, Fig. 6. 72. Hint. See Appendix VI, Appendix VI, Fig. 8. 75. Hint. See Appendix VI, Fig. 19 78. Hint. See Appendix VI, Fig. 23. 80. Hint. See Appendix VI, Fig. 9. 81. Hint. See Appendix VI, Fig. 9. 82. Hint. See Appendix VI, Fig. 10 83. Hint. See Appendix VI, Fig. 10. 84. Hint. See Appendix VI, Fig 11. 85. Hint. See Appendix VI, Fig. 11. 87. Hint. The period of the function 5 sin 2x with amis T 2njn. 89. Hint. The desired graph is the sine curve y plitude 5 and period n displaced rightwards along the x-axis by the quantity 5.

Fig.

1

Hint.

71.

Hint.

Fig. 7. 73.

See

See

90. Hint. Putting

.

a=A cos

cp

and

b= A

sin

--- V

In our case,

cp,

we

will have

y=A

sin (x

cp)

,

A

where

=-

2

Hint. cos x

V& + b

2

and

(p

=

arctan(

4 = 10,

cp=0.927. 92.

= -jr-(l+cos2jc). 93. Hint. The desired graph the sum of the graphs = sinjc. 94. Hint. The desired graph the product of the graphs = sinx. 99. Hint. The function even For x>0 we determine and 3) y =\. When x at which >+<, # = 0; 2) y = is

i

yl yl

=x =x

the

and and

points

t/

is

2

is

i/ 2

y+\.

1)

\\

101. Hint. See Appendix VI, Fig. 14. 102. Hint. 103. Hint. See Appendix VI, Fig. 17. 104. Hint. 17. 105. Hint. See Appendix VI, Fig. 18. 107. Hint. Hint. 18. 118. Hint. See Appendix VI, Fig. 12. 119.

Fig. Fig. Fig.

15.

Fig.

12.

120.

See See See See

Appendix VI, Appendix VI, Appendix VI, Appendix VI, Hint. See Appendix VI, Fig. 13. 121. Hint. See Appendix 132. Hint. See Appendix VI, Fig. 30. 133. Hint See Appendix VI, Hint. See Appendix VI, Fig. 31. 138. Hint. See Appendix VI, Hint. See Appendix VI, Fig. 28. 140. Hint. See Appendix VI,

VI, Fig.

13.

Fig. 32. Fig. 33.

134. 139.

Fig. 25.

141. Hint.

398

Form

a table of values:

Constructing the points (x, y) obtained, we get the desired curve (see Appendix VI, Fig. 7). (Here, the parameter t cannot be laid off geometrically!) 142. See Appendix VI, Fig. 19. 143. See Appendix VI, Fig. 27. 144. See Appendix VI, Fig. 29. 145. See Appendix VI, Fig. 22 150. See Appendix VI, x 2 It is 1^25 Fig. 28. 151. Hint. Solving the equation for y, we get now easy to construct the desired curve from the points. 153. See Appendix VI, Fig. 21. 156. See Appendix VI, Fig. 27. It is sufficient to construct

y=

the points

(x, y)

to the abscissas x

corresponding

.

= 0,

a.

,

157. Hint.

-^

*]

the equation for x, we have * 10 \ogy Whence we get the y points (x, y) of the sought-for curve, assigning to the ordinate y arbitrary values (*/>0) and calculating the abscissa x from the formula * Bear in oo as y -+ 0. 159. Hint. Passing to polar coordinates mind that log y -* (

Solving

.

(

haver=

cos 8

>

q>

+ sin

r=" x = 0.4

:2.9,

0002;

x

x

f)

= -2,

*2

2.5;

=i=

n

166.

= 0.86. =2

= 2, ^ = 5; * = 5, a) *! ^ = 2,^ = 3;^ = 3, y = 2; c) 2

A^=9; 0.0002.

n

a)

3.6,

4; b)

I.

172. 1.

169.

A^

c)

Xl

/i

>

=

N

logAr<

a)

2

Use the formula 182.

0.

192.

oo.

183.

193.

oo.

~~.

173.

!

+2

184. 2.

194.

174.

2

0.

3.1;

y,25s

A-

=.,

yi

10; c)

n

^=2.7, y z

2

=

^

2

= 2;

x,=--2,

^2

9;

Sit

**= 167.

32.

/i

T

^

>

e

W = 99;

b)

^

f/

4

168. 6

when

x>X(N); c)\f(x)\>N when \x\>X(N). 171.

8C

l,

i

163.

will

//

165. ;

^^=

d)

> -~ V

a) c)

=

32)

= rsincp, we 161. F = 32 + ^=

32)/

Fig. ft

e)

'

Appendix VI, Fig and

.),,-'

/2

f

when

y,

_1^ 2 1=JV.

=15

*);

^=-2;

*,=_;

x

VI,

Appendix KK

(

r = e? (see = rcosq,

have

will

coordinates

polar

x=1.50;

e)

we

,

(p

1M.

.1

b)

3

= 0.6* (10

162.

b)

tancp=~

Hint. Passing to 3sin cp cos cp

160.

d)

and

YX* -f- y*

r

}

1.

175. 3.

+i = -g 2

-f

.

.

.

185. 72. oo.

195.

186. 2.

~

196.

(

170.

1.

+ l) (2/i+ 3a a

-

188. 197.

a)

a)

002;

>N

when

(e
0<*<6(AO;

176.

187. 2.

=

0;

177.

b) 2*

b) 1; c) 2; d)

J

.

178.

j.

~

.

oU

Hint.

180.0. 181.

1).

179. 0.

oo.

189. 0. 190. 1. 191. 0.

3* 2

.

198.

-1.

199.

^-

1.

1

200. 3. 201.

i-

202.

.

.

-1

203.

204.

.

399

12.

205.

-- i

206.

.

207.

.

1.

-| 208.

=:. 209.

.

x 215. 0. 216.

a)

228.

A.

1.

4o

236.

217. 3.

b) 0.

224.

A

218.

225. cos

ji.

A:.

244.

245.

251.6.252.

e*.

ji.

i-

239.

a)

e" 1

Solution,

1.

-m

2

233.

).

a)

y

241.

1.

.

247.

e

*

=

lim (cos*)

2

I

221.

0;

lim [1

1.

1.

242.

-i 4

.

249.

"*.

cos x)]

(1

X-0

X-+0

.

1.

b)

234.

.

e" 1

248.

.

.

Z

227.

240.

.

Jt.

-

214.

4

246.

0.

2

I(rt

4

-|. 250.

238.

--JL

226.

232.

I.

237.

.

2 19. 4". 220. o

.

2

230. 0. 231.

Jt

243.

22'

---.

213.

-

2

~:j4=. 235.

212.

-

j/jt

sin a.

229.

211.

3

-^s\n2\ z

222. cos a. 223.

--

210.

.

2

3

2sin a

"m

=

Si nee lim V

=

2 lim

2-1- lim

=e

*

that lim (cos*)

1.

b)

*->o

(see

follows

it

*

the

in

preceding

lim

lim (cos x)*

a),

As

Solution.

"/==-. e

7=0,

V i

case

^n

v

X-+Q

=e

.

Since lim

X-+0

= _J_, 2

2 lim

:

= Put

lim (cos*)*

1.

256.

1.

--

257.

6*^1=0,

1.

where a-^0.

Hint. Put

263. a)

267. a) 0; b)

1.

1;

b)

268.

M

= a,

-i a)

.

259.

264. a) 1;

b)

a

-1;

b)

1.

258

'

K

Utilize the identity a

Ina. Hint. -*

where

2

=
Hint

-

z

e

260. In a

262.

that

*-).o

lOloge. 255.

253. In 2. 254.

.

y

follows

it

269.

(see

1.

a)

Example

265. a) 1;

b)

-1; 1.

b)

259) 1.

270. a)

= e na a

261.

266. a)

.

6.

1; b) 0.

oo; b)

+00.

400 271.

Solution.

but

if

x

=

when x=l;

whenx = 0;i/ =

l<x< + oo. 280

t

e

x=l

x>

when

when

y

276.

281.

.

kn (fe=0,

then cos 2

=

/

&

x

If

/Swi,

x>0. 277.

^g.

1

6=0;

H

1 1

compound c)

>

x\

|

1000. 289. |x

1|<0.005;

b) |x

= 0.01;

b) 6

c) 3.

293

299.

3.

6)

a)

300.

x |

;

1

1

~

kn, where k

316.

a)

/(0)

x+ A*

= n;

b)

|

|

x

\

\

when

^* 2^

-

-

b)

A:)

is

<,\

we

/(0)1;

310. a)

an integer I

c) /(0)

313.

= 2;

x

7=

301.

y

292.

a)

1.

= 0.1;

1;

|

2;

b)

298.

1.

Hint.

0.98(09804);

1)

d)

;

Take

~

6

3.167(3.1623)

c)

have

Hint.

< 0.05;

|

a)

15. 297.

296.

b) 4; c)

2;

309. |

.

,

3.875(3.8730);

4)

1

jx

1 ~

0.985(0.9849);

Ax

<| Ax

No

295.

a)

(

U|>100;

b)

-^-=6;

10.954(10.954).

d)

y^zT

curve

a)

2|<

|x

Second, b) third.

a)

the

|*|>10;

a)

,

\&x\<x

cos

285

=x

279.

si.

"

0<e
d) 2; e) 3.

when

(x-f- AJC)

j/

proportionality factor (law of

6

290.

0.0005

l^"x)<| Ajc|/V^x.

integer; b) x inequality

1

;

-5-

i

U|>

0.0095(0.00952);

inequality |cos

1

<

+1;

1

3)

Ax+

the

is

when

291.

c)

=

when x<0; # =

= 4" of

asymptote

0.043(0.04139). 303,

then

Ax|/(V A:+

288.

.

<~

1

x

y

7)

the

is

1.03(1 0296);

=3

> 0,r

|

-i

a)

072(0.7480);

1

= 0.001.

b)

1;

c)

1.03(1.0309);

If

=

6

c)

)/"lO= l/"9Tl 2)

Q t = Q
interest);

#

1;

-!

2

where k

)

0<x <

278.

lim <4C n n-Mo

284.

nj

v

y

* .

y=x

line

=

j/

x 2 -*oo.

fL

straight

Q/^^Qo

287.

274.

y=x

when

0=1 when 0<x
275.

+

o

e

the

= |x|.

*,-*---;

282.

4-

1

J/

cos*x
then

...),

272.

/=!.

273.

1

2,

1,

and

1.12(1.125);

5)

j

.

Hint.

307.

=

+kn, where

the

of

k

is

an

311. Hint. Take advantage of the l. 315. No >l=4. 314. f(0)

= 2; kind. 318. x = d) f (0)

e)

= = 0; / (0)

f)

/(0)

=

1.

317. x 2 is a discontinuity of the second 1 is a removable dis2 is a discontinuity of the second kind; x=2 is a removable continuity. 319. x= 320. x is is a discontinuity of the first kind. 321. a) x discontinuity a discontinuity of the second kind; b) x=0 is'a removable discontinuity. 322. x=0 is a remo vable discontinuity, x kn (k=\, 2, ..) are infinite discontinuities

=

=

=

323.

x=2jifcy

(fc

= 0,

.

1,

2,...)

are

infinite

discontinuities.

=

x^kn (=0, is a 1, 2, ...) are infinite discontinuities. 325. x 1 is a removable discontinuity; discontinuity of the first kind. 326. x 1 is a point of discontiauity of the first kind. 327. x 1 is a discon-

324.

x=

=

=

401

x=0

is a removable discontinuity. 329. x=l tinuity of the second kind. 328. of the first kind. 330. x 3 is a discontinuity of the first is a discontinuity is a 1 kind. 332. discontinuity of the first kind. 333. The function is continuous. 334. a) is a discontinuity of the first kind; b) the function kn (k is integral) are discontinuities of the first kind. is continuous; c) x 335. a) k (k is integral) are discontinuities of the first kind; b) x is integral) are points of discontinuity of the first kind. 337. No, since (k the function y E(x) is discontinuous at x=l. 338. 1.53. 339. Hint. Show that when * is sufficiently large, we have P ( x ) P (x ) <0.

=

x=

x=0

=

=

x~k

=

II

Chapter 341. 344.

3;

a) a)

0.21;

b)

1560;

624;

2ft

c)

+A

0.01;

b)

2

342.

.

0.1;

a)

100; c)

3;

b)

c)

0.000011. 345. a) abx; b) 3x*&x

1;

-.;

2*(2-_l)

e)

g^'-i); A*

;

jt+Ax-f-J^x

2;

354.

a)

~-

lim

A!?

A/-H)

At

=

lim

*

"^ d)

Q

^ =E =

1

AJ^-^O

sin 2 x

a)

12 "~

c)

]im f(x A*-*O

the

-

L^

0.16;

/ (x)

time

at

time

t.

at

time

t.

A =^

b)

t.

at

substance

6

21 -

Ax

temperature of

347>

+ Ax)

turn

of

angle

quantity

0238;

21 2

357.

Ax cos x cos

2

"

is

"" h;

^

(x)

f

the

is

l;

_ _ 2

"

35l

0.25.

= J^- = sec x. cos x (x+ Ax) 1 Solution. 359.

-.

T

the

is

356.

Ax

lim Ax->o cos x cos

cp

where

,

Ax

0.249;

lim

_

lim AJC-M)

201

'=

where

Ar

-

>

Ax

where

,

b)

l;

/(*+Ax)~ /(x) 2,

lim Af-M>

b)

\

Ax

^

c)

X

b)

;

rX = dt

355.

dt

_

353. a)

JP=

b)

Ar

~~

a)

IT)

348. 15 cm/sec. 349. 7.5. 350. 352. a)

346>

*

AX

+

-

c)

(x-f-

Ax)

a)

3x 2

358.

n8)=

;

Solution.

sec x.

AX b)

lim

Ax

*o

-4 x* ^ (8

+

c>

;

7=

2 V^

A ^~^

v

(8)

Ax

A*-M>

+ Ax 8 = hm Ax AA:^O Ax 8 8 [ J/ (8 + Ax) + ^/ (8 + Ax) + j/ ] = Hm = 1. 360. r(0) = -8, r(l) = 0, = 12 Ax-^o 3/(8+ Ax) + 2^/8 + Ax+4 = = = For the given function the equation x Hint. 3. 0. 361. 0, x, /'(2) = x 362. 30m/sec. 2. 364. 363. /'(*)=:/ (x) has the form 3x = 365. nx ^. 366 L 2 tanq>=3. Hint. Use the results of Example 3 8

..

.

_

2

_-

1

2

,

2

2

8

-

)

xo

and Problem

1.

1,

.

365.

3

367. Solution, a)

f'(Q)^ lim

Ll

!_=

=

lim

Ax AJC+OJ^/

AX

4- oo;

402

1^15^= Ax

/'(!)= lim

b)

r

=

lim

=

2fe+l

lim

sinAxl

l

1

=1.

Ax

A*->+o 5x

*+A

2

,

2

371.

'

,

,

3x

-l

/

.

+n-1

6ax

= x x* = x

9y2 ^ A

sin/.

y'

= 0.

390.

sin 2

xe x

391.

xV(x-f-7). x 2 e x 396. e x

sin x). 395.

^-x

cot x

387.

.

/^arc sin

.

\

3x 2 lnx.

398.

402

2

2

_ tanh2x

403

toco.h^dnh* cosh x 2

2x 2 405.

y

x 407.

x 411.

x

,

Vx

12fl6+18^

1

-|-2x arc

e

2

(1x 412.

//.

x

K

fex2 .

>X

16x(3

+ 2x

2 8 )

"~

*

420. 2

419. 2

2 sin x

V

15 cos

xln 2

sin

x sinx.

_2cos^

*

sin>

3cosx)

arc sin

....

+

414.

^

2(l+Jt )V^arctanjc

_3jari!*)l /"l

x*

/

.

8

x2

/I 1-tan

2

x

+ tan

4

x

COS 2 X

.Hint. x 3 cos x

^

cos*x

Ssin^^

c

\

I)

418.

.

I

I

c

.

y.

2 3a fax-}-b\ fax b \

424>

3

cos 4 x

L !

1

-f-

tanhx

421.^^^---

423> (1

427.

xsmh 2 x

sin 8 2/

422.

425.

2

cot x

x

+ x coshx.

sinh x

^

)

2

-3(xln* + sinh*cosM

413.

.

In

401.

.

*~ 1) .

x /

1

.

x

* (21n

397.

.

2 2 )

-.1/j/l;-!. X f

416.

8 2

^

2

v5

*>y*

393.

1

(2x

41B

O

y

^f-.

.

l

"

X2

j/~i

.

2

arcsinx-f-

;

*\ arc coshx

2

2

cos x)

(sinx

arc sinh x

406.

.

2

I)

384.

2x

388.

. '

4()4 1

X

*

1 ^1 x In 10 x

1+1H/-A.400. x x x

399.

on

.

2

.

4***

x 2 (2x

392.

x+

wy 1*

2

-

383. sin

386.

394. e* (cos x

3 sinx.

2)

T

380. OOU.

.

+ 5)

5x

377.

.

!

2

5 cos x

8

fW-4-9ft VJA -7- z-o

(x

ji

374.

JL

z

Hint. y

.

.

5

373.

.

i

9

2x 8 370.

3

2

2. 382.

7=

V

x arc tan x.

389.

rn

/2*-fl

,

l + 2x

369.

l

2 (1 2

+ 2.

o*7A OIS7. 379.

.

r,

|

1 x

376. -*

+ dx)

(c

12x 2

8 .

I

+ b(m-\-n)t

U*U.

.

2

Y 385.

3x- 4

l/x r^

1.

2

5x

/'-

c)

sin Ax ~ L^iJLJ=-l; ,.

lim

5x 4

368.

-JL 2x

375.

==00;

\\

*JL '=

mat m

372.

.

^_

lim

= sin- + 2

1

+ 2 sinx

2 V^lSsinx

lOcosx

1 '

2 1^1 > '

x 2 >^1-|- arc sinx

^ '

2

(1

+ x ) (arc tan x)

2

403

429.

X

430.

.

cos(x

5

-.

5x+l)

2

2*

x

1)

-}-

asin(ax +

433.

(2x-5) X

4 32.

,

3

434.

p).

sin(2/+q>).

X 2 COS 2 X

""*

?21i

2

435.

436.

.

x

sin

n2

s

x cos 2x 2 sin 3x 2

437.

.

^

.

438.

.

Solution.

a l

(2x)'

,-

2

V^l

"~ 2

2

=

-

lA

(2x)

439.

.

4x

2

x/V

2

442.

10xe~*

443.

-

-p^jf

T

440.

446.

sin2

+2

^

450.

.

f

t /

cos 2 In

L

453. (1

-j-

+

In x) x

3

455. Solution, y' 3 H- sin

5x2 cos 4x

456.

+

(1 -j-x

(sin

)

5x)' cos

2 -

\

sin --

f

tanx x 6

10).

cot x log

4 arc sin x)

-}-5sinx

e.

VI

+

454.

.

x\

x f -^

449.

.

]

arc

+ x In

(1

2

-

x (e

J

+x

(

452.

.

xlnx

x

1 .

1

2xi0 2 *

448.

^^

451.

x2

1

x

441.

2

2

447.

2.

.

445.

In 5.

pX t

"~

1

2 J^jc

1

2*5~* f

444.

.

.

2x \ In x-jx \ 2

1

'

+ sin

3

/

5x

I

cos 2

= 15 sin

= 3 sin

-=-

6 J)

\

x2~sin

5xcos5xcos 2

2

x

5x cos 5x5 cos 2

-

xx 3

8

5xcos

-r-

sin --

.

)

x 2 -f4x

3 457.

6

x

x' .

458.

1

459.

.

463.

.

x

4x 3 (a

465.

2x 3 ) (a

5x 3 ).

474. sin 3 x cos 2 x.

sin

479. .

4

476.

x cos 4 x

10tan5xsec 2

-

3cos*cos2*. 480. tan'*. 481.

-- 1

.

484.

sin

arc sin x (2 arc cos x -

r

2

-f

]/"l

x2

4

. '

478.

.

3/

1

sin

2/.

=

482 .

.

x

arc sin x)

xcosx 2

477.

5x.

2 . OB 485.

K a

x

sin 2

x

+p

2

AQC

-

V^x

cos 2 x

.

2

486.

'

1+x 2

1

V

487.

491.

,

1^2x

xa

492.

arc sin

,

...

V

1

25x 2 arc

sin

5x

_

404

,

494.

- -

495.

.

.

1+cos 2 * a w *lna. 502. f

e-*cos3x. 505.

504.

(1

Psinhpjf)'. 539.

542.

/?*

499..

2

506.

^

548.

-

l(/tanx

-^-

544.

.

545.

.

sinx

/

y+

I1^3 -. 5

V (0)= =^;

/'+

(0)

tion.

or

567

do not

We

sin ' x .

+ l/Tosxln a).

1 ,' =

549.

.

when x

1

have

c)

6n. 554.

553.

/I (0)=

exist. 555.

y'

xy'=y(\x)

= e" x (1 566. *

/_

/+

d) /'_ (0)

=/'+

(0)

+ (0)-0;

1,

1

a)

/'

x.

556. 2

+ ^-.

=^

557.

(0)=1; (0)

< 0;

y' (0)

does

I

f'W-

= -1,

/

xarctanhx

546.

.

=

y'

550.

^L

x2

1

{ 552.

(1

2coth2x. 541.

540.

y'=l when x>0;

a)

'-,*,.

b)

exist;

tanh 2 2x)

(1

cos2x

1

547. xarcsinhx.

1

sin 2xe

500.

x p sin pO- 503. e" sin p*.

e*(acos p/

2xMna).

+ 4 sin*

_)/.

6tanh 2 2x 543.

x y In 2 x

not

5

"""

1+*-

+

^-'a-^fn

496.

.

2xcosa + xt

1

= 0, 1.

b)

/.

(0)

e) /'. (0)

558.

=

2 ,

and

561. Solu-

= -(lx) (1 +2x)(l +3x) + 2(l+x) (1 +3x) + 3(x + l) (1 +2x). ~ x).

Since e' x

t

it

follows that y' 2

40S

-24 573 '

>-

574.

.

'n*.

575. *

.

,

2!li

+ cos x Inx

/

*

SIn

578.

.

576.

'

- sin * tan x)

(cos * In cos *

(cos x)

)

J

58

-

'

581.

r

"

.-- 4'<(2 ""
586.

.

-

590.

594. tan^. 596.

601.

tity.

.

,;

603.

2

-. 2

Yes, since the equality 604.

-

2j/)

arc tan 2

=^63

(-2, -12).

45.

623.

26'.

626. (1,

-3).

620

.

624.

627.

arc tan

.

b) d)

y

y-5 = 0;

^2y = x

x

+ 2 = 0.

l=0;

34 = 0. 13

= 0,

GA;

/

(3,0):

t/

2x

y=l

1;

x\

616.

xy

a) 0; b)

~^r 36

1=0; y = 0.

x

the

for

625.

21'.

633.

(0,20);

= 0;

+

6x Jt 2// x 2f/-fl=0

f-i o

point

(

1,

1).

6;

;

x

+

t/

2

= 0.

at the point (2, 0): (/=

639.

14;c

638.

634.

At

x-f-2; y

(1,

15);

-1^ lo

\

2^

for the point (1, 1);

7x

= 0. 637.

45-.

y=2

a)

635.

5y-f 21=0.

622.

0.

c)

;

+y

2

= **- *+ 1.628. k = ^.629.

+ y2 = 0; c) 3 = 0; e) 2x + y

1=0

= -Ilf

= 2x

632.

3 cosy

10

11

631.

r 10

x

615.

y,

1

6 08

*

8

an iden

is

-

605.

.

_=_

\nx-xx

-*-.

593.

2

y

-.

J

l-f3ju/ -f4(/

yVx*+y*

-

589.

f.

a

={

599. No. 600.

oo.

-.

602.

5

ex

tan

~

-tan^.592.

597.

1.

fir-*- ra.

588.

W.

l

a);

+x*)arctanx

585.

(arc tan x)X

-

=x

0;

1

636.

the 2\

point

(1,0):

at the point

41

=0.

-406

640.

The equation

Hint.

of the

tangent

Hence, the tangent

^-{-^-=1. 2y Q

is

ZXQ

crosses the x-axis at the point A (2x 0) and the y-axis at B (0, 2f/ the midpoint of AB, we get the point (* y Q ). 643. 40 36'. 644. at the at bolas are and intersect point (0, 0) tangent

The para-

,

arctan-i=^88' 648

652

'

'

the

at

= 2jia

/1+4JT 2 t

--

~.

arc tan

=a fl

-^-in/sec.

x

2

The

.

2u g/sina

+g

To determine system. The range on

velocity

/

the

;

Sf =

'

S,

= a;

cm /sec;

0;

656.

2jt.

q>

.

is

the

of

slope

3

657.

-

The equation

660.

range

2 8

5,,= -^;

of the trajectory *

sin

\sy=x tan a

2a

The

.

vector

velocity

cm /sec

9

is

velocity,

cosa

v

axes

the

~ '

~

are

and

The magnitude

j-

the velocit y vector

of

the velocity

CO8

1

cm

40

of

he 2

360

at

B

the

g,

density

x2

1

v

681.

y

5* g/crn,

is

+~~210,Y

4 .

+

=

^

u = 5;

x*) .

A

2

671

+ 2).

i*

cosh

682.

a vl //

=

6 79. y"'

= 6.

680.

a

~64sin2x

684. 0;

1; 2;

=

2.

of the rod

the density 2 cos 2x 669.

2 arc tan x

672.

.

f

is 0,

X

674 (1

The mass

666.

the density at x* 668. e (4^

2

3(1

_

y cm/sec

m'/'sec. 665.

56x 6

667.

60g/cm. 2(1 ~"* 2) .

673.

JT

M

at

is

670.

Q ,

- '6. 8 cm/sec, the area, at a rate diagonal increases at a rale ol 2 664. The surface area increases at a rate of 2ji m /sec, /sec

the volume, at a rate of 0.05 is

is

directed along the tangent to the

is

~ 663.

.

the trajectory, eliminate the parameter t from the given is the abscissa of the point A (Fig. 17). The projections

Hint.

of

=

655

'

I

tan|i=

2_|_p2.

2

uj

tan^i

;

u

g

angle

S,

-y +2(p

2

}/~l-|-4ji

2

658. 15 cm/sec. 659.

K

654.

--

rt

;

~ n -~

>

647.

(1,1).

an

T

'

653.

*

point

Finding

).

,

685.

+

/'"(3)

The velocity

The acceleration, a 0; 0.6. 686. The law 0.006; the is x acoscof; the velocity at time / is point M, 2 acD cos(of. Initial velocity, 0; awsincof; the acceleration at time / is aw 2 velocity when * initial acceleration: is -aco; acceleration when The maximum absolute value of velocity is aa>; the maximum * is 0. 2 n (n) (n+l value of acceleration absolute is aw 687. y n\a 688. a) n\ (1 \ x)~ is

of

4997; motion of

4.7.

=

;

=

.

b)

(

_ ir +

~

i .

689

.

= =

.

407

d)

2 n jc

691

2

^(OJ^-tn

692.

1)1

a) 9/

8 ;

b) 2/

2

+ 2;

XU+*-);

b)

sin

693. a)

---. $< - - =. 696. . 708. -o 711. x=\ 718. 721. 725. ^b) 3 a) and = AJC No. y 719. dy= 45 ^~^- 2 =l- 697. f- .712. Ar/ 714 d// = - cf(/ = 0.009001; =. A: M+ 2 (AA') 720. . 7,0. . x d(/ 8 ) = when 717. For * . . 16 713. cf(l ^5 = 2^Ax, A5 = 2^ Ax + 722. "J 256 // d/-0.009. ^=^-0.0436. 0.0698. ~ 1 ' 709. 2 2 ft-4-. ' v 726. . -j; 2 4o 1 4 " 707. 9 694 a > 0; b) 2e>al 695 - a) 777 4a /"T^?. c) = 0. = ;=i 0.00037. 723. I 2xe~**dx. 727. \nxdx. 728. x y*xe 735. 738. 740. ~dx. IJ 737. a) 0.485; b) 0.965; c) 1.2; d) 0.045; e) 5"=^ 2. 25; 8 739. 565_cm ^^4.13; ^70^=8.38; 200=5=5. 85. 741. a) 5; /70=^4.13; J/ ^/10=^2.16; . """ d) 0.9. 742. 1.0019. 743. 0.57. 744. 2.03. 748. ~(^) sin,ln, + fc). 753. 2 . -.754. (d **' iy (1 750. ~ + 0. 025^0 ^)*/i ~ . ^640=^ 25 b) 749. 3.2" sin 3. l.l;c)0.93; *( dx ? . (1 (dx)*. 752. 751. 81. X2 ) -e /a __ Answers 408 e 755. xcosa sin(*sina + na)-(d*) 758. No. The point 763. (2, + 765. 4). 2( 3U' C<8,<1. 0 0<6< c* 770. s sin - ~ error ^ / x x ( _ = H l \ x \ (1 -- / I 2!L_ 8 = 77:. get: f x \ , 2 -f where |2 where J,, = 8 *, 2 ?-6*, n 775. = both cases g We Solution. 6jc; have factors in powers of #, e in __L ^ 2 ^ + -_x -- x. 1 1 I , 1 ^1-f-) i ; 40 Expanding both . J . = 0. ~ sin x-^- + sin g 2 , -L . & _i_ * ( [ we *= 1 than =7 5! i 762. x=(x- l)-i (*-l) - A b)' ; * less is exist. ( j_ a In not function. 769. sin 1. ~ + ^- 3 The 773. 768. 0<0< x = x- the of does (2) /' =i+x + J+|J. + ...+ ^lL.+5jeS 772. Error: a) 1. b)g = -~. 1 1 since No, discontinuity |= +6 (x- 1), ^ = 8,*, 0<9 <1; where 757. . a is =^; a) where 1)8> x = ~x /l I ; y . -^~^1H //T" , y --- -x \ 8 1 x J y* ^~9~2- Tne n expanding x x powers 778 oo 788. 1. 1 we , 779. ~. 788. same polynomial get the 780. 3. 1 n for 802 of 789. 793. 0. ~ 781. 790*. 1. 795. -1 - 804. 1 Find lini , -^ u ~>0 791. a. 0. 796. 805. . where S ~ 2 hh 783. 792. 797. . 806. oo 1. IT . sin a) (a 807. > n 799. 1. 1; a n for 800. 808. 1. the exact is . 784. 0. 785. oo. for 1 ^77. 810. 1. expression for = 801. e*. e e R2 = -~- x x =^rlH---r-n~-2 \2 e S a 782. 5. o 803. e the l; 1. Hint. area 3"^ of the segment (R is the radius of the corresponding circle). Chapter (00, III 812. ( increases; (2, oo), decreases. oo, 2), decreases; 813. 814. increases. (2, oo), increases. oo), 0) and (2, oo), increases; (0, 2), decreases 815. 2) and (2, oo), decreases. 816. 1), 817. increases; (1, oo), decreases. 2), 8) and (8, oo), decreases. 818. (0, 1), decreases; (1, oo), increases. 819. ( and (1, oo), in00, 1) 811. 2), (00, (00, (00, creases; creases; ( 1, f , 1), decreases oo 820. increases. 822. j, ( (00, (00, (2, 00, (2, oo), 0), increases increases. 823. 821. ( (0, oo,2), ], de- decreases; 409 Answers (0, 824. increases. oo), (2, oo), increases decreases; (1, 1), and 00, a) ( (a, = *=- when j 0) and 828. No (00, 825. decreases.' oo), 827. ymax . = 1296 when x = 6. extremum. 830. t/m in = when x=Q; t/ mln =0 when x= 12; t/max when x= 1; t/ mfn =^ 0.05 when 0.76 when x=5=0.23; i/ max = 831. t/min^ 2 833. #max = 832. No extremum. x=^1.43. No extremum when x = 2. - when *=0; |/ = 3/3 when x = i/ 840. [/ n =2 x= when 837. j/ 2, = 845. y mln when x = 2 value is m= when x = Af = 75- ^= 12 = for x l fjkJ:^ 1 c f/ = ~;// min ==0 x min for when x = 1; j- (fc^O, for x 2, 1, is is (2fe + r ". (fe-^0, 1) JT when x = 844. f/ m?n -. -l when 4 when x = 0; = No extremum. M = 7rwhen value, = 5. //max--2 m^=0 when 852. 1; M = 27 when x = 3. m = 1579 when x = Each side whose x 863> -f- 1) = l; M=JI 854. -j- when m- a) ; 6 M = 3745 when be equal to -- 10; 864. The Isosceles. 865. m 850. \. The side The altitude must be 866. . x terms must of the Smallest 849. when x = (2fc m=~ 851. must be twice the other of tlit half the -g- altitude -^L y3 the radius of the given sphere. R mln -7-- Altitude of the cylinder, where f/ =6 = -- x=\ square with side a cut-out square must be equal to 867. That x 848. ...). b) 861. 4. <7 side adjoining the wall 868. 846. = g when x=l. x~ greatest = 10; M = 5 2, when 842. y mln = /2 s/"3 when y mta = 839. max when x = 2K K^3" x=1 when 1. = y min r/ max = *, The rectangle must be base. = 0. x t/ 5 when y min=l Ji; 835. 836. ^==^ 9rr ^ n; ^/max^^cos l when 853. /n=s when x=l; M=-^2o6 when x = 5; 862. when fcjT p= when x = 3.2. yg l x== _l. 856. =9 r x= x=12. #max = 3/3" min ymln ==0when 2 and x f/ (/ = 847. 834. = V"3 when x = 2^3; *=1; ma x = when x = w ^n t/max^^- =2 max 841. ...). x ^L; yO *=12 Scos^when 843. when m ax-= 5 when 1, x i mm -0 when 838. = m is ; to equal radius 869. the radius of the given sphere. of the its Altitude diameter base of R ]/ the of the base ro , where cylinder, 870. Altitude of the cone, i< __ RV'2 -- 410 __ Answers 4 where R is the radius of the given sphere. where R is the radius of the given cone -jr-r, 3 where r 871. Altitude of the cone, Radius 872. sphere. 876. . = The altitude of the cylindrical + *yo = 1. ^* XQ -^- 878. 879. The the vertices of the rectangle which / 3 \ -j J. 1 881. ~Y=., ( arc cos 4- 885. a) and t-y-. P min = y%aqQ. spheres, the impact with m*V +w b),= =; If a = n= - "I/ (if r this ^ is battery , number ( the physical points ^"3") points 3), oo, 6) 2, J, . *. ^_ 884. . 886. not an integer or is meaning M, M lj2 concave up; ...); oo, -4r m ,= obtained is as as close as possible to the concave 2), 0), ( (4/e +3) 0) 1/~3, ( (|/* 0) and -^, (4 Af 2 ( 6, down; 0(0, + 5)~ l)it, 2&Ji), f(2fe+l)y, concave down(fe=0, sas of the points of inflection are equal to xkn. 0). , 895. |-V and (1^3, points of inflection, concave up; ((2k not a divisor of (2, oo), 892. (6, --|Vo(0, ^3), concave up; (0, is of the solution must be of inflection of P ^L inflection. (00, oo), concave up. 12), point concave down, ( 3, oo), concave up; no points of inflection. and (0, 6), concave up; ( 6, 0) and (6, oo), concave down; inflection and -jr- 1, M (2, oo, of I/ ^M- a divisor of TV). Since the internal resistance is = ( 2 (-~-a; of l follows: the internal resistance of the battery 2 external resistance. 889. y h. 891. ( o 893. 894. coordinates \f~Mm. Hint. For a completely elastic impact of two imparted to the stationary sphere of mass m after sphere_of mass m 2 moving with velocity v is equal to v concave up; where 887. 2 the 6J/~2, 8 equal to the greatest of the numbers AM=a 883. 7V,we take the closest integer which of is The d velocity 888. . ^ 880. that zero; = \l* a^Tand on the parabola lie The angle 882. arc tan of the ellipse. the crossthe sector is, must be part sides of the rectangle are a and b are the respective semiaxes /HI jr, that angle of the vessel should be in the shape of a hemisphere. 877. h is, R, of the base of the the radius of the base of the given cylinder. 873. That is whose altitude is twice the diameter of the sphere. 874. q> section of the channel is a semicircle. 875. The central /~2 -^ oo), 896. concave down J oV (00, concave down; (& = 897. (2/m, 1, 898. i2, [0, \ ...); ^ the abscis), V&J concave Answers [7^, down; ; 899. oo V/> concave M up; 0), concave up; 900. (00, 3) _JL ^= f (0, oo), and (1, ) W) (Y* J oo, ( inflection. Y 411 a point of inflection. is concave down; 0(0, 0) concave up; (3, a point is oo), of concave 1), = 901. of inflection are M ( 3, T and M ( = = = 0. 902. *=1, x^=3, = x= 903. 904. x. 2, y= *, 905.0 = right 907. jc-= = x, right. 906. = left, left, y=-x, left, = 2, left, y-=2x 2, right. 909. 0^2 910. jt-=0. = *, right 908. 911. * = 0, 0=1. 912. i/^O. 913. *-=!. 0=1, left, = 0, right. = x n, left; y = + right. 915. y a. 916. ma x = when x 0; 914. 4 when x = 2\ point of inflection, Af,(l, ~~ 2 917 ^/max^ when min = r = 0; points of inflection Ai l|f when jf=V 3; m ~) == 4 when x = 918. y min = Q when je=l, point of inflection, M, (0, 2). m ax 919- f/max^ 8 wnen ^ = 2, m n = when = 2; point of inflection^ M (0, 4). when A'^0; of inflection 920. M lt2 (Y5, 0) and points //min^ down; points A' 1, 2 l J. j 1. iy 1, 1, l, y JC JT, )- r=: in 1 - jc 1, f 1; // Jt i ! \, totes, x x 923. 924. (farm x 0=-xl. ~ 1, = 0. = 0. = ~ when 1 \ Ai J2/"3, lf of point = --J; x Ai(5, \ of 0(0, inflection, jc^= *--- 1; 0min and /^(4, B (8, A ( and 4) 3, Q V 940. + 938. = 0; m n= x s 2 x asymptotes, x x 0, =4 = when x = V% 0max= 935. 2. when jg= ^nax^ 936f ~YH max = 4 when x = of points = 0. when Asymptotes, 941. x = 3. jc= 1; 1; 4; max j ym n =y \ 942. 2 and max \\ 4r J 0. y =l when x-^4; 929. Point 0. --97 930. nnx = 4 when max = '16 , 931. = Q and =2 f/ = 3v 933. ;c 0(0, /I 0). 932. (8, 934. A (0, End-point, A(Y$* 0), 0(0. 0) inflection, M (1^3 -f 2 f^J,

point of

M

l

(Q

x t

m = M 1>2 (1,

= 0,

points

of

and M f (1, (when x = 0)

0);

1)

1

in

~~

m

i

n

=2

= 0,

2)

and

4)

End-points,

when

1

-

inflection,

asymptote, 939. max -=2

K 2); asymptote, 4 wh'en x==4; point of inflection, 0(0, ~~ ^ 3 / 4 4 when x when x 2, 0mi n

inflection,

when x =

J

'

and

= x

asymptotes, r

asymptote,

0);

M lt (\ \ x--2 and

= 0.

and

2

and

937. Points of inflection,

0).

asymptote,

943.

0);

l

V M, t(l, when

2

=2

/

x~\\

vvnen

m in^=

0);

x.

=4

928.

2 V 2 when 2) are end-points; ma x are end-points. Point of inflection,

B(Y$* 0=^ 0); asymptotes, -; = =- = points of inflection, 0(0, 0) asymptotes, x 77-]; y T2); asymptote O. asymptote, AT 1/2, when x--0; asymptotes, asymptote, 8 when when x-=2; asymp- inflection, m jx ==l whenx=l; -^)'. inflection, M( of points 2 926. # ma x-~ ^2 mm inflection, = 0, * min (l, f/ of /y inflection M, lt 4 when i=l; of 1; point wnen 0max~'Qo 925. asymptote, and Points 922. when jc=l; when x^O; 2 ^max^ - when x ^ max =3 927. y m n 921 - = when 944. x = 0; 0min^ = 4; =K asymptote, T7= 1/2 0. 0); when x=2. Answers 412 J/max 1/~JT = TT= when V 2 -o- M of inflection, of M, x 3a, ( = 4; 949. 6 =2 rnax of point /min= 953. =g min f/ y = 2 2, I of = 0; M when x = g; *- as 1+0 x -- ,0 jc points = j n+ = 2Jfeji; t/ ( ~+ fare cos On the *=n; M, 4 nal 2fen ( [0, (fe = 0, = 0, # = 1; n + kn, -j- ~^-J+2fejx, m in = 2 20, 2it], ji, [ ...); i/ max = 1 1 = when 2A5Ji (fc t/ max = Odd of / . when of x= V Asymptote, x oo). 2; points defined on \ max the = 2, ...); function with 1, Periodic inflection, ~; q = V$

M k (kn,

M

/

l

2

ml n=0.71

f

when * =

86;

^=0.37); J

e

when and

0)

961. Periodic function with period

j

1;

956.

m in =

(0.57,

periodic function with* period

when x = 0;

^=

2n. y m n

inflection,

*

x=

= 0,

/

39

# = 0.

"""472)-

1

x (as

period

j Ji-f2/5Ji;

,

-y

not

is

1

asymptote, A:-

1^0.63;

with

960.

points

yg )^l5j.

962.

jc

the function

*=

when

when ^ = 0; points 0.95).

when

max

and

(-rr==

~V,

,

(e

t/

^e ^ 7

=

*

M

= 0.

2

when

74

e

2,

ji],

-

J.

=

Oj.

_

1,

950.

1,

y m{n =

j+

wn en y

asymptote,

);

2

f\ e

function

f

interval i/

#

inflection,

2

+

^3

y mln ==

2ji.

Ai

inflection,

of

948. f/max 1111162

g

f lf 2

= -F ^0-54

max = ^2" when A:=

Mk

of inflection,

= 0.

asymptotes, x=*l. (when x-+ oo) and y

,

o

period

max

Points

'

~\

M

inflection,

959. Periodic

.

point

1.33);

=

Asymptotes,

when

Nk

of

when x=l;

inflection,

(limiting end-point). 955.

M

interval

x=

f/

inflection, 1>2 (1.89, 957. Asymptotes, r/ Q.

958.

of

point

point

point

asymptotes,

point of

0)

when x = 6;

/2 max =

t/max^

0.70);

0(0,

3 = -0-7=

947.

M

\

-n-l,

J

,

951.

^14.39,

954.

D

*/

M^J

^ = -4=.,

wnen

x-*0.

946.

points of inflection,

(e"'*

i

3,

[

^

/ 8 ^ 2 i/"~ ^2

when ^ = 0.

=

m

asymptote, y

;

)

inflection,

>:

j-

a,

2 (

/

Af,

= Q.

asymptote, y

;

)

M

and

)

\

-j-

when * = 0;

=

-^

of

/

w min

when

y-+Q

y

/iooy

inflection,

952.

945.

j^r=V, asymptote, x = 2

when

jc=l;

when

x=l

/12,

points

//

/min

5-

points of inflection,

I

V

M

inflection,

3

asymptotes,

;

J ')

Q

f

=

\

(3 3,

*

*.

2;i.

2

2xc.

when

0.13) and

On

inter-

i/max=l when

=

*

Ji

T

^nin

'

=

When * =

1

=l

M

2

(1.21,

when

= 2n;

M,(2.36,

0.86);

A4 e (5.50, 0).

*

of

points

with

function

when * = 3

asymptotes, x=-r-Ji

Even

965.

= c

^+

Mk

inflection,

+ ^Jt kn,

^min

(

skill:?, function

A:

= arccos

^zr

;

(/

=

lO^ with m

n

i

M

;

J^

(jt^arcsin

1

On

2it.

x

the

when x^Ji;

0(0,

x

0);

M k (kn, function,

//

and

Odd

/

m in =

n + 1 +2^n when

*-*+oo).

inflection (centre of 974. Odd

Oj

;

Even

966.

-

when

^min=--F-

n]

t/

(/j

ma x

M!

t/max^

f

-y

,

Oj;

-i/g)-

0,

2,

1,

57

1

Odd

969.

when

...). A:

=

function.

1

t/

JI

A;

(/

+-fr

,

+ oo). Point of inflection, + ?JT when max = -^-

when

972.

(/

m

JC==1;

symmetry)

|

of

points

/JJT;

2,

1,

(as

(

l

inflection,

1

2fcrc);

1

n

=

when x =

^max=

(0, Ji);

"

1

-^-^

--

1

asymptotes,

...).

(as

971.

x-*

Even and

oo)

(node); asymptote, y

Whe "

X=S ~" I;

y = x + 2n

(left)

=

P oint

and

l.

J

y=x

= 1.856 when -= M of 0, inflection, + (when asymptotes, point -^y and # = * In2. (as ^-^+00). 975. Asymptotes, # = oo) and y = -^

(right).

l;

function.

t/

m in=1.285 when f

;

j

^-^

(1,

function.

[0,

^= when

O

2k 4-

=^.^1

#==

q

oo)

970.

j JI+/JJT.

vvhen

0.34).

= -J + 1 asymptotes, x^-y -^ (6 = 0, = 0; asymptotes, = t/mln^ when x

+ /en;

-j-

1,

= -

M

of

points

interval

inflection,

57)

1

t

(

n;

...);

1

[0,

fejt)

(fcjt,

'

asymptotes, x\.

2,

1,

the

;

of

points

(2

graph

= 0,

^-)

,

when

\

inflection,

interval

function. Points of inflection, M* Even function. End-points, 4, 2 83, Af J(1.54 (cusp); points of inflection, lf of

0.86);

(- (-/I)-=

-.(/S-i/S)' Odd points

of

-=.

= arccos

967. 968.

Limiting

(4.35,

5

= 0whenx-n;t/ mln =

8

when

=

max

0.86);

period

On

when

1

Ai^O.36,

asymptotes, *

...);

2n

wnen ^==0; points

period

= -F=

(fc

=

#min

V~2 ~ ym n = L

2n.

+ 2foi

2,

1,

period

t/

arccos-^;

= arccos --7=)'

(arc periodic

with

when ^ =

rr^s

Af t

= 0,

(fc

-y- J

5

M

0.86);

period

ji

5

f

inflection,

964. Periodic function with

function

periodic

*=

wnen

l

M 4 (3.51,

0);

Periodic

963.

= ^- 7

ymax

;

413

JC=1;

t/

t/

max

ji

414 976.

j/

*=1;

m n =1.32 when i

2jt.

period

t/

=

min

asymptote, x

when

x=^-ji

-g-

= 0.

+ 2&JT;

t/

...); points of inflection,

2,

1,

+

arc sin

and

^- +

2;ri

i/^ + 2/m, UlA

M^ Urcsin

2

wnen * =

max ^=e

/

= 0,

(fc

with

977. Periodic function

2

e

I

g

978. (2fe+l)n,e 4(0, 1) End-points, ^ B(\, 4.81). Point of inflection, M(0.28, 1.74). 979. Points of inflection, 4.81 (as *-*-|-oo). 0.21 (as x -+ (0.5, 1.59); asymptotes, # oo) and r/ 980. The domain of definition of the function is the set of intervals (2kn t where fc Periodic function with period 2ji. ... 0, 1, 2, 2/jJt-f-ji),

A^ and

I

=

M

=

t/

when * =

=

max

The

981.

domain

of

Points

-^-

+

=

(/e

r/=1.57;

4

lf

2,

1,

2/jJi

i/-^ 1

y mln

x

=

as

-*

x

Domain

of

(limiting

period

=

+ kn.

i/

min

Limiting

asymptotes,

...);

definition,

increasing

<~

2/jn

]x

|

y mm =l when

2jr

984.

Asymptote,

985.

End-points,

^ 0.69

-

2ji.

period

monotonic

0;

with

986.

987.

>

end-point).

when ^^0.

^-^ +

2,

1,

definition,

with

function

Periodic

x

kn.

of intervals

= 0,

asymptotes,

...);

x

asymptotes,

...);

the set

function

Periodic

1.57

1

(fe

983.

as

-+

0)

(2fcji,

0.

2,

p0.37; y

integer.

x

1,

57);

an

is

of

...).

= 0,

,(1.31,

*=

Mk

asymptote,

(fc~0,

is

2,

1,

Domain

982.

2fejt.

0,

(&

definition

k

inflection,

function;

x

2&Ji

of

where

(2fc+-jrjJi

x=

+

when

4 (+0,

end-point,

Q);

i

t/max^ 6

e

^ 1-44

when # = e=^2.72; asymptote,

(/

=

of

point

!;

inflection,

when / = (0 = 3); y mm = 1 M, (0.58, 0.12) and M 2 (4 35,1 40). 988. Jc inln = when / = (x^3) 989. To obtain the graph it is sufficient to vary / from to 2jt. a (cusp) when ^min^ awhen^-ji(// = 0); ^max = awhen / = (#-0); min 1

1

1

t/

*

=

+-2-

.

3T

^

inflection

t

when

f

=

fl

(cusp)

when

<

7--

V'Se \

when

*

=

8 ,

-y

U = 0);

points of inflection

7jT

,

-j-

*=

1

I/" 2,

i.e.,

to=

^)l

^max =

XJ.

f

j

i.e.,

=

44

,

r wnen

m in=

Jf

+

ma x 5jl

r4

,

4

1

990.

/

3jT

= -j-

.

when

(x;

= 0);

-TT=;

V

1

i/

= 2x

when

t

/

_ /"^"t

asymptotes, jc=0 and y

(cusp); asymptote,

wn ^n

-*

= 0.991.

+ oo.

992.

=

l(jK:

= e);

and when

points of

t=

j

xmin =l i/

m in =

andymln =i when

*

= 0.

$93. ds = d u cosa = dx, 415 ; y <*> where = x sin tanh a 998. ds -. 997. . = 2asin~rff; cos ds a sin--; cosa= a cos/; sin sin a =a sin/. 1000. ds "(p;cosp= cos dx: Z Z d^; = cosh cos 1002. a= -~- . 999. ds = cosp = Z V^l +(p - ds^ 995. . tit I/ (X ~3asin ^ cos/ 62 Ka 2 c=- 2 d ds== dcp; tp* cos* 1003. . l = an2z 1 zzz K^36. 1006. 4^ /(-= 1010. ^"3^2" 1007. 1012. . both at 1005. . = " vertices. &2 ; ' ds== KB = ~ 1011. ( ^).,0,3. (-!"*.. ,= . a sin 2t 1016. . . , !023. . | ^a -ilfl, ( 2 3) 1019. -8. -^- a cos 3 1017. . -pK 2 = ~(X 8 p) ) R and sin p = cos 29. K^ [ -f- , R^- = |a/|. 1018. 1020. - ^ 1026. 1009. 1014. -^- f k2 d(p; . 3 1015. 1004. cos-|-. . = /? = (2,2). l025 - - (semicubical parabola). 1027. _ 8 ^c 8 , where c 2 =-a 2 6 2 . Chapter IV In the answers of this section the arbitrary additive constant ted for the sake of brevity. 1031. y aV. 1032. f / 2x 3 + 4* + 3*. 2 C is 1033. omit- + 416 Answers 1037. '-+-- --' a 2 x---a'x 1038. J. j/ <* ''---~. '*'--- JC . 3x 2 x 3x* fi 1040. ' 13 4//2 1042. + 1043.3= - In -4= 2/10 1044. = sec 2 -x; b) x . 1045. ln(x+ 2 /4 + x 1048*. tanhx. Hint. Put tanh z b) x 1; /"lO -^~ln (x+ /x ^). 1047. arc sin x-cothx. ^j~ 1050. a ' _i I x a J Solution. ^ numerator the |. ^ 1063. 2 1065. --^-. | cotx 1052. we 1057. c--l|. Inlx 1059. |. ~~' a 3*- 2 ]/ _Jiy(a_- 1062. y. 1066. 1 ^= 2 ). In /7 + 2 + b + x V ab + V7 + 8x 1064. 2 1 ~. . j fll ,, . -L arc tanx = get 1054. . 1 I06li :ln In 1049. a) . Solution, f In (2 1073. 2/2 * 1 . 1 1067. Put denominator, ~+ (x+W /x +l. Hint. Solution. . 1056. 1060. " r 1058. the by -I' + T j . sin- r-j . Dividing . arc c 1053. 1055. x. a = tan arc 1046. ). x=l 1051. a in . 2 tanx a) { s + 1009. / 1068. x 1072. /"2 arc -~= tan arc sin x / Answers 1075. . 5 ^ In (a * + 2 1079. y V^= 31n|x + 2 1077. 2 ft =l n 5 - 2 1078. 1080. -i arc sin ^. ^* 1). 1076-. ( -ln|jt arc tan )-}- 417 ^- 2 4 l 1081. . are tan 1082. 4 1083. 1|. ^(arcsmx) 8 V 2 1086. \-e~ a l 1090. . 2 1092. - 1084. . ,/ -H- r? 2 , 3 Ina i 1095. In 1097. | e* 1| i + 3) In (2- 3 In 2 1102. ~ 1098. . - ~ 1109. Xcos(logA:) Hint. See arc sin HOI. x __ !. Hint. Put in 1109 lilt. sin 2 jc= ~ tan sin (fljc + cos 2). tan(x ). 1117. -^-cos(l ^'.1119. Xln sin - . 1122. 5 In sin I 4- 1125. In ] 1120. . | x -^ cot x ln|sinx|. 1121. 21n|cos/"x|. 1123. 1. tan* 1118. r). ln|cosx|. a + ft 1 1 2 -- 1110. tan 1 lOx 1112. ft). a In 1116. + ^)- In 1108. /"x. ~(1 008(0 o ax 1113. 4 arc tan (a*). jJ 1104. e* ^-cos2ax. 1107. 2 1106. hint ' !n~5" 1100. _^L_J. 1103. 2sni -^r. 1096. . 4 Hint. -Jr V x ~ 1099. (a r 1105. e 2 1126. ^. |sin 1124. K (a 2 ft)X ^ 1127. 1 1129. - 1128. 4a - 1130. l o sin 1132. 1131. 4 -^-tan -^. U33. -| 5 1134 . _3^_lJL. 14-1900 M55.1 U 36. tan ( j). + 2au.) l(ln tanf | | 23 Answers 418 4- In 3a | ocot3x| 6 tanh 1140. In 1144. In I -r- sinhxl - ln(x 1152. . lr 1155. 1153. . In tan x | fls n n49 + | - 1 1147. . arc tan J/ ~ 1150. In ln|x*-4x+ 1 Incoshx. 1143. ln|tanhx|. 1146. "- 1148. Xarc tan 1142. */(5=x^. 1145. 1139.- +- . D 2arctane*. 1141. . . 1138. . x _. ------- 1 1137. . | )- ]/ 1151. . x - . sec 1156. . 1) ' 1157 . *. 1158. arc sin (x 2 ). 1159. . & a in 1160. taniw-x. a /(^ Vx 2 In cos 1165. | 1172. e sln 1168. . 1169. 1170. ^ -l^arcsin^A 1173. 1175. . \n(e* + ^21n 1171. . tan 1167. tan^-l. | K. =. -i-ln 1166. 1 x a f 1181. /"l <40 - 1187. - . 2 jc . - 1=r: 7 1182. 1185. arc sin ln(secx+ . arc tan /tanx\ I == } . 1183. -2cot2x. 1 2 l/~sec x-i- 1). 1186. 1 .- 2). = In -p=. + 2arctanx. ln|x| 1174. Ve*x tan jr ln|tanox|. arc cos -^ 1184. V 5 -f sin 2x sin2x . Hint. 2 v = x r J 1189. s inh(x . 1190. -Latanh*. 1191. a) -~=arccos^ when x > ^2; b) In (1 +*-*); Answers (5*'-3)'; c) - d) 2arctan \e* 1195. . V 2x4-1 aarccos 2 arccos 1204. if , x > 0, and arccos In* 1201. -. -~ V2~x -^ V2-x 1202. . 1196. 1. Putx = Hint. . TT arc sin x. ln(sinx+- J/"l+ sin**)- e) 1 1194. In 419 In 21n Inx-f | -- | 2 . ( ~J x<0*) if Hint. ' Put x = ~- substitution x = be used in place may ~ x arc 1216. sin x-}- V x2 1 1217. ^r~- 1214. . ^T^Y- the of ^ (9x 2 f In (1 -|-x x--^ * xcosx. sinx 12 18. x arc tan 3* st 1215. Note. The substitution. trigonometric 1212. x. . ^ o + igf y ^ of undeter- xV*dx = (4x 2 + Bx + C) e*x = (Ax + Bx -f C) 3e + (2Ax -f B)e9X 2 3Jf . * and Cancelling oute equating the coefficients of identical powers of 3 122 B= = whence ^ ; -^-; =Q (x)e ax where P , polynomial Hint. *) is ). after differentiation, X 2e* x a 2 6x-f- 2). Solution. In place by parts we can use the following method of repeated integration mined coefficients: or, xlnx 1211. a|. 1213. 1206. . good See rt of degree Problem c= 2f = 3C + In the x, we get: , C 6eneral ^ orrn \ p n to e ax dx = is the given polynomial of degree n and Q n (x) is n with undetermined coefficients 1219. e~* (x 2 -f-5} (x) X 1218*. 1220. -3e T "(x-h9x 2 + 54x+ 162). Hint. See Henceforward, in similar cases we shall sometimes give an answer thai for only a part of the domain of the integrand. Answers 420 Problem cos 2x mined - 1221. 1218*. Hint. It where P n (x) polynomials px dx (x) cos = Q n (x) cos PX + R n (x) sin X arc sinx + In | sin 3 -|- x | 1236. x8 f Inx l O^2 x -^ (x+ V^l+x Ay2 ^Ar x ) tan ~^ 2 1 arc sin x -j 4 2, 2 2 e* (sin x 1232. X xcotx-f 1230. . ' 2 1228. . x . sinx - cos x) x2 V x ft , +2 V , - get + -^ In (1 + x rt 2 2 1246. 1+ - -- -- , In cos 2x I x 1 ]/T=Ixx x "" xcos . 2 ^ 4 ' x2 x 1 1 1 \ and +2 V x + . 1248 ' (2 In x) + ' 2xsin (2 In x) 10 a=x 1 du = dx 2 X x (arc sinx) 2 1244. ). xtan2x ._ 1247. V X 2 ^ V In x 2 In x ~^-^-- ^ arc tan 3x-^ + ^o In(9x +l). 1243. ~-^X 1242. . / cos2x -2sin2x 1240. 1 x. fy-* 2 x 1 2 1238. -1). Inf-^-x. +X 1245. 2x. 1237. 1, 1239. x arc tan x _ , x2 -/m -_ _ + -3x. 1). arCMn * Xarcsinx Xarcsin 1234. (x*+ [ln(lnx)-l].lnx. X (arc tan x) 2 we 1225. i . 1250. 1218*). " cos (In x)J. X x In 1229. x*. 1238. 1241. (x) are ln * x arc tan x 1227. . 1231. . and R n Problem (x) (see ij(, 2 l/T -|- undeter- of | Qn 2xlnx + 2x. xln x 1224. 4 /"xlnx 2 method the apply the given polynomial of degree rc, and degree n with undetermined coefficients is of ^-Inx--^. y o 1226. 1222. form 2 1223. 2x also advisable to is coefficients in the Pn sin 4^8 = * C Whence . ^ ( x ~r x dx and x*dx d = 2 2(x +l) 1) 125K Hint. xy a 8 ^x* + -s- a=- ^ Utilize arc sin . the 1 identity Solution. Put j.. x; we have f }/aa a ~2 2 ~ 1252. = Ka x and -x'dx = x ^a -^- \ r xa - f -/-^2 \ J Ka =. ,f x 2 dx+a2 l dx , J r x whence 2 - -; a* x 2 ]. ^a2 Answers Consequently, 2 y~a*x \ ~ l257 - = - Fn arctan x* 2 -fa arc Problem 1258 TTT 2 -ln 1259. . 2 ~. sin 1252*. Problem 1252*. 1255. Hint. See . Va See Hint. . arc sin -| o 2 dx = x 2 421 4A ^~- . / f ~ ln( * a 2 1260. x 2). - y arc tan 7=-. y 7 Larcsin^P^. ^ ,262. x-f31n(x 1261. l 2 arc tan 7 1263. arcsin(2jc 1). 2 n (jc z 3 --T ~ 1256. 2t arc tan (x (x -~ 1254. arc tan T ' ~ V& +** 1253. + (A:3). 1264. In 1/2 2 1265. -2 V -x-x* -9 arc sin 1266. 4jc-f5. 1 =^ . 1 In r 1267. '71 5/"1 1268. In . ?=. 1270. arc arc sin 1269. x arc sin 1271. . "* ,273. 9 arc sin - ( + 2* 2 In 1272. + sin - y$

+5).

1

1274.

8

2x4-1 -

arc tan

1276.

o

o

1278.-ln|cosx + 2+ ~'

y\

1279.

1281.

jc

In 2 x

4 In x

+ 31n|x

1283. In

l

2 arc sin

l!L~.

1280.

o 3|

.

31n|A;

2|

5*

+ In

1284.

^

1282.

*

2

(x-4)'

- -b

x \

+a

I

.1285.

1+*

(*-!)' 1286.

2

30

27 1288.

1290.

11

^

4-j 1289.

=

2U-3)

-^-r .

49

U

5)

49 (x

+ 2)^4*

x~5 x-f-2

x

l

1292.

1293.

ln|

x-3|-

In

U~l + [

In

X

422

X arctan(* + 2). x*-x V

x

,

arc tan

2

+

In

1300.

+ji-i*-

_|arctanx- TT^-jT -^

1305.

x

1304.

1

2

5=41.

l

f

arc tanx.

In (x

5

x

1310.

y ln|* +H 7

ln|x|

1

1

o

2

FT

^/(a* +

6)']

1315.

56

^

X -6

-3

In

In

1324.

(A:

1 |

2 arc tan

-J-

~

x7

1)

.

tan x

_ J_ J

'

1

Sx^Sx*

x

1

2 arc tan

1317.

-^=.

1322.

I)

7

arc +1)-^o

.

^3

Xarctan]/|-.

=il.

3 In

7

7(jc

3

(l+/l.

3^/1-6 ^/"x 1320.

l)"

(A:

._-

1

s

4(x

9(jc_l)>

= (*

1

arc tan

1312.

1

1313.

1

1 . i

.

Put

Hint.

.

1

1

*+l

tan (x-^1).

4

.

1311.

2

1306.

2

X[2

1303.

4r (8

+ 21n

-

|

1

'+4^

arc tan

1298.

1299.

.

1302.

x

arc tan x

,,., 1297.

In

1295.

.

1

K"-\

+ arctan(*+l). -

= arc tan

+

In

1294.

-

1316. 1318.

1319.

+ J/l + 6 arc tan |

*/l.

arc tan

^1-x.

1323.

-

(jf_2)+-I

In

/

where

423

=7+l).

1327.

-

*-li

.

-+i

1329.

-arc sin -^

1330.

1331.

.

4/7=1

-* + l).

-retan

.

*

-2

-cos.,.

(*

1340.

1334.

where

.

1337.

1332.

+1)

-

,336.

.

2

1 .

1338.

sin

x

1339.

g-sin'x.

-cosx+-jtos>

x

1342.

*J^-**. |841.4eo.|-^|.

>-.-.+ --^=

cot*

i

_ 7 i_sin6x. 144

1347.

__-. 35

,849.

-1

21

._. ._

, 344 .

3 In

|

tan x

cot

jc

Jt

1348.

.

tan x

tan + -^ o

+2lnt.n.

|

+ To tan

x

5

x

1351.

^.

1350.

8

1353.

.

COS 2

"[

ln

,357.

3 cos*

*+i

t,n|

_S2*lf_in|sinx|.

.

1354.

-

1358.

4

cot'

x

sin 4 A:

In

8

8siu*x

+ cot x + x.

1361.

.

1362.

xlB

-1 J/55?x + -| J/cSS*-^ J/cw

SrRj{"^ .

-cosx.

,366.

.369.

-

arc tan

^' .

.

1367.

,370.

11

-tan! --f

1359.

-

!. 1363. 2 J^twix. 1364.

where I

" iain

-+3-.

-

1365 '

.368.

lE_^M+
J-J^

cos-

137 ,.

!!H +

424

5*

sin

sin

7*

^ Jo

,

tan

-!-

1374.

+

[

tan

In -f *

1373.

1376

--

2

tan|

cos 4* --5- cos 2*. ^ cos 6* -^ z4 o lo

1372.

.

'

* In

1377.

1378. arc tan (l \

.

-

tan

We

+

3 cos x Solution. Whence 4- P (2 sin* -{-3 cos*)' 12 5 .,, Q

a= !3' r

I

*

iiC

-f-

arc tan

1381.

5>

f

have

lO

-^^

fraction by ccs 2 x

,

= j2^_

<.

In ||*--^ lo id

+

2 cos x

a

===

+ 2p = 2 p3sinx + 2crs*

= 3,

3^

(2 sin x

3a

and, ,

2^77+3^7^

J rfx

Si

J

1379.

/

3 sin x

put 2u

We

P== -!3-

+ tan 4V ^

+

2sin *

|

+

+

3 cos x) consequently,

12

==

13

5 2

ln

.

|

1380.

In |cosx

situc(.

1*3

Hint. Divide the numerator and denominator of the

)

-=- arctan

1382.

.

See

Hint.

.

V*

V

Problem

See

Hint.

Problem 1384.

1381.

1381.

4 In x

1

1387.

V2

2

/" 2 -f

sin

Vi

sni2*'

~~"

/\

r~~~

/"

t388.

4-ln^^^.

arc ian

1 .

1

Use the identity

sin*

2

sin*)

sin

In

cos*

-.

r cosh * !

1

1396.

.

,

2coth2*.

.

12 or

-j=. O

arc tan

/

3

1394.

5)

.

+m+ 1401.

I

r

-JT-

In (cosh

1397.

V

If

sinh*

*

*)-^^ 2

1399. arc tan (tanh*). 1400.-~=.arctan

* iA-?\l

(f V

Hint.

2

'

* ,

.

cos*

* *

-j-sin

cnth 8 *

coth*

1398. *

identity

+ 21n '

~

x

tan

1393.

tanh

the

arc tan

x+cos*

1-j-smx

^>

1395.

*

1390.

sin*

3

Sm 1392.

Use

Hint.

1

1

sit

(2

=

1389.

2

sinh * ^ *

sinh 2x r *

*

^^

5 ,

A

.,

..

..

...

Hint. Use the identity

J

,

cosh*

-^ (sinh* %

-J*