658 97 15MB
English Pages [497] Year 1970
Table of contents :
Title Page
CONTENTS
PREFACE
Chapter I INTRODUCTION TO ANALYSIS
Sec. 1. Functlcns
Sec. 2. Graphs of Elementary Functions
Sec. 3. Limits
Sec. 4. Infinitely Small and Large Quantities
Sec. 5. Continuity of Functions
Chapter II DIFFERENTIATION OF FUNCTIONS
Sec. 1. Calculating Derivatives Directly
Sec. 2. Tabular Differentiation
Sec. 3. The Derivatives of Functions Not Represented Explicitly
Sec. 4. Geometrical and Mechanical Applications of the Derivative
Sec. 5. Derivatives of Higher Orders
Sec. 6. Differentials of First and Higher Orders
Sec. 7. MeanValue Theorems
Sec. 8. Taylor's Formula
Sec. 9. The L'HospitaiBernoulli Rule for Evaluating Indeterminate Forms
Chapter Ill THE EXTREMA OF A FUNCTION AND THE GEOMETRIC APPLICATIONS OF A DERIVATIVE
Sec. 1. The Extrema of a Function of One Argument
Sec. 2. The Direction of Concavity. Points of Inflection
Sec. 3. Asymptotes
Sec. 4. Graphing Functions by Characteristic Points
Sec. 5. Differential of an Arc. Curvature
Chapter IV INDEFINITE INTEGRALS
Sec. 1. Direct Integration
Sec. 2. Integration by Substitution
Sec. 3. lntegration by Parts
Sec. 4. Standard Integrals Containing a Quadratic Trinomial
Sec. 5. Integration of Rational Functions
Sec. 6. lntegrating Certain Irrational Functions
Sec. 7. Integrating Trigonometric Functions
Sec. 8. Integration of Hyperbolic Functions
Sec. 9. Using Trigonometric and Hyperbolic Substitutions for Finding Integrals of the Form $\int R (x, \sqrt{ax^[2}+bx+c) dx$
Sec. 10. Integration of Various Transcendental Functions
Sec. 11. Using Reduction Formulas
Sec. 12. Miscellaneous Examples on Integration
Chapter V DEFINITE INTEGRALS
Sec. 1. The Definite Integral as the Limit of a Sum
Sec. 2. Evaluating Definite Integrals by Means of Indefinite Integrals
Sec. 3. Improper Integrals
Sec. 4. Change of Variable in a Definite Integral
Sec. 5. Integration by Parts
Sec. 6. MeanValue 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. 11. Moments. Centres of Gravity. Guldin's Theorems
Sec. 12. Applying Definite Integrals to the Solution of Physical Problems
Chapter VI FUNCTIONS OF SEVERAL VARIABLES
Sec. 1. Basic Notions
Sec. 2. Continuity
Sec. 3. Partial Derivatives
Sec. 4. Total Differential of a Function
Sec. 5. Differentiation of Composite Functions
Sec. 6. Derivative in a Given Direction and the Gradient of a Function
Sec. 7. HigherOrder Derivatives and Differentials
Sec. 8. Integration of Total Differentials
Sec. 9. Differentiation of Implicit Functions
Sec. 10. Change of Variables
Sec. 11. The Tangent Plane and the Normal to a Surface
Sec. 12. Taylor's Formula for a Function of Several Variables
Sec. 13. The Extremum of a Function of Several Variables
Sec. 14. Finding the Greatest and Smallest Values of Functions
Sec. 15. Singular Points of Plane Curves
Sec. 16. Envelope
Sec. 17. Arc Length of a Space Curve
Sec. 18. The Vector Function of a Scalar Argument
Sec. 19. The Natural Trihedron of a Space Curve
Sec. 20. Curvature and Torsion of a Space Curve
Chapter VII MULTIPLE AND LINE INTEGRALS
Sec. 1. The Double Integral in Rectangular Coordinates
Sec. 2. Change of Variables in a Double Integral
Sec. 3. Computing Areas
Sec. 4. Computing Volumes
Sec. 5. Computing the Areas of Surfaces
Sec. 6. Applications of the Double Integral in Mechanics
Sec. 7. Triple Integrals
Sec. 8. Improper Integrals Dependent on a Parameter. Improper Multiple Integrals
Sec. 9. Line Integrals
Sec. 10. Surface Integrals
Sec. 11. The OstrogradskyGauss Formula
Sec. 12. Fundamentals of Field Theory
Chapter V Ill SERIES
Sec. 1. Number Series
Sec. 2. Functional Series
Sec. 3. Taylor's Series
Sec. 4. Fourier Series
Chapter IX DlFFERENTIAL EQUATIONS
Sec. 1. Verifying Solutions. Forming Differential Equations of Families of Curves. Initial Conditions
Sec. 2. FirstOrder Differential Equations
Sec. 3. FirstOrder Differential Equations with Variables Separable. Orthogonal Trajectories
Sec. 4. FirstOrder Homogeneous Differential Equations
Sec. 5. FirstOrder Linear Differential Equations. Bernoulli's Equation
Sec. 6. Exact Differential Equations. Integrating Factor
Sec. 7. FirstOrder Differential Equations not Solved for the Derivative
Sec. 8. The Lagrange and Clairaut Equations
Sec. 9. Miscellaneous Exercises on FirstOrder Differential Equations
Sec. 10. HigherOrder Differential Equations
Sec. 11. Linear Differential Equations
Sec. 12. Linear Differential Equations of Second Order with Constant Coefficients
Sec. 13. Linear Differential Equations of Order Higher than Two with Constant Coefficients
Sec. 14. Euler's Equations
Sec. 15. Systems of Differential Equations
Sec. 16. Integration of Differential Equations by Means of Power Series
Sec. 17. Problems on Fourier's Method
Chapter X APPROXIMATE CALCULATIONS
Sec. 1. Operations on Approximate Numbers
Sec. 2. Interpolation of Functions
Sec. 3. Computing the Real Roots of Equations
Sec. 4. Numerical Integration of Functions
Sec. 5. Numerical Integration of Ordinary Differential Equations
Sec. 6. Approximating Fourier Coefficients
ANSWERS
Chapter I
Chapter II
Chapter Ill
Chapter IV
Chapter V
Chapter VI
Chapter VII
Chapter VIII
Chapter IX
Chapter X
APPENDIX
I. Greek Alphabet
II. Some Constants
Ill. Inverse Quantities, Powers, Roots, Logarithms
IV. Trigonometric Functions
V. Exponential, Hyperbolic and Trigonometric Functions
VI. Some Curves (for Reference)
INDEX
MIR PUBLJ:SHF.I(S
r. c.
5apaHeHK06. 5.
n
DeMUiJOf1U't,
B. A
Ef/JuMeHfCO,
C. M. Kocan. r Jl Jlynu,, E lP flopumeea, E fl. Cwteea, C. B. l/>po/108, P. fl. lllocmaK, A. P. fl nnollbCKUiJ.
3AJIA4M 11 YnPA)I(HEH11S1
no MATEMATM4ECKOMY AHAJ1113Y
lloiJ penaKl{Ut>iJ. B.
n.
A EM 11 A 0 B H q A
TocgiJapcmBeHHO!! U3iJameAbCmi>O t/Ju3uKOMameMamu 'leeK oil. 11u mepamypw MocKtHI
G. Baranenkov, B. Drmidovich V. Efimenko, S. Kogan, G. Lunt~. E. Por!:.hmva, E. ~!;t:hrta, S. frolw, R. ~ho!>tak, A. Yanpolsky
PROBLEMS IN MATHEMATICAL ANALYSIS Under the editor.,hip of B. DEMI DOVICH Translated from the Russian. by G. YAN KOVSKY
Mil{ PUBLISHEI{S Moscow
TO THE READER
Ml R Publishers would be glad to have your opinion of the translatwn and the design of thls book. Please send uour suggest tons to 2, Pervy Rtzh:.ky Pereulok, Mo:.ww, U. S. S. R.
Second PrintiR/l
Prlntecl ln the Unlon of Soviet Socialist Republics
CONTENTS Preface • • . • , •••••
9
Chapter I. INTRODUCTION TO ANALYSIS Sec. Sec. Sec. Sec. Sec.
l. Functions . . . . . . 2 Graphs of Elementary Functions 3 Limits . . . . . . . . . . . . 4 Infinitely Small and Large Quanltties 5. Continuity of Functions . . . . . .
11 16 22
33 36
Chapter 11 DIFFERENTIATION Of .fUNCTIONS Sec 1. Calculating Derivativl's Directly . . . • • . . . . . • • Sec 2 Tabular Differentiation . . . . . . . . . . . . . . Sec. 3 ThE.' Derivat•ves of Functions Not RE.'presented Explicitly Sec. 4. GE.'oml'trical and Mechanical Applications of the Derrvative Sec 5 Denvallveo; of Hig1er Orders ..... Sec 6 Differentials of First and Higher Orders Sec 7 Mean VahlE.' Theorems . Sec. 8 Taylor'5 Formula Sec 9 The L'HospitalBernoulli Rule for Evaluating Indeterminate Forms .......•. . . . . • . .
42 46 56 60 66 71
75 77 78
Chapter Ill THE EXTREMA Of A FUNCTION AND THE GEOMETRIC APPLICATIONS Of A DERIVATIVE Sec. Sec. Sec Sec Sec.
l. The Extrema of a Function of One Argument
2 3 4. 5.
The Direction of Conravity Points of I nllection Asymptotevolution , 166 Sec 11 JVorrents Centres of Cravity CJuldm's Throrrms • 168 ~ec 12. Applyir.g I:ef.r.ite IntEgrals to the Soluhon of Physical Prcb· Iems 173
Chapter VI. FUNCTIONS OF SEVERAL VARIABLES Sec. Sec. Sec Sec Sec Sec. Stc. Sec Sec ~ec
Sec. Sec Sec. 5t>c Sec Sec Sec.
1. Basic Notions • , , 2. Contmu1ty 3 P.art1al Derivatives 4 Total Dlf'lerential of a Function 5 Differentiation of Comros1te functions 6, Dtnvat1ve m a G1ven Direct•on and the Gradient of a Function 7 H 1gH'e, Crder Cenvat1ves and Difterentials 8 Integration of lotal Dif:erentials 9 Dif.erentiation of lmpllc1t Functions . • . 10 Change of Variables • 11. The Tangt'nt Plane and the Normal to a Surface 12 1 aylor'~ Formula lor a function of ~everal Variables 13 The Extremum of a Functwn of Several Variables • 14 Firdlr.!l tht Creates! and ~n·allest Values of Functions 15 Sml'ular Pomts ot Plane Curves . 16 Envelope 17, Arc Length of a Space Curve • , .
180 184
185 187 190 193
197 202 205 211
217 220 222 227 230
2.32 234
Contents
7
2J5
&c. 18. The Vector Function of a Scalar Argument Sec. 19 The Natural Trihedrnn of a Space Curve Sec. 20. Curvature and Torsion of a Space Curve
2J8 212
Chapter VII. MULTIPLE. AND LINE INTEGRALS Sec. 1 The Double Integral in Rectangular Coordinates 246 Sec. 2 Change of Variables in a Double Integral 252 256 Sec. 3. Computing Areas • . • • • • • • Sec. 4. Computing Volumes . • • • . • • • • • • 258 Sec. 5. Computing the Areas of SurfacE's 259 Sec. 6 Applications of the Double Integral in Mechanics 230 Sec. 7. Triple Integrals . . . . . . . • . . . . . . . . . 262 Sec. 8. Improper Integrals Dependent on a Parameter. Improper Multi tie Integrals. . •• • 269 Sec. 9 Line Integrals . • • • • . . . . . 273 284 Sec. 10. Surface Integrals • . . . . . . Sec. 11. The OstrogradskyGauss Formula 286 Sec. 12. Fundamentals of Field Theory 288
Chapter VIII. SERIES Src. Sec. Sec. Sec.
I. 2. 3. 4.
Number Series Functional Series Taylor's Series • Fourier's Series
293 304 311 . 318
Chapter IX DIFFERENTIAL EQUATIONS Sec. 1. Verifying Solutions. Forming Differential Equations of Fami· lies of Curves. Initial Conclitions . . . • • • • • . . . . . . . . Sec. 2 FirstOrder Differential Equations • • • . • . • • . • . . . Sec. 3. FirstOrder Diflerential Equations with Variables Separable. Orthogonal Trajer.tories . • . . . . . . . . . . . . . . . . . . . Sec. 4 FirstOrder Homogeneous Differential Equ:ttlons • • • . . . Equations. Bernoulli's Sec. 5. FirstOrder Linear Differential Equation . . . . . . . • • . • • • . • . . . . . • . . • • • . • Sec. 6 Exact Differential Equations. Integrating Factor • . • • . • Sec 7 FirstOrdrr Differential Equations not Solved for the Derivative Sec. 8. The Lagrange and Clairaut Equations . . • . . . . . . . • Sec. 9. Miscellaneous Exrrcises on FirstOrder Differential Equations Sec. 10. HigherOrder DiiTerential Equations • • • • . • • • • • • . Sec. 11. Linear Differential Equations • . • • • • • . . . . • • . . Sec. 12. L1near Differential Equations of Second Order with Constant Coefficients • • • • • • • • • • • • • , • • • . • • . . . • • • •
322 324 327 330 3~2
335 337 339 340 345 349 351
Contents
8
Sec. 13. LinE>ar Differt'ntial Equations of Order Higher than Two . . . . • • with Constant Coefficients Sec 14. Euler's Equations . . . . . . . • • • , , ••• , ••••• Sec 15. Systems of Differential Equations • . • . . . . • . . . . . Sec. 16. Integration of Diffen•ntial Equations by Means of Power Se· ries . . . . . . . . . . . . . . • • . • • • . • , , , •• Sec 17. Problems on Fourier's Method
Chapter X. Sec. I Sec. 2. Sec. 3. Sec. 4 Sec. 5. Sec. 6.
APPROXIMATE. CALCULATIONS Operations on Approx.rnatc Numbers InterpolatiOn of FunctiOns . . . . . Computtng thc,Rcal Roots of Equations .•• , • , • NunwriLal, lntel!ration of Functions . . . . . • . . . . 1'\un er:ral lnteFlllion of Crd1rary Dd'ltrtntial Equations Appro,.tmating h urtu's Cceflictents
ANSWERS APPENDIX I. Greek Alphabrt II. Some Con~tants III. Inverse Quantities, Powers, Roots, Logarithms IV Trigonometric Funct 1011~ • • • • • • • • • • • • V. Exroret~tJal, Ilnerbolic and Trigono111etnc Functions VI. Some Curves . • . • , , . • . . • . . . . . • , , ,
356 357 359 361 363
• 367 372 • 376 382 384 3Q3 • 396 475
475 475 476
478 479 480
PREFACE This collection of problems and exercises in mathematical analysis covers the maximum requirements of general courses in higher mathematics for higher technical schools. It contains over 3,000 problems sequentially arranged in Chapters I to X covering all 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 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, the Fourier method, and approximate calculalions. Experience shows that the number of problems given in this book not only fully satisfies the requirelr:£11 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 ch01ce of problems in each section and 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 th1s will greatly simplify the work of the student. Answers are given to all computational problems; one asterisk indicate'> that hints to the solution are given in the answers, two astensks, that the solution is given. The problems are frequently illustrated by drawings. 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.
Chapter I
INTRODUCTION TO ANALYSIS
Sec. 1. Functlcns 1°. Rnl nurrl:ers. Rational and irrational numbers are collectively known as real numbers 1 he ab,otutf value of a real number a IS undtrstood to be the nonnegative numb~r I a 1 d~fined by the conditions· 1 a 1 =a if a :;;.o, and =a if a< 0. lhe following in(quallty holds for all real numbers a an b:
laJ
2°. DeHnition of a function. If to every value*) of a variable x, which belongs to son.e collection (set) E. there corresponds one and on I y one linite value of the quantity y, then y is sa1d to be a functwn (~1nglevaluedl of x or a dependent 1 artable defined on the ~et E. x is the a•gument or tndependent vanable The fact that 11 1s a lunct10n of x IS expressed 10 bnef form by the notation y =I (x) or y = F (~ ), and the like If to every value of x belonging to some set E there corresponds one or several values of the vanablt> y, then y is called a multtplevalued functiOn of x defined on E. From now on we shall u~e the word "funrt10n" only in the meanl!lg of a ~tnglevalurd functwn, 1f not otherw1~e stated 3° The domain of dellnition of a function. The collectiOn of values of x lor which the g1ven funct1on IS defined IS called the domatn of defirutwn (or the domain) of th1s functiOn. In the sm1pl~st cas~s. the doma1n of a function IS either a clo~ed tnterval [a. bl. which is the set of n•al numbPrs x that sat1sfy the inequalities ae;;;;;~~b. or anopen tntenal (a.b). wh1ch :s the set of real numhers that satisfy the 1nl'qual1t.es a< x 0,
that is, if I xI> l. Thus, the domam of the function Is a set of two intervals: oo < x 1eret a! ca~e. th rquation y = f (x) refines a multiplevalued mverse furct1on x=f•(y) wch that y==flf'(y)j for ally that are values of ihe functiOn f (x) l:.nn pIe 2. Ceterrnine the inverse of the lunchon
y= 12x.
(1)
Solution. Solving equation (I) for x, we have 2X=Jy and X=
log(ly) "') log 2 ·
(2)
tre
dorrain of c'efinition of the function (2) is oo < y < 1. 5°. Corrroslte and irr.plicit functicns. A function y of x defined by a series of equalitu~sy=f(u), whereu=q>(x), etc., is called a comoosite function, or a functwn of a fun:tio't. A function defined by an rqu11tion not solved for the derencent variable is call~cl an tmpllCll function. For example, the equation x1 y 1 = 1 defines y as an imp1ic1t function of x. 6°. The graph of a function. A set of points (x, y) in an xyplane, whose coordinates are connecttd by the equation y = f (x}, is called the graph of the given funct:on. Obviou~ly,
+
1**. Prove that if a and b are real numbers then
!!all b II~ lab I~ ial+lb 1. 2. Prove the following equalities: a)
Iab I= I a 1·1 b I; c)
b)
lal =a 2
2 ;
I: I= (:) (b + 0);
d) V£i2=~\aj.
3. Solve the inequalities:
a) lx11O; Y 1 tt=X l.
Vv,
43. Write, explicitly, functions of y defined by the equations: a) x'arccos y=n; b) lOX+ l()Y = 10; c)
x+IYI=2y.
Find the domains of definition of the given implicit functions. Sec. 2. Graphs of Elementary Functions Graphs of functions Y=f {x) are mainly constructed by marking a sufficiently dt>me net of points M;(x;. 1/;). where y 1 =/(x;) (i=O, I, 2, ... ) and by conuecti ng the pomts with a Iine that takes account of intermediate pomts. Calculations are best done by a ~hde rule.
y
...................... .}It
·.
\
X
Fig. 3 Graphs of the basic elementary functions (see Appendix VI) are readily learned through their construction. Proceeding from the graph of Y=f (x),
(f)
we get the graphs of the following functions by means of simple geometric construct ions: 1) y, =  f (x) is the mirror image of the graph r about the xaxis; 2) fla=/(;c) is the mirror image of the graph r about theyaxis;
Graphs of Elementary l'•mctwni
Sec. 2)
17
3) y1 = f (xa) Is the r graph displaced along th~ xaxls by an amount a; y4 = b +I (x) Is the r graph displaced along the y·axls by an amount b (Fig. 3). Example. Construct the graph of the function
4)
y =sin ( x : ) . Solution. The desired line Is a sine curve y =sin x displaced along the x·axis to the :right by an amount
y
1 (Fig. 4) !f• sin
(:r f) X Fig. 4
Construct the graphs of the
following
linear functions
(straight linPs ): 44. y=kx, if k=O, 1, 2, 1/2,1, 2. 45. y=x+ b, if b=O, 1, 2, 1, 2. 46. y = 1.5x + 2.
Construct the graphs of rational integral funetions of degree two (parabolas). 47. y=ax•, if a=1, 2, 1/2,1, 2, 0. 48. y = x• + c, if c = 0, 1, 2,  1. 49. y=(xx.) 2 , ii x.=O, 1, 2, 1. 50. Y =Yo + (x 1)2 , if Yo= 0, 1, 2,  1. 51*. y=ax•+bx+c, if: 1) a=1, b=2, c=3; 2) a=2, b=6, C=O. 52. y=2 t·xx•. Find the points ot intersection of this parabola with the xaxis. Construct the graphs of the following rational integral lunctiom of degree above two: 53*. y=X 3 (cubic parabola). 54. y = 2 + (X 1)3 • 55. !J=X3 3xt2. 56. y= x•. 57. y = 2x1  x•. Construct the graphs of the following linear fractional func· tl ons (hyper bolas): 68*.
Y=+.
Jntr~duct10n
18
to Analysis
1
59. y= 1x· x2
60. Y= x+2. m
m=6. x.=1, y.=1, 61*. y=y o +,if • x  Xo 2x3 62 *· y=3x+2'
Construct the graphs o[ the [ractional rational functions: 1 63. y=x+x· xz
64. Y=
65*.
X+
1'
y=:.. 1
66. Y=.. X 67*. y =
~}~ 1 (Witch of Agnes£).
68. y = x 2 ~ 1 (Newton's serpentine). 2
69. Y=
1
x+xz·
(trident of Newton). 70. y = x• + _!_ X
Construct the graphs of the irrational functions: 7t*. y =Vi J/
72. IJ = V X. 73*. y= V? (Niele's parabola). 74. y = ± x Vx ('>emicubical parabola). 2 75*. y= ± V25x (elltpse).
76. y~ 77.
f
± V X 1 I Y= V 1 x 2 . 2
(hyperbola).
78*. y= ~X v~ (cissoid of Diocles). 79. y= ± xV'25x•.
Conslruct the graphs of the trigonometric functions: 83*. y =cot x. 80*. y = sin x. 84*. y=sec x. 81*. y=cosx. 85*. y=cosec x. 82*. y=tanx. 86. Y= A sin x, if A= 1, 10, 1/2, 2. 87*. y= sin nx, if n= 1, 2, 3, 1/2. 3 88. y=sin(xq>), if q>=O, ~. ; . n, ~. 89*. y=5sin{2x3).
(Ch. 1
Graphs of Elementary f•mclions
Sec. 2]
19
90*, y=a sin x+ b cosx, if a=6, b=8. 91. y = Sin X j COS X. 96. y= l2cosx. . x I sm . 3x. 92*. y=co~•x. 97 . y= sm
3
93*. y= x+ sin x.
94*. y = x sin x. 95. y=tan 1 x.
I
98. y= cosx+ 2 cos2x. 1t 99*. y = cos. X
tOO. y=± Vsinx.
Construct the graphs of the exponential and logarithmic funetions: 101. y=a", if a=2, ; , e(e=2, 718 ... )*). 102*. y= toga x, if a= 10, 2, ; , e. 103*. y= sinh x, where sinh X= lj2 (e"e~. 104*. y=co~hx, where co'>hx=l/2(e"+e"). sinh x 105*. y= tanh x, where t an h X=h. COS
X
I
106. y= wx. 2 107*. y = e" (probability curve). I
1
108. y=2Xi.
113. y=log.
109. y=logx 2 • 110. y= log' x. ttl. y =log (log x). I 112. ·l j = og , . x
lt4. y=log(x). 115. y= log 2 (l + x). 116. y =log (cos x).
X
Construct the graphs of the inverse tngonometric 118*. y =arc sin x. 122. y =arc sin
f.
I 119*. y=arc cosx. 123. y=arccos . 7 120*. y= arc tan x. 124. y=x+arccotx. 121*. y=arccotx. Construct the graphs of the functions: 125. Y=lxl. I 126. y= 2 (x+lxj). 127. a) y=xlxl; b) y=logv!lxl. 128. a) y=sinx+lsinxl; b) y=smxlsinxj. 3X2 when 1x1~1. 129. y= { 1~1 when lxl> l. *) About the number e see p. 22 for more details.
functions~
Introduction to Analusis
20
(Ch. 1
130. a) y=[x], b) y=x[x], where [x] is the m~egral part x, that is, the grea~est in:eger less than or equal to x. Constn:ct the graphs of the following functions in the polar coordina.e system (r, 0 we have 6 = 6 (e) > 0 such that If (x) A I< e lor 0 N (e).
The following conventional notation ts also used: lim f(x)=co,
x..a
which means that If (x) I > E for 0 < I xa I < 6 (E), where E is an arbitrary posttive number 3°. Onesided limits. If x a and x ..... a, then we wnte x ..... a +0. The numbers
f (a0) =
lim
x~a~o
f (X) and J (a+ 0) = lim f (x) x+a+o
.are called, respectively, the ltmtl Ol'l the left ol the function f (x) at the point a and the limtt on the rtght of the function f (x) at the point a (if these numbers ex1st).
Sec. 3]
23
Limits
f (x)
For the existence of the limit of a function and sufficient to have the follcwmg equality:
as x ..... a, it Is necessary
f(aO)=f(a+O). If the limits lim x..a
I old: 1) lim [f 1 (x)
f 1 (x)
lim {1 (x)
+
x~a
lim [{ 1 (x)
f 2 (x)) =
x~a
3)
lim { 2 (x) exist, then the following theorems.
x.a
+ fz (x)] =
x~a
2)
and
lim {1 (x); x~a
lim {1 (x) • lim f a(x); x~a
x~a
lim [f.(x)/f 1 (x)J = lim {.(x)/lim {2 (x)
(lim
JC+a
JC+a
x +a
f 1 (x) :F 0).
The following two limits are frequently used: lim sin x = 1 o
X
(1 +..!_)x = Jim
(1
JC+
and lim X+
(I)
I
a ...... o
X
+
a)a =e=2
71828 .. ,
Example 2. Find the limits on the right and left of the functioii I f (x) =arc tanX
as x .... o. Solution. We have
f(+O)= lim (arctan ..!_)=2: X++0
X
2
and f(0)= lim X+ 0
Obviously, the function
f (x)
(arctan..!..)=~. X
2
in this case has no limit as x ..0.
166. Prove that as n oo the limit of the sequence I
I
1 • 4 • · · · • nz • · .. is equal to zero. For which values of n will we have the inequality I
nz
N fulfilled if N is an arbitrary positive number? Find 6 if a) N = 10; b) N = 100; . D c) N = 1000. 291. Determine the order of smallness A ~::;;...o~.....;;::::t~~o... 8 of: a) 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 cirFig. 9 cular 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 L1 ABD.
Infinitely Small and Large Quantities
Sec. 4)
x ___.. 0
293. For
35
determine the orders of smallness relative to
x of the functions: a) b)
2x
1 +x
d) 1 cos x; e) tan x sin x.
;
YxtVx;
c) Vx~
Vx';
294. Prove that the length of an infinitesimal arc of a circle of constant radius is equivalent to the length of its chord. 295. Can we say that an infinitt!simally small segment and an infinitesimally small semicircle constructed on this segment as a diameter are equivalent? Using the theorem of the ratio of two infinitesimals, find . sin 3x·sin 5x 298 • l'tm 1In x • 29 6 . Itm ( •)z • xx
x~o
arc sin
297. lim x~o
Y
x+~
x
299 • lim co~ xcos2x • x~o cosx
x 1 x•
In (1 x)
i
300. Prove that when x0 the quantities and VI +x~i are equivalent. Using this result, demonstrate that when I xI is small we have the approximate equality
Vl+x:::::;l+~.
(1)
Applying formula (1), approximate the following: a) VL06; b) Vo.97; c)
Vw;
d) Vl20
and compare the values obtained with tabular data. 301. Prove that when x> 0 we have the following approxi· mate equalities accurate to terms of order x•: a)
l 1 +x
:::::; 1 x;
b) Va jx;:::;;a+ta (a>O); c) (1tx)n:::::;1jnx (n is a positive integer); d) log(ltx)=Mx, where M=loge=0.43429 ... 1
Using these formulas, approximate: 1 1) l.0 ; 2) _197 ; 3) 1 ; 4) 2 0 105
v15; 5) 1.04'; 6) 0.93'; 7) log 1.1·
Compare the values obtained with tabular data. 2*
Introduction to Analysis
36
(Ch. 1
302. Show that for x.. oo the rational integral function
P (x) = a0X 11 + a,x" 1 + ...
+a
(a 0 =F0)
11
is an infinitely large quantity equivalent to the term of highest degree aox". 303. Let x. oo. Taking x to be an infinite of the first order, determine the order of growth of the functions: a) b)
X
2
l00x 1,000; xs
xj2
Sec. 5. Continuity of Functions
s
1°. Definition of continuity. A function f (x) is continuous when x = (or "at the point s"), if: l) this function is defined at the point 6. that is, there exists a number f (6); 2) there exists a finite limit lim f (x); 3) this limit is equal to the value of the function at the point lim f (x) =
x+1;
x .... ;
6.
i.e.,
f (s).
(1)
Putting
x=s+M. where As 0, condition (1) may be rewritten as lim Lll;>0
Af=
lim lf0
f~
lim
(0) =
t!.x >+o
AX
I Ax I= 1. Ax
4°. lnftnite derivative. If at some point we have lim
f (x + Ax) f (x) _
t!.x> o
Ax
co,
then we say that the continuous function f (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 xaxis. Example 5. Find f' (0) of the function
y= Solution. We have
f'
(0)= lim
t!.x>o
vx:
VAx 1 .= Jim v=co. ux 6.x>o Ax 2
341. Find the increment of the function y= x• that corresponds to a change in argument: a) fromx=1 to X 1 =2; b) fromx= 1 to X1 =I. I; c) fromx= 1 to X1 = 1 +h. 342. Find t!y of the function y a) x=O, t!x=O.OOI; b) X=8, L\x=9; c) x=a, t!x=h. 343. Why can increment t!y if t!x = 5, while for 344. Find the tions: 1
a) y=(x• 2). b) y= c) y =log x
Vx
= V x if:
+
we, for the function y = 2x 3, determine the all we know is the corresponding increment the function y = x• this cannot be done? increment t!y and the ratio ~Y for the funcx
forx=l
and t!x= 0.4;
for x=O forx= 100,000
and t!x = 0.0001; and t!.x =  90,000.
45
Calculating Derivatives Directly
Sec. 1]
345. Find 6.y and!~ which correspond to a change in argument from x to x+ 6.x for the functions: a) y=ax b) y=x';
c) y=~; .X
+ b;
Vx;
d) y= e) y=2x;
f) y=lnx .
346. Find the slope of the secant to the parabola y =2xx 2 , if the abscissas of the points of intersection are equal: a) X 1 = 1, X2 =2; b) X 1 = 1, X 2 =0.9; C) X 1 = 1, X 2 = 1 +h.
To what limit does the slope of the secant tend in the latter case if h +0? 347. What is the mean rate of change of the function y = x• in the interval 1 ~ x ~ 4? 348. The law of motion of a point is s = + 3t + 5, where the distances 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= 1 to t=S? 349. Find the mean rise of the curve y = 2x in the interval
2r
I~x~S.
350. Find the mean rise of the curve y = f (x) in the interval
(x, x+ 6.x}.
351. What is to be understood by the rise of the curve y = f (x) at a given point x? 352. Define: a) the mean rate of rotation; b) the instantaneous rate of rot at ion. 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 m = f (x) be the mass of a nonhomogeneous 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 6.x]; b) the Iinear density of the rod at a point x? at the point 356. Find the ratio ~~ of the function y =.!. X x=2, if: a) 6.x= 1; b) 6.x=0.1; c) 6.x = 0.0 1. What is the deri vative y' when x '" 2?
+
[Ch. 2
Dif!erentiation of Functions
46
357**. Find the derivative of the function y =tan x. 358. Find y' = lim ~ of the functions: 11x+ o
a) y = x'; 1
b) y = Xi;
c) y =
vx;
d) y =cot x.
Vx.
359. Calculate f' (8), if f (x) = 360. Find f' (0), f' (1), f' (2), if f(x)= x(x1)1 (x2)'. 361. At what points does the derivative of the function f (x) = x' coincide numerically with the value of the function itself, that is, f (x) = f' (x)? 362. The law of motion of a point is s = 5tZ, where the distance s is in metres and the time t is in seconds. Find the speed at t =3. 363. Find the slope of the tangent to the curve y = 0.1x1 drawn at a joint with abscissa X= 2. 364. Fin the slope of the tangent to the curve y = sin x at the point (tt, 0). 365. Find the value of the derivative of the function f (x) = _!_ X at the point x = X0 (X 0 =I= 0). 366*. What are the slopes of the tangents to the curves y = l. X and y = x• at the point of their intersection? Find the angle between these tangents. 367**. Show that the following functions do not have finite derivatives at the indicated points:
VV
a) Y= x• b)y= x1 c) y = l cos xl
at X=O; at X= 1; at
2k+l X=rtt, k=O, ± 1, ±2, ...
Sec. 2. Tabular Differentiation 0
I Basic rules for finding a derivative. If c is a constant and u = q> (x), v = 'ljl (x) are functions that have derivatives, then •
1) (c)'= 0;
5)
2) (x)' = 1;
6)
3) (u
± v)'=u' ± v';
4) (cu)' =cu';
(uv)'=u'v+v'u; =vu'v'u
(.!:.)' v,
7) (
~~....
~ )' = ~~·
(u
1: 0);
(u =F 0).
Sec. 2]
Tabular Dlf!erentiation
47
2°. Table of derivatives of basic functions I. (xn)' =nxn1.
1
II. ( Vx)' =
,r (x > 0). 2 I' X III. (sin x)' =cos x. IV. (cos x)' =sin x. V. (tan x)'
= COS \. X 1
VI. (cot x)' = Stn . 1 • X 1
VII. (arc sin x)' = .r (I xI < 1). r 1x• 1
(I xI < 1).
VIII. (arc cosx)' = .r r 1x• ,
I
IX. (.trCiar x) = 1 +x•· 1
X. (arc cot x)' = x•+ 1 • XI. (aX)'=axlna. XII. (ex)' =ex.
XIII. (In x)' =~ (x > 0). X
X IV. (log 4 x)' =
 1 = loga e x
x 1na XV. (sinh x)' =cosh x. XVI. (cosh x)' =sinh x.
(x
> 0, a> 0).
1
XVII. (tanh x)' =  h 2
X
COS
XVIII.
•
1 . sm x
(cothx)'=~h•
XIX. (.ncsinhx)'= XX. (arc cosh x)'
XXI. (arc tanh x)' =
~· 1
+x
I
,r
1
r x1 1 x 1~
(I xI> 1). (I xI< 1).
I
XXII. (arccothx)'=1(lxl>1). x 1 30. Rule for differentiating a composite function. H y=f(u) and u=q>(x), that is, y = f [q> (x)], where the functions y and u have derivatives, then , Yx=YuUx (1) or in other notations dy dy du dx= du dx' This rule extends to a series of any finite number of differentiable functions.
DtUerentiation of Functions
48
[Ch. 2
Example I. Find the derivative of the function
y= (x1  2x+ 3)5• Solution. Putting y=u', where u=(x1 2x+3), by formula (1) we will have y' = (u 5 )~ (x1 2x + 3)~ = 5u 4 (2x2) = lO (x1) (x 2 2x 3)'.
+
Example 2. Find the derivative of the function y=sin1 4x. Solution. Putting
y=u';
u=sinv;
v=4x,
we find y' =3u 2 ·cos v·4 = 12 sin 2 4x cos 4x.
Find the derivatives of the following functions (the rule for differentiating a composite function is not used in problems 368408).
A. Algebraic Functions I
368. y=x'4x'+ 2x3. 369. y=!
~ x+x'0.5x'.
~
375. y=3£i 2xl tx•. 376*. y=x·
Vx•.
v v· a
b
370. y= ax 2 + bx +c.
377. Y=
5x' 371. y=. a
a+bx 378. Y = c+dx.
372 y = atm + btm+n. ax'+b
X2
X
X
2x+3
379 · Y = x1 5x +5 · .11
2
I
373. Y= Ya•+b•'
380. y=  X . 2X1
374. y=..::..+1n2. X
381 . Y
382. 383. 384 .
_t+
vz
.r 1 t' z
B. Inverse Circular and Trigonometric Functions y=5 sinx+3cosx. 386. y =arc tan x + arc cot x. y =tan x cot x. 387. y=xcotx. =slnx+cosx 388. y = x arc sin x. y sin XCOS X'
385. y=2t sin t{t'2) cost.
389. y=
(1
+x
2
)
arc tan x x 2
Sec.
49
Tabular DiOerenttation
2)
C. Exponential and Logarithmic Functions 390. f/ =
X
7
·ex.
396. y =ex arc sin x. x•
391. y=(xl)ex. ex 392. Y=xz.
397. y=1 nx . x• . 398. y=x • lnx 3
xs
I
lnx
393. Y= ex.
399. y=+21nx. X X
394. f (X) =ex COS X. 391i. y=(X 2 2x+2)ex.
400. y=inxiogxina logax.
D. Hyperbolic and Inverse Hyperbolic Functions 401. y= X Stnh X.
405. y =arc tan x arc tanh x.
x• 402 · Y= cosh x ·
406. y =arc sin x arc st nh x.
403. y=tanhxx.
407 . Y
x 404 Y = 3coth \nx
408 ' Y 1x•
_arc cosh x X
•
_arc coth 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 funchons:
409**.
y=
(1
+ 3x sx•r.
Solution. Denote I+ 3x5x2 = u; then y = u••. We have: y~ = 30u 11 ; u~ = 3 lOx; ~~~
410.
c= 30u 21 ·(310x) =30 (I+ 3x5x 2) 21 ·(310x).
u= (axe+ b)'.
411. f(y)=(2a+3by)'. 412. y = (3 2x1 )&.
+
413 · y =56 (2:1) 7
vr::xr.
24
414. Y= 415. y= Va+bx'. 416. Y= (a''•x''•)'/•.
(2~1)1
40 (2x1)1 '
[Ch. 2
DifJerenttation of Functions
50
417. y=(32 sin x)'.
y' = 5 (32 sin x)'·(32 sin x)' =5 (32 sin x)' ( 2 cos x) = Solution.  10 cos x (32 sin x)'.
y=tanx ~ tan'x+ ~ tan'x.
418.
419. y=
 vcot a. Vcotx
1 1 423. y= ~.
3 cos' x cos x _ ,/3sinx2c osx
y= 2x + 5 cos• x.
420.
421*. x=cosee t +sec•
424 · Y
t.
V
5
'
425. y= V sin' x+ co!•x.
I
422 · /(X)=6(1 3cosx) 1
'
Vt
+arc sinx. 426. y= 427. y = V arc tan x (arc sin x)'. I
428 · Y =arc tan x · 429. y=Vxe"'+ x. 430. y= t/2ex2x 1 +In' x. 431. y=sin3x+ cos ~ +tanVx.
+
Solution. y' =Cos 3x·(3x)' sin I . x   sm 
5
5
I + :==
2 V x cos•
Vx
~
(
i )' +cos•
1 }" x (y.X)' =3 cos 3x
·
432. y= sin (x'5x+ l) +tan~. X 433. f (x) =cos (ax+~). 434. f (t) = sin t sin (t + =.!_. r
dr
r'
The tangent MT and the norma\ MN at the point M together with the radius vector of the point of tangency and with the perpendicular to the radius vector drawn through the pole 0 determine the following four segments (see Fig. 14):
X T Fig. 14
t = MT is the segment of the polar tangent, n = MN is the segment of the polar normal, S 1=0T is the polar subtangent, Sn =ON is the polar subnormal.
[Ch. 2
Differentiation of Functwns
62
These segments are expressed by the following formulas:
t= MT =!, Ir I
Yr + (r') 1
1
,z
;
St =OT = j?"j;
sn =ON= I r' I·
n = MN = y=,z+:(:,,::::)1;
621. What angles cp are formed with the xaxis by the tangents to the curve y = xx• at points with abscissas: a) X=O; b) X= lj2; c) X= 1?
r
Solution. We have y' = 12x. Whence a) tan
, if: a) y=x·e";
d) Y=
y=x•ezx; c) y = (lx 2 ) cos x;
c)
b)
691. Find [ (0), if f (x) =In
1 1
l+x
Yx;
y=x'Inx.
x
B. 1/tgherOrder Derivatives of Functions Represented Parametrically and of Implicit Functions
In the following problems find :~ . c) { x=arcsint 692. a) { x=ln t,. b) { x=arctant, u=t'; u=ln(t +t•>; u=Vtt•. . a) { x=acos t, c { x=a(tsint), 693 y =a sin t; ) y =a ( 1cost); x=acos't, { x=a(sinttcost), b) { y=asin't; d) y=a(cost+tsint). x=cos2t, { x=arctant, 694. a) { y __ si·n• t· 695. a) 1 t• ' y='i ; x=eat, { x=lnt, b) { at b) I y=e • Y=1t' .
d'x
696. Fmd d., Y
. {
If
x = e' cos t, · t y=et sm •
11 ,
[Ch. 2
DtUerentiation of Functions
70
+/
1
), . { X = In ( 1 y=tz. 697. Fmd dx for 1=0, 1f 698. Show that y (as a function of x) defined by the equations x =sin t, y = aet Y2 +bet Y! for any constants a and b satisfies the differential equation
day
.
2
2
d 2y
dy
(1x )dxz xdx=2y.
In the following examples find y"'
x=sect,
t 699. { Y= ant.
701.
= :a;a .
{ x=et, 3 y=f.
1
. { X=lnmt,
dny
.
x=e cost,
702. Fmd dn, 1f _ .• t 700. { y=t . x y = e 1 sm . 703. Knowing the function y = f (x), find the derivatives x", x'" of the inverse function x = f 1 (y). 704. Find y", if x2 + y 2 = 1. Solution. By the rule for differentiating a composite function we have y xy' d !/' =  X )' =~ X . 2x+ 2yy ' = Q; whence y ' =an y2 y X y Substituting the value of y', we finally get: 1 y2+x2 , !I =qs= LJ'.
l'
In the following examples it IS required to determine the derivative y" of the function y=f(x) represented implicitly. 705. Y2 ; 2px. xz
yz
706. (i2 + fj2 = 1. 707. y=xtarctan y. d2 x d2 11 . 708. Having the equation y = x +In y, fmd dx 2 and dya. 709. Find y" at the point (1,1) if
x' +5xy+y 2 2x+y6=0. 710. Find y" at (0, 1) if
x'xy+y'= 1. 711. a) The function y is defined implicitly by the equation x• + 2xy + y' 4x + 2y 2 = 0. Find
:~
at the point (1, 1).
b)
• d d'y Fm dx•,
'f
1
I
I
I
x t y =a.
Sec. 6]
DiUerentials of First and Higher Orders
71
Sec. 6. Differentials of First and Higher Orders 1°. Firstorder differential. The dif!erential (firstorder) of a function y = f (x) is the principal part of its increment, which part is linear relative to the increment t..x=dx of the independent variable x. The ditJerentlal of a
y
0 Fig. 19 function is equal to the product of its derivative by the differential of the independent variable dy=y'dx, whence t/=dy. · dx
If M N is an arc of the graph of the function y = tangent at M (x, y) and PQ=t..x=dx,
f (x) (Fig.
19), MT is the
then the increment in the ordinate of the tangent
AT=dy and the segment AN= fly. Example 1. Find the Increment and the differential of the function y=3x 2 x. Solution. First method: t..y = 3 (x+ flx) 1 (x+ flx) 3x2 +x
or
fly=(6x1) t..x+3(flx) 2 • Hence,
dy = (6x1) t..x = (6x 1) dx. Second method:
y' =6x1; dy=y' dx= (6x1) dx. Example 2. Calculate fly and dy of the function y=3x 2 x for x= 1 and flx=O.Ol. Solution. t..y= (6x1)· t..x+ 3 (6.x) 2 =5·0.01 +3·(0.01) 1 =0.0503 and dy=(6xJ) 6.x=5·0.01 =0.0500.
72
Differentiation of Functions
[Ch. 2
2°. Principal properties of ditferentials. 1) dc=O, where c=const. 2) dx= !J.x, where x is an independent variable. 3) d (cu)=cdu. 4) d(u ± v)=du ± dv. 5) d (uv)=u dv
6) d
+ v du.
v duu dv (.!!...)= v v
(v
2
¥= 0).
f' (u) du. 3°, Applying the dilferential to approximate calculations. If the increment Ax of the argument x is small in absolute value, then the differential dy of the function y = f (x) and the increment !J.y of the function are approximately equal: !J.y::::::dy, that is, f (x !J.x) f (x) ::::::: f' (x) !J.x, whence f (x !J.x) ::::::: f (x) f' (x) dx. 7) df (u) =
+ +
+
Example 3. By how much (approximately) does the side of a square change If its area increases from 9 m 2 to 9.1 m2 ? Solution. If x is the area of the square and y is its side, then
Y=Vx. It is given that x=9 and !J.x=O.l. The increment !J.y in the side of the square may be calculated approximatrly as follows:
!J.y::::::dy=y' !J.x=
2
~ry
9
·0.1=0.016m.
4°, Higherorder dilferentials. A secondorder differential is the differential of a firstorder differential: We similarly define the differentials of the third and higher orders. If y = f (x) and x is an independent variable, then
d 2 y = y" (dx) 2 , d'y=y'" (dx) 1 ,
dny = y (x), then
d 2y=y" (du) 1 ty' d1u, d'y= y'" (du) 1 +3y" du·d 2 u y' d 1 u
+
and so forth. (Here the primes denote derivatives with respect to u).
712. Find the increment ~y and the differential dy of the function u=5xt x• for x=2 and L\x=O.OOl.
Sec. 6]
DiOerenttals of First and Higher Orders
73
713. Without calculating the derivative, find 3 d (lX ) I for x = 1 and ~x = ;;... . 714. The area of a square S with side x is given by S=x•. Find the increment and the differential of this function and explain the geometric significance of the latter. 715. Give a geometric interpretation of the increment and differential of the following functions: a) the area of a circle, S = nx 2 ; b) the volume of a cube, V=X 3 • 716. Show that when ~x+0, the increment in the function u= 2x, corresponding to an increment ~x in x, is, for any x, equivalent to the expression 2"'"ln 2 t1x. 717. For what value of x is the differential of the function y = X 2 not equivalent to the increment in this function as ~x.. 0? 718. Has the function u=\X\ a differential for X=O? 719. Using the derivative, find the differential of the function Jl .1t y =cos x for x = 6 and ~x = 36 720. Find the differential of the function 2
Y=yx for x=9 and ~x=0.01. 721. Calculate the diiTerential of the function y ==tan x Jl .1t for x= 3 and t1x= 180 • In the following problems find the differentials of the given functions for arbitrary values of the argument and its increment. 722.y,.=}m· 727.y=xlnxx. 723.
lj
=
I
728. y=ln
X X •
. 724. !J= arC Sltl
a. X
1x 1
+x'
729. r =cot
0
X
. (Ieos x)·l . (Ieos x) cos x =lun Solution. lim (1cosx)colx=hm x.o
X>0
=lim sin x .0
X>0
SUI X
Sill X
=
'=0
CtJS \'
788. lim ( 1  x) tan ~x
792. lim x" sin !:. , n>U.
•
x
xr.
x.,.1
793. hmlnxln(x1).
789. lim arc sin x cot x. X>0
I ) 794. I .IITI ( \  
790. lim (x"e x), n>O.
x +l
r10
xI
In x
•
. 791. lim x sin !:.. X . x In xx +I = . ( x  I ) =hm Solution. hm x..1 (xl)ln\ lnx x.1 .\I I I x+lnx1 .\ lnx x = hm~~=2. I =lim I =Jim ~'+2 x.'ln.\ +lJnxt I x x x x
) 5 I . ( ·3795. I llTI . . 2 _ 6 .\ X X X >3
. [ 796. I ltll x+1
79~
I . 2 ( I  }I x)
3(1t/ x)
].
X  l1 ) ( • hm n cotx 2cosx
t>2
798. lim xx. x>0
Solution.
We
iny=xlnx:
lim lny=llmxlnx=x..o
x
lnx
= Jim1 = 0, lim== .¥+0 _1 X+0 _
X
_
X2
whence limy= 1, that .¥+0
ts, Inn.\"'= l • X+0
x..o
82
[Ch. 2
Differentiation of Functions 1
1
804. limx1x.
799. limxi".
f+1 I
800. limxHin
805. lim( tan
x.
X+0
X+1
7)
tan~ 6
•
1
806. lim (cot x)iii%. X+0
X+0
nx
807. lim( .!_)tan x •
802. lim (1x)cosz.
X+0
f+1
X
1
803. lim (1 + x2)"i.
808. lim (cot x)~ 10 %, X+0
X+0
809. Prove that the limits of l
1'
x• slnX
a} 1~Si"ii"X=0;
b) lim xs~n x x+ao x+smx
cannot be found by the L'HospitalBernoulli rule. Find these limits directly. /)
Fig. 20
810*. Show that the area of a circular segment with minor central angle a, which has a chord AB=b and CD=h (Fig. 20), is approximately
with an arbitrarily small relative error when a0.
Chapter Ill
THE EXTREMA OF A FUNCTION AND THE GEOMETRIC APPLICATIONS OF A DERIVATIVE
Sec. 1. The Extrema of a Function of One Argument 1°. lncrea~e O when x>2. it follows that x= 2 is the minimum point of the function, and Ymln=2. We get the same result by utilizing the sign of the second derivative at the critical point y"=2>0.
827. y=2+xx•. 828. y=x'3x'+ 3x +2. 829. y=2x'+3x1 12x+5.
Solution, We find the derivative y'=6x1 +6x12=6 (x'+x2). Equating the derivative y' to zero, we get the critical points x 1 = 2 and x1 = I. To determine the nature of the extremum, we calculate the second derivative u"=6(2x+l). Sinee y"(2)0; therefore, x1 = I is the minimum point of the function y and Ymln= 2.
830. y=x 1 (x12) 2 • 831. y=x (x1) 1 (x2)'. x'
832. y= x• + 3 .
840. y = 2 cos ~ + 3 cos
i.
841. y=xln(l+x).
1
833. y=
x 2x+2 xl
834. y = (x2)x~8x) • 16
835. y = x (4x•).
842. y=xlnx. 843. y=x tn• x. 844. y =cosh x.
4
836. y = vX~+B. x•
837. y= 838.
+
845. y=xe".
vx 4 . X
1
u= V (r 1)'.
848. y = x arc tan x. 839. y = 2 sin 2x +sin 4x. Determine the least and greatest values of the functions on the indicated intervals (if the interval is not given, determine the
88
Extrema and the Geometric Applications of a Derivative
[Ch. 8
greatest and least values of the function throughout the domain
.of definition). X
853. y = x• on the interval [ 1 ,3].
849. y= 1+x*. 1:;:0x~). 850. y = 4 851. y = sin x + cos 4 x.
v·x;:(
854. y=2x1 j3x1 12x+1 a) on the interval r1 ,5]; b) on the interval [ 10, 12].
852. y =arc cos x. 855. Show that for positive values of x we have the inequality I x+;;;z,2. X
856. Determine the coefficients p and q of the quadratic tri·nomial y=x• +px +q so that this trinomial should have a minImum y = 3 when x = 1. Explain the result in ~eometrlcal terms. 857. Prove the inequality 1
110
> 1+ x
when x + 0.
Solution. Consider the function f (x) =1110 (1 +x). Jn lhe usual wa"j we find that this function has a single minimum Hence, f (x) > f (0) when x :1: 0, and so e110 > l +x when x ::p 0,
f (0) =0·
as we set out to prove.
Prove the inequalities: x• 858. x 6 < sinx 1 2 x2
when x>O. when x+O.
860. x 2 ++"1.' 2)
1f the curve
IS
g1ven by an equation implicitly, f(x, y)o.=.O, then
f:ll
F~
F:x
F~y F~
F~
F,,
0
F;x
K= {F'z + f'2)'/• X • y 3) if the curve is represented by equation> in parametric form, x = y = ¢ (t), then x' yl \
K=
I
1
W('
dx
1
= rft'
IJ
(I),
x" y" ' (x1z y'2)'1•
+
where X
!p
d2 x
dy
=dt' x" =(ill·
In polar coordinates, when the curve is given by the equation r=f (:p), have
where
r
I
dr
=
dcp
an d
d"r ,. = d!p2.
3°. Circle of curvature. The ctrcle of curvature (or osculatitt~,? ctrcle) of a curve at the po111t M IS the limittn~ positton of a circle drawn throul(h M 01nd two other pomts of the rurve, P and Q, as P+ M and Q  • M. The r;1dius of the ctrcle of curvature is equal to the radius of curvature, and the centre of the circle of curvature (the centre of curvature) lies on the nortnal to the curve drawn at the point M in the dtrectton of roncavity of the curve. The coordinates X and Y of the centre of curvature of lh 0). a
Since
Solution.
y' =sinh!__ and y•=..!..cosh_:_, a u a
and, hence, a cosh 2
_:_
it follows that K=
X R =a cosh 2 !__. We have dR d =sinh 2a x a
Equating
a
the derivative ddR to zero, we get sinh 2 _::. = 0,
a
x
whence we
critical point x=Q Computing the second derivative
~x~
find the
~ole
and putting mto
o=~>O. wegetddx~~ X= o=~cosh2!_1 a a X= a 2
it the value x=O,
Therefore,
x=O is the minimum point of the radius of curvature (or of thP maximum of curvature) of the catenary. The vertex of the catenary y=acosh_::. is, a thus, the point A (0, a).
Find the differential of the arc, and also the cosine and sine of the angle formed, with the positive xdirection, by the tangent to each of the following curves: 993. X2 + Y2 = a 2 (circle). yz xz 994. {i:i jj2 =I (ellipse). 995 y 2 = 2px (parabola),
+
Di(Jerenttal of an Arc. Curvature
Sec. 5]
105
+
996. xzfa yzfa = a 2i 1 (astroid). 997. y =a cosh:__ (catenary). a
998. x=a(tsint); y=a(lcost) (cycloid). 999. x=acos't, y=asin't (astroid). rind the diiTerential 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 = aqJ (spiral of Archimedes). (hyperbolic spiral). tOOt. r=!!.. cp 1002. r=a sec'; (parabola). 1003. r =a cos' ;
(cardioid).
1004. r = aq> (logarithmic spiral). 1005. r 2 =a 2 cos2qJ (lemniscate). Compute the curvature of the given curves at the indicated points: I 006. y = X 4 4x'l8x 2 at the coordinate origin. 1007. x 1 +xy+y"=3 at the point (1, 1). xz 11 z 1008. 2 +ijz =I at the vertices A (a, 0) and 8(0, b). 1009. x=l\ y=t' at the point (I, 1). 1010. r 2 =2a 2 cos2qJ at the vertices .
Whence
+" \ . dx2 . v Jl x +4 A
.\•+4x' =(3Ax"+2Bx+C) Vx•+4+(Ax'+Bx"+CxjD)x +
v x•j4
}r.l. 2 +4
.
V.l. 2 j4
Multiplying by Vx 2 +4 and equating the coefficients of identical degrees of x, we obtain Hence,
s
2 x•JI x2 J4 dx= xs ~ x V x•j4 2ln (x+ Vx• +4) +C.
3°, Integrals of the form
J
(xa)n
::x• + bx +c.
{4)
They are reduced to tnll'grals of the form (2) by the substitution: _1_=1 .
.\a
Find the integrals: 1326.
• x•d~ J' y,,•~+I
1329.
1:327.
r I.\" Jl ,r;:_==.dX.
1330.
I :328.
r ,r Ix•·i x• dx.
.l
1
1331 .
4°. Integrals of the binomial differentials
~ xm (a+ br")P dx,
(5)
where m, n and p are rational numbers. Cheby~hev's conditions. The integral (5) can be expressed in terms of a finite combination of elementary functions only in the following three cases: I) 1f p i~ a whole number; 111 1 2) if + is a whole number. Here, we make the substitution a+ bxn = n = z8 , where s is the denominator of lhe fraction p; 111 1 + p is a whole number. Here, use is made of the substitution + 3) if n 3 axn+b=z ,
[Ch. 4
Indefinite Integrals
128 Example 3. Find
1
1
1
Solution. Here, m=  2 ; n=4; P=3;
}+t
1 m+ n
1
=2. Hence,
4
we have here Case 2 integrability. The substitution I
+x 4 =z'
yields x=(z 8 1) 4 ; dx=12z 2 (z 1 1) 1 dz Therefore,
t/x.
where z=VI+
Find the integrals:
1332. ~
X
3
+ 2x
(1
s 2
1333.
svdx .
1334.
.r
)
dx ltx•
1336. ,.
dx
S.x
4
I'
1 fx
, 5
1 f.x• dx
v
1335. ,x· \
'
2 dx.
Jx2 (2fx 3 ) 3
v + Vxf . J V.x• •
1337. ~
. 2
dx
1
Sec. 7. Integrating Trigonometric Functions 1°. Integrals of the form
~ sinmxcosnxdx==lm, n• where m and n are integers. l) If m= 2k + 1 is an odd positive number, then we put
I m, n = 
~ sin 2k x cosn xd (cos x) =  ~ ( 1cos2 x) 1l cosn xd (cos x).
We do the same if n is an odd positive number. Example t. ~ sln10 x cos1x dx = ~ sh110 x ( 1 sln 2 x) d (sin x) = sin 11 x sin 13 x
= 11 r3 +c.
(1)
129
Integrating Trigonometric Functions
Sec. 7)
2) If m and n are even positive numbers, then the integrand (1) is transformed by means of the formulas sln2 x=
I
2
cos• x=
(Ieos 2x), sin x cos x=
I
2
(I+ cos 2x),
1 2 sin 2x.
Example 2. ~ cos• 3x sin 4 3x dx = ~ (cos 3x sin 3x) 2 sin 2 3x dx = 2
6x dx= I  4 6x 1cos = sln 2 8
S ~
=
S(lc~s 12x
= _!_ ( ~ 8
s
(sin 2 6xsln 2 6x cos 6x) dx=
sin 2 6x cos 6x) dx=
sin 12x _!_ sins 6x ) 18
24
2
+ C°
3) If m =  ~t and n =  v are integral negative numbers parity, then
Im
'
11
= S
=
Sill
f'
,
dx
X COS X
=Scosec~'xsec••xd(tanx)= 
S(
of identical
~
~+v
V2
1 )" (l+tan 2 x)  2 I+ 1an 2 x \
1 2
2
x) t .• d(tanx)= s(l+tan an• x
d (tan x}.
In particular, the following integrals reduce to this case:
Example 3.
=S cos• x S...!!:_
sec• xd (tan x) =
=lan Example 4.
=..!_ 8
x+ 3I
SSin~dt x = ¥1 ~ sin
r
1

tan 1
S(I+ tan
x+C.
x xdx;· cos 1 2 2
BI
2
x) d
s
tan a
(tan x) =
s X d X 2 sec 2 x =
~ (I +tan• i sec 2 ..::_dx=~~ [tana ~+2 + t an 3
X
2
2
8
2
t an
X
2
1 ~ ;] +C. +tan ~] d( tan f)=! [  x +2ln j tan ~ j + 2tan• 131
2
51900
2
[Ch. 4
Indefinite Integrals
130
4) Integrals of the form ) tanm x dx (or ) cotm x dx), where m is an In· fegral positive number, are evaluated by the formula tan 1 x=sec1 x1 (or, respectively, cot• x = cosec• x 1). tan• x dx = tan• x (sec• x1) dx = Example 5.
s
5
5
ta~• x
5tan• x dx =
 x tan x + x +C. (sec• x1) dx= tan• 3 5) In the general case, integrals I m n of the form (I) are evaluated by means of reduction formulas that are usually derived by integration by parts. 2 s sln x+ cos• x d s dx E x= cos' x cos' x = xamp 1e 6· s dx = cos x 1 s dx+ 1 . dx s =smx·sin x slnx·dx+ cos x 2 cos• x 2 cos• x cos x cos' x I sin x o : : : :2  + 1njtanx+secxl+C. 2 COS X 2
 3x= tan•
=S
Find the integrals: 1338. ~ cos• xdx. 1339. ) sin'xdx. 1340. ~ sin• x cos' x dx. · · x cos ,xdx. 1341. ssm 2 2 cos' x d 1342. sin' x X.
s
1352. )
dx X X sin cos1 2 2
) sin ( x+ ~ ) sin x cos x dx. dx 1354. sin' x ·
1353.
5
1355. ~sec' 4x dx.
~ ~in' xdx.
1356. ~ tan• 5xdx.
1344. ) sin~ x cos• x dx.
1357. ~ cot• xdx.
1345. ~ sin! x cos• x dx.
1358. ~ cot• x dx.
1346. ~ cos6 3x dx.
1359.
dx4 1347. s sin x' dx 1348. s cos• x· d x cos• 1349. sin 8 x X.
1360. ) x sin 1 x• dx. cos• x 1361. stn• x dx.
1350. s sln1 xdxcos• x • dx 1351. sin1 x cos• x •
1363.
1343.
5
s
5\ tan• 3 + tan• 4 1
X
X)
5
1362. ) sin' X Vcosxdx.
1364.
sY
s
dx sin x cos• x •
dx Ytanx'
dx.
I ntegrattng T rigonometrlc Funct tons
Sec. 7]
131
2°. Integrals of the form ~ sin mx cos nx dx, ~ sin mx sin nx dx and
~cos mx cos nx dx. In these cases the following formulas arc usedl l) sin mxcos nx= {[sin (m+ n) x+sin (mn) x]; 2) sin mx sin nx = 3) cos mxcos nx=
I
2 I
2
[cos (m n) xcos (m + n) x); [cos (mn) x +cos (m + n) x].
Example 7. Ssin9xsinxdx=S
=~sin8x~sin
~
[cos8xcosl0x]dx=
lOx+C.
Find the integrals: 1365. ~sin 3x cos5xdx.
1369. ~ ros(ax ~ b)cos(axb)dx.
1366. ~sin lOx sin 15xdx.
1370. ~sin rot sin (rot +cp) dt.
1367. 1368.
s s·
cos 2X cos 3X dx.
1371. ~cos x cos• 3x dx.
.2Xd Sin 3X Sin 3 X.
1372. ~ sin x sin 2x stn 3x dx.
3°, Integrals of the form
~ R (sin x, cos x) dx,
(2)
where R ts a rational fttnction. I) By means of substitution
tan
X
2
=t,
whence .
l1 2
21
2dt
stnx=l+t•• cost'=l+t•• dx=l+t~' integrals of form (2) are reduced to Integrals of rational functions by the new variable t. Example 8. Find dx 1 l+sin x+cos x '
S
Solution. Putting tan ~ =t, we will have 2dt
I=
s•
S
l +t•
2t lt 2 l+ 1 +ta+ 1 +t 1
[Ch. 4
Indefinite Integrals
132
2) If we have the identity R (sin x, cos x) eaR (sin x, cos x), lhen we can use the substitution tan x = t to reduce the integral (2) to a rational form. Here, l t , cosx= , r slnx= ,r r I+ t• r l + t•
and
dt x=arc tant, dx=l+t', Example 9. Find
dx
S1+sin" x
Solution. Putting tanx=t,
we will have I=
r
1
t• sln'x= 1 +t•,
dt
J (l +tl) ( 1+ 1 ~ t•)
(3)
·
dt dx=1+t•'
s~='S d(ty'2) 1 + 2t• V2 1 + (t y2 )I
.r1 .r1 arc tan (t r 2)+C= Y"2 arc tan ( r 2 tan x)+C. = Y2" We note that the integral (3) is evaluated faster if the numerator and denominator of the fraction are first divided by cos 2 x. In individual cases, it is useful to apply artificial procedures (see, for example, 1379).
Find the integrals: 1373.
s3+~:osx'
1374.
dx Ssin x+cos x'
1375.
Sl+cosx
1376.
Slsin x
cosx
sin x
d d
1382* ·
dx S3 sin x +·,s=co;;s•:x ·
1383*
•
dx Ssin '+3sinxcosxcos x'
•
Ssin
X.
1384*
X.
1385.
1377.
s
1378.
Scos x + dx2 sin x + 3 ·
dx 84sinx+7cosx'
1386. 1387•
2
2
2
2
dx x5sin x cos x ·
sin x S(Ieos
x)'
dx
·
sin 2x 1 +sin• x dx.
S d cos 2x Scos x+sin x X. 4
4
1379**,
d x+2 3 s2smx+3cos x
1380.
l tanx dx 1tan x '
Ssin x6sinx+5 dx 1389*. S (2sin x) (3sin x) ·
5l+:~os•x'
1390*. sls~nx+cosx dx 1+smxcos x ·
1381*.
S~n
COS X
s+
X.
1388 '
COSX
d
1
X.
Integration of Hyperbolic Functions
Sec. 8)
133
Sec. 8. Integration of Hyperbolic Functions Integration of hyperbolic functions is completely analogous to the inte• gration of trigonometric functions. The following basic formulas should be remembered: 1) cosh 2 xsinh 2 X= 1; 2) sinh 2 x =
~
3) cosh 2 x=
2 (cosh 2x+ 1);
(cosh 2x1);
l
4) sinh x cosh x =} sinh 2x.
Example t. Find
~ cosh 1 x dx. Solution. We have
5
cosh 2 x dx=
5~(cosh
2x+ l)dx=
~sinh
2x+
~
x+C.
Example 2. Find
~ cosh 1 x dx. Solution. We have
~ cosh 1 xdx= ~ cosh 2 xd (sinh x)= ~ (l+sinh 2 x)d(sinhx)= . =stnh
sinh x x+3 +C. 1
Find the integrals: 1391. ~sinh' x dx.
1397. ~tanh 1 xdx.
1392. ~ cosh 4 x dx.
1398. ~ coth 4 x dx.
1393. ~sinh' xcosh xdx.
1399.
5sinh x +dxcosh x • 1400 · 52 sinh x ~ 3 cosh x '
1394. ~ sinh• xcosh 2 xdx. 1395 · 5smhxd:osh 2 x' 1396·
5
sinh 2
1401*.
:~osh 2 x •
1402 •
2
2
5
tan::1 •
s
sinh x dx • Ycosh 2x
Sec. 9. Using Trigonometric and Hyperbolio Substitutions for Finding Integrals of the Form
~R(x,
where R ls a rational function.
Yax
1
+bx+c)dx,
(1)
Indefinite Integrals
134
[Ch. 4
Transforming the quadratic trinomial ax 1 + bx+c into a sum or difference of squares, the integral (1) becomes reducible to one of the following types of integrals: 1)
~ R (z,
2) 3)
2
z 2) dz;
j R (z,
Ym Ym
1
+z 2 ) dz;
~ R (z,
Y Z2
m 2) dz.
The latter integrals are, respectively, taken by means of substitutions: 1) z=msint or z=m tanht, 2) z=m tan t or z=m sinh t, 3) z = m sec t or z = m cosh t. Example 1. Find
Solution. We have
x2 + 2x+ 2= (x+ 1)•+ 1.
Putting x+l=tanz, we then have dx=sec 2 zdz and I=
r
2
dx
J (x+l)'Y(x+l)'+l=
s sec2 zdz tan zsecz
s~dz= 2 sin z
= __1_ + sin z
Y x + 2x + 2 + 2
C=
x+ I
C '
Example 2. Find Solution. We have
x2 +x+1= ( x+ 21 )
Putting
. V3 t x+'2=2smht
we get
I=
2
¥3
and dx=coshtdt, 2
t) V32 cosh t. V32 cosh t dt=
S( V3.
 smh t2 2
3s
smh t cosh 2 t dt=3JF3s. 8 8
cosh 1 t dt =
1 ) 3 ( 1 . t cosh 1 V3 smh t cosh tj=32 t 8 2 3 8  
Since sinh t= and
3 +;r·
~ 3 ( x+ ; ) , cosh t=
t=ln
r,
Yx 2 +x+ 1
(x+ ~+Yx 1 +x+t)+ln
;
3
,
+ C•
Sec. 11]
Using Reduction Formulas
135
we finally have I
1 1 +x+l) 1 ~41 ( x+2I) /=3(x Yx 1 +x+1
~ In (x+ ~ + Yx 2 +x +1) +C. Find the integrals: 1403. ~ V32xx• dx.
1409. ~Vx•6x7dx.
1404. ~V2+x 2 dx. "
x• Y J
1405. I
dx.
9+ x 2
1407. ~ V x• 4 dx. 1408. ~ Vx• tx dx. Sec. 10. Integration of Various Transcendental Functions
Find the integrals:
+ 1)
1415. ~ (X2
1
e2 x dx.
5
1421.
dx
e•"+e" 2 •
sVe•" +e" +I
1416. ~ x• cos• 3x dx.
1422.
1417. ~ x sin x cos 2x dx.
1423.
1418. ~ ezx sin' x dx.
1424. ~ ln 1 (x
1419. ~ ex sin x sin 3x dx.
1425. ~ x arccos (5x2) dx.
1420. ~ xex
1426. ~ sin x sinh x dx.
COS X
dx.
dx
•
5x•ln : +;dx. + V1 +X
1
)
dx.
Sec. 11. Using Reduction Formulas
Derive the reduction formulas for the following integrals: 1427. In=
5(x•!xa•)n;
find I 1 and I,.
1428. In=~ sinn xdx; find I, and I,.
[Ch. 4
Indefinite Integrals
136
1429. In= Sco~~ x ;
find I 1 and I 4 •
1430. In=~ xnexdx;
find 110 •
Sec. 12. Miscellaneous Examples on Integration
1432. 1433.
Sxt~;:_ + 2 dx.
1449.
dx.
1450.
5
r
I
x'
J x2+x+2
1438. 1439.
3)" • I).
Sx _::a + 1 •
s·
1441.
S Vx~/ I)• dx.

4
x
(I2Y x)"
dx.
1442.
sY + x+
1443.
~ ~x2:; dx.
1444.
J(V x•+ V x ) .
1445. 1446. 1447.
sY
s
dx
I
1459.
.
dx
2
dx 2x+ I · (4x 2 2x+ l)a dx
xz
(x 1 1)3
r
dx
J x Yx +x+ 1
1
•
.
2
dx.
s /X
x•  I dx
x4
" jy
,
y I x• .
.j/
dx
I
+x•
JV +x 5 x
4
I
•
dx.
1460. ~ cos 4 x dx.
Vsx+ Y5x ·
JJl
2
1457. \ v x
1458 .
0 and q > 0. 1575*. Prove that the Euler integral of the second kind (gam
mafunction)
..,
r (p) =
~
Xp!8x
dX
0
converges for p > 0. Sec. 4. Change of Variable in a Definite Integral
If a function f (x) is continuous over a...;xc;;;b and xcp (t) Ia a function continuous together with its derivative 0).
0
Solution. We put
x=asint; dx = a cos t dt. Then t=arcsin~ a
and,
consequently,
we can
take a=arcsin0=0,
~=arc sin I= ~ . Therefore, we shall have lt
a
2
S x• Va
x"dx= ~ a 2 sin 2 t }"a•a•sin•tacostdt=
2
0
0
=a•
lt
lt
2
•
Ssln
2
2
t cos t dt =
0
lt •
a•s a4s 4 sln 2t dt = B (Ieos 4!) dt = 2
0
0
2
4
a =a
(
I sin4t t 4
)Ina' =w. 0
1576. Can the substitution x =cost be made in the integral
Transform the following definite integrals by means of the indicated substitutions: lt
I
1577. ~Vx+ldx, x=2tl.
•
S f (x) dx, x =arc tan t.
1580.
I
1578.
S~d=x=Ytx•'
x= sin t.
I
b
•
Sf (x) dx (b >a)
4
a
1579.
dx
SYx'+l' • '
1581. For the integral
a
X=
sinh t.
Definite Integrals
148
(Ch. 5
indicate an integral linear substitution x=at + ~. as a result of which the limits of integration would be 0 and 1, respectively. Applying the indicated substitutions, evaluate the following integrals: 4
1582.
dx S1+ v ,
0 19
1583.
X
.,
(x2) ' • dx, (x2) 1• 3
+
S
I
x2=z'.
lna
1584.
SVex1 dx,
ex1 =Z1 •
0 :It
1585.
dt
t
tan 2 = z.
S3+2 cost. 0 :It
tan x= t. Evaluate the following integrals by means of appropriate substitutions: 1
1587.
Ins
S ~· dx.
Yl
ex
1589.
S
X
Jle'i=T d
fx+3
X.
0
I
15S8.
srx;=t
I
1590.
dx.
1
dx
S2x+ Y3x+ 0
1•
Evaluate the integrals: a
1591.
1592.
Sx V x dx+5x+ 1 • 2
a
1593.
)Vaxx• dx.
I
0
I
1:1'1
S(1 !xx•)•.
1594.
I
dx
S'5  3 cosx • 0
1595. Prove that if
f (x) is an even function, then a
a
a
o
) {(x) dx = 2 ~ f(x) dx.
149
Integration by Parts
Sec. 5]
But if f (x) is an odd function, then a
~ f(x)dx=O. a
1596. Show that «J
«J
«J
0
0
5ex"dx=2 5ex"dx= 5e;;dx. 
ct>
1597. Show that
5 l
1
xd X   sin dx
5
arc cos x
0
x
·
0
1598. Show that lt l
.:! I
~ f(slnx)dx= ~ f(cosx)dx. 0
0
Sec. 5. Integration by Parts If the functions u (x) and v (x) are continuously differentiable on the interval [a, b), then b
~ u (x) v' (x) dx=u (x) v (x) a
b
b
a
a
J ~
u (x) u' (x) dx.
(l)
Applying the formula for integration by parts, evaluate the following integrals: 2
1599. ~ xcosxdx. e
1600.
~In xdx.
«J
1603. ~ xexdx.
"'
1604. ~ eax cos bxdx
(a> 0).
1
1
1601. ~ x'eudx.
"'
1605. ~ eaxsm bxdx 0
lt
1602. ~ e" sin xdx. 0
(a> 0).
[Ch. 5
Definite Integrals
150
1606**. Show that for the gammafunction (see Example 1575) the following reduction formula holds true: r(p+ 1)=pr(p) (p>O). From this derive that r (n + 1) = nl, if n is a natural number. 1607. Show that for the integral
•
•
0
0
In=~ sinnxdx = ~ cosn xdx
the reduction formula holds true. Find I n• if n is a natural number. Using the formula obtained, .eva! uate I, and 110 • 1608. Applying repeated integration by parts, evaluate the integral (see Example 1574) 1
B (p, q) =
SxP
1
(1 x)q 1 dx,
0
where p and q are positive integers. 1609*. Express the following integral in terms of B (betafunction):
•
In.m = ~ sinm X cosn X dx, if m and n are nonnegative integers. Sec. 6. MeanValue Theorem 1°. Evaluation of integrals. If f(x)s;;.F(x) for as;;.xs;;.b, then b
b
Sf (x) dx,.;;;. ~ F (x) dx. a
(I)
a
If f (x) and q> (x) are continuous for a,.;;;. x,.;;;. b and, besides, q> (x) :;;;z, 0, then b
m ~ q> (x) dx ,.;;;. a
b
b
~ f (x) q> (x) dx o;;;;;; M ~ a
q> (x) dx,
(2)
a
where m is the smallest and M is the largest value of the function the interval [a, b}.
f (x) on
MeanValue Theorem
Sec. 6]
151
In particular, if q> (x)"""" I, then b
m (ba) o;;;;
~ f (x) dx o;;;; M (ba).
(3}
a
fhe inequalities (2) and (3) may be replaced, respectively, by their equiva· ,ent equalities: b
b
~ f (x) q> (x) dx = f (c) ~ q> (x) dx a
a
md b
~ f(x)dx=f(~)(ba), a
where c and ~ are certain numbers lying between a and b. Example 1. Evaluate the integral
Solution. Since 0 l). 1646*. Find the area bounded by the cissoid y• = 2ax' x and its asymptote x = 2a (a> 0). x (x a) 2 1647*. Find the area between the strophoid y•= and 2ax its asymptote (a> 0). 1648. Compute the area of the two parts into which the circle X2 + !/ = 8 is divided by the parabola y2 = 2x. 1649. Compute the area contained between the circle x• + y• = 16 and the parabola X 1 =12(yl). 1650. Find the area contained within the astroid X= a cos' t; y= b sin' t. 1651. Find the area bounded by the xaxis and one arc of the cycloid x =a (t sin t), { y = a ( 1 cos t). 1652. Find the area bounded by one branch of the trochoid x=atb sint, { y = a b cos t
(O. 1660. Find the area bounded by Pascal's Jimac;on r = 2 +cos q>.
Fig. 48
1661. Find the area bounded by the parabola r =a sec 1 ~ and the two halflines q> = ~ and cp =
;.
1662. Find the area of the ellipse r= I+:coscp (e< 1). 1663. Find the area bounded by the curve r = 2a cos 3cp and lying outside the circle r=a. 1664*. Find the area bounded by the curve x 4 + y 4 = x1 + y•. Sec. 8. The Arc Length of a Curve 1°. The arc length in rectangular coordinates. The arc length s of a curve Y=f(x) contained between two points with abscissas X=a and X=b is b S=
~ VI +y'2dx. a
+
Example 1. Find the length of the astroid x•l• y•l• =a 211 (Fig. 49). Solution. Differentiating the equation of the astroid, we get
y
'
y'l. =,,. X a
The Arc Length of a Curve
Sec. 8)
159
For this reason, we have for th:! arc length of a quarter of tlie astroid:
Whence s=6a. 2°. The arc length of a curve represented parametrically. If a curve is represented bv equations in parametric form, x = cp (t) and y = 1jl (t), then the arc length s of the curve is t, 2 s= ~ +y' 2 dt,
Yx'
t,
where t 1 and t 2 are values of the parameter that correspond to the extremities of the arc.
y
y
s=F?a X
Fig 49
Fig. 50
Example 2. Find the length of one arc of the cycloid (Fig. 50)
x=a(tslnt), { y=a (Ieos t). dx
dy
Solution. We have dt =a (I cost) and dt =a sin t. Therefore, 2n
s=
zn
S Ya (Jcost) +a sln tdt=2a Ssin: dt=Ba. 2
2
2
1
0
0
The limits of integration 11 =0 and 12 =2n: correspond to the extreme points of the arc of the cycloid. If a curve is defined by the equation r=f (cp) In polar coordinates, then the arc length s is 13
s=~ Yr2+r' 1 dcp, a
where a and ~ are the values of the polar angle at the extreme points of the arc.
(Ch. 5
Definite Integrals
160
Example 3. Find the length of the entire curve r=asin 1 ~ (Fig. 51). The entire curve is described by a point as
lative to a rlane. If the massPs conllnuously fill the line or figure of the xyplane, then the 11tatic moments Mx and My about the x and yaxes are exoressed \respectively) a!l intl"grals and not as the sums ll). For the cases of geometric figures, lbe density is considered equal to unity.
Sec. 11]
Moments. Centrts of Gravity. Guldin's Theorems
In particular: I) for the curve x=x (s); y = y (s),
wh~re
Is the arc length, we have
L
Mx=
169
the parameter s
L
~ y(s)ds; My=~ x(s)ds
(2)
(ds= Y(dx) 1 +(dy)1 is the differential of the arc);
y
v b
a
X
l
X
t b '1'
Fig. 57
Fig. 58
2) for a plane figure bounded by th~ curve y=y (x), th.! xaxis and two verhcal hnes x=a and y=b, we obtam
b
Mx=;
5ylyldx;
b
My= Sxlyldx.
Example 1. Find the stat1c
(3)
a
a mom~nts
about the x and yaxe> of a triangle
bounded by th~ straight lines: : +)i=1, x=O, y=O (F1g. 57) Solution. Here, y=b ( 1 : ) . Applyir.g formula (3), we obtain
and
2°. Moment of Inertia. The moment of 111ertta, about an /axis, of a m1fc· rial point of ma!>s m at a d1stance d from th~ lax•~. •~ the number 1 1 ~md • Tht> moment of werlia, about an /axis, of a system of n matenal pomts with masses m1, m:., ••• , m,1 1s the sum 2
11
It'=~ mid:. i"1
Definite Integrals
170
(Ch. 5
where d1 , d1 ••• , dn are the distances of the points from the taxis. 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 xaxis, for its altitude, the yaxis (Fig 58). Divide the triangle into intlnitely narrow horizontal strips of width dy, which play the role of elementary masses dm. Utilizing the similarity of triangles, we obtain hy dm=b hdy and
Whence
3°. Centre of gravity. The coordinates of the centre of gravity of a plane figure (arc or area) of mass M are computed from the formulas 
My
X=M.
Mx

y=M.
where Mx and My are the static moments of the mass. In the case of geomet· ric figures, the mass M is numerically equal to the corresponding arc or area. For the coordinates of the centre of gravity (x, if) of an arc of the plane curve y = f (x) (a~ x ~b), connecting the points A [a, f (a)] and B [ b, f (b)), we have B
h
~xd~
~xYI+(y')•dx
B
~
a  A X= .s= :;.b _ _ _ _ __
~
Yl + (y')
2
b
yds
A

y=s
dx
a
~Y
Yl +(y')
2
dx
a b
~
Yl + (y')
2
dx
a
The coordinates of the centre of gravity ('X; if) of the curvilinear trapezoid
a.;;;;;x.;;;;;b, O.,;;;;;yos;;;;,f(x) may be computed from the formulas b
b
Sxydx a
x=s·
~ y=
b
Sy"dx
_a::..__
s
where S = ~ y dx is the area of the figure. a
There are similar formulas for the coordinates of the centre of gravity of a volume. Example 3. Find the centre of gravity of an arc of the semicircle x•+u"=a•; (y~O) (Fig. 59).
Sec 111
Moments. Centres of Gravity Guldin's Theorems
171
Solution. We have
y = va~X2; y' = o::=X=ccy a 2 x• and
Whence a
u
J'
'
xds=
My=
v
a
a
u
Mx=
.~
"
a
a
Vaaxx 2
2
d~:cO,
Va•r•~
J yds·.. .\.
. Va•x•
··2a2 ,
u
M
.r
:rra.
a
Hence, x=O;
2
y=Ta.
4°. Guldin's theorems. Theorem 1. The area of a surface obtamed by the rotation of an arc of a plane curve about some axis lying in the same plane as the curve and not intersecting it is equal to the product of the length of the curve by the circumference of the circle 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 fi~ure by the circumference of the circle described by the centre of gravity of the figure.
y
)(
Fig. 59
1727. Find the static moments about the coordinate axes of a segment of the straight line ~+1.=1 a b •
lying between the axes.
172
Definite Integrals
[Ch. 6
1728. Find the static moments of a rectangle, with sides a and b, about its sides. 1729. Find the static momen~s. about the x and yaxes, and the coordinates of the cen~re of gravity of a triangle bounded by the straight lines x+y=a, x=O, and y=O. 1730. Find the static moments, about the x and yaxes, and the coordinates of the centre of gravity of an arc of the astroid t
I
I
x• tY' =a', lying in the first quadrant. 1731. Find the static moment of the circle r =2asin
ition of the centre of gravity of 2
2
•
an arc of the astroid x3 + y> =a> lying in the first quadrant. b) Find the Cl'n:rc of gravity of an area bounded by the curves y' = '2px and x' = 2py. 17fiO**. a) Find the centre of gravity of a sen:icircle usin~~ Guldin's theorem. b) Prove by Guldin's theorem that the centre of gravity uf a triangle is distant from its base by one thtrd of its altitude Sec. 12. Applying Definite Integrals to the Solution of Physical Problems 1°. The path traver~ed by a point. II a point IS m motion ~lon~ some curve ar:d the ~hsolute value of the velocity u=t (t) is ~ known fumuon of the time t, then lhe path traversed by the po1nt in an mterval ol time (t 1 , t 1 ) is t, S=
~ f (t) dt. I,
Example t. The velocity of a point is o=0.11 1
m/~ec.
F'ind the path s covE>red by the point m the interval of lime T= to ~ec followIng the comme1rement ol mot10n. What is the mean velocity d mouon durmg this interval?
Definite Integrals
174
[Ch. 5
Solution. We have:
s=
S0.1t dt = 0.1 !•1: = 250 metres 0
1
0
and Vmean
s = y= 25
mjsec.
2". The work of a force. If a variable force X= f (x) acts in the direction of the xaxis, then the work of this force over an interval [x 1 , x,] is x.
A=~ f (x) dx. x,
Example 2. What work has to be performed to stretch a spring 6 cm, if a force of 1 kgf stretches it by 1 em? Solution, According to Hook's law the force X kgf stretching the spring by xm is equal to X= kx, where k is a proportionality constant. Putting x=O.Ol m and X= 1 kgf, we get k= 100 and, hence, X= IOOr. Whence the soughtfor work is 0,06 o.oo A= ~ 100xdx=50x2 i =0.18 kgm 3". Kinetic energy. The kinetic energy of a material point of mass m and velocity v is defined as
The ktnettc energy of a system of n material points with masses m 1 , m2 , ••• , mn having respective velocities v 1 , v 2 , ••• , Vn• is equal to n
K=
L m~v
I
i.
(I)
{. =t
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 ro. 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:
dm= 2nr ·Mdr. Since the linear velocity of the mass dm Is equal to v = rro, the elementary kinetic energy is
Sec. 12]
Applying Definite Integrals to Solution of Physical Problems
175
Whence R
K=
1t(J)
2
M
r'dr =
S
nw 2{)R 4 h
4
.
0
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
p=yhS, where y is the specific weight of the liquid.
X
dr
r
y Fig. 60
Fi!.( 61
Example 4. Find the force of pressure expertenced by a semictrcle of radiu> r submerged vertically in water so that it> diameter is flush with the water surface (Ftg 61). Solution, We partition the area of the semicircle into elementsstrips parallel to the surface of the water. The area of one such element (ignoring higherorder infinitesimals) located at a distance h from the surface Is
ds:..: 2xdh =2 Vr"h 2 dh. The pressure experienced by this element is
dP=yhds=2yh Vr 2 h 2 dh, where y is the specific weight of the water equal to unity. Whence the entire pressure is
1751. The velocity of a body thrown vertically upwards with initial velocity v0 (air resistance neglected), is given by the
171
[Ch. 5
formula
v = V 0 gt,
where t is the time that elapses and g is the acceleration of gravity. At what distance from the initial position will the body be in t seconds from the time it is thrown? 1752. The velocity of a body thrown vertically upwards with initial velocity V 0 (air resistance allowed for) is given by the formula
V=C·tan
(! t+arctan ~o),
where t is the time, g is the acceleration of gravity, and c is a constant. Find the altitude reached by the body. 1753. A point on the xaxis performs harmonic oscillations about the coordinate origin; its velocity is given by the formula
v=
V0
cos u>t,
where t is the time and V 0 , ro are constants. Find the law of oscillation of a point if when t = 0 it had an abscissa x = 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=teo.ott 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 1 = a~bt (a bt > 0), find the velocity at any instant of time t, if the initial velocity is zero. Find the altitude reached at tim~ t = t 1 • 1756*. Calculate the work that has to be done to pump the water out of a vertical cylindrical barrel with base radius R and a !tit ude H. 1757. Calculate the work that has to be done in order to pump the water out of a conical vessel with verlex 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 wateJ out of a semispherical boiler of radius R = 10 m. 1759. Calculate the w;)rk needed to pum;J oil out of a tank through an up,Jer 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 mass m from the earth's surface (radius R) to an altitude h? What is the work if the body is removed to infinity? 
Sec. 12\
Applqin[! Definite
lnte~rals
to Solution of
Phq~ical
Problems
1n
1761**. Two electric charges e0 = 100 CGSE·and e1 =200 CGSE tie on the xaxis at points x0 = 0 and x, = J em, respectively. What work wi II be done if the second charge is moved to point X 2 = 10 em? 1762**. A cylinder with a movable piston of diameter D=20 em and length l = 80 em is filled with steam at a pressure p = 10 kgf cm 2 • What work must be done to halve the volume of the sleam with temperature kept constant (isothermic process)? 1763**. De:ermine the work performed in the adiabatic expan· ~ion of air (having initial volume V 0 = 1 m' and pressure p,= 1 kgfjcm 2 ) to volume v, = 10 m'? 1764**. A vertical shaft of weight P and i radius a rests on a bearing AB (Fig. 62). fp The frictional force between a small part a of the base of the shaft and the surf ace of the support in contact with it is F = JlPO, where p =canst is the pressure of the shaft on the surface of the support referred to unit area of the support, while ll is the coefficient of friction. Find the work done by the frictional force during one revolution of the shaft. 1765**. Calculate the kinetic energy of a Fig. 62 disk of mass M and radius R rotating with angular velocity about. an axis that passes through its centre perpendicular to its plane. 1766. Calculate lbe kinetic energy of a ri~ht circular cone of mass M rotating with angular velocity =2+0.001 x• gjcm. 1773. According to empirical data the specific thermal capacity of water at a temperature t° C (0 .s;;; t .s;;; 100°) is 5 C= 0.99835.184 X I0 t 6.912 X IO' / 1 ,
+
What quantity of heat has to be expended to heat I g of water from 0° C to 100° C? 1774. The wind exerts a uniform pressure p g/cm• on a door of width b em and height h em. Find the moment of the pressure of the wind striving to turn the door on its hinges. 1775. What is the force of attraction of a material rod of length l and mass M on a material point of mass m lying on a straight line with the rod at a distance a from one of its ends? 1776**. In the case of steadystate laminar flow of a liquid through a pipe of circular crosssection of radius a, the velocity of flow v at a point distant r from the axis of the pipe is given by the formula
where p is the pressure difference at the ends of the pipe, f.l. is the coefficient of viscosity, and l is the length of the pipe. Determine the discharge of liquid Q (that is, the quantity of liquid flow,ing through a crosssection of the pipe in unit time). 1777*. The conditions are the same as in Problem 1776, but the pipe has a rectangular crosssection, and the base a is great compared with the altitude 2b. Here the rate of flow v at a point M (x,y) is defined by the formula v=
2:/ [b2 (b y)2).
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 xaxis, and the reciprocals of corresponding accelerations a, on the yaxis. Show that the area S bounded by an arc of this graph, by two ordinates v = t/ 1 and v = V2 , and by the xaxis is numerically equal to the time needed to increase the velocity of motion of the automobile from v 1 to V1 ( acceleration time).
Sec. 12) Applying Definite Integrals to Solution of Physical Problems
179
1779. A horizontal beam of length l is in equilibrium due to a downward vertical load uniformly distributed over the length of the beam, and of support reactions A and B (A= B = ~ ) , directed vertically upwards. Find the bending moment Mx in a crosssection 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 I is in equilibrium due to support reactions A and B and a load distributed along the length of the beam with intensity q = kx, where x is the distance from the left support and k is a constant factor. Find the bending moment Mx in crosssection x. Note. The intensity of load distribution is the load (force) referred to unit length. 1781 *. Find the quantity of heat released by an alternating sinusoidal current
I= 10 sin(:;
t0orx 2 +y 2 > 0 such that when 0 < Q < 6, when' Q = V (xa) 2 + (y b)1 is the distance between P and P', we have the inequality
Z=f (x, y) B > 0 there
lf(x, y)AI.zy·
Solution. Tile function will be m~aningless if the denomin1tor becomes zero. But )( 2 y=O or y=x 2 i~ the equation of a parabola. Hence, the given function has for its discontmuity the parabola y=x 2 •
1797*. Find the following limits of functions: a) Ii;n (x~ r y~) sin..!_ · J( ....
e) lin +x
xy'
o
X· .. OX
·. x+y . b) 11.11 z+2 .
.
d) li.n ( 1 +E..)";
y
X>"' X +
;
!/> 0
lj+.}
l/
y
x ~o>. lJ
y>k
00
x•!lz
f) It 11 .......+3 •
Xj
XtoOO
.....,.o
1798. Test the following function for continuity: {(x, y) = {
V1
X
0
2 
y
2
when when
X X
2 2
+ !l .,;;;; 1, +y• > 1.
1799. Find points of discontinuity of the functions: a) z =In Vx• I
+y
b) Z = , ; ,..ty 12
2 ;
c)
z
I
lx2 ya ·• I
d) Z =COS. >.y
Y
Sec.
31
Derit~atiues
Partial
185
1800*. Show that the function Z= {
2xy z+ Y 2
X
0
when xI + yI =I= 0, when x=y=O
is continuous with respect to each of the variables x and y separately, but is not continuous at the point (0, 0) with respect to these variables together.
Sec. 3. Partial Derivatives 1°. Definition of a partial derivative. If z = example, y constant, we get the derivative
f (x,
y), then assuming, for
f!!= . f(x+!J.t,y)f(x,y)= f'(x ) 1tm " " x , Y, ux .o.~ .... o u.X which is called the partial deriuatiue of the function z with respect to thevartable x. In similar fa~hion we define and denote the partial derivative of the function z w!th respect to the variable y It is obvious that to find partial derivative;, one can use the ordinary formulas of differentiation. Example 1. Find the partial derivatives of the function
z=lntan~. y
Solution. Regarding y as constant, we get
iJz
I
I
OX =tx ~X an cos y y
I
•
2
y = ~· ysmy
Similarly, holding x constant, we will have
i}z
I 1 ( a,=x·x Y tan cos2 y y
x)
z = y
2x . 2x" y stn1
u
Example 2. Find the partial derivatives of the following function of three arguments: u =x 1y2z 2x3y z 5.
+
Solution.
++
iJu
ax=3x 2y2 z+2,
iJu oy=2x 1yz3,
au oz =x•y•+t. 2". Euler's theorem. A function f (x, y) fs called a homo_;eneous function of degree n if for every real factor k we have the equality
f (kx,
ky)
=;
knf (x, y)
Functions of Several Variables
186
[Ch. 6
A rational integral function will be homogeneous if all its terms are of one and the same degree. The following relationship holds for a homogeneous differentiable function .of degree n (Euler's theorem): xf~ (x, y)
+ yf~ (x, y) =
nf (x, y).
Find the partial derivatives of the following functions: 1801. Z=X 1 +y 1 3axy. 1808. Z=XY. sin .!!... 1802. Z=xy. 1809, Z=e X x+y .
1803. Z= fL. X

1804. Z= Vx•y•. 1805, Z=
X
Y x"+ y•
, /x•y•
1810. z =arcsin J1 x"+Y". 1811. z=lnsin;J;.
, 1
1806. z=ln(x+Vr+y
1812. u=(xyy. 1813, U = zXY,
).
1807. z=arctanlL. X ,
'
1814. Find fx(2, I) and fu(2, I) if f(x, Y)= ,
,
,
x Yrxyf. y
1815. Find fx(l, 2, 0), / 11 (I, 2, 0), fz(l, 2, 0) if f (x, y, z) =In (xy + z). Verify Euler's theorem on homogeneous functions in Examples I8I6 to 18I9: 1816. f(x,y)=Ax"+2BxyCy 2 • 1818. t(x,y)=v~· ~+Y
,
1817.
Z= "+x 2 X ~
1820. Find
1819. f(x,y)=ln.lt.,
•
X
ax(~·
1821. Calculate
/It iJr
where r=Vx•+u•+z 1 •
ii ,
if x=rcoscp and y=rsincp.
iJip
az dz , 1822. Sh ow t hat xa;+Y ay=2, If z=ln(x•+xy+y•).
iJz
iJz
.
JL
1823. Sh ow t hat xa;+Y ay=xy+z, If z=xy+xex· h au au au . 182. 4 Sh ow tat ax+ay+a 2 =0, If u=(xy)(yz)(zx). 1825. Show that ~+aau+~=I, if QX
y
QZ
°
u=x+xy yz'
1826. Find z = z (x, y), it 0 =~. y x +Y 2
Sec. 4)
Total DiUerential of a Function
187
1827. Find z=z(x, y) knowing that iJz xz+y2  =and X 0X
z(x, y)=siny when x= 1.
1828. Through the point M(l,2,6) of a surface z=2x•+y1 are 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 crosssections) drawn at their common point M. 1829. The area of a trapezoid with bases a and b and altias
as
as
tude his equal to S= 1 / 2 (ajb)h. Find au, ab, iJh and, using the drawing, determine their geometrical meaning. 1830*. Show that the function
_I f( X, y ) . \
2xy y
~+2
X
,
·r
I
0 ' if
X
2
+Y =I= 0 , 2
X=tj=O
. ' has partial derivatives f~ (x, y) and f~ (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 \
z=
0
,
f (x,
Total increment of a function. The total increment of a function y) is the difference ~Z"i'lf(x, !1)=f(xj~x.
y+t\y)f(x, y).
2°, The total dift'erential of a function. The total (or exact) dtUerential of a function z = f (x, y) is the principal part of the total increment ~z. which is linear with respect to the increments in the arguments ~x and ~y. The difference between the total increment and the total differential of the function is an inlinitesimal of higher order compared with Q= Y ~x 2 ~y 2 • A function definitely has a total differential if its partial derivatives are continuous. If a function has a total differential, then it is called differentiable. The differentials of independent variables coincide with their incre· ments, that is, dx= ~x and dy= ~y. The total differential of the function z = f (x, y) is computed by the formula az iJz dz = ihdxf ay dy.
+
Similarly, the total differential of a function of three arguments is computed from the formula
au au au du = iJxdx+ ay dy+ oz dz. Example 1. For the function
f (x, y)=x1 txyy 2 find the total increment and the total differential.
u = f tx, y, z)
(Ch. I
Functions of Several Variables
188
Solution. f (JC+ l:!.x, y + l:!.y) = (x+ llx) 2 + (x + l:!.x) (y+ l:!.y)(y + lly) 2 ; l:!.f (x, y) = {( t'+ l\x) 2 + (x + l:!.x) (y + 1\y)(y lly) 2 )(x 2 xyy") = = 2X•I:!.X I:!.A 2 +X•I:!.y lj•l:!.x+ l\X•I:!.y·2!J·I:!.lj!J.y'= = [(2x+ y) l:!.x + (x2y) l:!.y) + (llx 2 + llx·lly 6.!/).
+
+
+
+
Here, the expression df = (2x + y) l:!.x + (x2y) l:!.y is the total differential of the function. while (.!\x' + b.x·l:!.yl:!.y 2 ) is an Infinitesimal of higher order compared with V llt 2 + i\y•. Example 2. Find the total differential of the function Z=
oz =
Solution.
ax
Vx
2
+y2•
x oz Y ; vx• + y• ay= v..~.•+y~· xdt"+ydy vl•+u'.
dz
3°. Applying the to!al dilfcrential of a function to approximate calculations, For suffirim'ly small 1.!\xl and I b.yl and, hence, for ~ufficiently small Q= y' .!\. 2 + b.y 2 , we have for a differentiable function z=ftx, y) the approx· imate equality .!\z:::::: dz or az az l:!.z:::::: Ax b.y.
ax +ay
Example 3. The altitude of a· cone is H = 30 em, the radius of tht> base R = 10 em. How will the volume of the cone change, if we increase H by 3 mm and diminish R by I mm? Solutlo11. The volume of the cone is V =
~ nR'H.
The change in volume
we replace approximately by the differential 1
AV:::::: dV=an (2RH dR+R 2 dH)=
= 31 n (2·10·30·0.1 +
100·0.3)= lO.n ::::::31.4 em•.
Example 4. Compute 1.02'·01 approximately. Solution. We consider the functiOn z == xY. The desired numbE.'r may be considered the increasE.'d value of this function when x= l, y=3, l\x=0.02, lly=O.Ol. The initial value of the function z=l'=l, llz:::::: dz= yxYJ b.x+xY In x b.y= 3·1·0.02+ l·ln 1·0.01 =0.06. Hence, 1.02a.o 1 :::::: 1 + 0.06= 1.06.
1831. For the function f(x, y)=x•y find the total increment and the total dillerential at the point (1, 2); compare them if a) 11x = 1, 11y = 2; b) L\x = 0.1, 11y = 0.2. 1832. Show that for the functions u and v of several (for example, two) variables the ordinary rules of differentiation holsired to reduce it by 1°. By how much should the radius of the sector be increased so that the area will remain unchanged, if theoriginal leng:h of the radius is 20 em? 1851. Approximate: a) (1.02)'· (0.97) 2 ; b) V(4.05) 1 + (2.93) 1 ; c) sin 32°·Cos 59° (when converting degrees into radius and calculating sin 60° take three significant figures; round off the last digit). 1852. Show that the relative error of a product is approximately equal to the sum of the relative errors of the factors. 1853. Measurements of a triangle ABC yielded the following data: side a= 100m±2m. side b = 200m±3 m, angle C=60°±l 0 • To what degree of accuracy can we compute the side c? 1854. The oscillation period T of a pendulum is computed from the formula
ry
T=2ttY  g ,
Functions of Several Variables
190
[Ch. 6
where l is the length of the pendulum and g is the acceleration of gravity. Find the error, when determining T, obtained as a result of small errors Ill= a and tlg = ~ in measuring l and g. 1855. The distance between the points P0 (x 0 , Yo' and P (x, y) is equal to Q, while the angle formed by the vector P 0 P with the xaxis is a. By how much will the angle a change if the point P (P 0 is fixed) moves to P1 (x +dx, y+dy)? Sec. 5. Differentiation of Composite Functions 1°. The case of one independent variable. If z=f(x, y) is a differentiable function of the arguments x and y, which in turn are differentiable function~ of an independent variable t, X=q> (/), y='ljl (/),
then the derivative of the composite function z= f [q> (t), 'ljl (t)] may be computed from the formula (I)
In particular, if t coincides with one of the arguments, for instance x, then the "total" derivative of the function z with respect to x will be:
(2) Example t. Find ::, if
z=e1"+ 2Y, where x=cost, Y=l 2 • Solution. From formula (I) we have: acost+2l' ' dz (4t3~int). dt=e1"+ 2Y·3(sint)+e1 "+ 2Y·2·2t =e'"+ 2Y(4t3sin t)=e Example 2. Find the partial derivative
Solution.
~=ye"Y.
aaz
x
and the total derivative dz II
dx'
z=e"Y, where g=q> (x).
From formula (2) we obtain dz dx
= ye"Y + xeXY q>' (x).
2°. The case of several independent variables. If z is a composite function of several independent vari~bles, for insta':'ce, z = f (x,y), where x= q> (u,u), y='ljl (u, v) (u and v are mdependent vanables). then the partial derivatives z with respect to u and v are expressed as
az=~~+~ay
au
dxilu
ilyilu
(3)
Sec. 5]
Differentiation of Composite Functions
191
az az ax az ay av =ax + ay av
(4)
and
au
0
In all the cases considered the following formula holds:
dz=
az
ax dx+ azay dy
(tile invariance property of a total dtfferential). Example 3. Find
~~
f (x,
z=
:~
and
y),
, if
u
where
X=UV, Y=v.
Solution. Applying formulas t3) and (4), we get: az • • 1 f X (X, y)' V + f y (X, !f) V
au=
and iJz _ • F' u (fv fx (x, y) u, Y (x, y) V2
.
Example 4. Show that the function z = cp (x1 + y 2) satisfies the equation
az
az
urrxx ay=O. Solution. The function cp depends on x and y via the intermediate argument x2 + y2 = t, therefore,
~ dt ~ CJxCj) at axand
I (
X
I+ Z) 2
y
X
aat =cp' (xz+ yz) 2y. a~=~ y dt y
Substituting the partial derivatives into the lefthand side of the equation, we get
az az y (fx  X ay = ycp'(x1 +y 1 ) 2xxcp' (x1 +y 2 ) 2y = 2xycp' (x1 +y 2 )2xy cp'(x1 + y 2 )oa0, that is, the function z satisfies the given equation.
1856. Find :: if z=f. where x=e', y=lnt. 1857. Find
:~ if .
u= l n sm 1858. Find
:~
X VY,
where X= 3tZ, y= Vf+l.
if
u =xyz, where X= t• + 1, y= In t, z = tant.
Functions of Several Variables
192
1859. Find du dt
u=
[Ch. 6
if
V xz+yz z ,
where x =
R cost, y = R sin t, z =H.
1860. Find :: if
z=u'O, where u=sinx, v=cosx. 1861. Find :: and :: if
z =arc tan.!L and y = X 1862. Find
x•.
~ and ~ if z = xY, where y = cp (x).
~
1863. Find :: and
if
z=f(u, v), where u=x•y•, v=exY, 1864. Find
~ and ~ if
z=arctan~. where x=usinv, y=ucosu. y
~ if
1865. Find :: and
z=
f (u),
where u = xy
+Y • X
1866. Show that if
+ y• + z•), where x = R cos
= 0 (coupllllf! equal ton). To find the conditiOnal extr£mum of a function (x, y), given the relationship cr (x, y) =(I we form the SO·Called Lagra,zge
fu net ion
F (x, y) ~ f (x, y) +A.·cp (x, y),
where A. is an undetermmed multiplkr, and we seek the ordinary extremum of this auxiliary function. The neces~ary conditions for the extremum reduce to a system of three equal ions:
iJF iJx
=::;
iJf +A. iJcp =O ax ax •
iJF= ~+A. iJljl =O iJy • { dy  iJy cp (x, y) =0
(2)
with three unknowns x, y, J.., from which it is, generally speaking, possible to determine these unknowns.
Functions of Several Variables
224
[Ch. 6
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: arp azp aap 1 2 2 d F(x, y)= axz dx t2axaydxdy+ ayr dy for the given system of values of x, y, A. obtained from (2) or the condition that dx and dy are related by the equation
~: dxj ~: dy=O
(dx 2 tdy 2 =F 0).
Namely, the funchon f (x, y) has a conditional maximum, if d'F < 0, and a conditional minimum, if d 2F > 0. As a particular case, if the discriminant t1 of the function F (x, y) at a stationary point is positive, then at this point there is a conditional maximum of the function f (x, y), if A < 0 (or C < 0), and a conditional minimum, if A> 0 (or C > 0) 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 undetermmed multipliers factors as there are coupling equations. Example 2. Find the extremum of the function z=64x3y provided the variables x and y satisfy the equation x2jy2= l
Solution. Geometrically, the problem reduces to finding the greatest and least values of the zcoordinate of the plane z=64x3y for points of its intersection with the cylinder .t 2 +y 2 =l We form the Lagrange function F (x, y) = 6 4x3y +A. (x 2 + y2 l). We have:.~ =4+21.x, ~~=3+2f.y. The necessary conditions yield the following system of rquatwns:
Solving this system we find 5
",=2· and
Since
It follows
tha~
3
y,=5·
If
225
The Extremum of a Function of Ser•eral Variables
Sec. 13]
x=s4
5
t..= 2 ,
and
3 y=s,
then d2 F
> 0,
and, consequenHy, the function
has a conditional minimum at this point. If!..=
i, x=: andy=~,
then d"F < 0, and, consequently, the function at this point has a conditional maximum. Thus,
6'. Greatest and smallest values of a function. A function that is diffe· rentiable in a limited closed region attains its greatest (smallest) value either at a stationary point or at a point of the boundary of the reg10n. Example 3. Determine the greatest and smallest values of the function
z = x• + y 2  xy + x + y in the reg10n x,;;;;O, y ,;;;;0, x+y;:::: 3 Solution. The indicated region is a tri· angle (Fig. 70). 1) Let us find the stationary po111ts:
f
z't""" 2xy 1· I =0,
\
z~ =
2yx+ I =0;
whence x= 1, II·= I; and we get the point M (1, 1) At M the value of the function z111 =l It is not absolutely necessary to test for an extremum :2) Let us inves!tgate the function on the boundaries of the region. When x 0 we have z = y 2 y, and the problem reduces to seeking the greatest and smallest values of thts function of one argument on thr interval 3.;;;;;y,;:;;O. Investigating, we lind that (Zgr)x=o=6 at the point (0, 3);
+
(z8 m)~=·=+
at the po111t (0, 
When y::..O we get
2
Z=X !X.
point (3, 0); (Zsmly=o=
i
1
/ 2)
Similarly, we find that (Zgrlv=o=6 at the
at the point ( 1/ 2 , 0)
When xfy=3 or y=3x we will have Z=3x 2 +9x+6. Similarly 3 3 . (\ 2· :~ ) ; (Z~tr>x+v=•=6 2 we find that (lsmlx+v=•=4 a t th e potnt
+
metres coincides with (Zgrlx=• and (lgr)~=·· On the straight line x y =  3 we could test the function for a coni:litional extremum without reducing to a function of one argument. 3) Correlating all the values obtained of the function z, we conclude that z11 r=6 at the points (0, 3) and (3, 0); Zsm=1 at the stationary point M. 81900
Functions of Several Variables
226
[Ch. 6
Test for maximum and minimum the following functions of two variables: 2008. Z= (x})1 +2y1 • 2009. z = (x1} 1 2y'. 2010. z=x1 +xy+y'2xy. 2011. z=x'y 2 (6xy)(x>0, y>O). 2012. z = X 4 + y4 2x• +4xy2y' . 2013.
Z=XY
x•
.. /
y•
V 1(jf"1)2 •
2014. Z= 1(x1 fy2 )'ia. 2015. Z=(x'+y•)ecx'+Y•J. 2016. z = 1+xy .
Yl +x1 +Y2
Find the extrema of the following functions of three variables: 2017. U=x•+y• +z•xy+x2z. y2
z2
2
y>O, z>O). ++(x>O, 2018. u=x+Z y 4X Find the extrema of the following implicitly represented functions: 2019*. x• + y• + z'2x+ 4y6z11 = 0. 2020. x'y 1 3x+ 4y + z• + z8=0. Determine the conditional extrema of the following functions: for x+y= 1. 2021. z=xy for x• + y• = 5. 2022. Z= x +2y
i + ~ = 1.
2023. z = x• + y•
for
2024. z=cos•x+cos•y
for yx=: .
2025. u=x2y+2z
for x• + y• + z• = 9.
2026. u = x• + y• + z•
x•
y•
z2
for £i2lfi2+cz= 1 (a>b >c>O). 8 2027. U=xy'z for x+y+z= 12(x>0,y>0, z>O). 2028. U=XYZ provided x+y+z=5, xy+yz+zx=B. 2029. Prove the inequality x+y+z ;;;;:. 3
if x;;;;;:.O,
y~O,
vxyz,
z ~0.
Hint: Seek the maximum of the function u=xyz provided x+u+z=S.
Sec.
141
Finding the Greatest and Smallest Values of Functions
227
2030. Determine the greatest value of the function z = 1 +x+ 2y in the regions: a) x~O. y~O. x+y~ 1; b) x~O. y~O. xy~ 1. 2031. Determine the greatest and smallest values of the functions a) Z=X 1Y and b) Z=X 1 Y1 in the region x'+y ~1. 2032. Determine the greatest and smallest values of the function z=sinx+siny+sin(x+y) in the region O~x~;. 1
:rt
O~y~2.
2033. Determine the greatest and smallest values of the function z=x'+y 1 3xy in the region O~x~2. 1~y~2. Sec. 14. Finding the Greatest and Smallest Values of Functions Example 1. 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, axy. We seek the maximum of the function f (x, y) =xy (axy). According to the problem, the function f (x, y) is considered inside a closed triangle x~O. y~O. x+y~a (Fig. 71).
y
X Fig. 71 Solving the system of equations f~(x, y)==y(a2xy)=0,
{
fu (x, y)=ax(ax2y)=0,
we will have the unique stationary point (
~
,
~)
for the intel'lor of the
triangle. Let us test the sufficiency conditions. We have
fn(x, Y)=2y, t:y(x, Y)=a2x2y, t;y(x, y)=2x.
a•
228
Functions of Several Variables
[Ch. 6
Consequently, A=
t:x ( ~ , ~ ) = {
B = t:u
i) = 
(~ ,
~ a,
C=f~v(;, ~)=fa £\ = AC 8
2
>
0, A
a,
and
< 0.
And so at ( ~ . ~ ) the function reaches a maximum. Since f (x, y) = 0 on the contour of the triangle, this maximum will be the greatest Vdlue, which is to say that the product will be greatest, if x=y=axy=!!_ , and the 3 aa greatest value is equal to 'if. Note The ploblem ran also be solved by the methods of a conditional extremum, by seeking the maximum of the function u=xyz on the condition that x+y+z=a.
2034. From among all 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 of a 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 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. 2039. Fi11d a point M (x, y), on an xyplane, the sum of the squares of the distances of which from three straight lines (x=O, y=O, xy+ 1 =0) is the least possible. 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 1 (x" !/ 1 ), P 2 (X 2 , !/ 2 ), Pa (X 3 , !/ 3 ) with masses mt> m 2 , m 3 • For what position of the point P (x, y) will the quadratic moment (the moment nf inertia) of the given system of points relative to the point P (i.e., the sum m 1 P 1 P 2 + m 2 P 2 P 2 + m 3 P 3 P 2 ) be the least? 2042. Draw a plane through the point M (a, b, c) to form 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 6 and capacity (internal) V so that the smallest quantity of material is used to make it.
Sec. 141
Finding the Greattst and Smallest Values of Functions
229
2045. At what point of the ellipse xz
y2
7+v= 1 does the tangent line to it form with the coordinate axes a triangle of smallest area? 2046*. Find the axes of the ellipse 5x 1
+ Bxy+ 5y
2
= 9.
2047. Inscribe in a given sphere a cylinder having the greatest total surface. 2048. The beds of two rivers (in a certain region) approximately represent a parabola y=X 2 and a straight line xy2=0. 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 the point M (l, 2, 3) to the straight line X
y
Z
T= 3=2 · 2050*. The points A and B are situated in different optical medta separated by a straight line (Fig. Z2). The \"elocity of
A
B
I
/3 Fu~.
7'2
c
I Fig. i3
light in the first medium is ua> in the second, U 2 • Applying the Fermat principle, according to which a light ray is propagated along a line AMB 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 I llows in an electric circuit containin~ a resistance R, then the quantity of heat released in unit time is proportional to J"R. Determine how to divide the current I into
Functions of Several Variables
236
(Ch. 6
currents I 1 , I 2 , I • by means of three wires, whose resistances are R1 , R1 , R 1 , so that the generation of heat would be the least possible? Sec. 15. Singular Points of Plane Curves 1°. Definition of a singular point. A point M
(x~,
y 0) of a plane curve
f(x, y)=O is called a singular point if its coordinates satisfy three equations
at once:
f (xo,
Yo)= 0,
f~ (xo, Yo)= 0,
f~ (X0 , Yo)= 0.
2°. Basic types of singular points. At a singular point M (x 0 , y 0), let the second derivatives A=
B
f:x (xo,
= t:y
Yo),
(xo, Yo),
C = f;y (xo, Yo)
be not all equal to zero and then: a) b) c) second
if 6. > 0, then M is an isolated point (Fig. 74); if 6. < 0, then M is a node (double point) (Ftg. 75); if 6.=0, then M is either a cusp of the first kind (Fig. 76) or of the kind (Fig. 77), or an isolated point, or a tacnode (Fig, 78).
Fig. 74
Fig. 75
When solving the problems of this section it is always necessary to draw the curves. Example 1. Show that the curve y 1 = ax 2 + x• has a node if a > 0; an isolated point if a < 0; a cusp of the first kind if a= 0. Solution. Here, f(x, y)=ax 2 +x'y 1 • Let us find the partial derivati· ves and equate them to zero:
f~ (x, y) =2ax +3x1 =0, f~ (x, y)= 2y=0.
231
Singular Points of Plane Curves
Sec. M)
This system has two solutions: 0 (0, 0) and N (  : a, 0) : but the coordinates of the point N do not satisfy th~ equation of the given curve. Hence, there is a unique singular point 0 (0, 0).
y
X
Fig. 78
Fig. 77
Fig. 76
Let us find the second derivatives and their values at the pomt 0:
t:x (x, y) = 2a + 6x,
t:u (x, y) = 0, t;,,(x, y)=2,
A= 2a, 8 = 0, C=2,
.1=AC82 = 4a. y
y
Fig. 79
F1g. 80
Fig. 81
Hence, if a> 0, then .1 < 0 and the point 0 is a node (Fig. 79); if a< 0, then A> 0 and 0 is an isolated point (Fig. 80); if a~= 0, then .1 =0. The equation of the curve in this case will be y 2 =x• or y= ± y7; y=exists only when x~O; the curve is symmetric about the xax is, which is a tangent. Hence, the point M is a cusp of the first kind (Fig. 81 ).
232
Functions of Several Variables
(Ch. 6
Determine the character of the singular points of the following curves: 2053. y• =x• +x\ 2054. (yx 1 ) 2 = x•. 205li. a'y2 = a• x' x 8 • 2056. x•y• x•y• =0. 2057. x 8 +y'3axy=O (folium of Descartes). 2058. y• (ax) = X3 ( cissoid). 2059. (x 2 y 1 ) 1 = a2 {x 2 y 2 } (lemniscate). 2060. (a+ x) y• =(ax) x• ( strophoid ). 2061. (x• y•) (xa)• = b•x• (a> 0, b > 0) (conchoid). Consider three cases:
+
+
1) a>b, 2) a=b, 3) aO) is defined by the equations
x=2t,
y=lnt,
z=t 1 •
Find the mean velocity of motion between times t = 1 and t = 10. Sec. 18. The Vector Function of a Scalar Argument 0 ) , The derivative of the vector function of a scalar argument. The uector function a= a (t) may be defined by specifying three scalar functions ax (t), ay (t) and az (t), which are its projections on the coordinate axes: a=ax (t) I +ay (t)J+az (t) k.
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
~ _ lim a (t + ht)a(t) di!J.tH i\t
dax (t) i +day (t) j dt dt
+daz (t) k dt
•
The modulus of the derivative of the vector function is
I I= ,v ~ dt
1
2
f(dax)• (da_v) dt + dt
+
(
daz ) dt •
The endpoint of the variable of the radius vector r=r(t) describes in space the curve r=x (t) l+y (t)J+z(t)k, which is called the hodograph of the vector r. The derivative
!~
is a vector, tangent to the hodograph at the corre
Idf I=dt"·
sponding point; here,
dr
ds
where s is the arc length of the hodograph reckoned from some lnlttal point, For example,
I:~I=
1.
Functions of Several Variables
236
If the parameter t is the time, then
r,
extremity of the vector
and
~=
~;~= ~~ ='W
'll
(Ch. 6
is the velocity vector of the
is the acceleration vector of the
extremity of the vector r. 2°. Basic rules for differentiating the vector function of a scalar argument. 1)
de
db
da
d
dt (a+bc)=dt+dtdt;
~~
2)
d~
(ma)=m
3)
d dt
(cpa)=dta+q>df, where q>(l) is a scalar function oft;
d 4) (if
5)
, where m is a constant scalar;
dqJ
da
da
db
(ab)="iitb+adf;
d
db
da
Tt (axb)=Ttxb+axdt; da
d
6) dfa(q>(t)]="'it(p·
7)
a~~
dq> .
lit'
=0, if lal=const.
Example 1. The radius vector of a moving point is at any instant of 1ime defined by the equation
r= i4t'J+3t 2k.
(I)
Determine the trajectory of motion, the velocity and acceleration. Solution. From (I) we have: y=41 2 , Z=31 2 • X= I, Eliminating the time t, we find that the trajectory of motion is a straight line: z y xl 0 4=3· From equation (I), differentiating, we find the velocity
!!:!_=8tJ+6tk dt and the acceleration
The magnitude of the velocity is
I~~
I= V K==
R
~s+o
lls
Sec. 20)
243
Curuature and Torsion of a Space Curue
where
O).
Solution. We have
dr dt =  l a sin t
+j a cos t +kb,
tFT
dt 2 =  l a cos t  j a sin t,
d3r dt•=  l a sin t ja cost. Whenct'
I
1 1 dr X ddt•= r dt a sin t
a cost
I
aJcost kb =lab sin tjab cos t+a 1k a sin t 0
244
Functions of Several Variables
and
[Ch. 6
I
a cost dr d 2r d• r ~a sin t . b 2 dtdt 2 dt•= ac~st asmt 0 =a b. a sm t a cost 0 Hence, on the basis of formulas (1) and (2), we get
1
a yaz~
a
R= (aa+bz)•J. = a2+b2 and
a 2b
1
b
Q= az (az+ba) = a2+b!. Thus, for a screwline, the curvature and torsion are constants. 3° frenet formulas:
2104. Prove that if the curvature at all points of a line is zero, then the 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 X=1+3t+2t\ y=22tj5i 2 , Z=l/ 1 is a plane curve; find the plane in which it I ies. 2107. Compute the curvature of the following curves: a) x=cost, y=sint, z=cosh tat the point i=O; b) X2 i\Z 2 =l, y 2 2x+z=0 at the point (1, 1, 1). 2108. Compute the curvature and torsion at any point of the curves: a) x = e1 cos t, y = e1 sin t, z = e1 ; b) x=a cosh t, y=a sinh t, Z=at (hyperbolic screwline). 2109. Find the radii of curvature and torsion at an arbitrary point (x, y, z) of the curves: a) X 2 = 2ay, X 8 = 6a 2 z; b) x• = 3p 2 y, 2xz = p 2 • 21 t 0. Prove that the tangential and normal components of acceleration w are expressed by the formulas dv
'Uil'
= dt l',
Wv=
v2
R V,
where v is the velocity, R is the radius of curvature of the trajectory, '\' and v are unit vectors of the tangent and principal normal to the curve.
Sec. 20)
Curvature and Torsion of a Space Curve
245.
2111. A point is in uniform motion along a screwline r = ia cost+ ja sin t + btk with velocity v. Compute its acceleration w. 2112. The equation of motion is r= ti + f 1 j+ t'k. DetermiRe, at times f=O and t=l: 1) the curvature of thetrajectory and 2) the tangential and normal components of the. acceleration. =
Chapter VII
MULTIPLE AND LINE INTEGRALS
Sec. 1. The Double Integral in RectanguiP.r 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 twodimensional integral sum
~ f (x,
lim
y) dx dy=
max ax1 ~ o max ay,, ~ 0
(S)
~~~(xi, i
Yk) Mt !lyk,
(I)
k
where !lxi=xi+sXi, !lyk=Yk+ 1 Yk and the sum is extended over tho~e values of i and k for which the points (x1, Yk) belong to S. 2°. Setting up the limits of integration in a double integral. We distinguish two basic types of region of integration.
y
y
c
B
Yz y
A
B
!It 0
:r,
D
:X
Xz Fig. 85
X
A
a
c X
Fig. 86
1) The region of integration S (Fig. 85) is bounded on the left and right by the straight lines x=x1 and x=x1 (x 2 > x1), from below and from abovE by the continuous curves y = q>dx)(AB) and y = q> 1 (x)(CD)(q>1 (x) ;;;;.q> 1 (x)l. each of which intersects the vertical x =X (x1 ...;;; X 1 (x) to y1 = q> 1 (x). The integral (1) rna~
Sec. 1)
The Double Integral in Rectangular Coordinates
247
be computed by reducing to an iterated integral by the formula
Hf (x,
X2
Cl'o (X)
~ dx ~ f (x,
Y) dx dy =
x1
(S)
y) dy,
cp 1 (X) Cl'2 (X)
~ f (x,
where x is held constant when calculating
y) dy.
IP1 (X)
2) The region of integration S is bounded from below and from above by the straight linE's y=y 1 and y=y1 (y 1 > y 1), and from the left and the right by the continuous curves x = 1j) 1 (y) (AB) and x = 11' 1 (y) (CD) (1j) 1 (y) ?11'1 (y) ), each of which intersects the parallel y = Y (y 1 .,.;;; Y,.;;; y 1) at only one poinU (Fig. 86). As before, we have
H
f(x, y)dxdy=
(S)
u.
,.
(Y)
'"
,p,
(1/)
~ dy ~
f(x, y)dx,
"'· (y)
~ f (x.
herl', in the integral 111.
y) dx we consider y constant.
(y)
If the region of integration doe~ not belong to any of the abovediscussed types, then an attempt is made to break it up into parts, each of which doe!'. belong to one of these two types. Example l. Evaluate the integral I
1
I=~ dx ~ (x+ y) dy. 0
y
Fig. 87 Solution.
Example 2. Determine the limits of integration of the integral
Hf (Sl
(x, y) dx dy
Multiple and Line Integrals
'248
(Ch. 7
if the region of integration S (Fig. 87) is bounded by the hyperbola y 2 x2 =1 and by two straight lines x=2 and x= 2 (we have in view the region containing the coordinate origin). Solution. The region of integration ABCD (Fig. 87) is bounded by the straight lines x= 2 and x=2 and by two branches of the hyperbola
y=
Yi +x
2
y= Yi
and
+x•;
that is, it belongs to the first type. We have: V1+x•
2
~ ~ f (x, y) dx dy = ~ dx (S)
 2
~

f (x, y) dy.
VJ+x•
Evaluate the following iterated integrals: 2
I
2113. ~ dy ) (x 2 + 2y) dx.
a
1
2117. ~ dy ) (x+2y)dx. a
4
2114
•
Y'4
a
llt
Sdx S(x~y 11 •
2118. ~ dc:p
~ r dr. a sin q>
l
2115.
S
dx
sl
a cos q>
2
x•dy 1
+u•.
~ dc:p ~ r• sin 2 right and if it does not SQ  rdr 0
0
1 +r 2
=rt
In (I
+Q2);
lim I (a)= oo,
II._....,
diverges. Thus,· the integral (3) converges for p
2273. Find
>
that
is,
271
the
integral
I.
f' (x), if "'
f (x) = ~ exu• dy
(x
> 0).
X
2274. Prove that the function
oo
satisfies the Laplace equation iFu iJ"u   0. + dt!"
OX 2
2275. The La place transformatiOn F (p) for the function f (t) is defined by the formula "' F (p)
=~eMf {I)
dt.
u
Find
d)
F (p),
f (I)= cos ~t.
if:
f (t) = 1;
a)
b)
f (t) =ear~
2276. Taking advantage of the formula I
Sxn
1
(n dX = _!_ tl
> 0),
u
compute the integral I
~ xnl}nxdx. t)
2277*. Using the formu Ia
s"'
ePI
evaluate the integral
dt
=
i
(p
> 0),
c)
f (t) =sin ~t;
[Ch. 7
Multiple and Line Integrals
272
Applying di!Terentiation with respect to a parameter, evaluate the following integrals: rt:J
2278. se·x~e;ixdx (u>O, ~>0). 2279.
"' S e•x :;e;;x sin mx dx
(a > 0, ~
>
0).
0
"' tan ax d 2280 . JC arc x(I+x2 ) X.
sIn I
2281.
(1a"x") dx x• Ylx 2
(/a/< 1).
0 00
2282. ~eaxsin}xdx (u~O). 0
Evaluate the following improper integrals: "' "2283. ~ dx ~
e(H!il
y>
I
dy.
X
2284. ~ dy ~eli dx. 0
2285.
~S >~~:~., where
S is a region defined by the inequali
(S>
x~
ties
I,
y~x·.
(' s
00
"'
2286* . .) dx 0
2287. The
dr (x•+y"·J·a•)• (a>O).
0
EulerPoisson integral defined by the formula
00
00
f =~ex' dx may also be written in the form 1 = ~ o
f!Y'
dy. Eval
0
uate 1 by multiplying these formulas and then passing to polar coordinates. 2288. Evaluate 00
s
dx
0
"'
rt:J
0
0
~ dy ~ (x• + y• ~ z• + !)•.
Sea. 9]
273
Line Integrals
Test for convergence the improper double integrals: 2289**. ~~lnVx•ty"dxdy, where Sis a circle x•+y 1 ~l. (S)
2290.
55 (x~~:.)~,
where S is a region defined by the ine
IS)
quality x• 1 y• ~ 1 ("exterior" of the circle). 2291*.
• dxdtt SJ V (x))
+ cp' (x) Q (x,
ql (x))
I dx.
a
C
In the more general case when the curve C is represented parametrically: t varies from a to ~. we have
x=q> (t), y=¢ (t), where
~ P (x, c
y) dx
+ Q (x,
~
y)
dy + ~ [P (
rtional to the distance of the point f10111 the orig111 and if the point of application of the force traces counterclockwise a quarter of the ellipse~+~~= 1 !yin~ in the first quadrant. 2346. Find t:1e p)tential fun~tion of a for~c R {X, Y, Z} and determine the work don~ by the force over a given path if: a) X=O, Y=O. Z=mg (force of gravity) and the material point is moved from pvsition A (x,, y., z.) to position B (x •• Y.. z.); b) X =~t:, Y =J.I~, Z =tt:. where f.l=Const and r
r
'
r = V x• + y• + z" (Newton attractive force) and the material point ;.....~
moves from position A (a, b, c) to infinity;
Multiple and Line Integrals
284
[Ch. 7
c) X=k 2 x, Y=k 2 y, Z=k 2 z, where k=const (elastic force), and the initial point of the path is located on the sphere x• +y 2 + z 2 = R 2 , while the terminal point is located on the sphere 2 2 2 X +y z• = r (R > r).
+
Sec. tO. Surface Integrals 1°. Surface integral of the first type. Let f (x, y, z) be a continuous function and z=~c:p (x, y) a smooth surfaceS. The surface integral of the ftrst type is the limit of the integral sum n
) ) f (x, y,
z) dS = lim
S
.~ f (x;,
Yi· z,) M;,
n+oot=l
where I:!S; is the area of the Jth element of the surface S, the point (x;, y., z;) belongs to this element, and the maximum diameter of elements of parti lion 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 projectton a of the surface S on the xyplane i~ singlevalued, that is, every strdight line parallel to the zaxis intersects the surface S at only one point, then the appropriate surface integral of the first type may be calculated from the formula
~ ~ f (x, s
y, z) dS
= ~ ~ f (x,
y,
c:p (x, y)]
V 1+rr: (x, y) +cp; (x, y) dx dy.
(a)
Example t. Compute the surface integral
~~ s
(x+y+z)dS,
where S is the surface of the cube 0.,;;;; x.,;;;; I, 0.,;;;; y.,;;;; I, 0.,;;;; z.,;;;; I. Let us computP the sum of the surface integrals over the upper edge of the cube (z = I) and over the lower edge uf the cuue (z = 0): I I
H(x+y+ 1) 00
I I
dx dy+
~~
I I
H(2x+2y+ 1)
(xty) dxdy=
00
dxdy=3.
00
The desired surface integral is obviously three times greater and equal to
~~
(x
+ y + z) dS =
!),
s 2°. Surface integral of the second type. If P = P (x, y, z), Q = Q (x, !J, z),
R = R (x, y, z) are continuous functions and s+ is a side of the smooth sur
lace S characterized by the direchon of the normal n {cos u, cos~. cosy}, fhen ihe corresponding surface tntegral of the second type is expressed a~ follows:
)
~
S+
P dy dz
+ Q dz dx + R dx dy =
H
(P
S
cos a
+Q ws ~ + R cos \') dS.
285
Surface Integrals
Sec. 10]
s,
When we pass to the other side, of the surface, this integral reverses sign. If the surfaceS is represented implicitly, F{x, y, z)=O, then the direction cosines of the normal of this surface are determ!ned from the formulas cos a =
I iJF
l5 ax
I oF
I oF
, cos ~ = 0 ay , cos y = 0 az ,
where
and the choice of sign before the radical should be brought into agreement with the side of the surf ace S. 3°. Stokes' formula. If the functions P = P (x, y, z), Q = Q (x, !J, z), R = R (x, y, z) are continuously differentiable and Cis a closed contour bounding a twosided surface S, we then have the Stokes' formula
p
P dx
+ Q dy + N dz =
c
=55 [(oR_iJQ)cosa+(oP _aR)cos~+(aQ_aP) cosvl dS a!/
s
az
oz
ox
a~
cJy
'
where cos a, cos~. cosy are the direction cosinl's of the normal to the surface S, and the directwn of the normal is defined so that on the side of the normal thl' contour S IS traced counterclockwise (m a rigiJI·handed coorclmate t>) stem).
Evaluate the following surface integrals of the first type: 2347. ~~(x"+u")d.S', where Sis the sphere x 2
+!J
2
/Z 2 =a 2 •
~
2348. .\
2
~~
s2
V x• + !/ dS
+·
11
Z
where S is the lateral surface of the
2
cone a2  12 =0 [O~z N and for any positive p the following inequality is fulfilled:
\a11+1+an+2+· · .+an+pl
I "+ 0
1
/
U11
y.
11=1
I
n
~1_1 2455. .....,.n Inn: 11
00
1 2451. ~. sm n•.
2452.
2456.
~n !~ 2 n •
11=1
11=2
00
00
~ ln ( 1 +~).
2457.
11=1
2458.
n=t
n=z
~nzl
n'
11 =z
00
~l:n •
L n·ln n·\n In n •
11=z rt>
"'
2453. ~ l n nt+l n• • 2454.
~a
00
2459.
Lo"" Y n (nI +I) •
11 = 1
+ •••
Sec. 1)
Number Series
301
QO
2460.
L
n=J
'
"'
2465.
Vn(n+l)(n+2)•
n=l
QO
246t.
L.
'
QO
2462. ~..,
QO
.
n=an Inn+ Vln'n
2466. L2nn1 nn • n=l ao
1
2467. L3nnl n" • n=l
~nt/nvii·
2463.
""
:E (2n 
n=1
:E~ nn •
Vn
ex,
v
1) ( 5 ~ n 1)
.
2468*.
~ennl
n=1
tln
•
ao
2464.
L. (1 cos ~ ) .
n=l
QO
2469. Prove that the series ~nPlnq '\...., ____!.___n •• n=2
1) converges for arbitrary q, if p>1, and forq>1, if p=l; 2) diverges for arbitrary q, if p < 1, and for q ~ 1, if p = 1. Test for convergence the following alternating series. For con· vergent series, test for absolute and conditional convergen4!:e. I
2470. 1 3 2471. 2472.
+ 5I 
... ·f
I I 1·;r+~r 2 V3 I I 3
2nl
... +
1;r+g ... + 2
(1)" 1
(1)"
Vn 1
n•
3
5
· r:22·3
2475. 2476.
2477. 2478. 2479.
+ 3.47 f( ... · 
+ ...
+ ...
(l)nln
2473 · 1 7+13 ... + 6n5 2474
+ ...
(1)" 1
+ ...
2n+l 1)nJ n(n+l)
+ ...
Series
302
L"" (1)n 1: n.
2481.
[Ch. 8
00
2482.
L (l)nJ tan n .~. r n
n=l
n=t
2483. Convince yourself that the d' Alembert test for convergence does not decide the question of the convergence of the
"" an, where series ~ 1!=1
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 converge absolutely: I
I
I
I
I
I
a) Y2IY2+1+Y3IY3+1+Y4 1Yf+l+ ...
y
( a2kt = I
b) 13 +
I I 233
k;
+
I I '
I 22  35 I
a2k = 
I
I
c) l3+~3 2
I
I
I + I) ;
2L1) ;
a2k =  3
I
+533+ · · ·
( a2k• = 2k I I' I
k;
+ ...
1 ( a2k1 = 2k• • I
y
I
I
a2k=
~k);
I
d) 3 1 1ys+rrg+ ... (a2k•=4kl
I'
a2k=4kl3).
Test the following series with complex terms for convergence:
L n (2+Q~ 2n • 00
2485.
00
2488.
n=t
L n (2tl)n 3" • 00
2486 ..
.Lv~ . n=• n+l
2490.
I.""
00
Ln(3~i)n' n=t
""
2489.
n=t
2487.
L.i~. n==l
I
n=• (n+i)
Yn ·
303
Number Series
Sec. I) ~
2491
'
~
2 2492.~[n( i)+ 1 J" .._. n (32i) 3i '
1 ~ "" [n + (2n I) i) 2 •
n=t
n=t
in
I
and y = 2X and to the ri~ht their point of intersection are constructed segments parallel the yaxis at 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 Problem 2493 be finite if the curve y X is replaced by the 2493. Between the curves y =
of to of
I
8 X
=;
I
curve y=? X ~
2495. Form the sum of the series
L n=t
t
1
rn
n and
L (l~:n. n===I
Does this sum converge? «>
2496. Form the difference of the divergent series £~ 211l 1
n=l
and
"'
L 2~
and test it for convergence.
Jl:;:;J
2497.
Does the series formed
oc
by subtracting the
series
00
~ 2rz 1 ""n=t
I
from the series ""~ n1 converge? 11=t
2498. Choose two series such that their sum converges while
their difference diverges. 2499. form the product of the series
L"' ll=l n
.~ and r
11
L"' 2}_
1 •
n=l
r
Does this product converge? + + ... + 2}_ 1 + ... Does 2500. Form the series ( 1 this series converge? 2501. Given the series 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
+; +
"
} + 2\ ( ~ ) + ~! ( ; ) + •••+ h(}) n + ••• 2
1
by the sum of its first n terms.
Series
304
[Ch. 8
2503. Estimate the error due to replacing the sum of the
5eries 1
1
1
1 +21+31+ ... +iii+··
by the sum of its first n terms. In particular, estimate the accuracy of such an approximation for n = 10. 2504**. Estimate the error due to replacing the sum of the series 1 1 1 1 + 22 + 32 + ... + nz + ...
by the sum of its first n terms. In particular, estimate the accuracy of such an approximation for n =I ,000. 2505**. Estimate the error due to replacing the sum of the f.eries 1 +2 + 3 ( + ... +n ( + ...
(! r
+r
+rn2
by the sum of its first n terms. (yJ
2506. How many terms of the series
L rr; (n+l)l and consequently lim Rn (x) = 0 for any x. This signifies that the sum of the n+cr;
sen~s
l3) for any x is indeed equal to cosh x. 2". Techniques employed for expanding in power series. Making use of the principal expansions
x
x2
xn
). r=l+n+2f+ ... +iiT+ ... (oo 0)]; e ~· 3) has no more than a finite number of points of strict extremum. Dirichlet's theorem asserts that a funct10n f (x), which in the interval ( n, n) satisfies the Dirichlet conditions at any point x of this interval at which f (x) is continuous, may be expanded in a trigonometric Founer senes:
finite limit on the right
0
f(x) = ~ +a1 cos x+ b1 sin x+a 2 cos 2xj b2 sin 2x+ .•. +an cos nx+
+bnsinnx+ ... ,
(I)
where the Fourier coeffictents an and bn are calculated from the formulas
s
s .
lt
I On=;t
lt
I f(x)cosnxdx(n=O, 1, 2, .•. ); bn=rr:
lt
f(x)smnxdx(n=l,2, ... ).
lt
If xis a poinf of discontinuity, belonging to the interval (n, J!), of a function f (x), then the sum of the Fourier series S (x) is equal to the arithmetical mean of the left and right limits of the function: I S(x)="2 [f(xO)+f(x+O)]. At the endpoints of the interval x=n and x=n,
S ( n) = S (n) =
I
2 [/ ( n + 0) + f (n0)].
2°. Incomplete Fourier series. If a function f(x) is even [i.e., f(x) = =f(x)), then in formula (I)
bn=O (n= I, 2, ... ) e~nd lt
an =
}
SI (x) cos nx dx 0
(n = 0, I, 2, •.• ).
Fourier Series
Sec. 41
319
li a !unction f(x) is odd [i.e., f(x)=f(x)), then an=O (n=O, I, 2 •.• ) and n
~
bn =
5
f (x) sin nx dx (n = 1, 2, ... ).
0
A function specified in an interval (0, :n:) may, at our discretion, be continued in the interval ( :n:, 0) either as an even or an odd function; hence, it may be expanded in the interval (0, l't) tn an incomplete Fourier series of sines or of cosines of multirle arcs. 3°. Fourier series of a period 21. If a function f (x) satisfies the Dirichlet conditions in some mterval (1, l) of length 2/, then at the discontinuities of the function belonging to this interval the following expansion holds: a f(x)= 0
:n:x
l'tx
2l'tx
. 2:n:x
1 +b2 Sln1+ ... 2 +a 1 cos y+b 1 sin T +a 2 cosn:n:x
nnx
•.. +ancos1+bnsln + ... 1
•
where l
1 an=y
5f
nnx dx (n=O, I, 2, ... ), (x) cos~
I I
I bn=y
5{(x)sm. nnx1dx (n=l, 2,
(2) ... ).
I
At the points of discontinuity of the function
f (x)
and at the endpoints
x= ±I of the interval, the stun of the Founer series is defined in a manner simllar to that whtch we lta~e in the expansiOn in threntiatmg the equation (5). we find the d1,Jen:nt1al equation of the lam1ly x1 '2yy' =0.
y
X
Fig. 106 Wht'nce, replarlng y' by  y'. orthogonal trajeclortes
we
get
the differential
equation ef the
2
2 x Y =0 or q'= Y. X 1/
Jntegrating, we have y = Cx1 llatmly of parabolas) (Fig. 106).
Sec. 31
DiOerential Equations with Variables Separable
329
4°. Forming dift'erential equations. When formmg differential equations tR geometrical problems, we can fr£quently make use of the geometrical meaning of the derivative as the tangent of an angle formed by the tangent hne to the curve in the pos•live xdirection. In many cases thts makes it rossible straightway to establish a relationship hetween the ordinate y of the desired curve, its abscts•a x, and the taneent of the angle of the tangent line y', that is to say, to obtain the diiTe•enttal equat10n. In other 10stances ~see Problems 2783, 2890, 2895), u~e is made of the geometr.cal signtlicance of the definite integral as the area of a curviltnear trapezotd or the length of an arc. In !hi' ca,e, by hypoth('sts we have a simple mtegral equation !since the destred function is under the 'ign of the mtegral); however, we can readily pass to a differential equatton hy dtflerenliatmg both ~ides. Example 3. Ftnd a curve passing through the pomt (3,2) for which the segment of any tangent line contained between the coordinate axes is dividl'd in half at the potnt of tangency. Solution. Let M (x,y) be the midpomt of the tangent line AB. which by hypothl'sis is the point of tangency (the potnts A and B are points of mter~ection of the tangent line with the y and xaxes). It is gtven that OA =2y and 08 =2x. The slope of the tangent to the curve at M (x, y) is
dy
OA
!1
([X= ua=;;
This is the difierential equation of the soughtfor curve. Transforming, we gel
d.\+~=0 X
y
and, consequently, lnx+lny
~lnCorxy=C.
Utilizing the inittal conditiOn, we determ111e C=3·2=6. Hence, the desired curve is the hyperbola AY = 6.
Solve the differential equations: 2742. tan x sin• y dx + cos 2 x cot y dy = 0. 2743. xy'  !I= !l. 2744. xyy' ~ 1x•. 2745. yxy' =a (I +x 2 y'). 2746. 3t>x tan y dx +(lex) sec• y dy = 0. 2747. y' tan x = y. Find the particular solutiOns of equations that satisfy the indicated initial conditions: 2748. (l +ex) y y' =ex; !J = l when x = 0. 2749. (xy•+x)dx+(x 2 yy)dy=0; y=l when x=O. 2750. y' sin x =yIn y; y ~= l when x Solve 3751. 2752. 2753. 2754.
= ~ .
the di!Terential equations by changing the variables: y' = (x + y) 2 • y = (8x + 2y + l )1 • (2x + 3y 1) dx + (4x + 6y 5) dy = 0.
(2xy)dx+(4x2y+3)rly=0.
Dtfferential Equations
330
[Ch. 9
In Examples 2755 and 2756, pass to polar coordinates: 2755. y' = 1
~x y
.
1
2756. (x +Y )dxxydy=0. 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 pa5ses through the point (3, 1), for which the segment of the tangent between the point of tangency and the xaxis is divided in half at the point of intersection with the yaxis. 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 yaxis is of constant length 2. Find the orthogonal trajectories of the given families of cur· ves (a is a parameter), construct the families and their orthogonal trajectories. 2766. xy=a. 2764. X 2 +if = a•. 2765. y• =ax. 2767. (xa) 1 r y 2 = a 1 • Sec. 4. FirstOrder Homogeneous Differential Equations 1°. Homogeneous equations. A differential equation P (x, y) dx+ Q (x, y) dy=O
(l)
is called homogeneous, if P (x, y) and Q (x, y) are homogeneous functions of the same degree. Equation (l) may be reduced to the form y'=f
(f);
and by means of the substitution y=~xu, where u is a new unknown function, it is transformed to an equatiOn wtth variables separable. We can also apply the substitution x= yu. Example l. Find the general solution to the equation
FirstOrder Homogeneous Dif!erential Equations
Sec. 4)
Solution. Put u= ux; then u +xu'= eu
331
+u or
dx e ad u=. X
Integrating, we get U=ln ln.£, whence X
c
y=xlnln. X
2°. Equations that reduce to homogeneous equations.
If (2)
l\=1::::1;c x.
y
HigherOrder DiUerential Equations
Sec. 10]
Solve the following equations: 2911. y" = _!_.
2920. yy" = y•y'
X
2912. y" =
2!' .

2913. y" = 1 y'
1
2914. xy" ·l y'
0.
=
+ y''.
2921. yy" y' ( 1 + y') = 0. 2922. y"= x•.
•
y
2923. (x+l)y"(xt2)y'+x+ + 2 = 0. 2924. xy" =y' In y'.
1
yy" = y' • X 1 yy" + y' = 0. + 1 ,,1 , ' (1 !x 2 )y"+y' 1 +1=0. 2925 · Y 4Y =xy. y'(l+y' 2 )=ay". 2926. xy'"+y"=1+x. 2 2 x•y" + xy' = 1. 2927. y"' + y" = 1. Find the particular solutions for the indicated initial condit ions: 2928. (1+x 2 )y"2xy'=0; y=O, y'=3 for x=O. 2929. 1 + y'• = 2yy"; y = I, y' = 1 for x = 1. 2930. yy" + y'" = y'~; y = I, y' = 1 for x = 0. 2931. xy"=y'; y=O, y'=O for x=O. Find the general integrals of the following equations: 2915. 2916. 2917. 2918. 2919.
2932. yy' =
vY + y'' y" y' y". 2
2933. yy" = y'' + y' Vy• + y' 2934. y'" yy" = y•y'. 2 2935. yy'' y'  y'"ln y = 0.
2 •
+
Find solutions that satisfy the indicated conditions: I
for x= 2 . 2 2937. yy"+y' =I; y=l, y'=l for x=O.
2936. y"y 3 =1; y=1, y'=1 2938. xy"= V1
+y'
2 ;
I
y=O for x= 1; y= 1 for x=e1 • I
2939. y"(1 +lnx)+x·Y'=2+lnx;y= 2 , y'=l tor x=l. 2940.
y"=~(l+ln~'); y=}, y'=l for X=l.
2941. 2942. 2943. 2944.
y"y' +y'(yl)=0; y=2, y'=2 for x=O. 3y'y"=y+y'''+l; y=2, y'=O for x=O. y'+y''2yy"=0; y=l, y'=1 for x=O. yy'+y''+yy"=O; y=l for x=Oandy=Oforx=1.
2
348
[Ch. 9
Differential Equations
2945. 2946. 2947. 2948.
2y' + (y' 6x)·y" = 0; y =0, y' = 2 for X= 2. y'y' + yy" y'' = 0; y= I, y' = 2 for X= 0. 1 2yy"3y' =4y 2 ; y= I, y'=O for x=O. 1 2yy"ty'y' =0; y=I, y'=I for x=O. 2
2949. y" = y'•y; y= }, y' 2
=; for X=
1.
2
2950, y" f ; 2 eY y' 2yy' = 0; !/=I, y' = e for X o= ;e, 2
2951. I+yy"1!1' =0; y=O, y'=I for x=l. 2 2952. (I+yy')y"=(l+y' )y'; y=I, y'=I for x=O. 2 2953. (x+ l)y" +xy' =y'; y= 2, y' =4 for X= 1.
Solve the equations:
2954. y' = xy"' + y" • 2 2955. y' =xy" + y"y" , 2956. y'""=4y". 3 2 2957. yy'y" = y' + y" • Choose the integral curve passing through the point (0, 0) and tangent, at it, to the straight line y+x=O. 2
2958. Find the curves of constant radius of curvature. 2959. Find a curve whose radius of curvature is proportional to the cube of the norma I. 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 yaxis 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 J..l.. (Hint. The frictional force is JLN, where N is the force of reaction of the plane.)
2966*. We may consider that the air resistance in free fall is proportional to the square of the velocity. Find the law of motion if the initial velocity is zero •.
Linear Di{Jerential Equations
Sec. 111
349
2967*. A motorboat 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 y 1 =q>,(x), y 1 =q> 2 (x), oo• y = 'Pn (x) are called lweor/y dependent if there are constants C1 , C1 , ••• , Cn not at1 equal to zero, such that 00.,
C1y 1 + CtYa + ... + C,y, """0; otherwise, these functions are called linearly Independent. The general solution of a homogeneous lmear differential equation y (a) =0, then Y=x'eaxQn(X), \\here r is the multiplicity of the root a(r=l or r=2). 2. f (x)=e0 x [P 11 (x) cos bx+ Q111 (x) sin bx).
352
Difjerential Equations
If q> (a
(Ch. 9
± bi) :p 0, then we put Y =e0 " [SN (x) cos bx+ TN (x) sin bx),
where SN (x) and T 1:1 (x) are polynomials of degree Nmax {n, m}. But if q> (a± b1) =0, then Y=xreax [SN(x) cos bx+TN (x) sin bx), whrre r is the multiplicity of the roots a
± b1 (for secondorder equations,
r= 1).
In the general case, the method of variation of parameters (see Sec. II) is usl'd to solve equatiOn (3). Example l. Find the ~eneral solution of the equation 2y"y'y=4xe2". Solution. The characteristic equation 2k 2  k  l =0 has roots k 1 =I and k2
=; .
The general solution of the corresponding homogeneous equation X
(first type) is lfo= C1e" +C,e 2 • 1 he right side of the given equation is f (x) = =4xe2"=t 0 "Pn(x). Hence, Y=e 2"(Ax+Bl, since n=l and t=O. Difierentiating Y tw1ce and putting the derivatives into the given equation, we obtain: 2e 1" (4Ax+ 48 + 4A)e 2" (2Ax+ 28 + A)e 2" (Ax+ B)= 4xe~". Can,crlling out e2" and rquating the coefficients of identical powers of x and the absolute terms on the left and right of the equality, we have 5A =4 and 28 4 1A+5B=0, whence A=s and B= . 25 Thus, Ye 2"
(:
x~~),
and the general solution of the given equation 1s I
y= C1ex + C2e .
'+ e·•x
(
4x 2528) .
5
Example 2. Find the general solution of the equation y"2y'+y=xe". Solution. The I we can take, for the limiting relative error, the number
where k is the first significant digit of the number a. And conversely, if it 1 is known that 6.,.;;; (k1+ ) ( lO , then the number a has n correct decimal 2 1 places in the narrow meaning of the word. In particular, the number a
)nt
r.
definitely has n correct decimals in the narrow meaning if 6.,.;;; ~ (I~ If the absolute error of an approximate number a does not exceed a unit of the last decimal place (such, for example, are numbers resulting 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.
Approximate Calculations
368
[Ch. 10
Hencelorth, we shall assume that all digits in the initial data are correct (if not otherwise staled) in the narrow sense. The results of inter· mediate calculations may contain one or two reserve digits. We note that the examples of this secliun are, as a rule, the results of final calculataons, 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 ol an algebraic sum of sevE'ral numbers is equal to the sum of the limiting absolute errors of these numbers. Therefore, in order to have, in the sum of a ~mall number of approximate numbers (all decimal places of which are correct), only correct dig1ts (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 summancl a reserve digit should be retained. Then add lhe 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, lhey should be rouncled off and one or two reserve digits ~hould be retained. Then be guided by thE' forego~ng rule of addition while rt>taining the appro· priate extra digits in the sum up to the end of the calculations. Example 1. 215.21 + 14.182 + 21.4 = 215.2tl) + 14.1 (8) + 21 4 = 250.8. The relative error of a sum of po~itive terms l1es between the least and greatest relative errors of these terms. The relative error of a d1fl'erence is not amenable to simple counting. Particularly unfavourable in this sense is the dilference of two close numbers. Example 2. In subtracting the a;,proximate numbers 6 135 and 6.131 to four correct decimal places, we get the difference 0 004. The limiting relative . ~ 0.001 + ; 0.1'01 I error IS 6= 0.00 =;r=0.25. Hence, not one of the decimals 4 of the difference is correct. Therefore, it is always advisable to avoid subtracting close approximate numbers and to tnmsform the given expression, if need be, so that this undesir~l;;le operation is omitted. 5°. Multiplication and division of approximate numbers. The limiting relative error of a product and a quotient of approximate numhers is equal to the sum of the limiting relative errors of these numbers Proceeding from lhis and applying the rule for the number of correct decin.als (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 d£cimals of the factors are correct, we find that the limiting relative error of the product is I 6=220.01
I
+ 420.01 ~0.003.
Whence the number of correct decimals of the product Is three and the result, if it is final, should be written as follows: 25.3·4 12= 104, or more correctly, 25 34.12= 104 2 ± 11.3. 6°. Powers and roots of approximate numbers. The limihng relative error of the mth power of an approximate number a is equal to the mfold limiting relative error of this number The limiting relative error of the mth root of an approximate number a I
Is the  th part of the limiting relative error of the number a. m 7°. Calculating the error of the result of various operations on approximate numbers. If Aa1 , ... , lla,. are the limiting absolute errors of the appro
Operations on Approximate Numbers
Sec. I)
ximate numbers a 1,
•• , ,
369
an, then the limiting absolute error ll.S of the result S = f (ar. ... , an)
may be evaluated approximately from the formula
=I :1.1
t:.S
t:.a 1 + ' '·
+I :L I
!:.an.
The limiting relative error S is then equal to 68
=
~; 1=I:!. I· fn +···+I :!J ~~~n =
= olnfl aa. t:.a. + ...
+ latnf/ dan !:.an.
Example 4. Evaluate S =In (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 ll.S in the general form: S=ln (a+
Jl.a=t:.b~ 210;
Yb).
ll.S= a+
vn = 2.0976... ; we
Vbb) .
We
have
leave 2.1, since the relative error of
the approximate number Y4.4 is equal is then equal to ::::::: 2 ~ =
IV b(t:.a +~
to:::::~ ·~=;0 ;
the absolute error
k; we can be sure of the first decimal place. Hence,
I ( I I I ) ll.S= 10.3+2.1 20 +lr. 20·2.1
I
= 12.4·20
(
I+
I )
13
4.2 =2604::::::: 0 · 005 ·
Thus, two decimal places will be correct. Now let us do the calculations with one reserve decimal: log (10.3 + V44) ~log 12 4 = 1.093, In (10 3 + YU):::::: 1.093·2.303 = 2.517. And we get the answer: 2 52 8°. Establishing admissible errors of approximate numbers for a given error in the result of operations on them. Arplying the formulas of 7° for the quantities ll.S or 6S given us and considering all particular differentials
1a~k 1 flak or the quantities 1a:~ 1~ti equal. we calculate the admissible ab~olute errors t:.a 1 , ••• , ll.an, ... of the approximate numbers a 1 , ••• , an, .•• that enter into the operations (the principle of equal effects). It should be pointed out that sometimes when calculating the admissible error~ of the arguments of a funcbon 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 !hi!. is possible) so that the overall total error does not exceed a specified quantity. Thus, strictly speaking, the problem thus posed is indeterm !nate. Example 5. The volume of a "cylindrical segment", that is, ·a mlid cut off a circular cylinder by a plane passing through the diameter of the base (equal to 2R) at an angle a to the base, is computed from the formula 2 V ='3 R1 tan a. To what degree of accuracy should we measure the radius
Approximate Calculations
370
[Ch. 10
R::::::: 60 em and the angle of Inclination a so that the volume of the cylindrical segment is found to an accuracy up to 1%? Solution. If LlV, LlR and .!la 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 6 =MR 2Aa ,.;;;;1 R sin2a 100 ·
+
We assume
MR
I
R,.;;;; 200
2Aa 1 and sin a,.;;;; . Whence 2 200
R 60 em AR ,.;;;;600::::::: 600 = 1 mm; sin 2a 1 d' , 6 a ,.;;;; 400 ,.;;;; 400 ra tan::::::: 9 •
Thus, we ensure the desired accuracy in the answer to I% if we measure the radius to 1 mm and the angle of inclination a to 9',
3108. Measurements yielded the following approximate numbers that are correct in the broad meaning to the number of decimal places indicated: a) 12°07'14"; b) 38.5 em; c) 62.215 kg. Compute 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) 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 1%; c) 592.8 for an accuracy of 2%. 3111. Add the approximate numbers, which are correct to the indicated decimals:
+0.49 + 3. I0 + 0.5; b) 1.2·101 +41.72+0.09; c) 38. I+ 2.0 + 3.124. a) 25.386
3112. Subtract the approximate numbers, which are correct to the indicated decimals: a) 148.163.871; b) 29.72II.25; c) 34.2234.21. 3113*. Find the difference of the areas of two squares whose measured sides are 15.28 em and 15.22 em (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.
Sec. I]
Operations on Approximate Numbers
371
3115. The sides of a rectangle are 4.02 and 4.96 m (accurate to 1 em). Compute the area of the rectangle. 3116. Find the quotient of the approximate numbers, which are correct to the indicated decimals: a) 5.684 : 5.032; b) 0.144 : 1.2; c) 216:4. 3117. The legs of a right triangle are 12.10 em and 25.21 em (accurate to 0.01 em). 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 1 ; b) 65.2'; c) 1.51 • 3119. The side of a square is 45.3 em (accurate to I mm). Find the area. 3120. Compute the values of the roots (the radicands are correct to the indicated decimals): a) V2.715; b) V65.2; c) ~/81.1. 3121. The radii of the bases and the generatrix of a truncated cone are R=23.64 cm±0.01 em; r=17.31 cm±0.01 em; l= = 10.21 em+ 0.01 em; n = 3.14. Use these data to compute the 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 em+ 0.1 em; one of the legs is 6.8 em± 0.1 em. To what degree of accuracy 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 em and altitude 11 em weigh~ 93.4 gm. The relative error in measuring the lengths is 0.01, while the relative error in weighing is 0.001. 3124. Compute the current if the electromotive force is equal to 221 volts ± 1 volt and the resistance is 809 ohms± 1 ohm. 3125. The period of oscillation of a pendulum of length l is equal to
T=2n
yT , g
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 I%, the lateral surface of a truncated cone whose bast radii are 2 m and 1 m, and the generatrix is 5 m (approximately). To what degree of
372
[Ch. JQ
Approximate Calculations
accuracy do we have to measure the radii and the generatrix and to 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 crosssection we use the formula I
I'P
E=4 'd 1 bs' where 1 is the rod length, b and d are the basis and altitude of the crosssect ion of the rod, s is the sag, and P the load. To what degree of accuracy do we have to measure the length l 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 b are known to an accuracy of 1%, l ::::;:, 50 em, s ~ 2.5 em?
Sec. 2. Interpolation qf Functions 1°. Newton's interpolation formu Ia. Let x0 , x 1 , ••• , xn be the tabular values of an argument, the difference of which h=flx; (fl.x;=x;+ 1 x;; i=O,!, , .. , n1) is constant (table tnlerval) and y0 , y,, ., y, art> the correspondin~ values of the functiOn y Then the valuP of the funct1on y lor an intermt>dlate value of the argumtnt x is approximately given by Newton's interpolation formula _ + .fl. +q(q1) fl.Z + +q(q1) .. (qn+l) fl.n (1) YYo q Yo 2! Yo · · • nl Yo.
where
xx q=T
and fly 0 =y,y0 , fl 2 y 0 =fly,fly0 ,
are success1ve finite
...
dil,erences of tht> furction y. \\hen x=x; (t=O, I, ... , n), the rolynomial (1) takes on, accorclingly, the tabular values Y; (t=O, 1, . . , n). As partie· ular cases of 1\:ewtc.n's formula we obta1n: for n =I, linear wterpolalwn; for n = 2. qupdratrr rnterpolatwn. To simplify the use of Newton's formula, it 1s advisable first to st>t up a table of finite clJfferences. If y=f (x) is a polynomial of degree n, then fl.ny;
= const
and fl.n+'y;=0
and, hence, formula (I) i~ exact In t11e general case. if f (x) ha~ a continuous derivative f Cn+n (x\ on the interva I ta. b), which includes the points x0 , x1 , ••• , xn and x, then the error of formula (l) is
R ()n
X 
Y
""...,q(ql) ... (qi+l)fl.' _ rl Yo
~
l=o =hn+t Q (qJ).' .(QM) f'B+Il
(n+ 1)1
m '
(2)
wht>re; Is some intermediate value between x; (t =0, I, ... , n) and x. For pracllcal use, the following approximate formula is more convenient:
Interpolation of Functions
Sec. 2)
373
If the number n may be any number, then it is best to choose it so that the difference ~n+'y 0 :::::: 0 w1thin the hmits of the given accuracy; in other words. the diflerences !J."y0 should be constant to within the g1ven places of decimals Example t. Find sm 26nl5' using the tabular data sin26°=0.43837, sin 27"= 0.45399, sin 28° _,_ 0.46947. Solution. We set up the table
I
I
x,
I '" I "'' · I o 42837 0 45399 0 46947
26°
0 I 2
27~
28°
I 15621 1548
fl21/,
14
26°15' 26° , 60 4 . Applying formula (I) and usmg the first horizontal line of the table, we have
Here, h=60', q=
sin 26°15' =0.43837 + ,} 0.01562+
_!_ 4
(!1 ) ~1
'
•
(0.00014) =0.44229.
Let us evaluate the error R 2 Usmg formula (2) and taking into account that if y = sm x, then 1y'no 1 ~ l, we will have:
i(i )(* )(:n)' IR J..;;; 2
1
31
1
180
7 1 l · =12o'5.1.33':::::4' 10 '
Thus, all the dec1mals of sin 2fl 0 15' are correct. Using Newlon's formula, it is alsc posstble, from a given intermediate value of the function y, to flntl the correspo.1d1ng value of the argument x (inuerse interpolation). To do lhts, first determtne the corresponding value q by the method of success1ve approxm1at10n, pulttng
q'"' =YYo !J.y.
and
q 10
(q 1' 1 l )
Ny q1' 1(qu 1l) .. . (q 111 n+ I) ~ny • • • lly0 nI !'l.rJ00 (1 =0, I, 2 .... ).
q''+''=q'"'.   · 
21
Here, for q we take the common value (to the given accuracy!) of two sue. cessive approximations q1rn' =·i'rn+•l. Whence x=x.+q·h. Example 2. Us1ng the table
• 2 2 2 4 2 6
I
I
11=•inh •
4.457 5.466 6.695
I "'" I
I
1.009 I 229
!J.>y
0.220
I
approximate the root of the equation swh x = 5.
Approximate Calculations
'874
[Ch
10
Solution. Taking Yo= 4.457, we have (0)_54.457 0.543_0 538· q 1.009 == 1.009 . • 2 (I)_ (O)+ qiO) (Jq 0 or f' (x) < 0 when a 0; f (2) f (3) 0. Hence, the conditions of 3° for Xo = 3 are fulfilled. We take a= ( 2 I ) • =0.6. 3
f (3) r (3) >
< 0;
We carry out the calculations using formulas (3') with two reserve decimals: X 1 =30.6 (2·3Jn 34) =2 4592; x 1 =2.45920.6 (2·2 4592ln 2 4592 4) = 2 4481; x, =2.44810.6 (2·2.4481ln 2.44814) =2.4477; x4 = 2.44770.6 (2·2 4477 In 2.4477 4) =2 4475.
At this stage we stop the calculations, since _the third decimal place does not change any more. The answer is: the root ,;=2.45. We omit the evaluation of 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 gtven degree of accuracy): f (x, y)=O, { rp (x, y) =0, and let there be an initial approximation to one of the solutions (;, TJ) of this system x = x 0 , y =Yo· This initial approximation may be obtained, for example, graphically, by plotting (in the same Cartesian coordinate system) the curves f (x, y) =0 and rp (x, y) ,=0 and by determining the coordinates of the points of intersection of these curves. a) Newton's method. Let us suppose that the functional determinant
a a
I= (f. rp) (x, y)
does not vanish near the Initial approximation x =X0, y =Yo· Then by Newton's method the first approximate solution to the system (6) has the form x1 =x 0 +a0 , y 1 =Yo+~o· where a 0 , ~ 0 are the solution of the system of two. linear equations I {
(Xo• Yo)+aof~(Xo, Yo)+~of~(Xo, Yo)=O,
rp (Xo, Yo)+ Cloq>: (Xo, Yo)+ ~orp~ (Xo, Yo)= 0.
The second approximation is obtained in the very same way: X 1 =x.+a.,
where a 1,
~.
Yz=Y.+~t•
are the solution of the system of linear equations f(x 1 , y 1 )+a 1 {~(x 1 , Y 1 )+~.f~.(x 1 , y 1)=0, { rp (x , y ) + a rpx (x., y ) + ~ rpy (xp y ) =0. 1 1 1 1 1 1
Similarly we obtain the third and succeeding approximations.
Approximate Calculat cons
380
[Ch
J(J
b) Iterative method. We can also apply the Iterative method to solving the system ol equations (6), by transforming this ~ystem to an equivalent one x=F (x, y), { y=ll> (x, y)
( )
IF~(x,y>j+I (Xo, 1/o), x 2 =f(x 1 , y1), y2 =(x 1 , y1 ), Xa= F (Xz, Yz), Ya = 11> (Xz, Yz),
If all
(Xn,
y11 ) belong to U. then
Jim
Xn
= ;.
n+oo
lim y11 =fJ.
n .....
.:x~
The following technique 1s advtsed for transformtng the system of equa· lions (6) to (7) with cond1tion (8) observed. We consider the system of equations a/ (x. y) + ~c:p (x, y) =0, { y/ (x, y) + bc:p (x, y) = 0, which is equivalent to (6) provided that
I~: ~
J
=1=
0. Rewrite it in the form
+
x=x+af (x, y) ~c:p (x, y) ~ F (x, y), Y=Y+Yf(x, y)+l~lp(X, y)c=;QJ(x, y).
Choose the parameters a, ~. y, tl such that the partial deriv11tives of the fum·ttons F (~. y) :nd !l> (x. y) will be equal or close to zero tn the tnitial arprOXIIlllliiO.Il; Ill Olher Words, We find U, ~. y, b as apprOXImate SOIUltOilS o the system of equattons 1 +at: (X0 , Yo) + ~c:p: (Xo, Yo)
= 0,
af~ (Xo. Y0 ) + ~c:p;, (Xo, Y0 ) = 0, Yf~ (xo.
{ 1
Yo)+ tlc:p: (X0,
+ vt;, (xo.
1/0) = 0,
Yol + tlc:pl/ (Xo, 1/ol =0.
Condition (8) will he ohserved in such a choice of parameters a, ~. y, b on the 11ssumption that the p11r!tal derivatives of the functwns f (x, y) and cp (.\, y) do not vary very rapidly in the neighbourhood of the initial approx· imation (x 0 , y 0 ). Example 3. Reduce to the form (7) the sy~tem of equations X {
2
+y 2  l =0,
3
x lt=0
,given the initial approximation to the root x0 =0.8, y0 =0.55.
Sec. 3]
Computing the Real Roots of Equations
381
f(x, y)=x2 +y 2 l, IJ'(X, y)=x'y; f~(x0 , y8 )=1.6, (xo, Yo)= 1.1; 'P~ (xo, Yo)= 1.92, 'P~ (Xo, Yo)= l. Solution. Here,
f,
Write down the system (that is equivalent to the initial one) 3
a(x +y l)+~(x y)=O, { y(x 2 +y 2 l)+t'l(x3 y)=0 2
in the form
2
(Ia,y, 6~~=~:o)
x=x+a (x 2 + y 2  l ) + ~ (x'y), Y=Y+Y (x 2 +Y2 1)+6 (x3 y).
For suitable numerical values of a, ~. y and t'l choose the solution of the system of equations 1+1.6a+l.92~=0, l.Ia~=0,
{
l.6y+ 1.92~ =0, I+I.Iyil=O;
I. e., we put a ~0.3, ~ ~0.3, y ~0.5, t'l ~ 0.4. Then the system of equations
x=x0.3 (x 2 + y2 l)0.3 (x1 y), { y=y0.5 (x 2 + y 2 l) +0 4 (x 3 y), which is equivalent to the initial system, has the form (7); and in a suffi. ciently small neighbourhood of the pomt (x 0 , y 0 ) condition (8) will be fulfilled.
Isolate the real roots of the equations by trial and error, and by means of the rule of proportional parts compute them to two decimal places. 3138. x• x ·I 1 = 0. 3139. x• + 0 5x1.55 = 0. 3140. x'4x 1 =0. Proceeding from the graphically found initial approximations, use Newton's method to compute the real roots of the equations to two decimal places: 3141. X 1 2x5=0. 3143. 2x=4x.
3142. 2xlnx4=0.
I
3144. logx=. X
Utilizing the graphically found initial approximations, use the iterative method to compute the real roots of the equations to two decimal places: 3145. x• 5x 1 0.1 = 0. 3147. x' x2 = 0. 3146. 4X= COSX. Find graphically the initial approximations and compute the real roots of the equations and systems to two decimals: 3148. x'3x 1 = 0. 3151. xln x14 = 0. 3149. x•2x 2 +3x·5=0. 3152. x•+3x0.5=0. 3150. x•+x1 2x2=0. 3153. 4x7sinx=0.
+
382
Approximate Calculations
3154. XX +2x6 = 0. 3155. ex+eax4=0. 3156. { x•+y•t=O,
[Ch. 10
3157. { xa+y4=0, ylogx1 =0.
x'y=O.
3158. Compute to three decimals the smallest positive root of the equation tan x = x. 3159. Compute the roots of the equation xtanh X= 1 to four decimal places. Sec. 4. Numerical Integration of Functions 1°. Trapezoidal formula. For the approximate evaluation of the integral b
~ f (x) dx a
[/ (x) is a function continuous on [a, b]) we divide the interval of integration [a, b] into n equal parts and choose the interval of calculations h = ba . I!
Let xi=x0 +ih (x 0 =a, Xn=b, i=O, l, 2, ... , n) be the abscissas of the partition points, and let Yi = f (xi) be the corresponding values of the integrand y = f (x). Then the trapezoidal formula yields b
Sf(x) dx~h (Yo~Yn +Y1+Ya+ ... +Yn1) a
with an absolute error of
hi
Rn ~ f2 (ba)·M 2 , where M 2 =~axJr(x)j when a~x~b. To attain the specified accuracy e when evaluating the integral, the interval h is found from the inequality h212e ._(ba) M 1
2
t l
That is, h must be of the order of Vi". The value of h obtained is rounded off to the smaller value so that ba h=n should be an integer; this is what gives us the number of partitions n. Having established h and n from (l), 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 in the notation of 1° Simpson's formula b
5f a
(x) dx ~~ HYo+ Yn)+4 (y, + Ya + •.. +Yn1) + +2(Ya+y,+ .. •+!lnaH
(3)
Numerical Integration of FunctiOfls
Sec. 4)
383
holds with an absolute error of
h'
Rn.;;;; lBO (ba) M,,
I/
(4)
I
where M 4 =max 1v (x) when a.;;;;x.;;;;b. To ensure the specified accuracy e when evaluating the Integral, interval of calculations h is determined from the inequality
h' i8o (b a) M,,..;;;; B. That is, the interval h is of the order
ve.
the (5)
The number h is rounded off
to the smaller value so that n = ba is an even integer.
n
Remark. Since, generally speaking, it is difficult to determine the interval h and the number n associated with it from the inequalities (2) and (5), in practical work h is determined in the form of a rough estimate. Then, after the result is ohtained, the number n is doubled; that is, h is halved. If the new result coincides with the earlier one to the number of decimal places that we retain, then the calculations are stopped, otherwise the procedure 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 Runge princtple, according to which
where 11 and l: are the result~ of calculations from formula (3) with interval h and H =2h, respectively.
3160. Under the action of a variable force F directed along the xaxis, a material point is made to move along the xaxis from x = 0 to x = 4. Approximate the work A of a force F if a table is given of the values of its modulus F:
_x+l_o.o~lo_.s~l_t.o~l1_.s~l_2.0~12_.s~l_3.o+1_3_.5~4.0
I
I
F 11.50 0. 751 0.50 0.7511.5012.751 4.50
I
6. 75 110.00
L
Carry out the calculations by the trapezoidal formula and by the Simpson formula. I
3161. Approximate ~ (3x1 4x) dx by the trapezoidal formula 0
putting n= 10. Evaluate this integral exactly and find the absolute and relative errors of the result. Establish the upper limit Ll of absolute error in calculating for n= 10, utilizing the error formula given in the text.
384
Approximate Calculations
(Ch. 10 I
5;~xi
3162. Using the Simpson formula, calculate
to four
0
decimal places, taking n= 10. Establish the upper limit ll of absolute error, using the error formula given in the text. Calculate the followmg definite integrals to two decimals: I
3163.
2
dx
5 +x .
3168.
1
0
5sl:x dx. 0
:rt
1
3164.
dx
5 +x 1
2
3169.
'
X
•
0
0
2
I
3165.
5sin x dx
5 :x''
3170.
1
5co:x dx. I
0
n
I
2
3166. ~ xlogxdx.
3171.
I
5cosx d
1+x X.
0
I
I
3167. 5lo~x dY.
3172. ~ ex• dx.
I
0
3173. Evaluate to two decimal places the improper integral tl>
5 !xx 1
2
by applying the substitution
by applying
Simps~~·s
is chosen ;;o that
5
1
x=+·
Verif_r the calculations
formula to the integral
5 ~x•, 1
where b
~\. < ~ · 10=.
b
3174. A plane figure bounded by a halfwave of the sine curve y = sin x and the xaxis is in rotation about the xaxis. 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 x• y• places the length ot an arc of the ellipse 1 (O.fi 222 ).= 1 situated in the first quadrant.
+
Sec. 5. Numerical Integration of Ordinary Differential Equations 1°. A method of successive approximation (Picard's method). Let there be given a firstorder differenhal equation y'=f(x, y)
subject to the initial condition Y=Yo when X=x 0.
1 1)
Sec. 5)
Numerical Integration of Ordinary Differential Equations
385
The solution y (x) of (1), wh1ch satisfies the given initial condition, can, generally speaking, be represented in the form
(2)
lim Yi (x)
y (x) =
I_,. "'
where the successHJe approximations Yi (x) are determined from the formulas
Yo (x) =y •. X
yj(x) =Yo+~ f (x, YtdX)) dx Xo
(l=O, 1, 2, ... ). If the right side
f (x,
y) is defined and continuous in the neighboe~rhood
R{lxx.lo;;;;a,
IYYolo;;;;b~
and satisfies, in this neighbourhood, the Lipschitz conditiOn
If (x, Y1)f (x, Y2 ) I,.;;;; L I y,y.l (l. is constant), then the process of successive approximation (2) definitely ronverges in the interval
lxx 0 jo;;;;h, where h=mJn(a.
~)and
M=m;xlf(x, y)l. And the error here is
Rn=ly(x)Yn(X)I~MLn
I xx. In+• (n+l)l
•
If
lxx 0 lo;;;;h. The method of successive approximation (Picard's 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 RungeKutta method. Let it be required, on a given 1nterval x0 o;;;;xo;;;;X, to find the solution y(x) of (I) to a specified degree of accuracy e. To do this, we choose the interval of calculations h=Xxo by dividing n the interval (x 0 , X] into n equal parts so that h 4 •
Where ft=f(xi, Yi) and h=f(x1,
Yj).
~
To check we calculate the quantity
1 8t= 29 YiYi
1 =I •
(6)
Sec . .5]
Numerical Integration of Ordinary Differential Equations
387
wm
If E; does not exceed the unit of the last decimal retained in the answer for y (x), then for Yi we take Yi and calculate the next value Yi+t> repeating the precess. But 1f E; > then one has to start from the begmning and reduce the interval of calculations. The magnitude of the initial interval is determined approximately from the inequality h 4 < 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 I. Given a differential equation y' =yx with the initial condition y (0)= 1.5. Calculate to two decimal places the value of the solution of this equation when the an;ument is x= 1.5. Carry out the calculations by a combined RungeKutta and Milne method. Solution. We choose the initial mterval It from the cond1tion h4 < 0.01. To avoid involved writing, let us take h=0.25. Then the entire interval of integration from x=O to X= 1.5 is divided into six equal parts of length 0.25 by means of points x; (i=O, I, 2, 3, 4, 5, 6); we denote by Yi and the corresponding values of the solution y and the derivattve y'. We calculate the first three values of y (not counting the initial one) by the RungeKutta method [from formulas (3)]; the remaining three values  y 4 , y5 , y 6  we calculate by the Milne method [from formulas (5)] The value of Yo 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 I and 2. At the end of Table 2 we obtain the answer. Calculating the value y~" Here, f(x, y)=X+!J, X0 =0, Y0 =1.5
wm,
wm.
y;
h = 0.25. l'ly 0 = ~ (k\ 0 )
+ 2k~o) + 2k~o) + k~0 )) = =
k\ 0 ) = f (x 0 , y 0) h = I
( 0 +
! (0.3750 + 2·0.3906 +2·0.3926 + 0.4106) =0.3920;
1.5000)0.25 ~ 0.3730;
k(o) )
k~0 )=f ( Xot],
!/d·+ h=(0.125t1.5000t0.1875)0.25=0.3906;
k~0 )= f
Yo++) h=(0 125+ 1.5000+0.1953) 0.25=0.3926;
k~o) =
(
x 0 f
~,
f (x 0 t ll,
k(O)
Yo+ k~ 0 )) h = ( 0.25 + 1.5000 + 0.3926) 0.25 =0.4106;
y 1 =Yo+ tly 0 = 1.5000+0.3920= 1.8920 (the first three decimals
In
thts
approximate number are guaranteed). Let us check:
8=
Iklk~o)0 >k~o) k~o)
I= I 0.39060.39261 I 0.37500.39061
20 0 13 156= · ·
By this criterion, the interval ll that we chose was rather rough. Similarly we calculate the values y 2 and y1 • The results are tabulated in Table 1.
13*
388
Approxtmate Calculations
[Ch. 10
Table /. Calculating y 1 , y,, Ya by the RungeKutta Method. f(x, y)=x+y; h=0.25
I.5000 I.8920 2.3243 2.8084
0 0.25 0.50 0 75
2
3
t( "i+i. k~l)) Yi1y
0 I 2 3
I .5703 I. 7323 I.9402 2.2073
Yt"'= =f (Xj, Yi)
Yi
X[
0 I
Value of i
t(xi+4,
'
Value of i
k(/) I
k(l)
k\1))
I
k(l)
'
y,12
1.5000 I.6420 I .8243 2.0584
0.3750 0.4I05 0.456I 0.5I46
I.5625 1.7223 I.9273 2.I907
0.3906 0.4306 0.4818 0.5477
Yt 1 k~1 l)
f (xi+h.
k(i)
!lyi
Yi+t
I.6426 I.825I 2.0593 2.3602
0.4I06 0.4562 0.5I48 0.5900
0.3920 0.4323 0.484I 0.5506
I . 8920 2.3243 2.8084 3.3590
0.3926 0.433I 0.4850 0.55I8
4
·Calculating the value of y 4 • We have: f(x, y)=xjy, h=0.25, x4 =1; · Yo= I.5000, y 1 = 1 .8920, y 1 = 2.3243, Ya = 2.8084; y~ = 1.5000, = 1. 6420, = 1.8243, = 2.0584.
u;
u;
u;
Applying formulas (5), we find y,=yo
+ 4h 3 (2y,y&' + 2y,)' = = 1.5000 + 4 ' 0325
(2·1.64201.8243 +2·2.0584) =3.3588;
!l, = f (x,, i.> = I + 3.3588 = 2.3588; u,=u. + 3h O and q>O. 1575. Hint.
f (X) =xP•ex.
r
0
arbitrary. 1576. No. 1577. 2
1580.
f (arc tan t)
5 l+t•
I
The first integral converges when p>O, the second when p i!
.r2 y
:t
s
s
1
n
a
...
(p) = ~ f (X) dx + ~ f (X) dx, wher~
.rt dt. 1578 y '
a
In a
dt
y' 1 + Sll12 t .
dt. 1581. X=(ba)t+a. 1582.421n3. 1583.8
n
rr
Jf
dt
.
In a
0
n
1579.
n
1584. 2 • 1585 . • r 1586. .r . 1587. 1 4 . 2 r 5 2 r I+ a2 , " 1589. 4a. 1590. 1 1u 112. 1...91. In 7+2Y1    . 1592. 1 2 9 5
1588.
+ 4n
.
9 2
.r
YJn. n
r 3 . 3 1 na 1593.
8
Answe,., 1594.
n 2 .
1599.
n
2
I.
42!}
e2 +3
1600. I.
1601. 
8
 .
1602.
I
2
(e"
+ 11.
1603. I.
«>
1604.
~b2' a+
r(p+I)=S xPe"dx. Applying
1605. /!b2' 1606. Solution.
a+
0
the formula of integration by parts, we put xP = u, ex dx = dv. Whence du=pxP• dx, v= ex and
.
r (p + 1) = [ xPe"J:' + p ~ xP•ex dx = pr (p)
(*)
lf p is a natural number, then, applying formula (•) p times and taking into account that
..
r
(!)=
~ eXdX= 1, 0
we get:
r "(t)(xdx+ydy) 2 +2cp'(t)(dx2 +dy 2 ).
x (ylnexdx+xln!._dy); y ey +2 (xu In ex y
1919. dz=
[(y ln
d2z= ( !._)"Y
2
\ y
In~+ In!._) dx dy ey y
+
(x ln ey.=__!._) dy J. y 2
1
!0.+ y
(;)"Y x Y )dx2 + x
2
2
1920. d2z = a2 f:u (u, v) dx 2 + 2abf:v (u, v) dx dy + b2 f~v (u, v) dy 2 • 1921. d2 z = (ye"f~+ e8Yf:u + 2ye"+Yf;w + y 2e2"f:v> dx 2 + +2 (eYf: + e"f~+xe2Yf:u +e"+ Y (I +xy) f;,v + ye 2"f:v) dx dy + + (xeYf: +x2e1Yf:u + 2xe"+Yf:v + e2"f~v) dy 2• 3 sin y dx 1 dy  3 cosy dx dy 2 + sin y dy 1).
1922. d 1 z = e" (cosy dx 1 1923. d1 z =  y cos x dx 1 3sinxdx 2 dy3cosydxdy 2 +xsinydy 1 • 1924. df(l, 2)=0; d2 f(l, 2) = = 6dx 2 +2dxdy+4. 5dy 2• 1925. d 2f (0, 0, 0) =2dx2 +4dy2 +6dz 2 4dxdy + ya 3
X
+Bdxdz+4dydz.
1926. xy+C.
1927. x•y +sinx+C. 1928.   +
+In (x + y) +C.
1929.  In (x 2 + y 2 ) + 2 arc tan+ C. y 2
1
x+y
X
X
1930.  + C. y
1932. a=1, b=1, z= ~+y 2 +C. 1933.x2 +y 2 +z 2 + X y +xy+xz+yz+C. 1934. x1 +2xy 2 +3xz+y 2 yz2zfC. 1935. x 2 yz3xy 2 z+ 1931.
Yx
2 +y 2 +C.
+ 4x2 y 2 +2x+y+3e+C. 1936. ~+..!L+~+C. 1937. Yx 2 +y 2 +z 2 + C y Z X 1938. "= 1. Hint. Write the condition of the total differential for the XI/
expression Xdx+Ydy.
1939.
t;=f~.
1940. u=Sf(z)dz+C.
1941.
~=
a
b2x. d 1 y b' d1 y 3b 8x . . . = az • d.= ""''i; d1 = 45. 1942. The equation defimng y ts the y
x
ay
x
ay
1943. ddyx = y" In Y
equation of a pair of straight lines. day1 _ _Y_1 dx  (ly)
1945. (dy) •
1946. ~=x+a!l.
d x axy ' 8 1948. oz x yz2 ih=xyz ;
xsinycosz cosxysinz
1xy" 1 '
=3
or
1;
dx x=t day (aa+ I) (xa+ yz) d.= x· (axy )a ~6y 2 3xz2
oy 3(xyz1 )
1950 . ~=l· ox
'
·
2
(d ~)
1944 dy _1!__ · •
=8
dx x=t dy y . 1947. d= x x'
dxy1 •
or
day~ d x z x 2 ·
~=zsinxcosy, 1949 . ox cosxysinz'
oz I ay=2·
8.
oz
a;,=
1951. iJz=_c 2x; ~=c"11. ox a z iJy b2 z • 2
Answers
439 1953. dz
dx=
x
1
x a +dy 1
y
y 2 a1 xy d2z = 1  dx 1  2 1 dx dy z z
1954. dz= dx   dy; z z
..p~ 21
4 1955. dz=O; d 2 z= l5(dx1 +dy 2).
1•
1956. dz=
z
1
_
2
+
(dx+dy);
2 z (d 1 2d d d 2 dy . dz 1 . d y . 'd z 4 d zz=(lz)• x + xu+ y). 1961. dx=oo, iiX=s· cJox•=oo, dx'=25 · y (zx) z (xy) a 2 1 1962. dy=()dx; dz= x (yz )dx; dy =dz =  a( x yz x yz )• X au au a•u a2u a•u x ((xy)z (yz) 2 (zx) 2] dx 2 • 1963. = = I; a"~ =aa = z=O; 0 2
+
~ ·
1'
ox
+
ov ay=O;
axz=2;
+ 1 +Ydy; d z
2v
oy• =0.
v
2
1
'lj); dX j qJ; dy

~'ll~,'ll~~
du=
1
~qJ~!j)~~
dv=
Wu Wv 1966. a) C)
~=cs~nv'
~~=cc~sv
b)
~={ !h..
Ya• y•
+S dy S
a
dy + 5 1x 5 f(x, y)dy.
o
II
1 1a
(x, y)
J
Vu• u•
Vu• 
5dy Sf
0
5dy S 0
1
2
a
2138.
f(x, y) dy= 5 dy 5 f(x, g)dx;
X
2a
12 2
5
x
o
1J
I
f (x, y) dx = 5 dx 5 f (x, y) dy + 5 dx 5 f
48
2136. 5 dy
5 dx
v..;;;
y + 2a
a
e)
f (x, y) dx; d)
I
y) dx.
Answers
444
(binormal). 2103. bxz 0 (osculating plane); ; : x+bz=O,} . (bmormal); u= 0
+ 3y+ 19z27 = 0.
~·
l+bk bl+k .r ; IS= V
1=
, 1 tb•

l+b•
¥6
} (principal
; \1 =
2107. a) V2; b)   . 2108. a) K = 4
j.
2106. 2x
e 1 ¥2 3
t
when
=2;
=
K=
1,
+v:~'
w~ =
;14 '
; T=
+
et
3:
(p~~;t>'
b) K = T = 2a c!sh•t . 2109. a) R = Q = (y +aa)• ; b) R = Q = av• 21tt. a•+bi. 2112. When t=O, K=2, Wn
normal)~
w~=O,
=
Wn
2
v~
.
Chapter VII 2
3 .
2tt3. 4 2120.
2ll4. In
25
24 .
lf' 2121. x=y: 1;
X=l;
x=3.
n
na1
9
!2 2116. 4 . 2117. 50.4. 2118. 2 . 2119. 2.4.
21111.
X=2y; y=6; y=2. 2122. y=x 2 ; y=x+9; X
2123. y=x; Y=IOx;
y=O; y=4. 2124. u=:r; y=2x;
X= I; X=3. 2125. y=O; y= V25x 2 ; x=O; x=3. 2126. y=x 2 ; u=x+2; I
x=1;
x=2. I
2127.
I
I
I
I
S dy S f(x, y)dx=S dx
Sf(x,
0
0
0
0
X
I
S dy Sf (x, y) dx= S dx Sf (x, y) dy. 2129.
2128.
y
0
1
•
0
1
I
= Sdx Sf (x, 0
0
y) dy
0
S f (x,
y) dx =
0
2
S f (x, y) dy.
I
y) dy.
1y
S dy
1X
+ S dx
I
2130.
2X+I
Sdx
S f(x, y)dy=
1
2X
0
.1!... 4
I
I
= S dy Sf(x,
I
7
y) dx+ S dy S f(x, y) dx+ 4
I
Sdy I
I
S f(x, y)dx. ~
I I
2131.
ss dy
y
0
I
+ S dx 0
Yl
II
f (x, y) dx+
s s
s
f (x,
dy
I
f (x, y) dy. 2132.
y) dx= S dx I
I
S dx S I
ax•
Ya x•
0
 Ya u•
Ya x• X
Yiu•
I
f (x,
y) dy = S dy 0
s
f (x, y) dy+
x
{+ S
v+
f(x,
y) dx.
Answers V~
1
2133.
V4
2 
x•

+ Sdx
V4
f(x, y)dy= 5 dy x•

2

VI
2134.
5 dx
f (x,
5
y)
I
S
5
dx + dy
y)
 1
dy
f (x,
5
y)
dx.
s vs dy

ftx, y) dy+
i"""'+X2
I
 Vy;:=j
f (x, y) dx+ 5 dy
2135.
1X
Sdx 5
a)
Va2 
I
f (x,
a  V a•  x' I+ VJ:4iji
o
2136.
5
f (x,
y)
dx; d)
J/~
y + 2a
e) S dy 5
a
~~
5dy
5
0
I
o
Sf (x,
1
X
f
5
a
y)
dy = 5 dy 5 f (x, y} dx;
a y)
V u'  y'
S
5 ftx, y) dx+ dy 2138. 5 dy o Vu• zuy a I
5tx5 f(x, tax 2a
o 8
1J
I
•
a
dy +
a
1
Sf (x, y) dx + 5dy 5 f (x, Y) dx,_
0
12
1
aa
2
f (x, y) dx. 2137. 5 dy
!1
1
aa
o
V x  x•
I
5 dx
x
.!!._
1
dx~
5
f(x,y)dx;c) 5dx
f (x, y) dx = 5 dx 5 f (x, y) dy + dx 5 f (x,
u Y
a
y)
vx::x:.
I
 IJ'
I
S dy 5 I
S f (x,
dy = 5 dy
y)
 V a'  y'
a
2
'{>
a
V a2
a
f(x, y) dy= 5 dy
.,.
1y
o
x>
S dx 5
b)
v.....:y;

I
f(x, y} dy~
5
8  !!'

v~
a
g). b...
Vu•1
1
5dy 5 f (X, y) dx+  V& v y•  I VI V8 •1'
f (x,
_y,_IJ.
S
_ I'
f(x.y)d~+
V 4  u• V4 u•

vi+Xi
dx 2
1
f (x, y) dx+ 5 f (x, y) dy = 5 dy 5 Vg:x2 VI Vou• V u  ,,. 1 V 8  u• Vi
a
+5
+5 
v~
t
dy
Sdy S 
f (x,
y•
f(x.gldy+
V~
VJ="iii
I
2
_ V ,.::x;
 a
Sdx S  I
f(x, y)dx+
S
dy
v ii""=Xi
2
f (x, y) dy + x>
y>
5
1


~~
1
V 4  ,,. v4 _,,.

I
+S
V4

I
5  V4
S
 I
V~
I
V~
1
f (x, y) dy + S dx
S dx S 
«5
•
V a•  u•
S
o
J!_
flx, y) dx.
y)dy.
Answers
446 a 'VI a
2139.
S dy S f (x, y) dx + S dy a 'V a• a 'VI a
S

y) dx.
~
za
a
a 'VQq
~ dy
2140.
f (x, y2
2
2
a
a
a
a
s
f(x, y)dx+ sdy
za
'VIa
2
s dy sf (x, y)dx.
f(x, y)dx+
!!:
a+ Va•y•
!!!_
4a
&a
2141.
I 1X
'VIX'
0
S dx
S
I
0
2..
f(x, y) dy+ Sdx S f(x, y)dy. 2142. 0
0
'V2X
2
S dx
S f (x,
0
0
y) dy
+
R'VI
+
'VI
I
'VI
'V;::x>
Sdx Sf(x,y)dy + S dx 'V2 n
1
S
f (x,
f(x, y) dx.
y) dy. 2143. y
0
arc sin y
1 f (x, y) dx. 2145. { . 2146. 6 . 2147. 2 a. 2148. 6 ll
ll
S
2144. S dy
arc •In y
2149. 6.
2150.
1
2
.
8 V2 5 2153. 2,p.
2151.
a
2154.
2152.
In 2
'VI (X
S
S dx

a) 2)'
4
3;
b)
2 15n16 150 ; c) 2 5
xydy=~.
·
~a V2a.
2155.
0
inR
2156.
mR 1 •
Hint. SSudxdy= S dx R
lit
=
5
R (Ieos t) dt
o
(S)
y=f(~l
ydy=
S o
(ICOS t)
5
y dy, where the last integral
Is obtained from
0
the preceding one by the substitution
X=
R4 2158. I . . t). 2157. SO. R (tsm
6
&
2159 •.
2160.
Sdq>
1
s
cos Ql
rf (r cos q>, r sin cp) dr+
0
0 It
1
:!..
2
1
sin Ql
"
cos Ql
11
0
+ S dq> T
5 rf (r cos q>, r sin q>) dr.
2161.
Sdq>
5 rf
0
0
(r 2 ) dr.
Answers
447
111
1
11
4
sin cp
4
2162. S dcp S 11
rf(rcoscp rsincp)dr.
sin cp ces• cp
Sf (tan cp) dcp S r dr +
2163.
1
•
0
0
111
sin cp ces• cp
1
sin cp
'
+Sf (tan cp) dcp 11
0
11
S r dr +
S f (tan cp) dcp S r dr.
0
111
0
4
~
a V cos
4
S dcp
2164.
1cp
111
a VCciS'ZiP
rf (r cos (~ u
v
1
u2v)dv+
u2
°) du +
Hint. After change of variables, the equa
v
tions of the sides of the square will be u=v; u+v=2; uv=2; u=v. 2 2 b ) arc tan ak ab] . Solution. The equation of the curve 2174. ab aJiiJii bh+hk
[(
Answers
448
~r(::
cos1 tp:: sln1 tp). whence the lower limit for r wi\1 be 0 and
y
.the upper limit, r= ~: cos• tp :: sin1 tp , Since r must be real, It az bz follows that hz cos 2 tp ill sln 2 tp;;;:. 0; whence for the first quadrantal angle
ak
we have tan tp..;;; bh . Due to symmetry of the region of integration relative
1
lo lhe axes, we can compute
of the entire integral, confining ourselves
4
r
ak arc tan /iii
'lo lhe
ftrst
55 dxdy=4 5
quadrant:
~171.
a)
4
Vy
a)
b)
{,
1 0, S =1 when x < 0; S=O when
"
n
nt1
2506. 99; 999.
2507. 2; 3; 5.
2508. S = 1.
Hint.
x=O. 2510.Convcrges absolutelyfor x>l, diver~esfor x..;;l.25tt.Converges absolutely for x >I, converges conditionally for 0 < x..;; I, diverges for x..;;;O. 2512. Converges absolutely for x>e, converges conditionally for I I, x.;;;;1. I 2 2520. x>3, x< 1. 2521. x::;:,l, x..;;1. 2522. x~5 , xI, x