M
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P.
PUBLISH KS
C.
T. C. BapaneHKoe. B. AeMudoeuH, B. A M. Koean, r Jl JJyHit, E noptuneea, C. B. ltPpo/iogt
P. fl. UlocmaK, A. P.
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H. Ctweea,
SAflAMM H VnPAKHEHHfl
no
MATEMATM H ECKOMV
AHAJHV
Iod
B.
H.
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rocydapcmeeHHoe
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G. Baranenkov B. Drmidovich V. Efimenko, S. Kogan, G. Luntsgtgt E. Porshncva, E. bychfia,
S. frolov, /. bhostak, A. Yanpolsky
PROBLEMS
IN
MATHEMATICAL
ANALYSIS
Under
B.
the editorship of
DEMIDOVICH
Translated from the Russian by
G.
YANKOVSKV
MIR PUBLISHERS
Moscow
TO THE READER
MIR
opinion
book.
of
Publishers would be the translation and
glad
the
to
have your
of
design
this
Please send your suggestions to , Pervy Rtzhtky Pereulok, Moscow, U. S. S. R.
Second Printing
Printed
in
the
Union
of
Soviet Socialist
Republic
CONTENTS
Preface
I.
.
Chapter
Sec.
INTRODUCTION TO ANALYSIS
Functions
Sec. Sec. Sec.
Graphs
Limits
of
Elementary Functions
Infinitely Small and Large Quantities Sec. . Continuity of Functions
Chapter II
Sec Sec
.
DIFFERENTIATION OF FUNCTIONS
Calculating Derivatives Directly Tabular Differentiation
.
.
Sec. The Derivatwes of Functions Not Represented Explicitly Sec. . Geometrical and
Mechanical Applications of the Derivative Sec Derivatives of Higier Orders
.
Sec
Sec Sec
Sec.
and Higher Orders Mean Value Theorems Taylors Formula
Differentials of First
for
The LHospitalBernoulli Rule Forms
Evaluating
Indeterminate
Chapter III
THE EXTREMA OF A FUNCTION AND THE GEOMETRIC APPLICATIONS OF A
DERIVATIVE
.
Sec.
Sec.
The Extrema of a Function The Direction of Concavity
.
of
One Argument
Points of Inflection
Sec Sec
.
Sec. .
Asymptotes Graphing Functions by Characteristic Points Differential of an Arc Curvature
.
.
Chapter IV
Sec.
INDEFINITE INTEGRALS
Direct Integration Integration by Substitution Integration by Parts
Sec Sec
Sec.
Sec. .
Standard Integrals Containing a Quadratic Trinomial Integration of Rational Functions
....
Contents
Integrating Certain Irrational Functions Integrating Trigoncrretric Functions Integration of
Hyperbolic Functions
Sec. .
for
Sec Sec
.
Sec.
.
Using Ingonometric and Hyperbolic Substitutions
Finding
is
integrals of the
tional Function
Form
f
R
x,
a bx c dx,
Where R
a
Ra
Sec
.
Integration of
Vanou Transcendental Functions
Sec
Sec.
Using Reduction Formulas Miscellaneous Examples on Integration
Chapter V
Sec.
.
DEFINITE INTEGRALS
The Definite Integral
Improper Integrals
as the Limit of a
Sum
Sec Sec
Evaluating Ccfirite Integrals by Means of Indefinite Integrals
Sec.
Charge
of
Variable in a Definite Integral
Sec. .
Integration by Parts
Sec
MeanValue Theorem Sec. . The Areas of Plane Figures Sec . The Arc Length of a Curve Sec
Volumes of Solids Sec The Area of a Surface of Revolution
Sec
Sec
.
torrents
Centres
of
Gravity
Guldins Theorems
Applying Definite Integrals
to the Solution of Physical
Prob
lems
Chapter VI.
Sec.
.
FUNCTIONS OF SEVERAL VARIABLES
Basic Notions
Partial Derivatives
Sec. . Continuity
Sec
Sec
Sec
Total Differential of a Function
Sec. . Derivative in a
Sec.
Differentiation of Composite Functions Given Direction and the Gradient of a Function
HigKei Order Derivatives and Differentials
Sec
Sec
Sec
Sec.
Integration of Total Differentials Differentiation of Implicit Functions Change of Variables
.
.
..
Sec
Sec.
Sec
Sec
The Tangent Plane and the Normal to a Surface for a Function of Several Variables . The
Extremum of a Function of Several Variables .... Firdirg the Greatest and tallest Values of
Functions Smcular Points of Plane Curves
Taylors Formula
.
.
Sec
Envelope
.
.
Sec. . Arc Length o a Space
Curve
Contents
Sec.
.
Sec.
Sec. .
The Vector Function of a Scalar Argument The Natural Trihedron of a Space Curve
Curvature and Torsion of a Space Curve
Chapter VII.
Sec.
Sec. Sec.
Sec.
MULTIPLE AND LINE INTEGRALS
The Double
Integral in Rectangular Coordinates
Sec.
Sec. Sec.
Change of Variables in a Double Integral . Computing Areas . Computing Volumes .
Computing the Areas of Surfaces Applications of the Double Integral in Mechanics
.
Triple Integrals
Sec.
.
Improper Integrals
Dependent
on
a
Parameter.
Improper
Sec.
Multifle Integrals Line Integrals
.
.
Sec.
Surface Integrals
Sec.
Sec.
.
The OstrogradskyGauss Formula Fundamentals of Field Theory
Chapter VIII. SERIES
Sec.
.
Number
Series
Sec. . Functional Series
Sec. . Taylors Series Sec. . Fouriers Series
Chapter
Sec.
IX DIFFERENTIAL EQUATIONS
.
lies of
Verifying Solutions. Forming Differential Equations Curves. Initial Conditions
of
Fami
Sec.
Sec.
.
FirstOrder Differential Equations FirstOrder Diflerential Equations with Variables Separable.
Differential
Orthogonal Trajectories
Sec.
FirstOrder
Homogeneous
Linear
Equations
Equations.
Bernoullis
Sec. . FirstOrder
Differential
Equation Sec. Exact Differential Equations. Integrating Factor Sec FirstOrder Differential
Equations not Solved for the Derivative Sec. . The Lagrange and Clairaut Equations Sec. .
Miscellaneous Exercises on FirstOrder Differential Equations Sec. . HigherOrder Differential
Equations Sec. . Linear Differential Equations
Sec.
.
Linear Differential Equations of Second Order with Constant
Coefficients
Sec. . Linear
Contents
Differential
Equations of Order
Higher
than
Two
with Constant Coefficients
Sec
Sec
.
.
Eulers Equations
Systems
of
Differential
Equations
Sec. .
ries
Integration of Differential Equations by
Means
of
Power
Se
.
Sec
Problems on Fouriers Method
Chapter X.
Sec.
APPROXIMATE CALCULATIONS
Operations on Approximate Numbers Interpolation of Functions
.
Sec. .
Sec.
.
Computing theRcal Roots
Numerical, Integration of
Sec.
Sec. .
Equations Functions
.
of
Integration of Ordinary DilUrtntial Equations Sec. . Approximating Ftuncrs Coefficients
erca
Nun
gt
ANSWERS
APPENDIX
I.
II.
Greek Alphabet Some Constants
Inverse Quantities, Powers, Roots, Logarithms Trigonometric Functions V. Exponential,
Hyperbolic and Trigonometric Functions VI. Some Curves
III.
IV
PREFACE
This collection of problems and exercises in mathematical analcovers the maximum
requirements of general courses in ysis higher mathematics for higher technical schools. It
contains over
, problems sequentially arranged in Chapters I to X covering branches of higher mathematics
with the exception of analytical geometry given in college courses. Particular attention is
given to the most important sections of the course that require established skills the finding of
limits, differentiation techniques, the graphing of functions, integration techniques, the
applications
all
of definite integrals, series, the solution of differential equations. Since some institutes have
extended courses of mathematics,
the authors have included problems on field theory,
the
Fourier
method,
and
the number of the requireiren s of the student, as far as practical masering of the various
sections of the course goes, but also enables the instructor to supply a varied choice of
problems in each section
to select problems for tests and examinations. Each chap.er begins with a brief theoretical
introduction that covers the basic definitions and formulas of that section of the course. Here
the most important typical problems are worked out in full. We believe that this will greatly
simplify the work of the student. Answers are given to all computational problems one
asterisk indicates that hints to the solution are given in the answers, two asterisks, that the
solution is given. The are frequently illustrated by drawings. problems This collection of
problems is the result of many years of teaching higher mathematics in the technical schools
of the Soviet Union. It includes, in addition to original problems and examples, a large
number of commonly used problems.
approximate calculaiions. Experience shows that problems given in this book not only fully
satisfies
and
the domain of a function is either a closed interval a. . then y is called a multiple. there
corresponds one and only one finite value of the quantity /. and a if a lt . Also possible is plex
structure of the domain of definition of a function see. Rational and irrational numbers are
collectively known numbers The absolute value of a real number a is understood to be the
nonnegative number a defined by the conditions aa if aO. if not otherwise stated. The
collection of values of x for which the given function is defined is called the domain of
definition or the domain of this function. of the function is a set of ooltxlt xgt . If to every value
of a variable x. which s the set of real numbers that satisfy the inequalities a a more comx b. .
Functions ana . Real nurrbers. and the ke If to every value of x belonging to some set E there
corresponds one or several values of the variable /y. lt lt lem Example . The function is
defined if x that vals is. if there is a function x gy such that y Hencetorth all values will be
considered as real. From now on we shall use the word quotfunctionquot only in the meaning
of a singlevalued function. the domain two inter and ltxlt oo . The following inequality holds
for all real numbers a aj as real b . x is the a r gument or independent variable The fact that y
is a Junction of x is expressed in brief form by the notation ylx or y F A. In the simplest cases.
for instance.e collection set E. Determine the domain of definition of the function Solution.
which is the set of real numbers x that satisfy the inequalities or an open intenal a.Chapter I
INTRODUCTION TO ANALYSIS Sec. if not otherwise stated The domain of definition of a
function. that is. then y is said to be a function singlevalued of x or a dependent tariable
defined on the set E. Prob ab. which belongs to son. Definition of a function.valued function
of x defined on E. if lgt. If the equation t/ /x may be solved uniquely for the variable x. Thus.
Inverse functions.b.b.
Corrposite and irrplicit functicns. is called the graph of the given function. Solution./. /
.o/plane. Solve the inequalities a lt. For example. Prove that if a and b are real numbers then
abltablta b. In He fereia case. or a function of a function. if/ /I.b. the equation y fx defines a
multiplevalued inverse function x f y such that y y for all y that are values of the function f x
Lxanple . where u x. A function defined by an equation not solved for the dependent variable
the equation xi/l defines is called an implicit unction. The graph of a function. is the inverse of
y fx. Prove the following equalities a b abHa. /. I then the function x gy. . c a a x . Find fx is /.
/. c b . Find . whose coordinates are connected by the equation y fx. if/ . f . xlgt. //T arc cos
log x. The function and / linear. . A function y of x defined by a seof equalities y /. /. . .
gfxsampx. d . Solving equation for x. . or. /. we have y l and logly log j w Obviously.
Introduction to Analysis Ch. y in an . y as an implicit function of x. in standard notation. A set
of points x. that is. Obviously. Definition . etc. Find / x /O. is called a comoosite function. the
function f x is the inverse of gx and vice vesa. . Determine the inverse of the function yl. ygx.
this function. d jtif Find /.llx . Log x is the logarithm of the number x to the base . ries of the
function is the domain of oo lt/lt. ./.. /.
. . / . Determine the domains oi definition of the following functions . Find l . / . if of . EL VV. x
A and . . y / x lo . if as a single formula using the absolutevalue sign.Sec Functions Find the
rational integral function fx of degree two. is called euen defined in a symmetric region / and
orfd if / x fx. . Given that f . t/ . b / . Prove that /lt. and / . Write the function . a. f fx x
Determine which of the following functions are even and which f x if A function are odd e . . .
fjc A SA.lt/ may any function be represented fx in defined in the interval the form of the sum
of an even function and an odd function. Determine the domain y of definition of the function
/sin x. linear linear interpolation of a function. . Approximate the value if we consider the
function / x on the interval /. a y x.
to lt/. if if . . m AM Find cpM and i jtpxJ. Fig. and the numbers jc lf xt x t form an arithmetic
progression. . d / x e sin . Find fx of l x of Show Let fn be the that sum n terms an arithmetic
progression. q z and q respecis equal t B B Fig. Find H/Um. . the graphs of these functions. /
x .. The linear density that is. sin J/. . . mass per unit length of a rod AB l Fig. a sin K b cos
tar. . AMN xAM MN AfsMquot M I f . and for the periodic functions find their least period T a /
x b c / sin . on the segments AC l CD and DB l / l t / . Express the mass rod as a function of
x. In t roduction to Analysis Ch. Construct the graph of this function. Show that if fx . if vx x
and qx . and that the product of an even function by an odd function is an odd function. A
function f x is called periodic if there exists a positive numter T the period of the function such
that fx Tfx for all valves of x within the dcmain of definition of fx. then the numbers J x f x t
and / xj likewise form such a prol gression. x of this of a variable segment tively. Determine
uhich of the following functions are periodic. /Ul. Prove that the product of two even functions
or of two odd functions is an even function. of the figure and the area S Express the length of
the segment y as a function of Construct Fig . .
x. jc . Show and p lt/ x ltp W ltp y y ty y ltp . if y . Find /I. y i/ c y COS . Let that ltp a f and t AT
a gt a.. if Write the given functions as a series of equalities each of which contains a simple
elementary function poweri exponential. / x Prove that if fx is an exponential function. and the
like member a b r. trigonometric. x t form an arithmetic progression. what regions will these
inverse functions be defined Find the inverse of the function / x. ioo if . . . /I arc sin arc tan if x
for x for lt c oo.. Functions metric progression. r . d y arc sin . . f and /jcj form a geo. and the
numbers x v . Determine the roots zeros of the rrgion of positivity of the region of negativity of
the function y if a and r/l fx. that is. Let Show that .Sec. /O. then the numbers /. a x a gt. d b c
yx y x ff x. Find the inverse of the function y In .
. Write as a single equation the composite functions repre sented as a series of equalities a
b sin. Graphs of the basic elementary functions see Ap pendix VI are learned through their
construction. if t/lt. Write. we . y y. Graphs of Elementary Functions Graphs of functions / are
mainly constructed by marking a suffiand / . arctan. Sec. / . . by connecting the points with a
line that takes account of intermediate points. Calculations are best done by a slide rule.
Proceeding from the graph of y readily fx. . ciently dense net of points Ai /. Fig. y defined by
the equations .. i i/ is the mirror image of the graph F about the axis.. T get the graphs of the
following functions by means of simple geometric constructions M js th mirror image of the
graph T about the axis. ugt y w y log. where /. b c . u Yv. functions of a x arc cos y n. Find the
domains of definition of the given implicit functions. . . . //. .. explicitly. Introduction to Analysts
Ch. ifugt. w.
. . an amount Fg. c . . // .Sec. y kx. xl . . y sinx displaced along the axis to the right by an
amount j Y Fig. t/ x. . y ax bx c. i/ x . c . . al. x . f . . . if al. . lt/ x . . y if lt/ if . . . a . . . b .c. . .
Graphs is is of Elementary functions th xaxis by an Uve /axis by i / amp / / the F graph
displaced along graph displaced of the function the F along amount a. Find the points ol
intersection of this . . . . . functions of degree Construct the graphs of rational integral two
parabolas. Construct the graphs of the following rational integral func tions of degree above
two x cubic parabola. . . y . following linear functions straight lines . Example. if . f x x. if c. ./.
the following linear fractional func . . . Construct the graph Solution. if . Construct the fc
graphs of the /. . y xx. . y Construct the graphs of tions hyperbolas . y. . t/ pa rabola with the
Acaxis. . . . yax . . . The desired line is a sine curve Fig. . if . /. .
. /. . y. x / x . //sinx / f if if / . Introduction to Analysis Ch.rj x x Newtons serpentine. if . r/
Construct the graphs of the trigonometric functions . y sinx cp. y lt/ Construct the graphs of
the fractional rational functions . . ll y sinx . l hyperbola. . /tanx. . y cosec x. yxfx f/ semicubical
parabola. ysmnx. nl. . . . . y sinx.. . . . y x y cissoid of Diodes. yj/ Vx x ellipse.gt. yt/x. t yh. . y.
/cotjc. . . . // . y sec x. JlA . . Construct the graphs of the irrational functions . n ft J. . r WtYc/i
of Agnesl. . . y/x Nieles parabola. Jl . t . . . rv /. . y H trident of Newton. . . j T . . . y cosx. i/ / . .
/ . arc tan. l e . . . . arccosx. y . y oga x. . whcn see p. where .Sec. t/ Construct graphs of the
inverse trigonometric functions . //logx . . where sinhxl/ex e x y coshx. for About the number
more details.. . . A . . arc cot x. y logl x f/ sin x sinx. . xx a y a b t/ sinA sin jt. ax a . sinhx. / arc
cot x. . Graphs of Elementary functions Construct the graphs of the exponential and
logarithmic functions . . . //loglogx. y // . . sin. if f if . f/ f/ . x arc cos. arcsin. Construct the
graphs of the functions . j/ . . . . ye probability curve. /yrVlog X the logquotcosx. a . lt l. .. . . .
yx. yarc sin. b x when jclt . . . x x. . . where coshx tanhx. quotquot . ylog A.. .
cos/. .straight line. that is.x . . . r . logarithmic spiral. / a hyperbola. .branch of y sin segment
of /t yt xa cos/ cos/. r al fcoscp agt cardioid. Construct the graphs of the following coordinate
system r. x t y . / . I . . f J/ rTT //wm Descartes. x . t/ asm / /cos/ y involute of a circle. jc cos f f
. Cjnstruct the graphs of the functions represented parametrically t semicubical parabola. f r r
ltgt spiral of Archimedes. where x is the integral part number x. . a of the . asin/ t t . / . x T y T
. jc acos/f / sin/. the greatest in. rsinp threeleafed rose . . sin / cardioid. / . . . secy parabola. . /
l. r . . xy hyperbola. . .eger less than or equal functions in the polar to x. / j/ t . cos /. i/ circle. r
a cosp agt lemniscate. sin / astroid. cp rO . xasfct. Introduction to Analysis C/i. b y xx.
hyperbolic spiral.aT astroid. ysin/ ellipse. . jc . t / Cjnstruct the graphs of the following
functions defined implicitly . jc parabola. a straight line. . / cosip circle. . .
Graphs of Elementary Functions . Construct the graph of the function obtained.jc. xl cot .
tion of the base x. Solve tjie systems of equations graphically a xy.c. y . . . Construct the
graph of this function and value. Derive the conversion formula Irom the Celsius scale Q to
the Fahrenheit scale F if it is known that corresponds e aquot C to F and C corresponds to
F.Sec. d e f I x x sin. x xr/. logJt xx ltjcltjt. Given a triangle ACB with BC a. Fig ACB x Fig. cos
ltxlt jt. /V y logarithmic spiral. . . Plot the graph . x f y b c d e j/ sinx. is a the area of the
rectangle y as a funcExpress Fig. . a x b x c x x . . Inscribed in a triangle base . . AC find its
greatest b and a variable A x. altitude h rectangle Fig. Give a graphic solution of the
equations . angle area ABC as a function of Express of this function and find its greatest
value. x x// folium of Descartes.
. there is a number NN e such that xn lte when ngt N. the limit on the left of the function f x at
the point a and the //mi/ on the right of the function / x at the point a if these numbers
exist.l/nfl. rtr Solution. Form the difference Evaluating the absolute value of this difference.
formula is true. for every gt we have for gt fjO such that fxA lte Similarly.A as x a A and a are
numbers. / Sec. or lim fx x a if gt A. x lt . hence. x gta lim lta and x a. The limit of a function.
ngt N Thus. then where E is number limits...a fa If x . ii an arbitrary . lt x alt. .. a is the limit of
a sequence xn n gt if oo for any a egt . . for is if f X Alt lim xgt /Ve. for every positive number
there will be a number Af such N we will have inequality Consequently. or The lim number a.
a which means that positive f x gt E for lt lt E. if lt e. The following conventional notation a /x x
also used oo. the number is that for n the limit of the sequence x n n. We say that a function /
x .. that Urn n Example Show L . we have e. lim gt . The limit of a sequence. X. Introdnction
to Analysis C/t. The numbers lim / fx and /a x a a o re called. similarly. Lfmits . .. a. Onesided
x . . respectively.. and x o then we we write conventionally write af.
. x f lim f. .. . x xlimaf . For which values of n will we have the inequal ity e is an arbitrary
positive number Calcula e numerically for a e . arc tan x J o fa and f Obviously. Prove that as
n oo the limit of the sequence is equal to zero. Solution. is For the existence of the limit of a
function and sufficient to have the following equality /a If / x it necessary O/ following
theorems. x. x lima jcJ /. lim o faictanlW / A in this case has no limit as x.o W/ JKJ lim /. /. the
function / x xgt. The following two limits are frequently used lim AP gt ILii and lim JL x lim l a
a left . Example . Limits as jco. . x and lim f x exist. We have lim x . Find the limits on the / x
right and of the function arc tan as x gt. the limits lim /. x Jt a . then the x gta x gt a I old x lim
a lim /. x a x a lim f . lim f t x. xl jc lim a a x lim f . c . xa f. x. Prove that the limit of the
sequence . b e . x..Sec.
CD n gt. oo... . . n /il quotquot . lim n CD . . / /quot. l oo Hm n J . c /. X gt e . How positive
should one choose.. X . . . . c following liai/x X gt . notations Give the exact meaning of the
Hoi log oo.. d ... some number so that the inequality lt b should follow from Compute . ..
Prove that liin x. a b V I Find the limits of the sequences i. Hm n lirn fl a l. .... . b lim . b e . c e
. .. ///quot. . a gt for a e . Find the . For which values of ngtN will we have Kllte c e is an
arbitrary positive number Find N for a e . quot ... for a given positive number e. as the
inequality Introduction to Analysis rt CH. .. limits . gtoo is unity.
r lim . o/ Hm Vn n i lt fn.Sec. ltxgt . then it is advisable to camel the binomial a out of the
fraction P Q x once or several times. or rrr. . lim J. lim gt jA Y L h . lirn . . . Example . lim J . it
seeking is the first limit useful of a ratio of two integral polynomials in n to divide both terms
of the ratio by x . O . many cases for fractions contain Example lim . lim lim v . lim . . V X . lim
When x oo. rrc are integral . lim Vx Vx If PA and Q x polynomials and P u or Q a then the
limit of the rational fraction lim is obtained directly. / . . where as n is the highest decree of
these polynomials. lim . . lim /T lim quotquot f xf Hm . Limits . JtfA Example .. A similar
procedure is also possible in ing irrational terms. But if Pa Qa. lim . lim / jc . Jquot . .
Putting we Mm have lim E /. .. lim . lim x aX aVx V a lim gt a jc f V a fi . l . . lim x . I .n / . lim .
. Q j . or vice versa. T Another way of finding the limit of an irrational expression is to transfer
the irrational term from the numerator to the denominator. . lim f. Introduction to Analysis . lim
quot . Um gt fl in Ch. li. . lim / Lquotquot . Example lim . lim . terms are many cases rational
Find lim Solution. . lim lt x . lim . from the denominator to the numerator. . . lim o . / t/x .. lim X
. Example . lim The expressions containing irrational ized by introducing a new variable.
sin a and lim cos granted that lim sin is taken for Example .n. . a b lim xsinl. . li. gt CO sill .
lim lim lim . . The formula llm X X i It r frequently used when solving the following examples.
K . lim x sin X x . . . lim lim . lim JC arc sin lim lim I crs tan . lim i .Sec. Limits . n I . lira lim . . X
. lim cot x cot f x. cos a. .. .cosn V sui . lim sin six . a lim. . . / sin JTX lim M sin BJIJC l . bli.
lim Jt ji lim n ncc sin. lim cosmx tan A . Hm/xfa JffCO . lim X lim / a xj. . . . lim JtM x tan . . sin
lim X sin .
I n . where a x as x a and. Find is Napiers number. nx ta . .. linipUl and we put qgtx ax. Hm a
x x Hm p x . quotquotr quotquotquot f p Jt When taking limits of the form lim lltpJt U X fl C
one should bear if in mind that there are final limits lim cp x A and lim x B. Here. if lim px is /l
and lim tyx oo.T. . then the problem of finding then Cquot. Example . lim x then co. lim JfO X
lim xo and lim hence.quot Introduction to Analysis Ch.ij ty x x a where e . We have lim r J x
end Hm . the limit of if solved in straightforward fashion. liene. m.. Find Solution. Example lim
Solution.
quotquot quot li . . lim . as indicated above. J . a lim cos x X .. lim X XKgt M x/ i . . . Find lim f
x t Solution. b H f I. lirnl . /i a gt Jill T / . V Hmfl Iiml / . lim / j .Sec Limits Therefore. . Example
.. T . it is useful to lim remember that . we have In this case it is easier to find the limit without
resorting to the general procedure Generally.CO i .o sinjc . Transforming. We have lim X X .
a lira . gt. ptX . . ital the . lX/ . . li . . Example tO. lim. a a lim lim . n llmn/a V pCLX . a lim b im
V . Prove that Solution. T a/lLutanh. p. lim In X . . o . see b x Problems and . fa Hm quot i
bJirn . . lim. b a lim limtanh. Hm. We have lim X ln X is XQ Formula frequently used in the
solution of problems. . . When solving the problems that follow. limit lim/x exists and is
positive. gt b Hm where tanh . . if is useful to know that the In Hm XQ f x. Find following limits
that occur on one side . lim lnjtl lt. li sin agt. a lira . Inx.ltgt quot limfjlnl/Ji. then lim In / xa
Introduction to Analysis it Ch.
Mn inscribed in a logarithmic spiral . .. n. x Construct the graphs of the following functions im
cos quot.. drawn x . M.Sec. y . will happen to the roots of the quadratic equation if the
coefficient a approaches zero while the coefficients b and c are constant. a limi. where xl. t/ lt
. Find the limit of the interior angle of a regular ngon gt oo. Find the limit of the perimeters of
regular ngons inscribed oo. a Hm.. . Limits .. Find the limit of the sum of the areas of the
squares constructed on the ordinates of the curve at the points as on bases. . Find the limit
of the perimeter of a broken line oo. and fc . as n oo.. Find the limit of the sum of the lengths
of the ordinates of the curve y ecos nx. ylim nc i xn . t/ im agto J/Vta li. it What as the limit of
the corresponding finite fraction.. n. in a circle of radius R and circumscribed about it as n . .
.. as n . .. . xO.. . . Regard . . Transform a the following mixed periodic fraction into common
fraction a . provided that n . y ngtoo .n li .. .
t the time interval Find Qlhi .Introduction to Analysis Ch. Fig in half.. on each of Determine
the limit of the area of the steplike figure . . I oo.C in half. . Assuming that the quantity of
substance at the n after determine the quantity of substance Q nth part of if the increase
takes place each the elapse of time t What initial time is Q . Fig. is the geometric meaning of
. A certain chemical process proceeds in such fashion that the increase in quantity of a
substance during each interval of time r out of the infinite sequence of intervals tr. and so
segment C. . qgtrt y Fig. A segment AB a length of AB despite the fact that in the limit the
broken line quotgeometrically merges with the segment ABquot. the point C divides on. i f lt
substance /. is divided into n equal parts. . Fig. The side a of a right triangle is divided into n
equal which is constructed an inscribed rectangle parts. .. thus formed if n ou. . The l l . .
each pnrt serving as the base of an isoscelos triangle with base angles u . . if the vertices of
this broken line have. ltPJ y . . divides a segment AB divides a segment AC in half. is
proportional to the quantity of the available at the commencement of each interval and to the
length of the interval. Determine the limiting position of the point C n when /ioo. Find the
constants k and b from the equation . divides a point C C C in half. as n the polar angles ltP.
Show that the limit of the perimeter of the broken line thus formed dilTers from the . the point
C. respectively. . the point C.
. xgta i. . If lim a x . The function u x is called an infinitesimal of order n compared with the
function p x if lim Qx quot C where If lt J C lt f oo. then we say that the function a x is an
infinitesimal of Higher order than p x. when taking the limit of a fraction of The sum two term
whose order is lim aPW . By virtue of this theorem. tanx x. lim i j/T lim a/ . infinitesimals of
different orders is equivalent to the lower. t gt as x a we can subtract from or add to where a
x gt. but if C . . Infinitely small quantities infinitesimals. Infinitely Small and Large Quantities .
similar fashion we define the infinitesimal a x K lt also infinitesimals as If The sum and
product of xa. then the functions a x and p A are called equivalent functions as x a For
example. xa P x where C is some number different from zero. The limit of a ratio of two
infinitesimals remains unchanged if the terms of the ratio are replaced by equivalent
quantities. ax and p a limited number x of infinitesimals as x a are x are infinitesimals as a
and lim SlJfUc. as x x a. ber Af there Infinitely large quantities infinites.. for x gt we have
sinxx. o x .Sec. lnlfx x and so forth. gt Example . Infinitely Small and Large Quantities Sec. if
axlte when as lt In infinitesimal oo. and p x the numerator or denominator infinitesimals of
higher orders chosen so that the resultant quantities should be equivalent to the original
quantities. then the function a x is an a fie. then the function fx is called an infinite as x gta.. If
for an arbitrarily large numa x exists a N such that when N we have the lt lt inequality
lfMIgttf.e.. then the functions ax and px are called infinitesimals of the same order.
b the volume of a sphere if the radius of the sphere r is an infinitesimal of order one. b e . x
is an infinite for x . Let the central angle a of a circular sector ABO Fig. is an infinitesimal as x
oo. the function . c b . . with radius R tend to zero. As in the case of infinites of different
orders. Determine the order of smallness the surface of a sphere.. What will the orders be of
the radius of the sphere and the volume of the sphere with respect to its surface . Determine
the orders of the infinitesimals relative to the infinitesimal a a of the chord AB b of the line
CD. we introduce the concept . c of the area of A/BD. Prove that the function ... For what
values of x is the ine quality /lt fulfilled if e is an arbitrary positive number Calculate numeri
cally for . analogous. The definition of an infinite f x as x of infinitesimals.. In what
neighbourhoods of x lt is the inequality lfxgtN fulfilled if N is an arbitrary positive number Find
if a . Prove that c e. is an infinitesimal for x gt.. Prove that the function Introduction to
Analysis gt co is Ch. b e. For what values of x is the ine quality l/WIlte fulfilled if e is an
arbitrary number Calculate for a e . . c e a e. o of a J ..
. Using this result. Prove that the length of an infinitesimal arc of a circle constant radius is
equivalent to the length of its chord.. lim x o lnl . lim arc sin .. c /lO. two infinitesimals. . Can
we say that an infinitesimally small segment and an infinitesimally small semicircle
constructed on this segment as a diameter are equivalent of Using the theorem si quot of the
ratio of . . . Prove that when x the quantities and Y xl x is are equivalent. Prove that when x
mate equalities accurate to terms of order x b c d loglx where Afx. approximate gt gt .. gt Ilt . .
approximate the following /L. Af log e . Infinitely Small and Large Quantities of .. . f . Using
these formulas. find . b c / . . we have the following approxi. demonstrate that small we have
the approximate equality when VTT Applying formula a .Sec. Compare the values obtained
with tabular data.. log .. For x determine the orders d e smallness relative to x of the
functions cos A x tan sin A. d /T and compare the values obtained with tabular data. lim s
quot . n x amp nx n is a positive integer. b /. . lim .
then it is said to be continuous in this region.x n a. Definition of continuity.. the function sin is
continuous when ooltxlt . f an is an infinitely large quantity equivalent to the term of highest n
degree a x .. Let xoo.e.. . there exists a number / g. determine the order of growth of the
functions . if this function is defined at the point g. If a function is continuous at every point of
some region interval. llmf fc /. . there exists a finite limit lim f x. Continuity of Functions .
Prove that the function y fs sin x x. cos f x si lim T T and it follows that for any x we have lim
A/ Hence. Introduction to Analysis Ch. this lim x it is equal to the value of the function at the
point g. a gt . c b Sec. continuous Solution. Show XOQ the P x a. condition lim may be lim
rewritten as g A/g l/g Agf . A function / x is continuous when x or quotat the point gquot. for
every We value of the argument have cos Ay sin Since lt Asin x sin x sin . Example . or the
function / x is continuous at the point g if and only if at this point to an infinitesimal increment
in the argument there corresponds an infinitesimal increment in the function. . Putting where
Ag .x n that for l rational integral function . i. Taking x to bean infinite of the first order. that is.
. etc.
a is . the nxQ is three numbers fx a discontinuity of the first kind. discontinuous This function
not defined at the point x and no matter how we choose the number /I. Q if called then is For
continuity ficient that called a removable discontinuity. If the function f x has finite limits
function / x will not be con Hm and not all / f . the redefined tinuous for .Sec. We say that a
function /xhas discontinuity at or at the point XQ within the domain of definition of the function
or on the boundary of this domain if there is a break in the continuity of the function at this
point. / f x In particular. a x Example when xl. of a function fx at a point JC Q. it is necessary
and suf . . The function is fx Fig. and Urn / / are equal. Continuity of Functions . Points of
discontinuity of a function.
obviously.e. . where Ex denotes the integral number x i. on all a.. at every integral all the
discontinuities are of the first kind. continuous. B. there is some number a. and a function cp
x is continuous in A B. if n is an integer. In t reduction to Analysis Ch. x is continuous for any
value argument . Indeed. there will be at least one value JC Y ltYltP such that fyC. The
function /jcjy lim has a discontinuity of the first kind . . At all /i /il and /i other points this
function is. . Indeed. i... J. matter what the if and fl altaltplt. In particular. These ane points
such that at least one of the onesided limits. then no ithat is. /ltgt or /o is equal to oo see
Example . A function f x continuous in an interval a. Example . here. E x is an integer that
satisfies the equality x Ex ltltlt. p. then the composite function cp/J is continuous in a. The
function cos Fig. number C between A and B.. Discontinuities of a function that are not of the
first kind are called discontinuities of the second kind. and a set of its values is contained in
the interval A. is and Fig. The iunction y Ex. When testing functions for continuity.. . then /i. b.
I Example at . the quotient of two functions continuous in some region is a continuous
function for all values of the argument of this region that do not make the divisor zero.
discontinuous point x q. lOc at the point x has a discontinuity of the second kind. b has the
following properis ties f x boanded on /ltM when / x has a / x takes altlt. since here lim cos L
X X both onesided limits are nonexistent and lim Jtgt cosi X . b. / L i and / of the lim jco x part
where . then the equation fa A /P intermediate has at least one real root in the interval a.e.
Properties of continuous functions. Infinite discontinuities also belong to discontinuities of the
second kind. Example . if a function t f x is continuous in an interval a. of the Show that the
function y x. bear in mind the following theorems of a limited number of functions continuous
in the sum and product some region is a function that is continuous in this region. if fa/plt. M
such that minimum and a maximum values value on b between the two given values. i.
. Prove that the rational integral tunction Continuity of Functions a. Prove that if the function f
x is continuous and non negative in the interval then the function is likewise continuous in
this interval. The right side of the equation that How fx is lx sin should one choose the value
/O jc meaningless for x . . Show Plot the function should one choose the value of the function
Af so the thus redefined function fx is continuous for Plot the graph of the function y fx. Is it
possible to define the value of / such a way that the redefined function should be continuous
jc . The function f is arctan in for meaningless for x. . For what values of x are the functions a
tan and b cotjc continuous that the function is continuous. is continuous for any value of x. . .
Prove that the function y Yx is continuous for xampzQ. Prove that the rational fractional
function is continuous for all values of x except those that make the denominator zero. . .
Prove that the absolute value of a continuous is a continuous function. . Prove that the
function y cos x is continuous for any x. A function is defined by the formulas . graph of this
function. .Sec. . How so that fx is continuous for .
. . not defined . Prove that the Dirichlet function x which is zero for and unity for rational x.
Introduction to Analysis Ch.. . . if for x . for quotT xgt. irrational x value of x. b c cos A / lnx d /
f / cot. b y xsn. lt/ l y yssi . The so that fix function fx is is continuous for x . . lt/ arc tan .
Define /O a fx / y y x / l I . is discontinuous for every t Investigate the following functions for
continuity and construct their graphs . . y arc tan A . yJ l . y . . . yncosx. in y lim x arc tan nx. y
. Pot the graph of this function. . Investigate the following functions for continuity . . . n is a
positive integer. . y . / / .
Continuity of Functions . c i/ sgnsinjt. . if if if gt. b y xEx. Prove least . Let a be a regular
positive fraction tending to zero ltaltl.Sec. a y sgnx. Give an example to show that the sum of
two discontinuous functions may be a continuous function. where the function sgn x defined
by the formulas . . Prove that the equation has an infinite number of real roots. Show true for
all values of a that the equation has a real root in the interval . a y xEx. that any polynomial
Approximate P x of odd this root. Can we put the limit of a into the equality which is l a l. . .
sgn x . .. is b y x I sgnx. . power has at one real root. . x . where E x is the integral part of the
number x. lt.
Example t. is called the mean rate of change of the function y over the interval x. For the
function MN . and Ai or / y f x. f Ax. If x and x l are values of the argument x. . . The ratio
ampxMA same interval jc. called the increment of the argument x in the interval xj. Increment
of the argument and increment of the function.Chapter II DIFFERENTIATION OF
FUNCTIONS Sec. f x f x A r ltn Fig. and y fx and t/ /jc are corresponding values of the
function y fx. Calculating Derivatives Directly . . is where called the increment of the function
y in the and by AN. then xx is l x x. Fig. s and the slope of the secant of the graph of the
function yfx Fig.
Calculating Derivatives Directly calculate a Ax and A. r V J Here. f. the limit of the ratio r
when Ax approaches zero.l . Find the derivative of the function is y Solution. to a change in
the argument fromxl from b x Solution. of the secant passing through the points Solution. AJ/
M . Example . graph of the function y of the derivative yields the slope of the fx at the point x
Fig. The expressions / lim AJO f / Ax and /x lim Ax . In the case of the hyperbola y N . /
tangent MT to the y tan qgt. I . .. . Ax and Ay U o U Hence. . lim Ax L lim AJCgtO Onesided
derivatives. corresponding to xl. ygt The magnitude lim AJC gt o A gt. x. Example . . The
derivative yf x is the rate of change of the function at the point x.Sec. Ax derivative. The The
derivative is yj of a function yfx with re spect to the argument x that is. We a have Ai/ b l. .l.
Finding the derivative usually called differentiation of the function. and .. find the slope . xi
From formula we have A Ay and Ax Ax Hence. Ax At/ Axl.. to x .
Find / of the function Solution. Example . AA b c .. the tangent to the graph of the function y
fx is perpendicular to the xaxis. the function Example Find / and / of Solution. Find the
increment of the function to a a change in argument to b c from x from x from A to xt xl A...
Uoo. it is necessary and sufficient that /. Find A// of the function a . the lefthand or righthand
derivative of the function fx at the point x. yi/xil AA. are called. In this case. for the function y .
If at some point we have lim //. . Differentiation of Functions Ch. . . Why can we. We have VV
im /llm Ax lt y x that corresponds .c .. . /. For / x to exist. and AJC . . a. while for the function y
x this cannot be done . Find tions the increment f by and the ratio A for the func a b c y Jg f
quot andAjc . .. and A. /Z. By the definition / we have lim L f lim Ato Ax . respectively. yl/quotx
y ogx forx for x . determine the increment Ay if all we know is the corresponding increment
Ax . h. Infinite derivative. then we say that the continuous function / x has an infinite
derivative at x. to .
What to be understood at a M . c xl. b the rate of cooling at a given instant .Sec. Let /X be
the mass of a non. . x lfc. Find to a change in argu a yax yx. mean rise of the curve j/ /x in the
interval fx by the rise of the curve y given point x . Ax . What is to be understood by the rate
of reaction of a substance in a chemical reaction . Find the x. b the instantaneous rate of
rotation.homogeneous rod over the interval . . What is to be understood by a the mean linear
density of the rod on the interval x. Define a the mean rate of rotation. What is the average
velocity of the point over the interval of time from t to in the interval .Ax for the functions .
Find the slope of the secant to the parabola y if x x t the abscissas of the points of
intersection are equal a x. b the linear density of the rod at a point x is xAx. where the
distance s is given in centimetres and the time t is in seconds.. What is to be understood by a
the mean rate of cooling.. A hot body placed in a medium of lower temperature cools off.
What interval is the mean in the lx rate of change of the function y x / . x. The law of motion
of a point is s / . x a . if Ax. Calculating Derivatives Directly which correspond Ay and ment
fromx to x. . Find the mean rise of the curve y . Find a the ratio b of the c function / at the
point is x . . Ax . d y e b y /x.l. xAx. limit does the slope of the secant tend in the latter case
To what if /igt . What the deriv ative y when x .
to the curve y Q. Show that the following functions derivatives at the indicated points a have
finite y l/xl y at at at b y c x x cosx jt. Differentiation of Functions C/t. . if / . cucu . Find the
value of the derivative of the function at the point i x x X Q x . Find /. What are the slopes of
the tangents to the curves y and y x at the point tween these tangents. Calculate f. then and
w ogtjc . Find the speed at . Tabular Differentiation .lx to the curve ysinjt at f / the point . t. of
the functions a t/ x f . At what points does the derivative of the function coincide numerically
with the value of the function itself. . i . r . Sec. xV . v . Find / lirn ianx. / that is. of their
intersection . The law of motion of a point is s /. . Basic v tyx are c rules for finding a
derivative.. /I. / / . Find the do not angle be . . Find the slope of the tangent ji. . if . where the
distance s is in metres and the time t is in seconds. If c is a constant functions that have
derivatives. fc . /.. . c y . . Find the derivative of the function y . Find the slope of the tangent
drawn at a point with abscissa x .
XVIII. Vx XXI. . lt .Sec. If y fu and u where the functions y and u have derivatives. . XI. cosx
cos. // composite function. arc X. sin. arc tanh x . XVI. joypj. ZL. x nx . sinx IV. VT arrdn
arccos i.quotquotquot . Tabular Differentiation . for differentiating a ltp A. Rule that is. VIII.
Table of derivatives of basic functions n n. I. cothx cosh x quot . XIV. XV. ltlt. XIX. XII. then y
or in other notations / dx du dx This rule extends to a series of any finite number of
differentiate functions. XIII. IX. arcsinhjt Jr. XXII. XVII. jcltl. III.
if . u x x Example . il Va . . i B.aHWquot. Find the derivative of the function y sin Solution. . .
by formula jt we will have y uu x . ouo. a of the composite following functions the rule for
function is not used in problems A. Putting / . y v X . / . Differentiation of Functions of the
function Ch. /quot f y X I f v A . yn n . Putting a . O Q Of i/. t Qv JA f O v A. Example . S.
where w jc . we find gt sin xcosjt.. . u jc. t/ sin cos x. . QQ oOy. jc .rJtpJT A V U. yarctanh f/
arc cot x. JC . . yax f ampA C. . Find the derivative Solution. . Find the derivatives
differentiating . quot quot. .OA QIST o/O . Inverse Circular and Trigonometric Functions . . u
sinu. tanx cotx. ltgt i/ v I quot .j//sincos. . Algebraic Functions s . y QO V A A V TEA Ov AA v
j.
.jAi r i jAt j Solution.a I x jc L. y xle. Exponential and Logarithmic Functions . w M jc . .Sec. . t/
Jtsinhjt. j . y . . yVcosh x // t/ arctanx arc J / arctanh t/ tanhA iiL Inx . / . y a Hi iv . earc sin x. . .
/ . y . . Denote jt w. jc. y nxogx In a log a jc. Hyperbolic and Inverse Hyperbolic Functions .
t/J/TJquot. y .. We have u . then t/ w jlt . . gt quot x I . . i/ . r/ J. use the rule for differentiating a
composite function with one intermediate argument. f/ gt . . . yKe. Tabular Differentiation C.
Find the derivatives of the following functions Q quotv u . . Composite Functions In problems
to . . wa. . y / . D. / x e cos . i / . y/ . A. // i x E.
sin x Ch. . Solution f cos sin f V / cos . . cosec sec . cos x Solution. . sin y t Jt sin x cos x . y
Solution. . arc tan. t/ sin sin x . /sin/si . .. j/ y . f . y arcsin. . y cos . arc sin x y J/arc tan /. y/ex
tart y sin cos . Differentiation of Functions . f fx cosx / / arc sin x. j/tanjc . t/sinx . y . quot lcos .
y x cos . . / sincos y y t/ . x . . t/ . /x . y .CGSJC . acot y arc sin x. . /x . /x cosct. . . y r/J/coU
/coU. arccosJ/. /Yx Sm sin x quot f .
az ijc x f/ . . . z . y n Inlnjc. . tsm . y lnl . j/ . Miscellaneous Functions . y . x . y arc cose. . .
ysinjccosy. . y . .Sec. . y /In xl In /. j/ j/ t . / . t/ x a J quot Jti . . y ne arctan lnA lnarctan. t/ . e. //
. AAA . Tabular Differentiation . t/ gt . . t/ F. X logsinjc. ft sin x arc sinx.
. lt/ arc tan t/ arc tan Vbxx.sinx. . F F . . y arc sinjc arccosA. . . j/sin . . / i/ . O . y/a sin p cos x.
CL .af c sin fx I/ a / . . y arc sin arc sin cos sin x . y arc sin Inx. y K x a arc sin f cotx. . ln/lelln/l
cosx cos . y V . . jcI y lnarcsinx. Differentiation of Functions C/t. . t/ / arc cot x. . w . . . t/jt/a . .
T a arc sin. . y O tan ianx x. y . tan ltan xHOtan fl . jc yarcsinl a . . . y arc sin arc cos jt. . tani .
yianx. . .. / sinAcos o A. . y . y ..
y tan In x.smx x lt cos . . . y . l l n z x . I . y x / In / lna /ax . Tabular Differentiation . y . t/ l . y
.Sec. . . / . i/ . xsinlnx COS X . / . . y . arctanln . l . y e. y K r I/cos a . l arc sln sin arc cos x ux
u sm . y / xquotaquot. y . . . / j/ i . ./ Uln arc tan n . ylnln . . . . ylnarc sinx f/ jc arc sin In x. m
cos bx . In .
e cosh px. r/ . / Find . . .In / . . In J In . Find f x if x for xlt. y. arc cosh In. e fx x sin le T . Vsin
arc tan x. e cos cos jc. fxlnlx arcsin. In Ji. f a . arc cot h sec. y arc tanh y arcsinh. y . . Find
and / of the functions b / x arc sin Y . . f / e x sin x e sin e x . . if Construct the graphs of the
functions y and . . t/ . C/t. t/ . y . . y . Find . a yx b . gj. t/ Aa . ytan . if . Find /I. smhx. Calculate
/ if / Solution. . yr. arc tanh tan x. Differentiation of Functions arc tan . y ianhx. cos . y
Insinhjc. Find y.
derivative of a periodic function the is also xy dxy . is fM first sometimes simplified by of the
taking logs of the func Example. satisfies lrlx the equa xy yyix G. y fx is the derivative of the y
Finding the derivative tion. that is. Prove that the a periodic function. Solution. or y u whence
. and the derivative of an odd function is an even function. Given the functions /xl . Show that
the function . Find / f xf of the function fxe. . Tabular Differentiation . x and cpjc sin r nna find
ff . satisfies the n . Prove that the derivative of an even function is an odd function. Given the
functions fx tarx and p Y x. Differentiate both sides of this equation with respect to x In y v In
u v In a. y xe . Taking logarithms we get In y v In u. xy that the function xy. Find the derivative
exponential function where u yx and v tyx. tion y x . Logarithmic Derivative A logarithmic
derivative of a function logarithm of this function. lnl xgt l find tind n q/or . Show Show that the
function y xe satisfies equation e ltl uati .Sec. . Find / x / of the function fx .
Ln the following problems find y after function y fx taking logs of the Sec. Find Solution. . .ui i
y y x x x . In sin x x cot x. I a function yfx has a is . y . The Derivatives of Functions Not
Represented Explicitly . In y . sin x x In sin x x cot x. or Differentiation of Functions Ch. first .
In y y / if y smx / x In sinx. .i . . . whence y / x x .In x In u x In x In sin x In cos x. cosx jx sin x
smx cosx . The derivative of an derivative y x inverse function. pj cotx tan xj . then the
derivative of the inverse function x/ t/ . Find y . if Solution.
Fx. Whence dx a sin / function. related to an argument x by means of a parameter t a
function then t or. if Solution. The derivative and y is of an implicit the relationship between x
given in implicit form. that is. x. The derivatives of functions represented parametrically. We
find dt a sin/ and r d acosf. JL tdx dt Example . If s . The Derivatives of Functions Not
Represented Explicitly or dy dy Tx Example . to put FA.Sec. we and equating it get y V a y xy
. ito Forming the derivative of the left side of zero. in other notation. and to solve the resulting
equation for Example . Find . taking y as a function of x to equate this derivative to zero. We
have y x x x t I . dx if x a cos t. Solution. Find the derivative yx if . with respect to x. x gt t .
hence. /. . If gt . Find the derivative x y . of the left side of equation . / y a sin Solution. then to
find the derivative I y y in the simplest cases it is sufficient to x calculate the derivative.f/ .y Q.
t. . . Calculate when f /t cn a sin at al sin/ if sin /. y . . ay axy xy f/ . arc cos arc sin y . whence
y . ///i y Solution. sin / V coslr x . sin gt t . . b sin x acos yb sin t. cos / Tnr . Differentiation of
Functions Ch. . find the derivative the functions y represented parametrically . a In tan cos
asin cosO. . x acosf. r al cosO cos/ . . cos/. ye . t / cos/. .dy In the following problems. Find
the derivative a if c y .
/a. . l/S / / y . x l/ K /quota. // . / sm. x / f lt lt . xy . . . . sin y . . in/ y x e cosf. a cos tan// x I s ty
a . Let y differentiation of Va . When x the following equation jc is true x. iflt i . x x f x. . . i dv
dx when / .Sec. . The Derivatives of Functions Not Represented Explicitly and S fdy quotT I .
Does it follow from this that when x . arctan l . Is it possible to perform termbyterm x In the y it
examples that follow of the is required to find the deriva tive y r x implicit functions y. Find . . .
. . f/ Prove that a function represented parametrically by the equations satisfies the equation .
Find when i f ji r if tlnt.
M Y . / . Equations of the tangent and the normal. . For the normal we have the equation .
we get whence y y xyy. /. . The angle between angle between the curves curves. The d and
at their common angle co point to M Z is the Q between these get M A and point Using a
familiar formula of analytic geometry. Differentiating. tangents curves at the . r/l. /x y care tan
y . . the we . The straight through the point of tangency perpendicularly to the tangent is
called the normal to the curve. Sec. Find the indicated points a derivatives y of specified
functions y at the y for for x b c xl ye e y for x and and and yl. . we obtain lf/. where y is the Q
line passing value of the derivative y at the point X Q y Q . Segments associated with the
tangent and the normal in a rectangular coordinate system. The tangent and the normal
determine the following four ..yQ at a point M . Putting x l and . x Find at the point A .
Differentiation of Functions Ch. From the geometric significance of a derivative it follows that
the equation of the tangent to a curve will be t/ y fx or Fx. . Geometrical and Mechanical
Applications of the Derivative . . if Solution. M M B y Q Fig.
Sec Geometrical and Mechanical Applications of the Deriiative segments Fig. /f Sn N X Since
KM y and tan y y Q . . If a en in polar coordinates /qgt. and the is . the segment of the polar
normal. ments see Fig. curve is givby the equathe then formed by the tangent radius vector r
Fig. is the polar subnormal. Af The tangent and the normal together with the radithe point of
tangency and with the perpendicular to the radius vector drawn through the pole determine
the following four segat the point us vector of M MT MN Fig. defined by the following formula
OM MT angle u. t MT is is is n MN S t OT S n ON the segment of the polar tangent. the polar
subtangent. . it follows that j/o . Segments tion r associated with the tangent and the normal
in a polar sys tern of coordinates. NM S n KN the subtangent. t TM is is is is the socalled S t
TK n segment of the tangent. St Fig. the segment of the normal. the subnormal.
. Find the points at which the tangents to the curve are parallel to the jcaxis. . These
segments are expressed by the following formulas to . t/ or whence the equation of the
normal x y . At Fig. c xl x. . Fig.. x y z f. It follows that the equation of the tangent is / or x
Since the slope of the normal must be perpendicular. We have y . c tan qgt qgt . qgt. y .
Determine the slope of the tangent to the curve xy Q at the point . At what point of the curve
y x is the tangent perto the straight line pendicular . have y f We k y x i T V . At what angle
does the tanianx intersect the gent curve y axis of abscissas at the origin tx . At what point is
the tangent to the parabola x y . N what angles do the sine and y smx interof sect the axis
abscissas at the origin . y K x with abscissa x . Differentiation of Functions Ch. b tan .. . Write
the equation of the tangent and the normal to the parallel to the straight line xy xy . Whence
Solution. x whence the slope of the tangent is Since the point of tangency has coordinates . b
xl/. At what angle does the curve y e intersect the line x straight . Find the equation of the
parabola yxbxc that is tangent to the straight line x y at the point . x ./ parabola at the point
Solution. . . x . l. ltp a tan cp . What angles cp are formed with the xaxis by the tangents x x at
points with abscissas the curve y a x .
Sec.. Write the equations of the tangent and the normal to the x at the point with ordinate y .
x y . Write the equations of the tangent to the curve at the origin and at the point j curve xy Q
at the point . Show that the segment of the tangent to the hyperbola xy a the segment lies
between the coordinate axes is divided in two at the point of tangency. Geometrical and
Mechanical Applications of the Derivative .. . a y b y arc sin at the point of intersection with
the Aaxis.. . . c d e y arc cos x at the point of intersection with the yaxis. x at the points of
intersection with the straight ye line y . x curve y x . Write the equations of the tangent point .
Form the equations of the tangent and the normal to the curves at the indicated points tanx
at the origin. Write the equations of the tangent and the normal to the at the point . Write the
equations of the tangents and the normals to the curve y x at the points of its intersection x ljt
with the axis.. Write the equation of the tangent to the curve x y / . y ln at the point of
intersection with the axis. . Write the equations of the tangent and the normal to the curve y x
xy at the point . Show that in the case of the astroid x yt a/ J the segment of the tangent
between the coordinate axes has a constant value equal to a. to the curve t and the normal
at the . . Find the equations of the tangent and the normal to the curve at the point . . .
an arbitrary point tt . of the subtangent of the Show that in the equilateral length of the
normal at any point is of this point. y x and y . What procedure of construction of the tangent
to the ellipse follows from this . Given a parabola y x. at y al . Find the length of the segment
at any point of it. tor r Find the angle between the tangent and the radius vecthe point of
tangency in the case of the lemniscate a cos qgt. . curve y . . Show that the length of the
segment of the subnormal a at any point is equal to the abscissa in the hyperbola x y a the y
hyperbola x to the radius vector equal of this point. Find the angle at which the parabolas y x
. Will at Show x x we have the same thing . and normal. At the point . the normal. of . x acost
t sin/. At what angle do the parabolas y and inter x Sect . . evaluate the lengths of the
segments of the subtangent. the subtangent. that the hyperbolas Show intersect at a right
angle. tangent.. Find the length of the segment ol the tangent. Show Differentiation of
Functions that the normals to the involute of the circle t Ch. and the subnormal of the cycloid
x atsmt. . . x i ntersect. x Show that the segments of the subtangents of the ellipse y jjrfrl an d
the circle x y a at points with the same abscissas are equal. . Find the angle between the
tangent and the radius vector of the point of tangency in the case of the logarithmic spiral at .
subnormal. y asinf cost are tangents to the circle . that the curves y x x and y are tangent to
each other at the point .
A Fig. on the axis is Find the velocity of the point at / in centimetres and / is in seconds. m A
is moving are sliding along at m/sec. subnormal. Geometrical and Mechanical Applications of
the Derivative . and normal. following laws of motion x With what speed are these points
receding from each other at the time of encounter x is in centimetres and / is in seconds . is
given by the formulas air resistance is . tangent and normal. The law of motion of a material
point thrown up at an angle a to the horizon with initial velocity V Q in the vertical plane OXY
in Fig. and t x is . and also the angle between the tanat an gent and the radius vector in the
hyperbolic spiral r . What is the rate of motion of B when A is at a distance OA m from the
origin . . tangent. and also the angle between the tangent and the radius vector of the point of
tangency in the case of the spiral of Archimedes at a point with polar angle ltp jt. The
endpoints of a segment AB the coordinate axes OX and OY Fig. Moving along the axis are
two points that have the t and l// where tO. Find the lengths of the segments of the polar
subtangent.Sec. Find the lengths of the segments of the polar subtangent. . . . The law of
motion of a point . subnormal. r rQ cp arbitrary point cp .
the rate of change of its ordinate when the point passes yx does the ordinate increase at
twice the rate of the abscissa . A derivative of the second order. AM. . of the function yfx is
the derivative of its derivative. increases with the square of the distance of the from the end A
and is gm when cm. The radius of a sphere is increasing at a uniform rate of cm/sec. . Find
the mass of the entire rod AB and the linear density at any point M. i. b. as or or . At what
point of the parabola becomes cm . The mass of a part of it. fquotx. . while the other
side.sec. lt/quot lt/. a cm. disregarded Differentiation of Functions Ch. is the time and
trajectory of motion speed of motion and . What is the linear density of the rod at A and S cm
vector is M AM Sec. Definition of higher derivatives. At what rate are the area of the surface
of the sphere and the volume of the sphere increasing when the radius through . or Ihe
second derivative. moving point. If // is the law of rectilinear motion of a point. Derivatives of
Higher Orders . increases at a constant rate of cm. g is the acceleration of gravity. A
nonhomogeneous rod AB is cm long. A point is in motion along the spiral of Archimedes a so
that the angular velocity of rotation of its radius constant and equal to per second. Also
determine the its direction. is of constant length. The second derivative may be denoted
lt/quot. At what rate are the diagonal of the rectangle and its area increas cm ing when .
Determine the rate of elongation of the radius vector r when r cm. . Find the and the distance
covered. then is the accel eration of this motion. that is. where / / igt /sin a . r/ A point is in
motion along a hyperbola so that its abscissa What is x increases uniformly at a rate of unit
per second. /cosa. One side of a rectangle.
Leibniz rule.Sec. then to evaluate the nth derivative of a product of these functions we can
use the Leibniz rule or formula up uv ltquotgt ultquot of . the ith t. Fnd w / . acosO aslnlt osln .
If the functions u qgtx and vtyx have derivatives to the nth order inclusive. Solution. Example
Find the second derivative y of the function n x. Generally. or f n x. Derivatives of Higher
Orders a derivative of order n derivative of a function y fx is the derivative . . f . quot quot
amp a cos. / . For the nth derivative we use the notation parametricaKy. / amp cos . We have
If . asm a LUl I and . Higherorder derivatives functions represented If i I qgt. /.JZL. . if
Solution. be of y v t or . can successively calculated then the derivatives y x r f/jc by the
formulas xt xt For a second derivative we have the formula Example .
Find the law of motion of i projection M. T and /quotO. y the function yC l equot for all
constants satisfies the equa y s x x . C e x C and C l satisfies th . tion yquot . if The equation
of motion of a poin along the jcaxis is fx . ysmx. . f . V I y arc sin x acosh u / . on the xaxis if
at time / the point is at Q a. Find t/ VI of the function ysinx. y . . e x sinx. Find /O. Find the
velocity and the ac celeration of motion of M. Show that the function y e x cosx satisfies the
differ ential equation y lv y Q. find the second derivative of given function. yy e the x e x
satisfies x the differen x . .OOU . Fig. . . n t/x v . . c t l y an . Find yquot if y x Find /quot if / . y
. . A. . . . XH/ the f t O. . x . . Find the velocity and the acceleration point . at the in What is the
velocity and the acceleration of tial time and when it passes through the origin What are the
maximum values of the absolute velocity and th A circle x y a point M M M l absolute
acceleration of Ai. . for times / . Find y v of the function lnlx. y j . HigherOrder Derivatives of
Explicit Functions In the examples that follow. y thlt x x e .. velocity CD. Differentiation of
Functions Ch. is in motion around with constant anguls Fig. tial Show that the function y
equation yquot . f function y e x smx . a Show that the function y OY satisfies the differ ential
equation . . . Show Show f that that equation yquot y . // .
/cos/. e . Find the nth derivatives of the functions js . Find n / . . HigherOrder Derivatives of
Functions Represented Parametrically and of Implicit Functions In the following problems
find . Find a a Derivatives of Higher Orders y yTx and b y . f al cos/. du B. where n the nth
derivative of the function natural number.e // . e b. y l . c y f yj yJ. b J arc tan/. . dy e b c if y x
A cos x . l x y iv c arc sin/ . J/ g h ysinjr. . . Find the /zth derivative of the functions j/sinx. cos.f
x y jc . x a sin/. quot cos/. Find . acos/f/ sin/. d /lnlx. find y a n if y x. Using the Leibniz rule. a
K nt. asin/. f .Sec.
Find / at the point . the . x fy.arc tan /. . . ax for /. Having the equation y x ny t find and j ..
Find at the point . b Find . x yl. if x i/ a . f y y l. Find Solution. . yelttnt. sin be iV y ae that y as
a function of x defined by any constants a the equaand satisfies the differential equation In
the following examples find . if x xy if x . . if Show t. Find yquot at . . y quotcos/. Find .
whence y Substituting the value of . y fx required to determine represented implicitly.. . tions
x I lnlf/ y for t . if . . differentiating By the rule for composite function we have i. x x . . jtquot
Knowing /. . a The function y is defined implicitly by the equation . a find the derivatives
xquot. . du . finally get y if it if is In the following examples derivative yquot of the function y .
the if of the inverse function y fx. tan/. j we and / . Differentiation of Functions Ch. lt r/ Find
function x e / . yquot f s sec.
.. MT is the PQ Axdx. x. whence u y If is an arc of the graph tangent at M. y Solution. y and
.. Ax . Find the increment x x. Example . x At/ . . then the increment in the ordinate of the
tangent and the segment AN by. ... First method A// and the differential of the function x Ax
At/ x Ax x x or Ax Ax . Solution. which part is linear relative to the increment Ax dx of the
independent variable x. . Calculate x and dy the function y x x for xl A/ lAx ltfy and jt Ax . dy
dx MN of the function y fx Fig. Differentials of First and Higher Orders differential. Example .
Second method t/ dy x Ax x dx. . df/ j/dx x of dx. The differential firstorder of a function the
principal part of its increment. Differentials of First and Higher Orders Sec. Firstorder is Fig.
The differential of a y fx .Sec. Hence. .. function is equal to the product of its derivative by the
differential of the independent variable dyydx. and Ax .
A secondorder differential the differential of a firstorder differential We similarly define the
differentials of the third and higher orders. then Example if its m It is given that At/ in and A .
is . . then dy yquot du yquot du du y du M. dv. A dy. Higherorder differentials. Find the
increment Ay and the differentia dy of the func f x for x and A . Principal properties of
differentials. Here the primes denote derivatives with respect to tion . Applying the of the
equal that is. If y fx and x is an independent variable. . . where w cpx. dc d.. then But if y /.Ax.
. Differentiation of Functions C/t. If x is the area of the square and y is its side. If the
increment argument x is small in absolute value. v du d uv udv v du. whence . and so forth.
m. The increment mately as follows the side of the square may be calculated approxi kyzdyy
Ax jz . where c const. where x is an independent variable. cdu. m area increases from
Solution. By how much approximately does the side of a square change to . dcu du A
differential to approximate calculations. then the differential dy of the function y fx and the
increment At/ of the function are approximately ..
lane. Differentials of First and Higher Orders find . S b the volume of a cube. . . . . y
problems find the differentials of the given functions for arbitrary values of the argument and
its increment. . Give a geometric interpretation of the increment and differential of the
following functions nx a the area of a circle. ymlt. equivalent to the expression In A. Show
that when Ax . and Ax . y e Find d// . x dy y dy y dx . and x . . . . arc sin .Sec. the increment in
the function corresponding to an increment Ax in x. Without calculating the derivative. In the
following . The area of a square S with side x is Find the increment and the differential of this
x. x if x xy y a. //arctan. . vx . given by S function and ex plain the geometric significance of
the latter. xnx x. Find the differential of the function for . Taking advantage of the invariancy
of the form of a differential. Calculate the differential of the function Ax for xJ and Ax. is.
Using the derivative. r .. find the differential of the function // X y cos x for x y and Ax . For
what value of x is the differential of the function y x not equivalent to the increment in this
function as Ax gt . cot s arc qgt f cosec p. Has the function y x a differential for x . we obtain x
dx Whence Solution. dlx for x . for any x.
.. . Derive the approximate formula and find approximate values .. . arcjr and Ax arc .
Differentiation of Functions Ch.. approximate V . Approximate the functions for for ..
Approximate tan quot.. . Find the approximate value of sin Solution.. In the following
examples find the differentials of the functions defined implicitly.. . c /x. /. . from formula see
we have sin sin cos . . . Y. jI/. Putting yy x. . Find the approximate value of arc sin . d In . .
tial. e arc tan .. Approximate . for j/TO. Replacing the increment of the function by the
differencalculate approximately a cos b tan c . . /..J.. X . if . ./quot ye l for d x for / x . What will
be the approximate increase in the volume of a sphere if its radius cm increases by . Derive
the approximate formula for ampx that are small compared to x mm A Using it. j/. e . Find dy
at the point .
/sinxlnx. Compute dy. Rolles b. MeanValue Theorems . relative error of / in A. may be A/ . //
dquotw. if y cox. due to a small change found approximately by the formula the resistance.
find e cosa sin. / . show in that a small change in the current. find du. . and FbFa axb altxltb t
f x and F x are have derivatives that continuous on the do not vanish then f a . . . find d y. .
Using MeanValue Theorems Ohms law. . If a function f is continuous on the interval and has
a derivative at every interior point of this interval. find dz. M sinjtf . .t sin u. d y yquot dx cos
dx . If has a derivative a function at / x the interval is continuous on f x every interior point of
this interval. and then the argument x has at least one value . determining the results in a
relative Solut ion. where a lt lt b. n Sec. . function fx Show that the g. length / of the radius. .
xx Rolle on the lltxlt and ltxltl satisfies the theorem. z . find dz. find dy. Find intervals the
appropriate values of . // x arccosx. z flnd d TJ.Sec. . theorem. Show that. . e . then where a
interval ltlt ft. Lagranges theorem. Cauchys theorem. a error of approximately in calculating
the area of a circle and the surface of a sphere. u . such that . If the functions and for
simultaneously. . find dy.
where . Show that this equation cannot have any other real root. Does the Rolle theorem
hold for the function takes on fx /x equal values the endpoints of the interval . Whence and
ltE lt fcJ/l. lto . The equation obviously has a root x . . find a point the tangent to which is
parallel to the chord AB. for which the is xx Whence. Given a segment of the parabola y x
lying between two points A . . Hence. / The function / j/ at the Rolle theorem hold for this
function on . . lt lt holds continuous and difTerentiable for we by the l/ E. that is. prove the
formula . the Rolle theorem is applicable on intervals and ltlt. / /E Test the validity of the
Lagrange theorem and find the intermediate point for the function fx x /s on the interval .
Solution. and . J/l where ilt g. Differentiation of Functions the The function / x is continuous
and different able for all values / /. and find the appropriate intermediate all Solution. .. Test
whether the Lagrange theorem holds for the function on the interval value of .h sin x h cos .
To find form the equation we and / .. Using the Lagrange theorem. Ch. .. Kxlt n l . The
function fx values of A. . JT Let Show that the equation / has three real roots. and / x x hve
and inequality /l/y gl. . Does on the interval . of x. Hence. Lagrange formula. appropriate sin x
. . the only suitable value is .
quotgt / /ltgt I. Taylors Formula . f A . and there then Tay lors formula fw f lto r o Va V w .
Sec. for . respect to / sin and Fx cosx on the interval . holds true on the interval. Therefore. n.
f a and a when ltlt. Hence. . .v x in A . . . . . xf r . f / jc x. .c in posi . Taylors Formula If a
function f x is continuous and has continuous derivatives up to the interval n is lth order
inclusive on the a finite derivative f n x at altxlt each interior point or of the interval. we have
Maclaurins formula W / jc. . test the interval and b do the same with To. . Expand the function
/x the term with x . . lnjt in powers of x up to .r . ampltlta.r .Sec. a For the functions /x whether
the Cauchy theorem holds on find E. nl where a jc In particular. or Jt x z x . where ltlt. .
Expand the polynomial /A tive integral powers of the binomial x Solution. n l x e for all n.
Whence H. p JL. where . /quot njt A . ltGlt. and Fx x . x Expand the function fxe powers of x l
to the term containing xf Solution.
Expand fx n l ing x . if the quotient . . to within J . a that If fx and qgt are both infinitesimals or
both infinites as x p and x is. in powers of x up to the term containing x e in powers of x up to
the term contain. xlt formula and evaluate their errors. . The LHospitalBernoulli Rule for
Evaluating Indeterminate Forms oo . is Show that the shape thread approximately expressed
by the parabola . and to the term containing x. Determine the origin . Show that for xlta. is
one of the indeterminate forms or oo i then lt . Let the singlevalued a lt/i. a heavy suspended
thread small x lies a catenary of the line y a cosh. we have the approximate equality Sec.
Evaluating the indeterminate forms functions / x and u oo . Show that sina /i differs from .
Evaluate the error in the . at x a. . lim / p x lim xa f qgt x provided that the limit of the ratio of
derivatives exists. . . Differentiation of Functions Expand / x sin x for Ch. sin a h cos a
approximate formulas by not more than l//i . the derivative be differentiate for lt of one of
them does not vanish. . . of the a b VTxamplx y yiiamplxx. xltl. in Due to its own weight.
Compute lim JL x gto cot form quot. JEfLllm pLr xocotx jcocot lim x We get the indeterminate
form jp however. one should transform the l appropriate difference /. it should be borne of in
mind that the limit of the ratio may see exist. oo M In the case of the indeterminate form oo f
x oo. .Sec. rule V HospitalBernoulli is Rule for Indeterminate Forms . /Tw The indeterminate
forms I. at the point x satisfy x all of one of the two abovementioned types and / x and the
requirements that have been stated for fx and q x. jt and KO. x then we re duce the
expression to the form the form . are evaluated by first faking loga rithms and then finding the
limit of the logarithm of the power f l x x which requires evaluating a form like oo. since Um
CfrO sint Hm H sin X X We thus finally get quot JCgt COt X . first into the product / t x L if / x j
and evaluate the indeterminate form . oo Solution. etc. To evaluate an indeterminate form
like oo. Applying the LHospital rule we have lim jco . a. into thequetient the form U TT /i X the
orm . transform the appropriate product fixf t x. r lim i X xa i . In certain cases it is useful to
combine the LHospital rule with tht finding of limits by elementary techniques. / lim/ gta oo.
Example . Other indeterminate forms. pass to the ratio of second derivatives. whereas the
ratios the derivatives do not tend to any limit Example . the quotient again yields an
indeterminate form. we can then However. .. wne re lim/. we do not need to use the
LHospital rule. The If also applicable / when a qgt f fix .
lim Jfgt x cos x e.zlf x x J xo x sin x z Before applying the LHospital rule. lim . Solution. lir m I
gtilsi n r x on . Hence. F ind the indicated limits of functions in the following x exam ples.
Example . is lim tan sin . in elementary fashion. G. we get lim In cos X J ncos X lim xQ X .
Compute MV sin limfJ. Reducing to a common denominator. . Sec. sin A. Xgt OJi r quot / .
we lim xo sin get v fonn JUlim. sin x x Example . lim quot . lim tj IU xcosx v r ji sinx .
Differentiation of Functions Ch.gt. we replace the denominator of the fraction by an
equivalent infinitesimal Ch. obtain ter lat We o The LHospital lim rule gives L lim A quot S a
m lim Then. x..x L x J form oo oo. Compute lim cos x XM form I Taking logarithms and
applying the LHospital lim p XQ rule. we find xltgt sin cos x . .
that is. . n x lim rr Solution. We r/. Ji . lim xgto cos x cot x . . Solution. lim A. Hm Sill AT cos A
.. tanx quot i . lim . In yx In Aquot. . lim y s . lim In t/ limjtln x s lim p lim j . vo . V a . ImiA l. lim
j A j lim A T MiiAl A A x gtl nx X A h A . lim x sin Solution. L HospitalBernoulli Rule for
Indeterminate Forms . f A nquot lr . lim.Vl A Ian . linrZiiJ . ngt. . liml . whence lim//l. lim xquot
sin XX . . lim .. limjc equot. . . lim arc sin x cot x. . A A lim n A . .lim Xgt Sill A . lim lim Ll y A /
AO J limf Vc cosx/ have . XQ ri hnilnxln lim i x .Sec. . lim cos x cot . . sec tan COS A n x x .
cot S .
Hmftanf / Xl . Differentiation of Functions Ch. Prove that the limits of X a sin cannot be found
by the LHospitalBernoulli rule. . . . . . . limx quot. lim cot x in . limlx X . Show that the area of a
circular segment with minor and central angle a. . . liml . is ABb CDA approximately with an
arbitrarily small relative error when a gt. which has a chord Fig. lim cot x ln XH cos ta . ltaI x /
xo quot. a . linue s/n gt . . limx. li VH tan . Find these limits directly.
These intervals are bounded by ciitic or f x does not exist. b and / xgt / AltOJ for alt .lt a lt/i
Fig /. hence. points x where /jc Example . amp. the function / increases in the interval
.Chapter III THE EXTREMA OF A FUNCTION AND THE GEOMETRIC APPLICATIONS OF
A DERIVATIVE Sec. Whence ooltxltl.ltb. Increase and decrease increasing decreasing l i ffxj
i. and . i/lt. From we have if oo. ed into In the simplest cases. then /A increases decreases on
the interval a. a Fifi. and. I fx is continuous on the interval a. y for xl. the function f x
decreases in the interval if lltAlt oo. any points x and x which belong to this interval. The
Extrema of a Function of One Argument of tunctions.ltA.gt/ A. from the inequality A. xz X Fig.
Tlu Junction y fx is called on some interval if. f oo. On number we get two intervals of monot .
Fm. . fo. and. then j/gt. hence. Test the following function for increase and decrease Solution.
. . b. the domain of definition of f x may be subdividincrease and decrease of the funca finite
number of intervals of tion intervals of monotonicity. We find the derivative t/ x a scale . then
onicity . . we get the inequality / . Fig.
is called the maximum point of the function and is the maximum of the function Fig. Here. we
get from f/ gt. increases Similar and the function in this interval we find that yltQ in the
interval . If xn is an extremal point of the function f then / . we can . use xl/ and ygt the
interval . retains its sign. ygtQ in the interval . of the func Example tion . for is a discontinuity
of the function and / . we can take ly. we Solving the equation x x . Similarly. as a check. and
. hence. v in the interval in Y I /A . the function y decreases in the intervals ooltlt Example
and Test the following function for increase or decrease /i y . The minimum fx. . . one has to
determine the sign of the derivative in each of the intervals. Since y can change sign only
when passing through points at which it vanishes or becomes discontinuous in the given
case. . f oo. If there . find the points x l Q. Extrema and the Geometric Applications of a
Derivative Ch. Determine the intervals of increase and decrease Solution. . /j point or
maximum point of a function is its extremal point bending point. then . . Hence. decreases in
the interval and again increases in the interval . the function being tested inThus. . the
derivative in each of the intervals . Extremum of a function. creases in the interval oo. x. inter.
at which the derivative y vanishes. while the number x lf the inequality any point xjx l of some
fltfx is fulfilled. here. the function under investigation is monotonic in each of these intervals.
exists a twosided neighbourhood of a point X Q such that for every point XX of this Q
neighbourhood we have the inequality fxgtfx Q then the point x is called the J Fig for
minimum point of the function is y fx. To determine what the sign of y is in the interval . . i o
ltQ . for this reason. v s y a Solution Here. or x. of the function y neighbourhood of the point
mum / x called the miniif fx. To determine in which of the indicated intervals the function
increases and in which it decreases. for example. y has no discontinuities. . l . it is sufficient
to determine the sign of y at some point of the val. taking x . and the minimum or maximum
of a function is called the extremum of the function.
Example . point of the function. Then. we /i.lt. let the first of the derivatives not equal to zero
at the function f x be of the order k. but I is the maximum point of the and //maxl. then the
point X Q is not an extremal point. r/ f /lt. It is also possible to test the behaviour of the
function at the point x by means of the second derivative whence is /Here. r If f XQ ltjc X Q lt.
one can If it is difficult calculate arithmetically by taking for h a sufficiently small positive
number. b. and //gt. . Consequently. find the critical point is x l . . The least greatest value of
a continuous function f x on a given interval a. if k is even.O is not n extremal point. of
positive number. function if f XQ f . . r/quotlt for is the maximum I x. The Extrema of a
Function of One Argument The sufficient conditions for the existence / x continuous function
and absence of an are given by the following rules extremum of a . then is the maximum
point of the function / . Q and fquot and /quot of then X Q is the maximum and /quotlt. . p to
determine the sign of the derivative y. and. of a critical point such that /xgt for X Q dltxltxQ
and /jtlt for xQ ltxltxQ d. if / ft x gt But if k is odd. / x. b is attained either at the critical points of
the function or at the endpoints of the interval a. sign if there for is some positive number
such is unchanged . hence. but if f . where for/ we have no derivative // For /i. then x that / x
retains its not an extremal point of the point. then /ygt. Greatest and least values. the
maximum point. we obviously have //lt. and it is the minimum point. if f k lt. Equating the
denominator the expression of y in to zero. From formula we have if xif where h then a
sufficiently small x function h. is the minimum point of the function y. If there exists a
neighbourhood X Q . namely. Find the extrema of the function More generally the point x
point X Q lt i/ j Solution. . the is an extremal point. Hence. we get Whence. gt. then A. we get
i/ e find the second critical point of the function A there .Sec. then x is the minimum point.
Find the derivative x V v Equating the derivative y to zero. . m jn Fig. and if / lt for ltltx is the
minimum point of the and / xgt for x ltxltxe function then / Finally.
intervals of decrease and increase i of the func yl / / . . . gt s . y A . . ml. Find the greatest
and least values of the function on the interval Solution. jf. A ll at point at the point xl at /i J
the minimum point. Extrema and the Geometric Applications of a Derivative Cfi. Determine
the ions . Comparing the values of the function at these points and the values function at the
endpoints of the given interval of the we conclude the Fig. x . . and the greatest value o at the
righthand endpoint of the interval. and Y Fig.. . that the function attains its least value. Since
P/ it follows that the critical points of the function y are . Example .
t/ . x f sin x. . citi sin . V / . y . . We find the derivative y x jc . determine the . y . rr. therefore.
Equating the derivative y to zero. yt/ I . The Extrema of a Function of One Argument . y e y z
. lt . and . Determine the indicated least if and greatest values the interval is intervals of the
functions on the not given. We get the same result by utilizing the sign of the second
derivative at the critical point yquotlt Solution. . y Vx. . g Tquot Test the following functions x .
We to for extrema of given function. . it follows that the minimum point of the function. y . f/ .
Solution. gt .Sec. . wJ/ it / xe. we calculate the second derivative quot . . O cin O v z sin ZA /ft
// oto. / . and max . it follows that x. I . we have the function y and t/lgt. x l is the minimum
point of i . y . v k. y find the derivative the the Equating zero. Since / lt. and ygtQ when gt . t/ .
value of the argument x . X OQQ ooy. M ar/ idii /tdrc fan t. y . Since i/lt when xlt is . arcsinlfx.
and min . / . is the maximum point of the function y. we get critical . je lnl. y xnx. . we get the
critical points x. lt. To determine the nature of the extremum. Similarly. . .
x f a . we set out to prove. ltlnl xltx . /gt/ and so e as gt x when x when x . V x on the interval .
Separate a given positive number a into two summands such that their product is the
greatest possible.. .. Prove the inequalities x sin x . Determine nomial yxpx q imum when t/ .
What is the optimum in the sense of area shape of the playground if / metres of wire netting
are available . greatest and values of the function throughout the domain of definition. .
rihamp. . xlOx. Explain the result in geometrical terms.. It is required to build a rectangular
playground so that should have a wire net on three sides and a long stone wall it on the
fourth. minimum / Hence. .. y y sin cos A. Solution. Consider the function In the usual way we
find this function has a single . b on the interval on the interval . cosgtl A when when . . arc
cos x. . . . . Show that for positive values of we have the inequality . Prove the inequality e
coefficients p and q of the quadratic triso that this trinomial should have a minJt . x lt o JL lt
when gt. the gt x lhat when x . y A. / Extrema and the Geometric Applications least of a
Derivative Ch. Bend a piece of wire of length / into a rectangle so that the area of the latter is
greatest. What right triangle of given perimeter p has the greatest area .
. It is required to make an open rectangular box of greatest capacity out of a square sheet of
cardboard with side a by cutting squares at each of the angles and bending up the ends of
the resulting crosslike figure. . slides along a horizontal straight line AB. The width of the
tower is Fig. About a given cylinder circumscribe a right cone of least volume the planes and
centres of their circular bases coincide. . What will the dimensions of the vessel be if a
minimum of material is used for a tical OB of the door of a ver. In a given sphere inscribe a
cylinder with the greatest volume. given capacity ABCD MN dltl . A sheet of tin of width a has
to be bent into an open cylindrical channel Fig. What size will it be if the least amount of tin is
used . . What should the central angle cp be so that the channel will have maximum capacity
D N I M Fig. The Extrema of a Function of One Argument . . . . Out of a circular sheet cut a
sector such that when made into a funnel it will have the greatest possible capacity. the end
of which. Which of the cones circumscribed about a given sphere has the least volume .
Determine the least height h so that this door can pass a rigid rod tower of length /. . In a
given sphere inscribe a cone with the greatest volume. An open tank with a square base
must have a capacity of v litres. Inscribe in a given sphere a right circular cone with the
greatest lateral surface.Sec. . In a given sphere inscribe a cylinder having the greatest lateral
surface. . the walls are of constant thickness. An open vessel consists of a cylinder with a
hemisphere at the bottom. M. Which cylinder of a given volume has the least overall surface .
. to bending width of the crosssection by the square of sion its height. A homogeneous rod
AB. M . determine the angle at which the is k times messenger has to cross the river so as to
reach B in the shortest possible time. . A linear centimetre of the rod weighs q kilograms. .
Extrema and the Geometric Applications of a Derivative Ch. square of the distance from the
light source. . On a straight line ABa connecting two sources of light A that of intensity p and
B of intensity lt/. . which can rotate about a point A Fig. . A lamp of radius so that r. The width
of the river is h and the distance between A and B along the bank is d. . suspended above
the centre of a round table At what distance should the lamp be above the table is an object
on the edge of the table will get the greatest illumination The intensity of illumination is
directly proportional to the cosine of the angle of incidence of the light rays and is inversely
proportional to the square of the distance from the light source. . . A point x lies in the first
quadrant of a coordinate plane. It is required to cut a beam of rectangular crosssection ont of
a round log of diameter d. The resistance of a beam to compresof its crossis proportional to
the area to the product of the section. On the curve y Xt . find the point receives least light
the intensity of illumination is inversely pro M portional to the . Inscribe in a given ellipse a
rectangle of largest area with sides parallel to the axes of the ellipse. . Determine the length
of the rod x so that the force P should be least. What should the width x and the height y be
of this crosssection so that the beam will offer maximum resistance a to compression and b
to I i/J bending Note. Draw a straight line through this point so that the triangle which it forms
with the positive semiaxes is of least area. and find P mln Fig. . find a point at which the
tangent forms with the Agtaxis the greatest in absolute value angle. Inscribe a rectangle of
maximum area in a segment of the parabola y px cut off by the straight line x a. Knowing that
the velocity along the bank that on the water. is carrying a load Q kilograms at a distance of a
cm from A and is held in equilibrium by a vertical force P applied to the free end B of the rod.
A messenger leaving A on one side of a river has to get to B on the other side.
N identical electric cells can be formed into a battery in different ways by combining n cells
in series and then combining the allel..Sec. The centres on a single straight locity v strikes
strikes C of mass fi. m. What mass should B have so that C will have the greatest possible
velocity . is of least value the principle of least squares. if for for the tangent drawn at any
point of the interval a.. b lt above or of the interval a. the electromotive force of one cell. For
what value of n will the battery produce the greatest lt diameter y of a circular opening in the
which the discharge of water per second Q will be greatest. Points of Inflection . where h is
the depth of the lowest point of the opening h and the empirical coefficient c are current . the
arithmetic mean Sec. of three elastic spheres A. M in pargroups the number of groups is The
current supplied by this battery is given by the formula resulting . Fig. is of Prove that the
most probable value of x the measurements. amp. irwgtoj. . .b concave the arc of the curve is
below or up in the interval a p . having acquired a certain velocity. Determine the body of a
dam for constant. The concavity of the graph of a function. If x lf . . . We say that the graph of
a differentiable function y fx is concave down in the interval a.. The Direction of Concavity.
then its most probable value will be that for which the sum of the squares of the errors .
Points of inflection. NnS where resistance. Points of Inflection .. A sufficient condition for the
concavity downwards upwards of a graph y fx is that the following inequality befulfilled in the
appropriate interval a. line. . is The Direction of Concavity. A point f jc at which the direction
of concavity of the graph of some function changes is called a point of inflection . . .ltxltb altxlt
rwlto Fig. B C are situated Sphere A of mass moving with vewhich. x n are the results of
measurements of equal precision of a quantity x. r is its internal and R is its external
resistance. if Q cy Vhtj.
F V VeJ r are points of inflection Fig. point of inflection x of the graph of a function derivative
f or /quot x . X and Equating the second second kind derivative y to zero. I i We have bx Fig.
a. Example . lt lt . we take one point in each lively. Points at which Q or f x does not exist are
called critical points of the second kind. The signs of t/ will be. It will be noted that due to the
symmetry of the Gaussian curve about the axis. Fig. x . And it is not a point of inflection if the
signs of f x are the same in the aboveindicated intervals. II j. and III x . fthis is obvious if. . .
the curve is concave down when ltxlt The points . respec . Extrema and the Geometric
Applications For the abscissa there is of a Derivative Ch. V F lt x ltf oo.d of the y f x no
second x where is some positive number. Determine the intervals of concavity and convexity
and also the points of inflection of the yfx Gaussian curve I Solution. for example. provided
these signs are Jc ltxlt lt lt . xj. the curve is concave up when oolt x lt V and F . it would be
sufficient to investigate the sign of the concavity of x this curve on the semiaxis alone. fx The
critical point of the second kind x is the abscissa of the point of inflec an the intervals x tion if
Iquot x retains constant signs in . opposite. b. we o find the critical points of tHe r and T
These points divide the number scale OOltAlt OO into three intervals . of the intervals and
substitute the corresponding values of x into y Therefore .
x. is the point of inflection Fig. The tangent at this point is parallel to the axis of ordinates.
Vertical asymptotes. . since the first derivative y is infinite at x . . . then the straight line x a
Inclined asymptotes. then this straight line is called an asymptote of the curve. yi/x . is an
asymptote vertical asymptote. . Definition. Solution. . we find that yquot does not exist for x .
ff X i . In x. obvious that yquot does not vanish anywhere. a then the straight line y k l xb l will
be an asymptote . Asymptotes . it follows that . Find the points of inflection of the graph of the
function y/. We have It is right of . . arc tanx y lxe. there are limits llm X gtgt If K and lim /X Ml
i. If a point . . of inflection . Sec. a right horizontal asymptote.Sec. y x . Since yquot for and
f/quotlt for . If there is a number a such that y Jim /v . . . X yx yx // sin./ is in continuous
motion along a curve fx in such a way that at least one of its coordinates approaches infinity
and at the same time the distance of the point from some straight line tends to zero. y x l y r
o . asymptote or. . . x. Equating to zero the denominator of the fraction on the gt gt xlt Find
the of the intervals of concavity and the points graphs of the following functions x x . when
right inclined If there are limits llm . Example Asymptotes . y cosx.
Equating the denominator lotos to zero. line a left Solution. and Urn then the straight or.
Testing a curve for asympconsideration the symmetry of the curve. Example Fig. S hence.
when oo. For x oo we obtain kl lim x lim ogt b l l. k zx y b is an asymptote a left inclined
asymptote horizontal asymptote. Extrema and the Geometric Applications of a Derivative Ch.
Find the asymptotes of the curve asymptote if y x we take into . / Fig. . gt seek the inclined
asymptotes. The graph of the function y fx we assume the function is singlevalued cannot
have more than one right inclined or horizontal and more than one left inclined or horizontal
asymptote. the totes is left simplified . we have fc a Hm lim ACgt . v x z x lim // lim X y . Find
the asymptotes of the curve when fe . we get two vertical asyinp x We and xl. Example . the
straight line y x is the right asymptote. fc Is Thus. Similarly.
If ltpU olt and lim then the curve has an inclined asymptote y kx b. j/ spiral r . is a vertical the
straight line x asymptote lower. Let us now curve only for the inclined right asymptote since
xgt. x quot /. When jf oo and pV c. . . If the curve is represented by a polar equation r /cp.
Find the asymptote of the hyperbolic . When p/ oo and tyt c. Since Asymptotes lim t/ oo. .
yYl. . nx oo. Hence. the curve has a vertical asymptote x c. u . . r sin ltp / qgt sin p. there is
no inclined asymptote. y Find the asymptotes of the following curves .Sec. while the other
remains finite. y. If a curve is represented by the parametric equations x cpi / then we first
test to find out whether there are any values of the parameter / for which one of the functions
cp t or gt / becomes infinite. . We have test the k b lim y XOD X x lim lim gtlt . . lt/ r . y x e . y .
i/ . then we can find its asymptotes by the preceding rule after transforming the equation of
the curve to the parametric form by the formulas x r cos cp /p cos qgt. . rr. Solution. the curve
has a horizontal asymptote y c.
yquot in each of the indicated intervals. a The function . bl lim X gt oo . and so the critical
points are only those at which y and yquot vanish. in each of the intervals constant sign l. We
have . . J/T. etc. oo. hence. lim . The function exists everywhere except at the points x . .
derivatives y and f are nonexistent only at that is. respectively. that is. yQ r/quot when x when
x V and x and . y retains a l. . d We find the critical points of the first and second kinds. From
and it follows that The xl. From the symmetry of the curve it follows that there is no lefthand
asymptote either. is odd. the straight lines x VM O are vertical asymptotes of the Xgt graph.
Thus. and therefore the graph is symmetric about the point are This simplifies construction of
the graph b The discontinuities x and jc . intervals . there is no right asymptote. periodicity. .
Graphing Functions by Characteristic Points In constructing the graph of a function. points at
which the first or. it is sufficient to determine the sign of y or yquot at some one point of each
of these intervals. Example . / in each . first find its domain of definition and then determine
the behaviour of the function on the boundary of this domain. . . asymptotes. . . . V and V t .
These elements help to determine the general nature of the graph of the function and to
obtain a mathematically correct outline of it. points of inflection. only at points where the
function y itself does not exist. . To determine the signs of y or. respectively. . such as
symmetry. Construct the graph of the function Solution. Sec. constancy of sign. of V. lim
gttoo y thus. the . etc. and find . and . the second derivative of the given function vanishes or
does not exist. bending points. It is also useful to note any peculiarities of the function if there
are any. c We seek inclined asymptotes. Extrema and the Geometric Applications of a
Derivative Ch. and lim J/ oo and t/oo. Then find any points of discontinuity. monotonicity. .
. It will be noted that due to the oddness of the function r/. the lefthand half of the graph is
constructed by the principle of odd symmetry. it is enough to calculate only for JcO. of the
investigation. we construct the graph of the function Fig / Fig.Sec Graphing Functions by
Characteristic Points It is convenient to tabulate the results of such an investigation Table I.
Table I e Usin the results . calculating also the ordinates of the characteristic points of the
graph of the function.
a The domain of definition of the function is ltxltfoo.. lim xgtlt . . . We Inx y y and yquot exist
at all points of the domain that o of is. definition of the function and y Q when lnl. we find the
point of intersection of the curve with the axis of abscissas. j/ The right asymptote is the axis
of abscissas find the critical points. b There are no discontinuities in the domain of definition.
we construct the graph of lunction Fig. In addition useful to find the points of intersection of
the curve with the coordinate axes. since x cannot tend to oo k . is that is. but as of the
domain of definition we have approach the boundary point limw lim JCgt we JL X oo X
Hence. Graph the function In x x Solution. Putting / . / when Inxy. and have d . when xel. lim
Xlt X . We form a table. c We seek the right asymptote there is no left asymptote. including io
the characteristic points the it characteristic points Table . Example . when x lt.Extrema and
the Geometric Applications of a Derivative Ch. the straight line jc ordinate axis is a vertical
asymptote. the curve does not intersect the axis of ordinates the e Utilizing the results of
investigation.
u non .h CM CM CO O quot gt gt C a c.
. Extrema and the Geometric Applications of a Derivative Ch. . discontinuities. y graph. . .
points of inflection of the direction of concavity. . . x x. . . y . . . quot . . . . y x l/xl l z . . interits
vals of increase and decrease. . its Graph the following functions and determine for each
function domain of definition. u x y y y / / lt/ . . extremal points. . . and also the asymptotes.
x /. y cos arc tanln arc sin In . y ./ . I agt. . Sec. Differential of an Arc. . tan A. then ds V F F .
f/ then ds d . . . Differential of an arc. . if J/d is a is plane curve expressed by dy . y lsin. . and
. . Construct the graphs of following functions represented parainelrically. ltgtarcsin K gt . The
differential of an arc s of represented by an equation in Cartesian coordinates x and y the
formula dshere. . c then ds qgt.Sec. . sin/ agt. Curvature . y fl / g. y sin .sin . /gt Differential of
an Arc. y arc cosh / I . xt /// t. i/ l ya .. f/ x arc tan .. .when when . y lnsin. . arc tan. . exercise is
in A good to graph the functions indicated the Fxam ples . te . . . y . / sin. jc /e. y arc sin /F.
xacob/. lnAarc tan. . . In cos x. . x i// ea cosh / a sinh/ ./ je. /. . . V F . l /. Curvature . . y arc
cot. / COSCOS. the equation of the curve of the form a // /. i/ . .Urt. garcun n sin x j . yx ..
then ds b /. Af .
Curvature of a curve.vaxis. K hore Au o As AS . . we get cos a dy r dx . The radius
curvature. i. of curvature R is the reciprocal of the absolute value of the e. ds Denoting by p
the angle between the radius vector of the point curve and the tangent to the curve at this
point. of Denoting by a the angle formed by the tangent in the direction increasing arc of the
curve s with the positive direction. where a is the radius of the circle and the straight line /C
are lines of constant curvature.. Extrema and the Geometric Applications of a Derivative Ch.
The circle f K . ds . rfs a is point M the angle between the positive directions of the tangent jt
the arid the . .V M Fig. The curvature K of a curve at one of its is the limit of the ratio of the
angle between the positive direcpoints and N of the curve angle of contintions of the tangents
at the points M gence to the length of the arc M MNs iim when . that is. sina In polar
coordinates. we have a P of the dr / sin p .
dt In polar coordinates. r dr dcp . equation /q. jfr the . then . The radius of the circle of
curvature is equal to the radius of curvature. If L . Find the equation of the evolute of the
parabola // x . quot where dx dy x y y . and the centre of the circle of curvature the centre of
curvature lies on the in the direction of concavity of normal to the curve drawn at the point M
gt M M M the curve. Circle of curvature. if the curve itself is represented by equations
parametric form z Example . Fx. as P and Q M. when the curve r z is given by the rrquot we
have r where . The circle of curvature or osculating circle of a is the curve at the point limiting
position of a circle drawn through v and two other points of the curve. f /. .. P and Q. dtp . The
coordinates X and Y computed from the formulas of the centre of curvature of the curve are
XxThe evolute curve. y fx. j of a curve for is the locus of centres of curvature of the in the
formulas determining the coordinates as the current coordinates of of the centre of a ture
lute. y . Flx F Fxy Fx F y Fyy x Fv /j / j if the curve is represented by equations in parametric
form. we regard X and Y then these formulas yield parametric equations parameter x or y or
/. then Fxx. curvapoint of the evoof the evolute vith in . Curvature We have the following
formulas for computing the curvature in rectangular coordinates accurate to within the sign if
the curve is given by an equation explicitly. then if the curve is given by an equation implicitly.
and dquotr .Sec Differential of an Arc.
M. we form the expression of the curvature K and find its extremal points. curvature of the
point The vertex of catenary acosh thus. x a and. lengths of thread. . l ellipse. Solution. / a
circle. J acosh we X . a We y have j. of the evolute equal to the corresponding increment the
radius of curvature CC. In place of the curvature K we can take the radius points if of
curvature R I and seek its extremal I Example y a cosh the computations are simpler in this
case.. by the tangent to each of the following curves . hence. The vertex of a curve of the
curve at which the curvature is a point has a maximum or a minimum. in AfC. we find the
equation of the evolute in explicit form. and also the cosine and sine of the angle formed. A
cosh a a curvature or of the gt . Equating M Iquot the derivative j critical ax to zero. get sinh a
. . Fig. obtained by unwinding a taut thread wound onto P.C. . Vertices of a curve. that is why
the involute P is also called the evolvent of the curve P. a. . y px parabola. Y o lTquot is The
evolute. whence we and find the sole point Q Q. . Y Eliminating the parameter x. Find the
vertex of the catenary a gt a . To determine the vertices of a curve. Computing the second
derivative putting into it the is value x the we of minimum the a point of the radius of
catenary. a cosh . Find the differential of the arc. X . Since // J sinh a / and J /quot cosh a rf/ it
follows that tf . Extrema and the Geometric Applications of a Derivative Ch. A . To each
evolute there corresponds an infinitude of invowhich are related to different initial lutes. with
the positive direction. sinh dx a X . get ry . the f/ Therefore. maximum is. Solution. involute of
a curve is a curve is for which the given curve to an the The normal length of the arc MC CC
of the l involute P is a tangent the evolute P.
. . Find the differential of the arc. Find the vertex of the curve ye Find the radii of curvature at
any point of the given x cubic parabola. r . . . . at the point . Compute the curvature of the
given points . y . x atsnt y alcost cycloid. x acost. acos y asnt astroid. is equal to . . .
hyperbolic spiral.parabola. . acosM sin . . the curvature x . Find the least value of the radius
of curvature of of the parabola y px. . b. ltp . ratp spiral of Archimedes. rza. x / f / t/ Differential
of an Arc. . and n. the point . r ae k v logarithmic spiral. A / /osO involute of a . Curvature / a
astroid. y lines . r . .S . r asec. circle. a eosqgt at the vertices cp . . S ellipse. . y acosh
catenary. . . r acos cardioid.v logarithmic spiral. f/ . /. a.al fcoscp cardioid. coordinates of the
centre of Compute curves at the indicated points given the curvature of the . a . and also the
cosine or sine of the angle formed by the radius vector and the tangent to each of the
following curves . y asmt astroid. Prove that the radius of curvature the catenary y acosh is
equal to a segment of the normal. . r.Sec. x curves at the indicated x x ISA xy y at at the
coordinate origin. y asnt /. r a cosq lemniscate. At what point of the parabola t/ at the vertices
A and .
. asin // / /cos is the involute of the circle . . ay Write the equations of the circles curves at the
indicated points of curvature of the given . Find the evolutes of the curves . . . . y px parabola.
a. xyl at the point . x . J gl ellipse. asm/. Show x logarithmic spiral with the same pole. x at the
point a. Prove that the evolute of the cycloid xat is sin/. y Gjc at the point . . . .c acob/. y e at
the point . . y al cost a displaced cycloid. Prove that the evolute of the logarithmic spiral r is
also a . Extrema and the Geometric Applications of a Derivative Ch. that the curve the
involute of a circle / a cos / / sin /.
If F A.T a C av fC yS . V. r a . Basic rules of integration. aO.a dx C A r . fc x d . a I arctan a C
a v arc cot a C a . . gt. . dx VII.r . dx. f f x dx. where A UH f is a constant quantity.MA. f f xdxF
fC and /cf then In particular. f dA III . VI. r J i ax a integrals. Direct Integration .Chapter IV
INDEFINITE INTEGRALS Sec. dx r aQ. IV s J X f. then where C is AfxdxA If an arbitrary
constant. ft v l dv fj / i f f O v. Table of standard II.
J XII XIII. . dVd. IX. VIII. x sin cotxC. COS X XI. cosxdxsinx C. smx cosx f tan x HC In cosec
x XIV.. and the formulas of integra find the following integrals . f Indefinite Integrals sinxdA. .
bx c dx f Applying the basic rules tion. . . XV. n . . T yZpxdx. . ljc/xdx. j/ . . x f .
sinhxdxcoshxC. Ch.dx v . . r x x i a x b dx. V J V x . cosx C. . . XVL XVIL Example f ax dx . J
nx quot dx. J a bx U clx. cot x fC.
Example x e xl dx e d jc ie C by virtue of Rule and tabular integral VII. Using the basic rules
and formulas tegrals integration. . Rule considerably sign expands the table of standard
integrals by virtue of this rule the table of integrals holds true irrespective of whether the
variable of integration is an independent variable or a differentiate function.e xdx . In
examples . .. .Sec Direct Integration . which were used in Exam dxdax b a . f dx We implied u
. where we put u . Vdx. Jcothjtdx. jc . differential. . of differentials. of transformation Some
common transformations are called integration under the differential sign. . and use was
made of Rule and tabular integral V. a b tan dx. a b Jtanh d. This type ples and a . We took
advantage of Rule and tabular integral Exa m p. Integration under the of the Example . . b
xdx of dx and so on. and form before making use of we reduced the given integral integral to
the following a tabular / ltP ltp dx is I u du t where a p x. find the following in .
.
.
i . f fl t JlanaArdx lt H.rfy. . ff. J tan gt . JVL l ltioo sinx sin ccs A J i/c V e .. Indefinite Integrals
C/i.. xedx. U. xl/ xl ilax I . . lt integrals x / dx. Jx cos . ftan/JC jcdx x djt i .. . f. . . sin SA f v quot
v . J H. Jld. . J Ad . quot Ti . cothxdx. dAtgt lt . . a cosxdx. f sinajtcosax J si . J tanhxdx. . v i A
f cos JCT . J I A f cos x dx.. J cosh x r /f v J sinh x cosh x Find the indefinite .. quot M j . M.
fd. J Ktanx ii. . . j j . . J to fi. i I . sinh x J . ff . f.. jsin Jdx.
and dx tdt. Putting where t is a new variable and cp is a continuously differentiable function.
Integration by Substitution J l/T Sec. Change of variable in an indefinite integral. . whence A /
. Integration by Substitution . Find a way that the right Solution. Sometimes substitutions of
the form are used. Example . we will have The attempt is made to choose the function qgt in
such side of becomes more convenient for integration.Sec. . It is natural to put t Vx . Hencu.
Suppose we succeeded in transforming t the integrand fxdx to the form fxdxgudu where u
qgtjc.
w jc s . cfw cx. . . Trigonometric substitutions. Sec. du xdx. then Actually.. . It is sometimes
more convenient to make use of hyperbolic substitutions. the usual thin is to put whence If an
integral contains the radical a . u . Find i dx. which are similar to trigonometric substitutions
see Example . x . see Example . u x. may be solved as follows Example . For more details
about trigonometric Sec. that is. du x dx x dx . Ch. If gudu is Indefinite Integrals known. d c .
we put xawct. dx du. we have already made use oi this method in Sec. an integral contains
the radical fa z V . whence It should be noted that trigonometric substitutions do not always
turn out to be advantageous. we put atan/. . and hyperbolic substitutions. . xdx . Examples .
Example . whence /x If a a tan an integral contains the radical VV a . Example . a sin If /.
integrals find the following c f e COS A d J Applying suitable substitutions..... . J cos si quot lt
J sin cos / dt / . f J iZEldx.cos J cos / sin M /cos/ J cos f / cos/ / J sin In tan sec J C In tan / V
jtan / tanlt . . n. J J lA. quot f f / . f yxl f Vian J tan dt sin / J / dt cos / i . J . f JxKtl .Sec. r . .
tegrals find the following in . Applying the indicated substitutions. K l . . gt. . / / Therefore. lt . .
Put xlant. find the following integrals csinA quot Sxquotdjc. Applying trigonometric
substitutions. Integration by Substitution J Solution. dx f sec t cos rr / . .d.
. Solution. If H ltp and u iare differen udv uv vdu. Find C putting x a cosh/. Sec. Find by
applying the hyperbolic substitution x asnt. Evaluate the integral r J dx / by means of the
substitution xsin /. then by parts. Integration by Parts A formula for integration on tiable
functions. . . . fidx x . .U . We have Aa x /a a suih /a cosh and dxa cosh /d/. f ff J x y x . J
Indefinite Integrals Ch. Whence ya z x dx a cosh ta cosh f ctf Since x and a we finally get
where C C In a is a new arbitrary constant.
JjVd. . x sin x cos x d a . . x Jjcsiiucr/. V Tx dx. Integration by Parts Find x . djc. Find e cos x
dx.v. jt xcos nxdx.vjecos x V e x cos x dx. . e cos x dx e x sm . one has to apply the fcrmula
of integration by parts several times. . . H x In xdx. .v f C. . Jjcarctanjcdjc. . Applying the
formula of integration by integrals parts. find the following . whence cos v dx sin x f. .cos .
Example . Jarcsin Arfjc. Example . In certain cases. dv xdx we have da dx v Whence dx x
Sometimes. Urfjc. J . . . Jarclanjcdx. Jlnxd. K A Jjcarcsmxdjc. Putting u In.Sec. . e x d cos x e
sin x e x cos x e cos x dx. JjccosAJx. We V have e cos x dx e xd sin x e sin x e x sin AT dx e
x sin x Hence. n x Jxrfx. . . fd. to reduce a given integral to tabular form. nxdx. integration by
parts yields an equation from which the desired integral is determined. . J . x A dA.
Sec. it is best to take the perfect square out of the quadratic trinomial. . j // x dx. xndx.
Applying various methods. . sinlnxdx. . . Standard Integrals Containing a Quadratic Trinomial
. fdx. z . dx. v e dx. reducing the quadratic trinomial to the form the tabular integrals III or IV
see Table. cosjtdx. J Vx jttan d. . J cos In xdx. f jc arctanjcdA. fare sin x J . j x dx. Integrals of
the form f mx n . eax mb dx . arc tan A . . xe dx. then. The following substitution may also be
used If m. . . Jesinxdx. . . fllLd. a . . arc sin jc . . the J The principal calculation procedure the
form is to reduce z quadratic trinomial to ax z bx c axk l. . . . JisFr. .t. To perform the
transformations in . Indefinite Integrals C/i. . we get . where k and / are constants. f x x nxdx.
. find the following integrals . dx. f n/ir .
Example . Qi o arc o tan TJS If mampQ. we can take the derivative f J a.v d J fquot A xi IV .
Example dx H g .Sec Standard Integrals Containing a Quadratic Trinomial .v .n i . and thus
we arrive at the integral discussed above. f J gt . The methods is of calculation are similar to
those analyzed above. if a The gt integral finally reduced to tabu lt . Integrals of the xl form I
d. dx f jt Example r . . dx Example . J quot f yVx J x JC . if a lar integral V. and VI. . then from
the numerator out of the quadratic trinomial quotSc.
By taking the perfect square to out of the quadratic trinomial. . . Integrals of the form T axbx
cdv. Find dx Solution. Integrals of the form f J mx . / these integrals are reduced to integrals
of the form . c/ gt . n i Vax bx c By means of the in verse substitution mxn Example . We put
whence .c S P xdx J .J Indefinite Integrals Ch. Example sin Find the following integrals . J Vl
. the given integral is reduced following two basic integrals sec examples and one of the J V
a A dx fax arc sin a C.
a. J f. . Integration of a rational function. . . and a. The method of undetermined coefficients.
v dy O f t J . Integration of Rational Functions t. i lt lt c r I dx . of the nume Qjr .. v dK. . after
taking out the whole part. V d r I . J f /quotl/quoto J X X / .H. V xdx V gtJ x . Integration of
Rational Functions j i oKn oJ.. . .. /. T . and then the coefficients of like powers of the variable
x are equated llrst method. t g x cJJt J . n ..quot dx.Sec.A/ / are real where a.. distinct roots of
the polynomial Q x. K are natural numbers root multiplicities. i J x ilnA. and the degree lower
than that of the denominator Q A. . . .. These coefficients may likewise be determined by
putting in equation or in an equivalent equation x equal to suitably chosen numbers second
method. n m K Sec. f sin x dA J TPTTTTlTInjfdi JYKI .. . reduces to integration of the proper
rational fraction where P rator If x and Q is P x A are integral polynomials. then
decomposition of into partial fractions is justified To calculate the undetermined coefficients A
lt A t . . both sides of the identity are reduced to an integral form.
Indefinite Integrals C/t. Applying the second method.. fl Cl. we get . and solving this example
When get . We get . We have quotquot xx tind I x x JC A it I is Bx x Cx. l. a F/rsf method of
determining the t coefficients.. i. i .. We the form xA.B xA ents of identical powers of x we get
B l B rewrite identity in Equating the coeffici Whence . Further. app lying the first method.B x
A. Then. we equate the coefficients of x in identity . Find Solution. we get C. Example .. we
will have Hence.. putting . . B and . B l /. Further. Putting Second method of determining the
will have coefficients. B . Find xdx Solution. Hence. / . putting jcl. i . i. we put in identity . b lt.
quot We have Whence t x.e. .e. Putting x .e. TJ v A T t x J A f AT Example . x in identity we .
advisable to combine the two methods of determining coefficients. i.
A k by the methods given above For k. Example Find Solution.Y A where Q. Bk are
undetermined coeflicients which are determined and A lt B lt . A and Q x. Example . AJ z. . .
respectively. dzz r j r Hit iini j jrc tan . . putting x z.Sec. A is the greatest common divisor of
the polynomial Q x and its derivative Q A.. The Ostrogradsky method. here. Since A x i then.
then partial fractions of the form will enter into the expansion .arc tan z . undetermined
coefficients. the fraction is integrated direct. Here. r wo got r . X arc. dx p dx . JC In JC rf C. .
. / If f dx J X f J r. ln jc J l . . use is made of the reduction method. . Px If Q A has multiple
roots. then p Av quot r . for kgt. A the polynomial Q x has complex roots a ib of multiplicity k..
ly. whose degrees by unity than those of Q. In t egration of Rational Functions Consequently.
Find A and Y x are polynomials with less C dx U . it is first advi z sable to represent the
quadratic trinomial x pxq in the form x f q and make the substitution . The undetermined
coeflicients of the polynomials X x and Y x are computed by differentiating the identity ..
. P B F . L x MX x A N A . Dx Ex F ax . r dx i i p dx r yS J o J x V J j Q X ll and J ln x Find the
following integrals r l mian C . consequently. dx Ax z Bx C we . ltj Equating the coefficients of
identical degrees of x on the . Differentiating this identity. have D . we we find . F fl. Equating
the coefficients of the respective degrees of x. D C . E whence . . we decompose the fraction
into partial fractions x lhat is. . D . right and left or Therefore. and. get Dx z Ex xAx Bx AxBx C
Dx Ex Fx we will i. get L. Solution. Putting . O F O C x x l dx J x To compute the r integral on
the right of .of . C . Indefinite Integrals Ch.
. q z are whole numbers.Sec. Applying Ostrogradskys method. find the integrals l I rf l l Sec.
Intagrating Certain Irrational Functions . . Integrals of the f. I . .. I C R V Av t Q J . dx C Q J J f
j Iijt dx ltJ dx c J d rggdx. l . quot quot I . C C/ rt . .rm i / lts a rational function and p lt q v p a
..Tjpfc JSf. . Jd. . . Integrating Certain Irrational Functions . find the following integrals
Applying different procedures.
.dx. Integrals of form are found by the substitution where n is the least Example .. J /Ti fdjc. .J
Indefinite Integrals Ch. f dx J VquotjcT... . J o A jv A V x U. Find common I multiple of the
numbers a dx lt/. Integrals of the form quot r ai t. . v . where P n Put x is a polynomial of
degree n dx where Q n ix cients and A. . /vl .x leads to an integral of the form ri quot J Find
the integrals . JZdjc.J/lquotquotj A The substitution f zdz z . The coefficients of the polynomial
Q n ix and the number K are found is a by differentiating identity . . is a polynomial of degree
n with undetermined coeffi number.. I . x c . Solution.vf c . J . J . . q . . / r dx .. /vl . . .
fv Vli v rfjc. FA Xl . n and p are rational numbers. fdx. . Multiplying by V we obtain and
equating the coefficients of identical degrees oi Hence. Integrating Certain Irrational
Functions . D. use is made of the substitution . They a n V ax substitution are reduced to
integrals of the form by the A a Find the integrals . Example Whence A. . Here. is m if p a
whole number. . quot if is a z n whole number.Sec. . X . Integrals of the form dx i. . The
integral can be expressed in terms of a finite combination of elementary functions only in the
following three cases if p is a whole number. Chebyshevs conditions. Here. Integrals of the
binomial differentials x t where m. we make the substitution a bx n s t where s is the
denominator of the fraction p. j.
Here.n r . Integrals of the form m. x. where e f e . . . sin x cos dx f sin x sin sin x .p . then we
put x cos quot xd cos n s lZk cos x k cosquot xd cos A. integrability. J . We do the same
Example . J i r y Sec. m.v d sin x sin . we have here Case The substitution yields z I . . dxz z
Y dz Therefore. an odd positive number. Indefinite Integrals Ch. JC . Integrating
Trigonometric Functions . . is m/jl . where m If and n are integers..feJef y J/T .v T . Find
Solution. f . if n is an odd positive number.. r integrals z T dx G f . Example . T . Hence. Find
the . .
sin x dx f / gt J x cos dx . I is trans y cos . the following integrals reduce to this case Example
. . then formed by means of the formulas sin . x f cos sin . sin JCGOS cos y cos . x sin dx .
Zc . c GA c dx cos / sin AC If m dx i and n v are integral negative numbers of identical parity.
Example f cos p sin . J COS X f f C J sec xd tan J ltanx tan xp.Sec. f sin A . then C J cos v In
particular.x .tan C. sin C cos fi ax x o cos n sin . sin . the integrand If m and n are even
positive numbers. Integrating Trigonometric Functions v . o jc d tan x r quotquot tan .
. . . . . . J sinxi/cosxdx. secx ldx tanx xC. . are evaluated by the formula or. J ltan . Jsinxdx.
tanxdx tv V tan x secx . x cos x dx. . f J V sin x cos . J sin x . COS X . JsecMxdx. jjcotd. x jr .
sin COS T . dx tanxdx x P. . . . J cos x dx. C J cos x dx. Indefinite Integrals Ch. tan x J In the
general case. cot x cosec x . dx . . J . . integrals mgtn of the form are evaluated by means of
reduction formulas that are usually derived by integration by parts. respectively. . Example
tan . J cos dx quotj c C cos x . sin A . smxcosxdx. Jcosxdx. . . where m is an in tegral positive
number. Example sin . Jtan jcdA. Integrals of the form f tan w xdx or V cot w xdx. J sin sin . .
f cos x sin x Find the integrals . JcosSxdx. J cos x A. f. .
f cos f sin cos djc. sin mx cos nx mx sin m n x sin m m n n x . sin Integrating Trigonometric
Functions . of substitution . J cosaA f sin co/ fcp . x Solution. . are reduced new variable
Example . J sin jc cosSxdx. In these cases the following formulas are used. tan whence
integrals of form t. By means x t. V Integrals of the form V sin mx cos nxdx. Integrals of the
form f / sin . I sin s m nfcos mf/i Example sin xdx cos cos lOjt dx Find the integrals . cosmcos
. sin agt/ dt. A . . . Find f J to integrals of rational functions by the sin x cos we wil E /. where
R is a rational function. . .Sec. cos x dx. . sin djc. dx. sin nx cos c x cosm n x x. Putting tan pr
have f J . sin sin d. J cos x cos x dx sin x sin sin . sin rnx sin nx dx and cos mx cos nx dx.
dt dx dt jz r we will have dt p i arctan K that / /quot C arc tan F is /T tan if C. sin sin sin I o.i. .
n P COS S pjrfx . f iooyi or J sin sin cos cos cos . dx. tan/. . J. dx J sin cos . use the
substitution tan to reduce the rational form. mi. dx . T ..gt. We note the integral it denominator
example. for Find the integrals J cos . cos s/ sin. sin jg t COS . . . P dx X r. integral to a we
can Here. . Indefinite Integrals Ch. If we have the identity R Ihen sin. / cos. . cos Sinx ltsi n . jj
Is. J sin C cos . Find Solution.. JiHiJT slncos P J T . T . j t . Putting Jnnsin lt gt tan f. to apply
the numerator and In individual cases. useful artificial procedures see. and arc Example ...j .
of the fraction are first is evaluated faster divided by cos .
. sinh cosh . integrals Jsinhjtd. . Integration of Hyperbolic Functions Integration of hyperbolic
functions is completely analogous to the integration of trigonometric functions. . C J sinh A
cosh . We f have xcU cosh xd sinh x J l sinh JcdsinhA . . V Example Solution. Find cosh . .
.Sec. . . cosh . Integration of Hyperbolic Functions Sec. The following basic formulas should
be remembered cosh X sinh l. A. . . . Using of the Trigonometric and Hyperbolio Substitutions
for Finding Integrals Form R is a rational function. . . sinh cosh Y cosh x Find cosh x . cosh
djt. sinh x cosh x sinh x. Sec. We have f cosh x djcrcosh x dx jsinh x Example Solution. smh
x Find the . coshxdx.
Example Find dx /. respectively. m z dz z z. We have Putting l tan. We have Putting . and f
tan zdz dz zsecquotJ Example . m cosh/. xY Solution. Find Solution. taken means of
substitutions msin/ mtan/ msec/ . The latter integrals are. R R z. /. t of integrals Transfor
Transforming the quadratic trinomial ax bxc into a sum or difference integral becomes
reducible to one of the following types by CH. we dx then have dx sec seczdz f cos x . or z or
or z m tanh msinh/. Indefinite Integrals the of squares. Vz m dz.
Using Reduction Formulas Derive the reduction formulas for the following integrals . /
slnquotxdje. . J flnjdjc. . Sec. .. . jc . . . . . . Sec. find /. SJCC/A. / J x snxcosxdx. .. .Sec.
smxsinhxdx. I e sin x jcdjt. .. quot . Integration of Various Transcendental Functions Find the
integrals . /. . Using Reduction Formulas we finally have Find the integrals . arc cos .. find /
and /. x . J J In x jc /FT djc. J I cos e x dx.. e Ae smxsmSxdx. . S/jtxd. cos x dx. . and /.. J Wd.
.Jj.
. . f COS . . f . l JxA dx CL . dx . Indefinite Integrals Ch. cosecSxdA. . J/ Jl ai ax . jd. n x n edx
find Sec. . . . J jt . r P y dA.. f J VTX . xdx C .. j yi T r t dx dv J dx dx . . IHBT sin A.. find I and /
. P dx dx. Miscellaneous Examples on Integration dx . /w Jh. xdx JOjt x f KIJ j x J . . . dx dx P
J x r dx i t/ f A y l/v r . .. J .
.jz . fare sin . e rfx. . Jxedjc. gt d. gt sin s cos quotquot. x nYTxdx. f lt aX sin . gt dv i J x x sin
rfjc. J sin x cos x . J coslnxdA. dx.v quot C. C . . .sinhquot . JV . . J e . . f fsinhlA J f X dc l J
sinx cosx quot dx f. . Jx cfx. . sinh cosh A dx. Miscellaneous Examples on Integration . M.J. J
I . a CJiLdx. i y Jxarctanxdx.. wJ A . .. Varc sinyd.f. / / gt src tdn x . . . J r Y . V dx.v. . Vdjt. . i j
sin J j jc sin x . dx . J A J/Tirfx. sin x dx.Sec.
Chapter V DEFINITE INTEGRALS Sec. Integral sum. . . Geometrically. .. b. n . is from xa
limit of the sum S nt provided that the to infinity. Let a function f x be defined on an interval
and axc ltXt lt lt x n b is an arbitrary partition of this interval n subintervals Fig. . The definite
integral. that is. is is called the integral sum of the function f x on a. Fig.. . and the largest of
them. max Aj gt o . . .. Fig. . Ax/. where / . S. . The Definite Integral as the Limit of a Sum into
. the algebraic area of a steplike figure see Fig. called the definite integral of the function f x
within the limits to amp. The number of subdivisions n tends to zero. A sum of the form .
and t Solution. Find the area bounded by an and the ordinates . in which the areas of the
parts located above the axis are plus. Geometrically. b. we get the area of the steplike figure
Fig. x t the Solution. . the limit of exists and is independent of the mode of partition of the
interval of integration a. i. b. b into subintervals and is independent of the choice of points / in
these subintervals. Example . The definitions of integral sum and definite integral are b.. The
Definite Integral as the Limit of a Sum it is integrable on a.e. Whence c/ n n . jcaxis. What is
the lim S n equal to n CO J Ax. minus Fig. Here. eralized to the case of an interval a. the
definite integral is the algebraic sum of the areas of the figures that make up the curvilinear
trapezoid aABb. arc of the parabola . . . Fig. Form the integral sum S n for the function gt
naturally gen on the interval ing x it by dividing the interval into n equal parts and
chooscoincide with the left endpoints of the subintervals x il . points / that Hence . Using the
formula for the sum of the squares of integers. those below the jcaxis.Sec. where a Example
. we have The areas of the rectangles are obtained by multiplying each yk by the base A Fig.
Summing. b If the function / x is continuous on a. . lim n gtgt oo S n L. gt nnn . parts the base
a into n equal of y Choosing the value the funcwill tion at the beginning of each subinterval.
and x a a gt Partition .
Find the area of a the hyperbola curvilinear bounded by by two ordinates x a and x b . The
NewtonLeibniz formula. S . Find X ltaltamp. passing to gnnlnl H gt the limit. Evaluating
Definite Integrals by Means of Indefinite Integrals . trapezoid . then the function If a function f
t is a is the aritiderivative of the function F x f x that is. j i x dx. we obtain Sn n S lim n lim co
w Dl al n gt Evaluate the following definite integrals. regarding them as the limits of
appropriate integral sums b i . . T . Sec. and the xaxis. Definite Integrals Ch. sintdt. b. xdx.
and g are constant. . o altlt. . F / th n . A definite integral with variable upper limit. continuous
on an interval a. a dx. J t glttt. f x for If . Jdx. we find S and.
fjcTd/. find the limits of the sums . f . Find the integral Find Find the derivatives X of the
following functions ntdt . Find the points of the X extremum of the function y jlild/ X in the
region gt. lim n . . / . l .Sec. Using definite integrals. Evaluating Definite Integrals by Indefinite
Integrals x is Rl The antiderivative F computed by finding the indefinite integral Example .
find the integrals . Jcos/lt. Applying the NewtonLeibniz formula. X J Jdt. . . .
f rfjt J cosh In . J sinhxdjc. . Evaluate the integrals A . Definite Integrals Ch. l iL . JF Jl .. Jl
cos a da. f dx JFF . . . sincpdcp. Jjt d. sec a da.L j coshAdA. . . Ins . . I Jl . JK d. . P dx .
Integrals with infinite limits. . e are finite. J I I/WKW /WrO then and the integral pxdx a
converges. then the infe gral converges as well. Improper Improper Integrals Integrals t . is is
a finite limit or not on the called convergent or divergent. right of . If a function f x is not
bounded any neighbourhood of a point c of an interval a. mltl the inte gral diverges.Sec. for /x
when oo. b oo b fxdx lim a oo J fxdx and f fxdx a lim agt oo fxdx.. then for mlt the integral If
converges. b such that Fx fx when x c generalized antiderivative. . mgt the integral
converges. lt the function / x continuous when oo. e. otherwise it is divergent. Sec. A . i. the
improper inteIf the limits on the right side of exist and a or c thj b.e. b and is continuous and
cltxltgtb then by definition we put for in altltc b J a CB lim C b fxdx a fxdx e lim C fxdx. then
we assume fxdx J lim b gt oo fxdx J and depending on whether there the respective integral
Similarly. and lim XC x c x m A oo.. then Fa. fx C I when x c. If /ltFx when altlt f and f a Fdx
converges. Integrals of unbounded functions. If and for lim f x xm Ajtltxgt t AQ. i. When c
definition is correspondingly simplified. then the in tegral also converges comparison test. . If
there is a continuous function F x on a. is for mgtl the integral diverges. gral is called
convergent.
Example . QO Example . the integral converges. Example . since ex lte not an improper
integral. we have Since the integral converges. f J x amp lim e Kgt J i f quotfquot eoJ e M lim
egtoVe lim IlU eoV e ll / and the integral diverges. Test the following integral for convergence
w r j y J Solution. Definite Integrals C/i. our integral likewise converges. Test the
convergence of the probability integral r Solution. Test for convergence the elliptic integral dx
. Example . We put C e x dx C ltr dv is C e x dx. while when x and l dx lim e b e e l t hence.
Example . The first of the two integrals on the right the second one converges. lim rT lim
ampgtlt ftoo J arc tan arc tan . When x ampgt.
for I we have / V /T Since the integral i xj ff nry converges. the given integral v dx converges
as well. The point of discontinuity of the integrand the Lagrange formula we get xl. i .
Applying where lt. or establish their divergence Evaluate the improper integrals xlnx a P AY f
. Improper Integrals is Solution. Jcotjcdx . .Sec.lt. Hence. J. I a L a . oo .
tan d . f j / / . q converges when mafunction pgt and qgt./ /ljc . . Ch.. kx eGO dx gt.. . . . Sec. .
Test the convergence of the following integrals . where aa and ampcpP.. Change of Variable
in a Definite Integral If a function f x is continuous over and qgt It a function continuous
together with its derivative cp t over altlt. Definite Integrals . . Prove that the Euler integral of
the fiist kind beta function BP. arc f . . Prove that the Euler integral of the second kind gam
converges for pgtQ. . C J . and /ltp is defined and continuous on the intervtl altxltb .
We put t dt. we shall have C x VV dx jl a sin J/a a sin JL Z a cos t dt IL t a f sin / cos / d/ f sin
/ d/ C Jtfl . fxdx. .Sec. IL consequently. . P a arc sin y x . ly. Can the substitution A COS/ be
made in the integral Transform the following indicated substitutions definite integrals by
means of the . we can take a arcsinO . i . x arc tan/. Therefore. For the integral b . Find x
Vaxdx asin a gt . Solution. Change of Variable in a Definite Integral then Example . Then arc
sin l and.
dx . Prove that if fx is an even function. . then . indicate an integral linear substitution Definite
Integrals Ch. as a result of respectively. C J xV y x e x . x Ve dx. J tyax o xdx. Evaluate the
integrals a . Applying the indicated integrals substitutions. . f J . Evaluate substitutions the
following integrals by means of appropriate . J fid. n o n . a . which the limits of integration
would be the and . x f A i . i J fgf x cos o . evaluate following . .
b. xcosxdx. evaluate the . then u x and v x are continuously differentiate on the u x v x dx u x
vx for t. o . . Integration by Parts But if fx is an odd function. oe . e . Applying the formula
following integrals L integration by parts. .aJC cosAdA flgt. then . Jesinxdx. Show that arc
cos o x J o x .Sec. o . JVdx. u x dx. J e. Integration by Parts functions If the interval a. . .
Jlnxd. Show that T L dx . Show that T C fsnxdx T f Sec.
function Express the following integral terms of B beta m if sin w x cos n m and n are
nonnegative integers. and M is the largest value of the function / on the interval bgt .
MeanValue Theorem . Sec. in where p and are positive . b If fxFx b for altltamp. Evaluation of
integrals. lt integers. Show that for the gammafunction see the following reduction formula
holds true Definite Integrals Ch. evaluate the integral see Example Bp. fx and ltpx are
continuous a a a a for altlt and. x m is the smallest a. then mJqgtxdltJ/dxltM where j a W . .
Applying repeated . if n is a natural Devaluate I and . integration by parts. besides. . number.
Find /. that for the integral n if n is a natural number. then fxdxF If X dx. Show T n n. Using
the formula obtained. Example From this derive that . p. sinquot / sm xdx n f cos the
reduction formula n n n holds true.
Determine the signs of the integrals without evaluating them n b . MeanValue Theorem if In
particular. ltpsl. lt/lt b . . we Jl have lt i/l K .fxdxM a a. that is. then b m ba lt. b respectively.
The mean value of a function. a where c and are certain .. Example Evaluate the integral L M
Solution.Sec. gt The inequalities lent equalities and b may be replaced. The number is called
the mean value of the function / x on the interval . Since Xsinxl. numbers lying between a and
b. by their equiva a and b a.
f . /x . x a . b . jfdx. JI . prove that JI . Definite Integrals Ch.. and V . fxsm x. Find the exact
value of this integral. . Integrating by parts. . r djf J gqpjj. Evaluate the integrals n . fx . JT . .
integrals i Determine without evaluating which of the following is greater i a Vx dx or J dx b x
sin x dx or J x sin x dx c e xdx or Find the mean values vals of the functions on the indicated
inter . Prove that J o dx / r lies between . ltxltl.
The soughtfor area is expressed by the integral . The Areas of Plane Figures Sec. lines x
Compute the . . The Areas of Plane Figures . by two vertical lines at the a Fig. .v area
bounded by the parabola and the xaxis Fig. Area in rectangular coordinates. and . the area of
the rectangular coordinates by the equation y curvilinear trapezoid bounded by this
curve.Sec. Example straight . If a continuous curve is defined in fx fxQ. b X a and x and
points x is given by the formula by a segment b of the jcaxis Fig. Solution. y X the b X yfi Fig.
Fig. .
we will then have Example . . By virtue simultaneously. Here. Solving the Jiniits of
integration x. X Fig. Evaluate the area bounded by the curve axes are changed / y and /axis
Fig. we obtain If then the curve is defined by equations in parametric form x qgt/. we find the
of formula . Definite Integrals Ch. Solution. by two . Example the . set of x and l / JC
equations and . . Solution. . In the curves / /iXM wn M more general case. Evaluate the area
contained between the curves y ltFig. the roles of the coordinate soughtfor area is expressed
by the integral and so the where the limits of integration y l and i/ l are found as the ordinates
of the points of intersection of the curve with the t/axis. y yt the area of the curvilinear
trapezoid bounded by this curve. if the area S is bounded by two continuous and y f x and by
two vertical lines x a and x b. a Fig. where n altltamp Fig.
and then x we get the limits of integration /. and by two radius vectors OA and OB. and of
Plane Figures of the axis by the vertical lines x integral by a segment is expressed where a f.
Find the area of the ellipse Fig. by using p/j /. Due to the symmetry. a. . y and ji iS b sin a sin
/ dtab sin / dt and. If a curve is defined in polar coordinates by the equation rf ltp. Find the
area contained inside Bernoullis lemniscate Fig. c i /.. and its parametric equations I c x y l a
cos bsint. Fig. If in the equation jc aco we / first put x Therefore. then the area of the sector
AOB Fig. . The Areas a and amp. p. S nab. . a and cp. Fig. The area in polar coordinates.
Solution. hence. .Sec. . Example . are determined from the equations and ltp / on the interval
/ lt /. bounded by an arc of the curve. quadrant and then multiply it is sufficient to compute the
area of a the result by four. to which integral correspond the values cp. is expressed by the
Example .
Compute the area contained between the curve y ianx. the straight line y x. yx.taxis and the
straight line x x x . Agnesi y Compute the area contained z between the witch of and the
parabola f. determine first one Whence S a . Find the area bounded by the curve y x. the
straight . Compute the area of the figure bounded by the curve and the yaxis. Definite
Integrals By virtue of the symmetry of the quadrant of the soughtfor area curve we Ch.
Compute the area bounded by the parabola y x x and the xaxis. Find the area bounded by
the curve y and the xaxis. the e. . y . . and x a agt and the xaxis. Compute the area bounded
by the curves yegt straight line and the xl. . x py. Compute the area bounded by the curve y
nx. . Find the area bounded by the parabolas ypx and and the straight line f/ . sinusoidal
curve . x . the vertical lines . between the Compute the area contained parabolas y and y .
Solution. x . Compute the area bounded by a single halfwave of the and the Jtaxis. line yl
and the vertical line x . /Y and . the straight line y . Find the area contained between the witch
of Agnesi xa y u x. Compute the area line . Compute the area contained between the
parabolas yx. a and the xaxis. . z ysmx the xaxis and the straight line x . x of a segment cut
off by the straight x from the parabola y x . . Find the area contained between the hyperbola
xy m. Evaluate the area bounded by the parabola y x x. .
the area contained between the circle x y acos /. cos/. Find the area bounded by one branch
of the trochoid tefr/and a tangent to it ltPltlt at its lower points.Sec. Compute the area
contained within the curve a . the xaxis. . Find the area between the equilateral hyperbola x y
the xaxis and the diameter passing through the point . . Find the area between the catenary y
a cosh the /axis and the straight line f/ . Compute the parabola xly . Find the area between
the strophoid its o X ax X y x x a and the Jfl its asymptote agt. Find the area contained within
the astroid jc the area of the two parts into which divided by the parabola if x. Find the area of
the figure l The Areas of Plane Figures bounded by the hyperbola f and the straight linex a.
Find the entire area bounded by the astroid . e x . y b sin /. . Find the cycloid the area
bounded by the xaxis and one arc f of x at al sin/. Find the area between the curve y i.c . . . . .
Find the area bounded by the curve ay . . . . Compute circle jt r/ is and . bounded by the J
cissoid r/ y and asymptote x a agt.. . and the ordinate . xl the Find area xgtl.
Definite Integrals Ch. r acp Fig. Find the area bounded by the cardioid cos/. Find the length
of the astroid / a Fig. Differentiating the equation of the astroid. . Find the area of the ellipse r
y. Solution. Find the entire area bound a sincp. . lying outside the circle r . Find the area of
the loop of the folium of Descartes rTquot a f coscp. . Find the area bounded by Pascals
limagon Fig. r r e cos r eltl. . ed by the curve r . v cp . The Arc Length of a Curve . sin. . Find
the area contained between the first and second turns of Archimedes spiral. r cos cp. . Find
the area bounded by the curve a. . . we get . Find the area bounded by the curve x acoscp
and y x y. . The arc length in rectangular coordinates. Find the area bounded by the parabola
r a sec and the two halflines and . Find the area bounded by the curve r asincp. Find the
entire area of the cardioid r . Find the area of one of the leaves of the curve r acoscp. yfx
contained between two points with abscissas x The arc length s of a and xb a is curve
Example I. . Sec.
sin/. If a curve is ltp and y represented by equations in parametric form. of the polar angle at
the extreme points of . The Arc Length of a Curve For this reason. Example . C Therefore. If
a curve is defined the arc length s is slnd/ and ji r correspond to the extreme points by the
equation P /cp in polar coordinates. . x at j/al Solution. Find the length of one arc of the
cycloid Fig. The arc s quot where t l and t are values of the parameter that correspond to the
extremities of the arc. /a The limits of integration of the arc of the cycloid. then the arc length
s of the curve is Whence s a. tyt. Fig Fig.Sec. then where a and p are the values the arc. we
have for th arc length of a quarter of the astroid length of a curve represented parametrically.
cos/ We have al cosf and asin/.
Definite Integrals
Ch.
Example
.
Find the
is
length of
the
entire
cp
curve
r
asinto ji.
Fig. .
The
entire curve
described by a point as
ranges from
Fig.
Solution.
We
have
r
a sin
cos
o
o
,
therefore the entire arc length of
She curve
is
JI
JI
s
J J/a
sin
o
sin
cos
f
a dltp
the
sin
j
f
rfqgt
.
y
x
.
from the coordinate origin to the point x . Find the length of the catenary y
Compute the
arc
length
of
semicubical
.
parabola
acosh
from the
vertex
A
to
,a to the point
.
x
Compute
Bb,h. the arc length of
of
xl.
the
the parabola y
/quotx
from
. Find the arc length the points , and l,e. . Find the arc length
to
curve y
e
lying between
of
the curve y
lnx
from x
/
from
lying
K.
. Find the
to
arc
length of the curve y
arc length
.
arc sin e In secy,
jcl.
.
Compute the
t/
of the
curve x
between
e.
and
j/
o
. Find the arc length of the curve x
y
ny
from
to
Sec,
Volumes
of Solids
. Find the length of the right branch of the tractrix
avVt/
y
from y a to
t/ampltlta.
xx
to
. Find the length of the closed part of the curve
a
a
.
. Find the length of the curve
t/
ln coth J
from
xb ltalt.
. Find the arc length of the involute of the circle
f
to
/
. Find the length of the evolute of the ellipse
. Find the length of the curve
x a cos/ y a sin/
. Find the length of the
cos /
f
I
sin /.
first
/
turn of Archimedes spiral
. Find the entire length of . Find the arc length of
r asec
the cardioid r that part of
line
al coscp.
the
parabola
y
which
is
cut
off
by a vertical
passing
through
the pole. . Find the length of the hyperbolic spiral rqgt from the point ,/ to the point C/,,. ae m
v, . Find the arc length of the logarithmic spiral r r a. lying inside the circle
j
. Find
the arc
length
of
the curve
ltp
g
r
from
rl
to
r.
Sec. . Volumes of Solids
.
. The volume of a solid of revolution. The volumes of solids formed by the revolution of a
curvilinear trapezoid bounded by the curve yampf x fhe AT ax is and two a and x vertical
lines x b about the x and /axes are
J
t
Definite Integrals
Cft.
expressed, respectively, by the formulas
b
b
Vx ji
Ja
ydx
VYn
J
a
xr/dx.
Example . Compute the volumes of solids formed by the revolution of a bounded by a single
lobe of the sinusoidal curve sinx and by the segment OltltJT of the xaxis about a the xaxis
and b the j/axis.
figure
Solution.
a
Vji
b
Vyn
xsinxdxjt
xcosx sinxJc ji t
.
The volume of a solid formed by revolution about the t/axis of a figure bounded by the curve
xgy, the /axis and by two parallel lines y c and
t/
d,
may
be determined from the formula
obtained x and y.
from formula
,
given
above, by
interchanging
the
coordinates
If the curve is defined in a different form parametrically, in polar coordinates, etc., then in the
foregoing formulas we must change the variable of integration in appropriate fashion. In the
more general case, the volumes of solids formed by the revolution about the x and /axes of a
figure bounded by the curves / / x and y f z x a and x b are, respectively, where fxf z x and
the straight lines x equal to
t
a
and
b
Example
circle x
.
y
amp
Find the volume of a torus formed by a a about the xaxis Fig. .
the rotation of the
The solid is formed by the revolution, about the /axis, of a curvilinear a, x trapezoid bounded
by the curve y b, fx and the straight lines x and . For a volume element we take the volume of
that part of the solid formed by revolving about the axis a rectangle with sides y and dx at a
distance x from the /axis, Then the volume element dVynxydx, whence
b
V K JI
xydx.
Sec.
Volumes
of Solids
Solution.
We
have
I/,
Ia
x and y
tgt Vax
Therefore,
the latter integral
is
taken by the substitution
xasi
K
a
xa
Fig
The volume of a solid obtained by the rotation, about the polar axis, of a sector formed by an
arc of the curve r Fp and by two radius vectors ipra, ltp P may be computed from the formula
Vp
C
JT
r sin cpd
qgt.
a
This same formula is conveniently used when seeking the volume obtained by the rotation,
about the polar axis, of some closed curve defined in polar
coordinates.
r
asinp
Example
.
Determine the volume formed
by the rotation
of
the curve
about the polar axis.
Solution.
.n
.L
rsinqgtdp
yJia
C
sin
p sin qgt dcp
Jia
f
sin
ltp
cos
cp
dcp
J
jia
.
lOo
. Computing the volumes of solids from known crosssections. If S Sx the crosssectional area
cut off by a plane perpendicular to some straight line which we take to be the xaxis at a point
with abscissa , then the volume of the solid is
is
Definite Integrals
Ch.
where , and x are the abscissas of the extreme crosssections of the solid. Example .
Determine the volume of a wedge cut off a circular cylinder by a plane passing through the
diameter of the base and inclined to the base at an angle a. The radius of the base is R Fig. .
Solution. For the axis we take th diameter of the base along which the cutting plane
intersects the base, and for the /axis we take the diameter of the base perpendicular to it.
The equation of the circumference of the base
is
R
j/
.
The area
of the section
ABC
at
a distance
x from the origin
r/
is
Sx area A ABC ABBC yy tana tana.
,
Therefore, the sought
for
volume
of the
wedge
is
R
R
y
fy
tanadtana
R
ltamp y tana R
.
. Find the volume of a solid formed by rotation, about the xaxis, of an area bounded by the
xaxis and the parabola . Find the volume of an ellipsoid formed by the rotation
of the ellipse r
l
about the xaxis.
of a solid
. Find the
the xaxis, of
xraxis,
formed by the rotation, about an area bounded by the catenary y acosh the
volume
and the straight lines . Find the volume of a solid formed by the rotation, about a the xaxis, of
the curve j/sin x in the interval between x
xa.
,
and x n. . Find the volume of a solid formed by the rotation, about the xaxis, of an area
bounded by the semicubical parabola if x s the xaxis, and the straight line x . . Find the
volume of a solid formed by the rotation of the same area as in Problem about the /axis. .
Find, the volumes of the solids formed by the rotation of an area bounded by the lines y e, x ,
y about a the xaxis and b the yaxis. . Find the volume of a solid formed by the rotation, about
the t/axis, of that part of the parabola j/ ax which is cut off by the straight line x a.
,
Sec.
. Find the volume of a solid formed by the rotation, about the straight line x a, of that part of
the parabola yax which is cut oft by this line. . Find the volume of a solid formed by the
rotation, about the straight line y p, of a figure bounded by the parabola
Volumes
of Solids
t/
p
and the straight
line
.
. Find the volume of a solid formed by the rotation, about the xaxis of the area contained
between the parabolas y x L
and
y Y.
. Find the volume of a solid formed by the rotation, about the xaxis, of a loop of the curve atf
axx j
. Find the
of the cyssoid y
volume
A
of a solid
its
generated
by the rotation
x
about
asymptote x a.
. Find the volume of a paraboloid of revolution whose base has radius R and whose altitude
is //. . A right parabolic segment whose base is a and altitude h is in rotation about the base.
Determine the volume of the resulting solid of revolution Cavalieris quotlemonquot. . Show
that the volume of a part cut by the plane jc a off a solid formed by the rotation of the
equilateral hyperbola x tfc about the axis is equal to the volume of a sphere of radius a. . Find
the volume of a solid formed by the rotation of a bounded by one arc of the cycloid xa / sin t,
figure y a cos/ and the xaxis, about a the xaxis, b the yaxis, and c the axis of symmetry ot the
figure. . Find the volume of a solid formed by the rotation of acos /, y bsmt about the //axis.
the astroid . Find the volume of a solid obtained by rotating the cardioid r al hcostp about the
polar axis. . Find the volume of a solid formed by rotation of the acos ltp about the polar axis.
curve r . Find the volume of an obelisk whose parallel bases are rectangles with sides A, B
arid a, ft, and the altitude is h. . Find the volume of a right elliptic cone whose base is an
ellipse with semiaxes a arid amp, and altitude h. a which are . On the chords of the astroid / //
to the axis, are constructed squares whose sides are parallel equal to the lengths of the
chords and whose planes are perpendicular to the Aplane. Find tfte volume of the solid
formed by these squares.
Definite Integrals
Ch.
of
. A circle undergoing deformation the points of its circumference lies on
is
moving so that one the yaxis, the centre
describes an ellipse , and the plane of the circle is perpendicular to the jq/plane. Find the
volume of the solid generated by the circle. . The plane of a moving triangle remains
perpendicular to the stationary diameter of a circle of radius a. The base of the triangle is a
chord of the circle, while its vertex slides along a straight line parallel to the stationary
diameter at a distance h from the plane of the circle. Find the volume of the solid called a
conoid formed by the motion of this triangle from one end of the diameter to the other. . Find
the volume of the solid bounded by the cylinders z z a. a and y . Find the volume of the
segment cut off from the ellipquot
z
tic
by the plane x a. paraboloid . Find the volume of the solid bounded by the hyperbo
u
loid of
one sheet
f
rj
and
the planes
XU
Z
and z
li.
. Find the
volume
of the ellipsoid
quot
the
is
Sec.
. The Area of a Surface of Revolution
The area
of
a surface
arc of the curve y the formula
formed by the rotation, about a and x fx between the points x
b,
xaxis, of
an expressed by
b
b
Vlydx
ds
is
the differential ol the arc of the curve.
Zfta
Fig.
If
the equation
surface
x
is
of the curve cbtained from formula
is
represented differently, the area of the by an appropriate change of variables.
Sec.
The Area
of
a Surface of Revolution
rotation, about
the
Example . Find the area of a surface formed by xaxis, of a loop of the curve i/ t x Fig. .
Solution. For the upper part of the curve, when
Xlt,
X
we have
Fromfor
/
x
yx. Whence
the differential of the arc
ds
mula
the area of the surface
V
rdx.
x
n
J
of a surface
V
Example
of
.
Find the area
x
the
cycloid
a
t
snt
y al
formed by the rotation of one arc cost about its axis of symmetry
Fig. .
Solution. The desired surface is formed by rotation of the arc OA about the straight line AB,
the equation of which is x na. Taking y as the independent variable and noting that the axis
of rotation AB is displaced relative to the axis a distance na, we
will
have
Y
Passing to the variable
/,
we obtain
da
na
d
n
na
nsin
/sin
y sinf sin
j
dt
Fig.
y
of
in Fig. .
of a parabolic mirror AOB are indicated required to find the area of its surface. . Find the area
of the surface of a spindle obtained by rotation of a lobe of the sinusoidal curve about the
.
The dimensions
It
is
ysmx
axis.
. Find the area of the surface formed
a part
of
the
tangential curve
t/
tan
by the rotation from Jt to
A,
about the
jcaxis.
. Find the area of the surface formed by rotation, about the xaxis, of an arc of the curve y e
from x to x
Definite Integrals
Ch.
. Find the area of the surface called a catenoid formed
about the xaxis from by the rotation of a catenary acosh x to x a. . Find the area of the
surface of rotation of the astroid
x/i
//
a
about the yaxis.
area
of
. Find the
JC
the surface of rotation of the curve
/
y
In /about the axis
from
y
to
y e.
circle
. Find the surface of a torus formed by rotation of the x b a about the axis ltbgta. y . Find the
area of the surface formed by rotation of the
ellipse
about
the axis, the yaxis
agt.
. Find the area of the surface formed by rotation of one arc of the cycloid x cos t about a the
at sin/ and r/ al xaxis, b the yaxis, c the tangent to the cycloid at its highest
point. . Find the area of the surface formed by rotation, about the jaxis, of the cardioid
x
t/
a cost asin/
cos
/,
/
sin/.
of the lemniscate r
. Determine the area of the surface formed by the rotation a cosltp about the polar axis. .
Determine the area of the surface formed by the rotation of the cardioid r a coscp about the
polar axis.
Sec. . Moments. Centres of Gravity. Guldins Theorems
point
. Static moment. The static moment relative to the /axis of a material A having mass m and at
a distance d from tha /axis is the quantitv
static
Mimd.
The
,
moment
,
with masses m,, m ..., m n lying d lt d ..., d n is the sum
relative to the /axis of a system of n material points in the plane of the axis and at distances
n
Mi
/
mA
where the distances of points lying on one side of the /axis have the plus sign, those on the
other side have the minus sign. In a similar manner we
define the static moment of a system of points relative to a plane. If the masses continuously
fill the line or figure of the xplane, then the static moments about the x and /axes are
expressed respectivex and not as the sums . For the cases of ly as integrals and geometric
figures, Itie density is considered equal to unity.
M
My
Sec.
II
Moments. Centres of Gravity. Guldins Theorems
is
In particular the arc length,
for
the curve
s
we have
L
yys whare
t
the
parameter
s
L
y
s ds
MY.
x
s
ds
ds
Vdx dy
is
the differential of the arc
Fig.
Fig.
for a
vertical lines x
plane figure bounded by th curve y a and y b we obtain
yx,
ihz Jtaxis and
two
t
b
b
xydx.
a
a
Example
.
Find the
static
moments about
the x and
//
/axes of a triangle
bounded by the
straight lines
fab
l,
x
,
Fig.
Solution. Here, y
b II
.
Applying formula
ab
,
we
obtain
and
of inertia. The moment of inertia, about an /axis, of a imfe point of mass m at a distance d
from the /axis, is the number l t tnd . an /axis, of a system of n material points The moment of
inertia, about with masses m lt m t ..., n is the sum
rial
. Moment
m
dy and Whence . we get an appropriate integral in place of a sum. Solution. f b.i.. where MX
and My are the static ric figures. d n are the distances of the points from the /axis. we obtain j
L dm b h n . Divide the triangle into infinitely narrow horizontal strips of width dy t which play
the role of elementary masses dm. Utilizing the similarity of triangles.n. Qyltfx may b of the
centre of gravity x.. The figure arc or area of mass M coordinates of the centre of gravity of a
plane are computed from the formulas moments of the mass. where d lf d .ydx f X b O where
y a . For the base of the triangle we take the xaxis. Sf y dx a is the area of the figure. Find the
centre of gravity of an arc of the semicircle . f x axb. Example . Centre of gravity. y of the
curvilinear trapezoid be computed from the formulas b xydx a . In the case of a continuous
mass.. y . the mass M is t we have B tgt B b y y dx yds A S a b dx The coordinates xb. In the
case of geometnumerically equal to the corresponding arc or area. the yaxis Fig . y of an arc
of the plane curve y connecting the points Aa f a and B . for its altitude. Definite Integrals Ch.
Find the moment of inertia of a triangle with base b and altitude h about its base. Example
..in pig . For the coordinates of the centre of gravity x. a There are similar formulas for the
coordinates of the centre of gravitv of y volume.
Guldins theorems. .. The area of a surface obtained by the plane curve about some axis
lying in the. ydsHence. xJ . Theorem . yV.. Find the static moments a segment of the straight
line x o about the coordinate axes of y o lying between the axes. to circle rotation of an arc of
as the curve and not the product of the length of the curve by the described by the centre of
gravity of the arc of Theorem . Centres of Gravity Guldins Theorems Solution. same plane it
intersecting is equal the circumference of the curve.v y a J a . We have and Whence f. . The
volume of a solid obtained by rotation of a plane figure about some axis lying in the plane of
the figure and not intersecting it is equal to the product of the area of this figure by the
circumference of the circle described by the centre of gravity of the figure.Sec Moments. Fig.
.
x Q. y a cos/.and /axes. x x at sin/. Find the centre of gravity of a homogeneous hemi sphere
of radius a origin. and the coordinates of the centre of gravity of a triangle bounded by the
straight lines x y Q. of an area of radius a . Find the static moments. r/ al cos/ and the jcaxis.
Find the centre of gravity of a hemisphere lying above the . about Definite Integrals Ch. Find
the static moment of the circle r asinltp about the polar axis. . . Find the static moments. .
Find the coordinates of the centre of gravity of an area bounded by the ellipse l and the
coordinate axes . . . Find the coordinates bounded by the curves of the centre of gravity of an
area . .q/plane with centre at the origin. . lying above the jo/plane with centre at the . . about
the x. sides. Find the centre of gravity of an arc of a circle of radius a subtending an angle a.
Find the static its moments of a rectangle. and the coordinates of the centre of gravity of an
arc of the astroid t i If/ aT gt gt lying in the first quadrant. Find the coordinates of the centre of
gravity of an arc of the catenary y a cosh from a to x a.and t/axes. with sides a and amp. a.
about the x. and y . Find the coordinates of the centre of gravity bounded by the first arch of
the cycloid x at sin/. Find the centre of gravity of a homogeneous right circular cone with
base radius r and altitude h. Find the coordinates of the centre of gravity of the arc of one
arch of the cycloid .
Example . The path traversed by a point. then the path traversed by the point in an interval
of time is . a Determine the position of the centre of gravity of M an arc of the astroid x i/ T a
lying in the first quadrant. t/ fgt. If a point is in motion along some curve and the absolute
value of the velocity o// is a known function of the time t. b Find the centre of gravity of an
area bounded by the curves px and x py. . of radius a and of mass . Find the moments of
inertia of the area of the ellipse JC U about its principal axes. Applying Definite Integrals to
the T t Solution of Physical Problems . Find the moment of inertia of a homogeneous sphere
about its diameter. . Find the moment of inertia of a right parabolic segment with base and
altitude ft about its axis of symmetry. / m/sec. The velocity of a point is o . Find the moments
of inertia of a rectangle with sides a and b about its sides. . Find the moment of inertia of a
homogeneous right circular cone with base radius R and altitude H about its axis. . the
moment of inertia with radii R about the axis passing through the centre of the ring and
perpen dicular to its plane. the interval of time sec followis the mean velocity cf motion Find
the path ing s the commencement this interval covered by the point ol motion. b Prove by
Guldins theorem that the centre of gravity of a triangle is distant from its base by one third of
its altitude Sec. in What during .ampc. Find the its moment of inertia of a circle of radius a
about diarneler. . Find the polar moment of inertia of a circular ring and R t R ltRJgt that is. .
a Find the centre of gravity of a semicircle using Guldins theorem. . . Find the surlace and
volume of a torus obtained by rotating a circle of radius a about an axis lying in its plane and
at a distance b bgta from its centre. Applying Definite Integrals to Solution of Physical
Problems .
by x m is equal to v. Kinetic energy. /co. the latter is appropriately partitioned into elementary
particles which play the part of material points. v n is equal to . . We have p . kgm point of
mass . Solution. The kinetic energy of a material velocity v is defined as m and mv The
kinetic . kinetic energy of a solid. Definite Integrals Ch.. A x dx j x . then by summing the
kinetic energies of these particles we get. The work of the xaxis.. if kgf stretches it by cm
Solution. X . and. Solution. lf v .Jo. hence. in the limit.irto. What work has to be performed to
stretch a spring cm. We have To compute the Since the linear velocity of kinetic energy is the
mass dm is equal to t. an integral in place of the sum . of a force. we get and l kgf.. . Example
. Putting x Whence the soughtfor work is Example a force of X m X . then the work If a
variable force Xfx acts in the direction is of this force over an interval x ly x z A ..i and T
m/sec. the elementary . mv m mn energy of a system of n material points with masses
having respective velocities t. where k is a proportionality constant. Find the kinetic energy of
a homogeneous circular cylinder of density with base radius R and altitude h rotating about
its axis with angular velocity CD. t metres . According to Hooks law the force X kgf stretching
the spring kx.. 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.. .
The pressure experienced by element where y is Whence the specific weight of the water
equal to unity. We partition the area of the semicircle into elements strips parallel to the
surface of the water. Pressure of a liquid. To compute the force of liquid pressure we use
Pascals law. a flush semicircle of with the water surface Fig Solution. radius Example . is
given by the . Find the force of pressure experienced by r submerged vertically in water so
that its diameter is . The area of one such element ignoring higherorder infinitesimals located
at a distance h from the surface is ds xdh this V is h dli. initial C J The velocity of a body
thrown vertically upwards with velocity V Q air resistance neglected. Applying Definite
Integrals to Solution of Physical Problems Whence r dr nco / fc .Sec. the entire pressure is r .
which states that the force of pressure of a liquid on an area S at a depth of immersion h Is
where y is the specific weight of the liquid.
Calculate the work that has to be done in order to pump the water out of a conical vessel
with vertex downwards. What work has to be done to raise a body of massm from the earths
surface radius R to an altitude ft What is the work if the body is removed to infinity . Find the
altitude reached at lima the / initial velocity through an upper opening the tank has the shape
of a cylinder with horizontal axis if the specific weight of the oil is y. co are constants. . The
velocity of motion of a point is v tequot Qlt m/sec. . the length of the tank H and the radius of
the base R. Calculate the work to be done in order to pump water out of a semispherical
boiler of radius R m. Considering that when the rocket thrust is constant. the radius of the
base of which is R and the altitude H. the acceleration due to decreasing weight of the rocket
increases by the law / if / a ftf gt. formula . . What is the mean value of the absolute
magnitude of the velocity of the point during one cycle . is zero. and c is a constant. The
velocity of a body thrown vertically upwards with initial velocity v air resistance allowed for is
given by the formula where t is the time. be in t seconds from the time it is thrown . point on
the xaxis performs harmonic oscillations A about the coordinate origin. where t is the time
that elapses and g is the acceleration of gravAt what distance from the initial position will the
body ity. g is the acceleration of gravity. . ff. .ITS Definite Integrals Cfi. Find the altitude
reached by the body. . Calculate the work needed to pump oil out of a tank find the velocity
at any instant of time /. A rocket rises vertically upwards. Find the path covered by the point
from the commencement of motion to full stop. r . . Calculate the work that has to be done to
pump the water out of a vertical cylindrical barrel with base radius R and altitude H. Find the
law of oscillation of a point if when / it had an abscissa . its velocity is given by the formula
where t is the time and t.
y . A vertical dam has the shipa of a trapezoid. while pi is the coefficient of friction. whose
specific weight is y. Calculate the kinetic energy of a right circular cone of mass rotating with
angular velocity CD about its axis. on the Two What work cm . Find the work done by the
frictional force during one revolution of the m ft shaft. Find the water pressure on a vertical
circular cone with radius of base R and altitude H submerged in walei vertex downwards so
that its base is on the surface of the water. Find the pressure of a liquid. Determine the work
performed in the adiabatic expan of air sion and pressure having initial volume u l m p l
kgf/cm to volume u. A vertical shaft of weight P and radius a rests on a bearing AB Fig. . on a
vertical ellipse with axes a and whose centre is submerged in the liquid to a distance h. const
is the pressure of the shaft where p on the surface of the support referred to unit area of the
support. s/cm . the lower base m. length electric charges CGSEand e l CGSE xaxis at points
and .Sec. Calculate the water pressure on the dam if we know that the upper base m. A
vertical triangle with base and altitude h is submerged vertex downwards in water so that its
base is on the surface of the water. . What work must be done to halve the volume of steam
with temperature kept constant isoihermic process . . What work has to be don to stop an
iron sphere of meres rotating with angular velocity w . Applying Definite Integrals to Solution
of Physical Problems lie . disk of mass . The frictional force between a small part a of the
base of the shaft and the surface of the support in contact with it is Ffipa. . cm. . M M . rpm
radius R j about its diameter Specific weight of iron. will be done if the second charge is
moved to point A cylinder with a movable piston of diameter / and the cm D cm pressure is
filled with steam at a pkgfcm . . a m. if the radius of the base of the cone is R and the altitude
is H. respectively. and the height h . while the major axis a of the ellipse is parallel to the
level of the liquid hb. Find the pressure of the water. Calculate the kinetic energy of a and
radius R rotating with angular velocity Ggt about an axis that passes through its centre
perpendicular to its plane. .
The wind exerts a uniform pressure p g/cm on a door width b cm and height h cm. on the
/axis. . Find the moment of the pressure the wind striving to turn the door on its hinges. and /
is the length of the pipe. x g/cm. According to empirical data the specific thermal capacity of
water at a temperature is . What is the force of attraction of a material rod of and mass on a
material point of mass a straight line with the rod at a distance a from one length / M m lying
on of its ends . . . but the pipe has a rectangular crosssection. . use is frequently made of
special types of diagrams the velocities v are laid off on the Jtraxis. Definite Integrals C/i.
Show that the area S bounded v t and v l and v by an arc of this graph. The conditions are
the same as in Problem .xlO/ / . In the case of steadystate laminar low of a liquid through a
pipe of circular crosssection of radius a. In studies of the dynamic qualities of an automobile.
What from of of /C /lt . and the reciprocals of corresponding accelerations a. . C quantity of
heat has to be expended to heat to C g of water . . i is is the pressure the coefficient of
viscosity. Determine the discharge of liquid Q that is. Find the mass of a rod of length / cm if
the linear density of the rod at a distance x cm from one of its ends is . the quantity of liquid
flowjng through a crosssection of the pipe in unit time. by two ordinates v by the axis is
numerically equal to the time needed to increase the velocity of motion of the automobile
from v l to v accelera tion time. Here the rate of flow u at a point Mx. Miscellaneous Problems
. and the base a is great compared with the altitude .xlO.y is defined by the formula
Determine the discharge of liquid Q. the velocity of flow v at a point distant r from the axis of
the pipe is given by the formula where p difference at the ends of the pipe.
Sec. Find the sinusoidal current quantity of heat released by an alternating during a cycle T
in a conductor with resistance /. . . that is. and support reactions A and fiylyj t directed
vertically upwards. Mx in crosssection x. . Find the bend M ing moment Note. of The intensity
load distribution is the load force referred to unit length. Find the bending moment x in a
crosssection x. A horizontal beam of length / is in equilibrium due to support reactions A and
B and a load distributed along the kx. the moment about the point P with abscissa x of all
forces acting on the portion of the beam AP. A horizontal beam of length a downward vertical
load of uniformly / is in. where x is the distance length of the beam with intensity q from the
left support and k is a constant factor.equilibrium due to distributed over the length of the
beam. Applying Definite Integrals to Solution of Physical Problems .
if nate system X. From geometry we know that the volume of a cone is where h is the
altitude of the cone.y a.Chapter VI FUNCTIONS OF SEVERAL VARIABLES Sec. if to each
set of their values x. Solution. that is. of the function zfx. Z a surface Fig. y. Find/ . . Fig.b or f
P Generally speakthe geometric representation of a funcz f tion like x. y. Functional notation.
variable quantity is called a singlevalued function of two variables jc. and/.b. when xa and y
at a bt ing.y in is a rectangular coordi. denoted by / a. Faiic Notions . Substituting r and t/ .
define functions of three or more arguments. we Fxample . But h y y Hence. The concept of
a function of several variables. Similarly. Y. // in a givm range there corresponds a unique
value of z The variables x and y are called arguments or independent variables. Express the
volume of a cone V as a function of its gen eratrix x and of its base radius y Solution. we find
. Example . This point is is The value P the desired functional relation. The functional relation
is denoted by A /.
for a function of three variables u fx. two the when or is when . the domain of definition of a
function U a finite or infinite part of the jo/plane bounded by one or several curves the
boundan. Similarly. of the domain. y we understand a set of points . The latter gt lt The Fig. r/
in an jq/plane in which the given function is defined that is to say. Find the domain of
definition of the function domain of x function has real values if or x y . /l. in which it takes on
definite real values In the simplest cases. i. we will have SL thai is.The domain of definition of
entire function shown in Fig. Fiy Example . of definition of a function. y. . y inequality is
satisfied hy the coordinates of points lying inside a circle of radius with centre at the
coordinate origin. z the definition of the function is a volume in At/zspace. . Basic Notions SI
Putting and replacing y by .Sec. Find the domain z of definition of the function arc sin / xy is if
Solution.. and includes the boundaries of the domain. Example . The first term of the function
defined for lt in or cases The second term has real values xr/O. Domain f. The domain of
definition oi the function is the interior of the circle Fig .e. By the domain of definition of a
function fx. . Solution.
if f. Find fx. Construct the the function z xy. The level line of a function fx. y if .x. y. . z Find y
fy. Functions of Several Variables Ch. if . . we get a family lines Fig. Example . level surface
of a C.. points of which the function takes value uC.. . Fig. y is a line / .. . Express the volume
V of a regular tetragonal pyramid as a function of its altitude x and lateral edge y. f x. . The
equation of y t z on is a sura constant level lines of the level lines has the form x y Putting of
level C . i . Find //. . Solution. Find the values assumed by the function at points of the
parabola y z . and construct the graph of the function . y. . . C . at the z function of three
arguments ufx. yR if . Determine fx. . The iace / x. or y j . Find the value of the function Z rrr
at points of the circle x .y x . Express the lateral surface S of a regular hexagonal truncated
pyramid as a function of the sides x and y of the bases and the altitude z. /I. Level lines and
level surfaces of a function. yC in an r/plane at the points of which C usually labelled in the
function takes on one and the same value z drawings.
Let z if z Basic Notions x when . V/l/x . Find the level lines of the following functions a z b /y
a. the and z if . Find the level surfaces of the functions pendent variables x yz. r . a u b u c u
J of three inde . zY x zV y x //. Construct the level lines of the given functions and determine
the character of the surfaces depicted by these functions a zx y. . Find and sketch lowing
functions a domains of definition of the fol b c z d e f g h . Find the domains y of the following
functions of three arguments a ux z c b u lnxyz warc sin jcarc sin d u Vx arc sinz. ar O d z .
lnf/. y z . b c z y e z xy.x arc cos zl/l z arc sin X V IT .Sec. d z . . Letz /V Determine the
functions f and when x. Determine the functions / y .
In this case we write lim XK yb yAlte. functions . fx. Continuity Functions of Several
Variables Ch. Continuity and continuous at a point P points of discontinuity. the given
function has for its discontinuity the parabola y x . . cli n M. Find the discontinuities of the
function gt Solution. A b. But The function r/ or y will be meaningless if the denominator
becomes x is the equation of a parabola. as A number A is approaches the point P o. Hence.
y A. Example . elinL. Test the following function for continuity ft x / / Vx I if when r// ltl. zero. .
y there e gt is . A function that is continuous at all points of a given range is called continuous
over this range A function /AT. c e . y may cease to be continuous either at separate points
isolated point of discontinuity or at points that form one or several lines lines of discontinuity
or at times more complex geometric objects. we have the inequality the point x. Find the
following limits of functions a liinsinl. a. Sec. y I P called the limit of a function ft. when x //gt .
Find points of discontinuity of the a z ln. where Q /x the distance between P and P. if for any
a U/ /. The limit of a function. is a gt such that when lt Q lt . y is called b if lim fx xa /a. /. . .
function zfx.
A function if for every real factor k f x. x y cos .Sec. Sec. y z y . Partial Derivatives . x y .
example. Definition of a partial derivative. continuous with respect to each of the variables x
and y sepabut is not continuous at the point . Example . ky . y constant. degree n Eulers
theorem. Show that the function SL is when when jt lt/ . we will / have x dz y y Example
arguments . Find the partial derivatives of the function y Solution. . rately. In similar fashion
we define and denote the partial derivative of the function z with respect to the variable y It is
obvious that to find partial derivatives. Find the partial derivatives of the following function of
three a Solution. one can use the ordinary formulas of differentiation. with respect to these
variables together. // f kx. we t get dz dx tan .y. Regarding y as constant. then assuming. we
get the derivative If z fx. y fs called a we have ttf homogeneous function of the equality x. y x
I/sin y y Similarly. . holding x constant. for which is called the partial derivative of the function
z with respect to the variable x. Partial Derivatives t.
. . . all its A rational integral function will be homogeneous for a if terms are of one and the
same degree.X . . if ln if Show Show Show x gfjS.y . / J . Exam Ax BxyCy . . . and f . on
homogeneous functions t . fx. y. Show that that that that . z . y if fx. Find /. . . . . . y
homogeneous differentiable function yfy x. Find . ggg. . if u xyyzz. y.y t f xy if L ... y. it if JJ. .
.z lnxyz. /il. axy. . Find /. . Functions of Several Variables Ch. Find the . . J. in Verify Eulers
theorem ples to . z arc tan . Calculate and y . where r . The following relationship holds of
degree n Eulers theorem xfx x. l. z partial derivatives of the following functions . y nf x. . /..
y at the point . Through the point Ml. Determine the angles formed with the coordinate axes
by the tangent lines to the resulting crosssections drawn at their common point M.Sec.fl amp
A. . y is the principal part of the total increment Az. The total increment function AzAf x. For
the function fx t yx xyy total find the total increment and the differential. y and fy x. Total
Differential of a Function z /i . has partial derivatives f x. . of a y.dz. Aj/f .. . The total or exact
differential of a function z fx. X J/ . If a function has a total differential. z du du dx . The area
of a trapezoid with bases a and b and alti. ampx function definitely has a total differential if its
partial derivatives are continuous. dy Similarly. determine their geometrical meaning. using
the drawing. The total differential of the function z /x. are tude h is equal to S l /. . y slny when
. Construct the geometric image of this function near the point . that is.d y dy . then it is called
differenThe differentials of independent variables coincide with their incret table. .and zx. dz
dy . which is linear with respect to the increments in the arguments Ax and A//. of a surface z
xy drawn planes parallel to the coordinate surfaces XOZ and YOZ. Sec. . dz . x although it is
discontinuous at this point. ments. du . . dx j du . Total y is increment of the difference a
function. y is computed by the formula the function A d . / x. The difference between the total
increment and the total differential of total differential of is an infinitesimal of higher order
compared with Q At/. y. r. g. dX . The / Ax. the total differential of a function of three
arguments u is computed from the formula . dz . Find zzx y t J Total Differential of a Function
knowing that g and. .. dxkx and dyky. a function. Example . Show that the function . Find g.
for sufficiently small we have for a differentiate function z fx t y the approxor .. while A f AJC
comared with VAx Ay compared Example . compare them if . f Ay . y Ay x f Ax f x A y f Ay y f
Ay x x y A x . Ax AA x Ay y Ax Ax Ay y Al y Ay Ay a A AAgt Ay Ay y Ay is the total differential
of Here.. fxAx. The altitude of a cone is // cm. s l l... Ay .J Solution. two variables the ordinary
rules of differentiation holcb v du udv . small AA and Ay and. b A .. The change o in volume
we replace approximately by the differential AV dV j n RH dR R dH lji . y .In . the radius of the
base the volume of the cone change.. The volume of the cone is V nR H. the expression d/ x
y A infinitesimal of higher order is an the function. . a Axl. .. We consider the function zx.. A//.
Compute . . Solution. For the function example. In x Hence. . Applying For sufficiently Q y Av
f Ay . imate equality Azdz the total differential of a function to approximate calculations..
Show that for the functions u and v of several for . . How will mm and diminish R fl Example .
. lOjiss . cm. s Example . considered the increased value of this function Ay . Find the total
differential of the function AyAy Solution. The initial value of the function z The desired
number may be when jcl.. Ajc . . and the fx. Functions of Several Variables Ch.
approximately.. dz cm.y xy find the total increment total differential at the point . . hence. if we
increase by H by mm Solution.
if the original lengh of the radius is cm . . if . if arctan. Approximate volume cm. By how much
should the radius of the sector be increased so that the area will remain unchanged. yx. . . .
the other amp cm. round off the C that the relative error of a product is approximasum of the
relative errors of the factors. Approximate of material used in making the box. Find dfl. take
when converting degrees into radius and three significant figures. A closed box with outer
dimensions .Sec. . The oscillation from the formula period T of a pendulum . . . b . z . a
diagonal / of the rectangle change if the side a is increased by and b is shortened by will One
mm mm the change and compare it with the exact value. it is desired to reduce it by . . side
fc angle . lnxy. w arctan. Find d/ How side of a rectangle is a cm. Approximate the a c . . /.
arc tan u u . z si . . K . . z. The central angle of a circular sector is . sincos sin calculating last
digit. . Measurements of a triangle yielded the following data side amm. mm. Find the total
Total Differential of a Function differentials of the following functions .quot tj . . and cm is
made of mmthick plywood.In tan i. x . cm. Show tely equal to the ABC . is To what degree of
accuracy can we compute the computed side c . lnli.
for instance then the quottotalquot derivative of the function z with respect to x will be x.
jZsyety. y is a differentiate function of the arguments x and y. Solution. where cos/. Find the
error. which in turn are differentiate functions of an independent variable /. y with if the Sec.
From formula we obtain of several independent variables. The distance between the points P
. ytyu t v u and v are independent variables. if ey t where y Solution. The case several
independent t dz dy . y dy point PP is fixed moves to P l x V g. . then the derivative of the
composite function z puted from the formula dz dt /ltpi ma X be com i quotdxdtdydt In
particular. then the partial derivatives z with respect to u and v are expressed as . The case
of one independent variable If z fx. while the angle formed by the vector P P the xaxis is a. if /
coincides with one of the arguments. Example Find the partial derivative z and the total
derivative . Find . is the length of the of gravity. From formula X we have e sinO . for
instance. y t. Differentiation of Composite Functions . dzdy Example . pendulum and g is the
acceleration when determining T. If z is a composite function of variables. a result of small
errors A/ . obtained as a equal to Q. z where qgt u v fx. where Functions of Several Variables
/ Ch. By how much will the angle a change dx. if y.y. is and Ag J in measuring / and and P x.
y. that is. y . Find if y if . . where / l. rt Substituting the partial derivatives into the lefthand
side of tion. f where x t y ytl. .Sec. . . t . and . . Differentiation of Composite Functions dzdzdx
. depends on x and y via the intermediate dzdzdt and z dzdt . y .y . if where xuv. The function
therefore. Applying formulas and we get and Example . Find j if u xyz. . Show that the
function z ltp x y satisfies equation dy argu dx Solution. I z . cp ment x yt. the equa we get t/ i
where x e t y nt. the invariance property of a total differential. yln. the dzdy In all the cases
considered the following formula holds dz j . Example p Find . dz . Solution.and du dv zfx.
Find u lnsin. the function z satisfies the given equation. dz dx dx dy dy .
if // a . then dz dz . . where x ip. Find g and if arc if tan and / . where x I. y /. if where Show
that where / is a differentiable function. cos cp sin e R cos gtgt sin . Find z if uv . and j z fu. if
where w x/ . Find t where y ltfgtx. Show that . Find g and xy . R cos . then . Find /u. ve . Find
if and a. Functions of Several Variables Ch. . ygt . Find and oy . Find if r. where u sin x. . ow
if where a A if t/ . v cos . . arc tan and z where x usmv. v.
Show that the function satisfies the equation x y j . . Their velocities are respectively km/hr
and km/hr. . The equations of motion of a material What is the rate of recession of this point
from the coordinate Two boats start out from A at one time. is increases at the rate of a
rectangle x the other side f/ decreases at m/sec.Sec. where u x at. The side m m angle . At
what rate does the distance between them increase Sec. y in a given direction / PP. Show
Derivative in a Given Direction that the function point are S w fu. the other in a northeasterly
direction. is . v v. origin . Derivative in a Given Direction and the Gradient of a Function of a .
one moves northwards. The derivative of a function function z in a given direction. y bt dw
satisfy the equation dT a . . The derivative /. dw Show that the function satisfies the equation
J . . What the rate of change of the perimeter and the area of the rect . dw dt . Show that the
function satisfies the equation A y xy of m/sec.
we get The minus sign indicates that the function diminishes at the given point and in the
given direction. P. The directional derivative characterises the rate of change of the function
in the given direction. In this case direction / for a t du dl du cos a du cos p du cos v.y Fig.
then the following formula holds where a is the angle formed by the vector / with the xaxis
Fig. The gradient of a function z fx. Find the derivative of the function z x in a direction that
makes a angle with the xaxis. . . Y e P. A dxjp Here. dx dz . Find the partial derivatives of the
given function and their values at the point P dz . Pl. at the point lt/ Example . Functions of
Several Variables / P. r T where a. dz dx dy . dz . . sina sin Applying formula . The gradient of
a function. where fP and If the function z are values of the function at the points P and
differentiate. Y are ne angles between the direction / and the corresponding coordinate axes.
. is Ch. y z. ij is vector whose projections on the coordinate axes are the corresponding par .
In similar fashion we define the derivative in a given function of three arguments u fx.
Solution.
.Sec. X Solution. du . . the derivative in a given direction is equal to the projection of the
gradient of the function on the direction of differentiation. thlft is. . du The gradient of a
function of three variables at each point is directed along the normal to the level surface
passing through this point. The direction of the gradient of the function at a given point is the
direction of the maximum rate of increase of the function at this point. equal to In similar
fashion u /. z we define the gradient of a function . The gradient of a function at each point is
directed along the normal to the corresponding level line of the function. du . y. Compute the
partial derivatives and their values at the point P. X J Fig. when /grad z the derivative takes
on its greatest value. . i dxjpdz Tquot . y s . dz The derivative of the given function in the
direction / the gradient of the function by the following formula is connected with That is. .
grad z tJ Fig. Derivative in a Given Direction tial derivatives of the given function dz . Hence.
Find and construct the gradient of the function zxy at the point PI. Example . of three
variables. .
y. The point at which the derivative of a function in any direction is zero is called the
stationary point of this function. . . .zx at quadrantal angle. p. Find the magnitude and . x
derivative of the function z xyy the direction that produces an angle . Find the derivative of
the function u x yz at in the direction that forms identical the point Afl. z the angles a. eP e n
e . if uxyz. Find grad u at the point . Find the derivative of the function u xy yz . if . Find grad z
at the point . y. . of direction grad u at the point . Functions of Several Variables Ch. Find the
derivative of the function z at the point Afl. angles with all the coordinate axes. taken to the
any point of the ellipse x y C along the ellipse is equal to zero. Find the stationary points of
the following functions a zx b z c x y y zxyyz u Show that the derivative of the function z .
Find grad z at the point . . if . . . in the direction of the bisector of the at first . in of with the
xaxis. Find the at the point Pl. . if at normal . in the direction from this point to the point NS.
respectively. Find the angle between the gradients . of the function lnj at the points A /. . . . in
the direction from x this xy xy point to the point tf . / and . . Find the derivative of the function
z lnYx y the point Pl. Find the derivative of the function u at the origin in the direction which
forms with the coordinate axes x. . . the point M.
Sec. f y y dz z . Construct a vector field of the gradient of the following functions a z jef y c z
x y z Sec. partial m derivatives J J . dxdy We note that the sosocalled el way. If the partial
derivatives to be evaluated are continuous. The second partial derivatives of a function z /.
HigherOrder Derivatives and Differentials . . Higherorder partial derivatives. . . For second
derivatives we use the notations d dz z Derivatives of order higher than second are similarly
defined and denoted. HigherOrder Derivatives and Differentials . then the result of repeated
differentiation is independent of the order in which the differentiation is performed. y are the
partial derivatives of its first partial derivatives. Find the second partial derivatives of the
function z arc tan y Solution. . First lind the dz first . Find the steepest slope of the surface x z
y at the point . namely quotmixedquot partial derivative may be found in a y different d /
dxdydydx dx . Example . j y dz dy Now differentiate a second time L d .
then d jt becomes identical with formula Example . then it is If sev . Differentiating we find x
and Differentiating again and remembering that dx and dy are not y. dx dj/ dx d d/ dy dxdx
dydy . formula If x and i/ are independent variables.t xyy Solution. dz fr Further we have djc
JL djcdquotquot l dy whence it follows that Second method. Functions of Several Variables d
y Q. function function We similarly define the differentials of a function z of order higher than
dz two. the following symbolic formula holds true formally expanded by the binomial law.
where x and y are independent variables. Find the total differentials the function z of the first .
The second differential of a of this y is the differential of the differential firstorder quot. z fx t
Higherorder differentials. /. If ddz l dz dd n z. and Ch. and second orders of . z f x. for
instance and. . generally. then the second the function z is computed from the formula
Generally. We have lt Therefore. differential of y. z /x. First method. we get dependent on z .
I. where the arguments x and y are functions of one or eral independent variables.
f xt/ Q. HigherOrder Derivatives and Differentials if .Sec. xy . Find if . Find if . Find if arc tan
A . Find f. if if u . . . Find all second partial derivatives of the function . l . if Show that arc sin .
Show that for the function . Find z sin xy. y Show that if . /O. if . . Find dxdy dz .
Find ampquot z fu. . provided that f. Show that the function / lny . Show that the function u
satisfies the arc tan Laplace equation . Functions of Several Variables Ch. where u x y. v. if
fx. v. // . x /. y. a are constants satisfies the equation of heat a/uU . that the function u
satisfies the x. . Show ux that the function i xx e yy n y z gt z gt gt where conduction . z. v . z
fu. we have r y . Find xy.. y. u g y. where r Yx Show a y b satisfies the Laplace equation .
where u ltx. t A sin akt cp sin Kx equation of oscillations of a string dt du a du dx .. where z
yx. .
. v xy. y. r . satiscp and the equation of oscillations of a string . Find dz if if . l x. that the
function where fies are arbitrary twice differentiable functions. Find d u . which satisfies the .
dv w dxdy y J y . y if dTSy . Show that the function satisfies the equation x . Find uux. dy n . if
. y . .rn l f /F if whpr vviitivx / tgt r jr y . Show HigherOrder Derivatives and Differentials . Show
that the function satisfies the equation d su . tion Show that the function z dz fx yy satisfies
the equa d dzd z dy dx dx dx dy ./y . Find dz . Find dz and dz z u v where u .Sec. Determine
equation the form of the function u ux.
Integration of Total Differentials t. . Find dz z if v. . x z Sec. v by fu. In the given case. where
u ax. where u xey v ye. The condition for a total differential. . it is necessary and suf aqap dx
dy Example a total t. ydx Q ydygt where the functions P x. to be some function u x. . P x y Q
xy. For an expression P x. therefore. Find d /. . pf/ we have . y. hence. . and. y are
continuous in a simply connected region in D the total ficient that D differential ol together
with their first partial derivatives. Make sure that the expression is differential of some
function. Find dz z e x cos y. whence qgt y y. Functions of Several Variables C/i. . Determine
all third partial derivatives. y and Q x. fx y. and t find that function. . that given jt . Find dz if
zfu. But on the other hand and Finally x y y x y. . Find dfl and dfl. . if v. Solution. if fx. y x xy y
if z nx Qny. . Therefore. where u It is is the desired function. Find the third differential of the
function z cos y y sin x.
Here. dP O dy . y. and find that function. hence. zdx y. have hence. y. We establish the fact
thai is Qz dQ dx and. Be sure that the expression the total differential of some function. is the
total differential of some function u x y. tion for total differential find qgt known and the condi
fulfilled. derivatives. f jc Solution. z. z if and only if the following conditions Qx. du dz dtp dz
whence of y z and P f/zl.Sec. Integration of Total Differentials case of three variables. where
P Rx. z. zdz. We that is. The Px. the expression y. z are. together with their first partial
functions of the variables x. yy. y and . x. continuous t are fulfilled dQW dRdQ dPdR dx dy dy
dz dz dx Example . . dR dQ r dy dz dP c. x. p . Qx. R yz. x y dx x xy x ltpy. Similarly. e C. And
finally. zdy Rx t y. ti . whose is The problem reduces partial derivatives are to finding the
function two variables qgt. rr OR r / dz dx where u is We the soughtfor function. z. V. u On
the other hand. y.
x y . differentials of Convince yourself that the expressions given below are some functions
and find these functions. Find the function u if du fxy ydx xdy. xyzyz zdxxy y xy dx zdy ldy x y
dz. . dxdy. Functions of Several Variables C/i. ydx xdy. What must the coefficient K be for the
force to have a potential . . y v to where A. x y xdx ydy zdz . . y satisfy for the expression fx.
find these functions. Given the projections of a force on the coordinate axes v . . . . What
condition must the function fx. y x dx y r z dy. Having convinced total are differentials yourself
that the expressions given below of certain functions. and find that total function. to be a total
differential ydx dy . . Determine the constants a and in such a the expression z ax xy y dxx xy
by dy manner that should be a total differential of some function z. cosxxydx xy . is a
constant.
Differentiation of Implicit Functions of f x. where differentiate function of the variables x and
y. The case of several independent variables. z z another way of finding the derivatives of
the function ating the equation F x. defines y as a function of x. y a that f x. dz TT. applying
formula . y Whence.Sec. rt Whence it is possible to determine dz. y. Find dx and dx if
Solution. Fz z dK F z x. . y. be found from the formulas y and x y.y found by successive
differentiation formula a Example . The case of one independent variable. y y . dF .
differentiate with respect to x the first derivative vhich we have found. . / If the equation fx.
and. z dl J F g x. y . generally speaking. y. z z. Similarly. then the derivative of this implicitly
defined function. dx and dy dz . provided . x. y. therefore. we get find the second derivative.
may be found from the formula dy dx Higherorder derivatives are f x y fyx. function can. . t of
the equation the variables defines z as a function of the independent variables x and y and
then the partial derivatives of this implicitly represented . we find Here is different dF . z.
Denoting the the partial derivatives lefthand side of this equation by we find f u x t yx z y z .
where F x. is Differentiation of Implicit Functions Sec. taking into consideration the fact that y
is a functiun of x To y J x dx y y J x y y J dx dx y J if if F x. y. . z is a differentiate function y. .
dF J .
dv y du /Q . G Du. y.dv . o . . z zl. . v dudv dGdG du dv then the differentials of these
functions and hence their partial derivatives as well may be found from the following set of
equations dF . A system of implicit functions. . dy dy . . . x y. If a f system of two equations
defines u Fx. First method. Whence we determine dz tion t that the total differential of the
implicit func// zdy dy we Comparing with the formula dz Q.dx z.dx and dy dx dy rr . see that
dz x dz yz y dx y z dy . F x x. F y x. . dx r l du dG .dx . /. y. z x. u t igt y. we obtain x dx f/ dy
zdz is.. y dz zdy dy .. Functions of Several Variables Ch. dv y dx . y. dF dF dF . u. t . z.
Example . //. z. y. x and y and the Jacobian and v as functions of the variables dF dF DF.
Find T and if j we Solution. z quot // xgt Second method. Differentiating the given equation.
Gx. z y dz dy Jy. Example . F z x. Denoting the left side of this equation find the partial
derivatives by F x.dy du . . The equations define u and v as functions of x and find w. z /
Applying formulas dz we get d Fxxgt F x z t y.
Knowing dx dz p and r the differential dzp d qdy. we find the partial derivatives dz . . be
found from the following system dx dx du du. Solving this system for the differentials du and
dv. u dx y dv v dy . Differentiation of Implicit Functions First Solution. dz dz . Parametric bles
x and y is dyx y z representation of a function. we obtain xy Whence dx x y tf/ x v x f y yfx dv
dxxy . dy .dv dx dy. dx dv . If a function represented parametrically by the equations z of the
varia zu. v and of then the differential equations of this function may .Sec. dv whence ff dx
Similarly u y x y v dv ux dx x y we find du y dv y find dy x y dy x Second method. xdu dv
dv.dv. method. Differentiating both equations with respect to we obtain du . . By
differentiation differentials of all four variables we two equations that connect the x du du .
First method. we wu u Whence jc au. Functions of Several Variables z Ch. and we obtain .
By differentiation connect the differentials of all five variables we find three equations that dx
du dv . dy u dx v u u w rfy Substituting into the third equation the values of du and have d dy
just found. Example . . The function of the arguments x and y is defined by the equations
Find . From the third given equation we can find Jf . dx dx Differentiate the respect to y first f
dy Jf t. From the first two equations we determine du and dv quot dx dy . dt/ TT wfy. dz . l dy
dy two equations first with respect to x and then with dx f . dy Solution. dz .and ox . v .f. dy dy
From the first system we find du v dv a u v dx v dx u From the second system we find da l v
dv dyu Substituting the expressions dyvu into formula . Second method.
The function y In is defined by the equation arc tan T dx z if dx and dx .. . i and . Find dx and
A dy z if cosy y cos zcosx . g K and g ax J ax jxi if Taking advantage of the results obtained. j
Flnd s // z . Find . . . Find yy x . z l. and g if y z A. show approximately the portions of the
given curve in the neighbourhood of the point . .Sec. Find . The function is defined by the
equation x y z xy . y a function defined by the equation Show that and explain the if result
obtained. Let y be a unction of x defined by the equation y and ana is Differentiation of
Implicit Functions Find ma dy dhj di d dxquot . // Find and j for the system of values . .q/z .
Find dx and . The the equation function z of the variables x and y r/ is defined by x c. . .
. y. equation ab . Show .z . if a function of x defined by the Find dz and d z. y bz Q. x y Show
that the function defined by the equation xqgt z ogt satisfies the equation dzf . . Fmd. dz . Let
the function be defined by the equation where ltp is an arbitrary differentiate that function and
a. The functions y and of the independent variable x are defined by a system of equations .
amp. .and secondorder derivatives of the function . What are the . Show that xf y z. /.
satisfies the an arbitrary differentiate function of two arguments. . KinA Functions of Several
Variables Ch. Find dz and d z for the values x . dZ . fy y . if if . f/. Find dz and d . px. x . is a
function of the variables x and defined by the equation x y z xz z . . tyx. . Show that the
function defined by the equation Fx where F is . Show is equation . f/ where y Find . c are
constants. . amp lgt d . if In zjc i/ first. y. . dz QR . .
The functions u and v of the independent variables and y are defined implicitly by the system
of equations x Find du. . . v of the variables x and defined implicitly by the system of
equations . . dv. .Sec. z Sec. du. . b du du du cty dxgt fy and g x M cos u. . v. v. b Find c Find
dz. The functions y and z of the independent variable x are defined by the following system
of equations Change of Variables H Find dy. find and . y b sincpcosij.v The functions u and y
are pw. Change of Variables When changing variables in differential in expressions.z uv.
where r and are functions of the variables . dz. . . . Regarding z as a function of x and y. cp
ltp x and y defined by the system of equations Find ax and dy . a Find . y u and v. dv. the
derivatives in differ them should be expressed terms of other derivatives by the rules of
entiation of a composite function. e Fr. x u v. y w sin u. The functions u and v of the
independent variables x and y are defined implicitly by the system of equations Calculate du
du du du dzu dy dv dv dx dy v dv dz v j d dxdy dy for xQ. . dy. if y l if jc . //iw. y . dz. if x a cos
q cos i.
derivatives. Express the derivatives of y with respect to x in terms of the derivatives of y with
respect to /. Functions of Several Variables Ch. Express the derivatives of y with respect to x
in derivatives of x with respect to y. Transform the equation x g yg . the derivatives get just
found into the given equation and replacing x by we or dt ruij quot Example . We have dy dt
dy dt dx dx dt t dt dt Substituting the expressions of r. Transform the equation dx putting x y y
Solution. Solution. taking y for the argument and x for the function. . will we have dx dj dx Lo
dy dy dy . terms of the dxJx dy quot dx dx dx dyl dx dx dy J fdx Ty dx dy fdx dy Substituting
these expressions of the derivatives into the given equation. Change of variables in
expressions containing ordinary Example .
as a function of r sin cp dcp. in Solution. and . Considering r rsinqgt. from formula cp we
have dx cos cp dr d// sin dr r cos r cp dcp.Sec. Solution. Take the equation of oscillations of
partial derivatives.r sin rcoscp cp rsincp or. where a . cp. r cos sin cp r dcp . or. whence dr .
du dp dudu da du dp didadTddT we get du d/ dadxdfidi Q T da du . . Change of Variables . //.
Change of variables in expressions containing Example .vrcoscp. du quot a du is du I dp dp
day dudu du . Applying the formulas for differentiating a composite function dudu da.f dltP r
cos cp rcos cpH rsm cp dr coscp . Let us express the partial derivatives of u with respect to x
and t terms of the partial derivatives of u with respect to a and p. a . we will have . . after
simplifications.f . a string and change it to the new independent variables a and p. Example
Transform the equation dx xy f/ by passing to the polar coordinates . . finally. d//sm cp dr f
cos cp dr J T . cos p Y cp dcp r sin cp dp T r cos dr cp r sin q Putting into the given equation
the expressions for sin cp x.v at.
dw dx or dw Whence . dz dx / dw dw J z xdv J du x . d u du dz u dxdxdx dadx d u du rtn /ft
du /ft /R Substituting into the iven equation.v. Example for . z g . du . oy dx dz .x . y taking for
z . Transform the equation x . Let us express the partial derivatives ypartial in terms of the
derivatives and . consequently. . du dw dv dv dz z dw du du . x . v the y x new independent
variables.dv x dv . and w x the new function. . same formulas dtdadt du d u. differentiate the
given relation ships between the old and new variables u x dx t v dy . Differentiate again
using the dt dt Functions of Several Variables C/i. i w fx m dy. dw . On the other hand. . amp
dw and. To do this.. and Solution. dw . dw Therefore. z dw . we will have d u du /d a d u d u
Jd u r d u .
taking y as the ar gument .Sec. Transform the equation putting A cos/. Transform the
equation x j putting xe. . T rm Transform the following equations. MT is The tangent of the
angle A formed by the tangent line and the radius vector OM of the point of tangency Fig.
Change of Variables and dz z dw dyy dv Substituting these expressions into the given
equation. expressed as follows tan u . we get or .
Transform Functions of Several Variables C/i. Express. X y rsmltp. . . Transform variables u
and v the following dz . Transform the equation dx taking u x y. Transform the equation y Ugt
dx ar/ putting uxy and y . Transform the equation dz dz yTx. . y r sin cp. gt and the new
function wnz x y. Transform the following equation to new independent dz variables u and v
dz if u x t v x y amp . p ltFu . in the polar coordinates x the formula of the curvature of the
curve r cos qgt.K jyy z by introducing new independent variables . . . Transform the Laplace
equation dM n quot z dxdy to the polar coordinates A rcoscp. v for the new independent
variables and gt for the new function. . this expression by passing to polar coordinates . n .
equation to new independent dz if x.
it . . which the equation of the normal. Z are the current coordinates the point of where ..
Example the surface z are the current coordinates of the point of the normal. Equations
representation implicitly. t/ f x the tangent plane. When the equation of a surface is
represented and F X Q . zo/i z .Sec. where wwu. . Transform the equation putting u Sec. y is
a differentiate z of the surface is the equation of the tangent plane at the point x f/ it M M M . .
in a rectangular coordinate system. r o. z function. F. oXo /io. The Tangent Plane and the
Normal to a Surface v. K. Write the equations of the tangent plane and the normal to y at the
point M . we will have z r/. . The Tangent Plane and the Normal to a Surface . z . the
corresponding equations will have the form . Solution. Z . The equations of a tangent plane
and a normal for the case of explicrepresentation of a surface.Y. xy. of i Here. The equations
of the normal are of the form and X. The tangent plane to a surface at a point to point of
tangency is a plane in which lie all the tangents at the point various curves drawn on the
surface through this point. is given in explicit form. The normal to the surface is the
perpendicular to the tangent plane at the point of tangency If the equation of a surface. y.
where f x. . which is the equation of the tangent plane and i. then f x. t/ . v x y. Let us lind the
partial derivatives their values at the point of the M given function and ltk v fo dx dxjM
Whence. . w xyz. of the tangent plane and the normal for the case of implicof a surface.
applying formulas or xf/ z is and .
we find the Applying formulas and . /sina.q/z . J/. . Thus. and XXQ F X . sphere at the xy z
Rz y . . we get or z a. which Functions of Several Variables is Ch. of the sphere x y Write the
equation of the plane passing through the tangents to the obtained sections at their common
point M. the tangent plane and the normal to a. z/ Solution. putting x . Show that the equation
of the tangent plane to the central surface of order two M ax by cz k . . the z a into the
equation of the surface . the point ffcosa. At what point of the ellipsoid y . which is the
equation of the tangent plane. a b c does the normal to it form equal angles with the
coordinate axes . . Find the zcoordinate of the point of tangency. x Qj/ ne a or r n i wn cn are
e q ua ions of the normal. Write the equations of s . Denoting by F x.j/ which are the
equations of the normal.. f. z a at a point for which x the surface . Example . y Fy gt . a
whence z a. . y. the equation of the tangent plane. paraboloid of revolution z xy point at the
point b to the c to cone the /. Yy Q ZZQ Z Fz . .and axes are drawn z . through the point .
quotquot f. point of tangency is M . a. z partial derivatives and the lefthand side of the their
values at the point Af equation. Planes perpendicular to the A. Write the equation of the
tangent plane tions of the normal to the following surfaces at points a to and the at the
equaindicated the . a.
Show that the surfaces xy z r t sphere. On the surface je find points at which the tangent
planes are parallel to the coordinate planes. pass through the coordinate origin. tj. z has the
form surface x to the plane x lt/ z parallel . The angle between the tangent planes drawn to
given surfaces at a point under consideration is called the angle between two surfaces at the
point of their intersection. OJ . cp. // . . .Sec. Draw the . Show / that all the planes tangent lo
the conical surface zxf at the point M .v . Show that the tangent planes to the surface / x/ y on
the coordinate axes. tf z tangent tangent planes . The Tangent Plane and the Normal to a
Surface at the point M x to . At what angle does the cylinder xy R and the sphere xR lt/z
intersect at the point Affy. segments whose sum if is constant. // a f/ / intersect the axis . Find
the projections of the ellipsoid / z xy any point of on the coordinate planes. . Draw off to the
ellipsoid afi Tl a plane which cuts z . where x . equal segments y m / zYa cut off. z . on the
coordinate axes. . Prove that the tangent planes to the surface jq/z a tetrahedron of constant
volume with the planes of the form coordinates. . Prove that the normal revolution z at the of
surface of rotation. are mutually ortho gonal. . which are the coordinate surfaces of the
spherical coordinates r. Show that the cone and the sphere are tangent at the points . b. .c.
Surfaces are called orthogonal if they intersect at right angles at each point of the line of their
intersection. r f . s . y xiany plane. and z yianty cone.
Jy /tfX if . The desired increment may be found by applying formula calculate the successive
partial derivatives and their values at . .. y rf / x. . l xx /. is called Maclaunns formula. Let a
function to the Functions of Several Variables Sec. y . y when passing from the values to the
values . the given point x f . . A Example. . Similar formulas hold for functions of three and a
larger number of variables. The particular case of formula . . ..il. .. byb where In other
notation.. . y First Solution.. Taylors Formula for a Function of Several Variables C/i. . .. Then
Taylors formula will hold in the neighbourhood under consideration up a fy a. of all orders f x.
/. when a b Q. f . or j df x. f f .//. y have continuous partial derivatives rclth inclusive in the
neighbourhood of a point a. b. Find the increment obtained by the function f x.
up to terms of order three inclusive . / to fx. Putting these results /i /z k. z Jcfy xy yz of zl by
Taylors formula in f x /. . y x xy y by Taylors formula in the neighbourhood of the point /y . t
the neighbourhood of the point .f fc y /i G /z fc. .Sec. tegral Expand of ft. up to terms of order
two inclusive /. y k in a series of positive integral powers of if .Y.y . Find the increment
received by the function xy when passing from the values xl. and x if / x. Expand the function
f x. // in a Maclaurins series up to terms of the third order inclusive sin//. Expand the following
function of order four inclusive / x. /ife f . fr. z / in a series positive in powers k. y. Taylors
Formula for a Function of Several Variables are All subsequent derivatives into formula . / / .
//J /. // Maclaurins series up to terms cos x cos y. Expand h and f x i fe ft. Expand the
following function in a Taylors series in the neighbourhood of the point . . Expand the function
. y. . . . Expand the following function /. Expand the following function in a Taylors series in
the neighbourhood of the point . we obtain identically zero. fx. in a . z // z jr// Ae yz.
Derive approximate formulas accurate to second order terms in a and P for the expressions
Functions of Several Variables Ch. . ft. then there no extremum P gt a t ft. fy a. xz b .
Definition of an extremum of a function. . Using Taylors formulas up to secondorder a terms.
and extremum at a minimum. . or df a. an implicit function of x and y defined by the y . //. y.
Then a if df a b lt for dx z dygtQ then /a. at the point if A . b be a stationary point of the
function fx. ft requires further investigation. . Let P a. P a. Sufficient conditions for an
extremum. Cgt. //. ft changes sign. f t/ x t yQ extremum. We form the Then I if Pa. We say
that a function fx. The Extremum of a Function of Several Variables . . In similar fashion we
define the extremum of a function of three or more t variables. The generic term for
maximum and minimum of a function is extremum. . ft. The case of a function of many
variables. c if d /a. b ft and Af xx a. then the if function or is has an maximum. System I is
equivalent to a single . / . that is. b lt f x y is fulfilled. b is not an extremum of /v. df a. y
different from P in a sufficiently small neighbourhood of P the inequality /a. which takes on
the value z for x equation and yl. . remains open if . ftgt for d . is O. ft . Write several terms of
the expansion of the function z in increasing powers of the differences and y . b if d z fa. then
the question which is to say. .di/ then /a. namely or A a gt . f a. b does not exist. Necessary
conditions for an extremum. z z x Sec. the function f x. the necessary conditions for the
existence of an extremum of it an extremum A. t necessary conditions for an t gt The
foregoing conditions are equivalent to the following let ft. b at the point P a.y has a maximum
minimum f a. b is the maximum of the function fx. amp. In the general case. y or. at the point
of the extremum equation. dfx. ft. Bf xy a. x . or df a. y may attain an extremum socalled
stationary points are found by solving the following system of equations t x. C /ci. if A lt Alt C
lt . approximate /T. then f a. if for all points P x. The points at which a differentiate function f
x. /a. b gt fx. of the function at P a. accordingly. y. b. For a function of three or more
variables.. b is the minimum of the function . if a and p are small compared with unity. b.
Cg p. . dy J dxdyjp. This minimum function for A .. Let us find tiie Pt d z adx P. B. A . .
generally speaking. coupling equation. b. At P of is equal the value the i For the point P no
extremum. X. In the simplest case. Example . There gt A lt . dxdy ry for r/. . there is no
extremum at the point P.v. C .. Test the following function for an extremum Solution. y A to gt
. . dz Tdz T P . To find given the relationship qgt A. and c . dy and form the discriminant A/C
B l each stationary point. Find the partial derivatives and form a system of equations or r xy .
/ Solving the system we get four stationary points . on //. B . For the point P . At the point . C
.l. FA. B fL . For the pomt P t A g . P. AC lt dx Jp . . is a maximum that its attained ltjr. For the
point P . A fi lt . P the function has a maximum or equal to ma x f Conditional extremum. . the
function has a minimum. The necessary conditions for the to a system of three equations
extremum reduce with three unknowns to x. this function which is related by the equation the
conditional extremum of a funcwe form the socalled Lagraltige i/ of minimum are function yf
multiplier. /I gt . C. from which it is. while the sufficient conditions are analogous to the
conditions a.Sec. Thus. The Extremum of a Function of Several Variables are similar to
conditions . .. t/. possible determine these unknowns.l. c x second derivatives c .. the
condition arguments w tion /A. and where X is an undetermined we seek the ordinary
extremum of this auxiliary function.. . the conditional extremum // of a function /A.
if dF conditional minimum. and a Namely. The question of the existence and character of a
conditional extremum is solved on the basis of a study of the sign of the second differential of
the Lagrange function yi for the given Functions of Several Variables Ch. As a particular
case. y. dx dxdy y dy system of values of x. the problem reduces to finding the greatest and .
then at this point there is a conditional maximum of the function / x. if d F . if A or C . if A or C
In similar fashion we find the conditional extremum of a function of three or more variables
provided there is one or several coupling equations the number of which. The necessary
conditions yield the following system of equations i Solving this system we find i quotquot
and X quot quotquotquot . y at a stationary point is positive. Geometrically. must be less than
the number of the variables Here. if the discriminant A of the function F x. X. h obtained from
or the condition that dx and dy are related by the equation Q. and a conditional minimum. We
have T gtjr. the function / x y has a conditional maximum.v plane z y for points of its
intersection with the cylinder ji f// l We form the Lagrange function least values of the
ecoordinate of the Fx. Find the extremum of the function t lt gt gt gt lt lt z y provided the
variables x and y satisfy the equation xy Solution. y xf/lM / . Example . we have to introduce
into the Lagrange function as many undetermined multipliers factors as there are coupling
equations. y. f Since dx it quot dxdy follows that . however.
the function at this point has a conditional minimum at this point. Greatest and smallest
values of a function. A function that is diffe Example . K Z . n ne straight line jc we could test
the Function for a conditional extremum without reducing to a function of one argument. and
the problem reduces to seeking the When A greatest and smallest values of this function of
one argument on the interval //. of and we get the point the function ZM M It .Sec. and .
consequently. Similarly. we find l that g r x at the point . 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 region.. consequently. // At A the value to test for . l When // we point get z x z x. z .r at
the point J . Similarly that . Determine the greatest and function smallest values of the in the
region Alt. sm v Investigating. y x . Correlating all the values obtained of the function z. . x yz
tri Solution. If and. x the function . z max i ll. at the point . Zsm y Tquot at thc A P oint V /
When we find xy z ampm x or // we will have z A . we conclude at the stationary that z gr at
the points . we have / ff/. Thus. Let us find the stationary points I z K J xy . The indicated
region is a angle Fig. /lt..and f/ O D and. whence x . fAj. . . x The Extremutn . .jfo. maximum.
o of a Function of Several Variables o and / thend / gt. has a conditional then d F ltQ. we find
that g r v at the . is an cxtrcmum not absolutely necessary Let us investigate the function on
the boundaries of the region. z sm point M. gr metres coincides with z gr x o anc r. .
. of three variables z f/ xy x gt. . x y. Test for . r f/ for for . z x xy tf y xyxgt z x y xxy y z x z . .
cos cos W y z u xtfz A A for// for A . x . . ygt. . zl . z . a x . uxyz provided xj/e. y z x if A . of the
following functions Determine the conditional extrema . zx . Find the extrema tions of the
following implicitly represented func . z xy for for for a f/ . xyyzzx. . z maximum and minimum
the following functions of two variables . x I y. zgt. of the following functions z. Functions of
Several Variables Ch. z . yx y z z . Prove the inequality if xzQ Hint Seek the maximum of the
function u xyz provided . .vfy z jcgt.. x y. x y z xy . z . Find the extrema . .
x yl b zlx y . Determine the greatest and smallest values of the func tions a z y in the region
xy l. we will have the unique stationary point TTI T We or ne of j the triangle. the function / x.
Determine the greatest value of the function the regions a xgt. a x We seek the maxi//. . xy
and b z . yzQ. Let us test the sufficiency conditions. function ya / x. Let the desired numbers
be x. It is required to break up a positive number a into three nonnegative numbers so that
their product should be the greatest possible. yy gt / . y is considered inside a Fig. Finding
the Greatest and Smallest Values of Functions Example t. Solution. Solving the system of
equations fx. yaxy. IXIf / .Sec. y xya to the problem. the closed triangle xO.sin x y in the
region Ojt. have fquot x. Determine the greatest and smallest values of the func x tion z xy in
the region y x. f x. Finding the Greatest and Smallest Values of Functions in . Fig. Determine
the greatest and smallest values of the func tion z sinxh sin yf. . y. x of mum According x y. .
Sec. y.
Find a triangle of a given perimeter p. x ya x // . the sum of . find the one whose total surface
is the least. P x z igt w ith masses m lf m m For what position of the point Px. Ch.P.
generates a solid of greatest volume. Draw a plane through the point c to form a. Determine
the outer dimensions of an open box with a given wall thickness and capacity internal V so
that the smallest quantity of material is used to make it. y. Functions of Several Variables so
at fg. by the methods of a conditional function u xyz on the condition . Find a point x. x .
.plane. / on the contour of the triangle. . Since fx. b a tetrahedron of least volume with the
planes of the coordinates. by seeking the maximum y z a. a greatest value is equal to . x lf .
Given in a plane are three material points P. x yl is the least possible. f/ . . M t . on an x.
which. . Find a rectangular parallelepiped of a given surface S with greatest volume. Inscribe
in an ellipsoid a rectangular parallelepiped of greatest volume. Consequently. find the one
that has the greatest area. For what dimensions does an open rectangular bathtub of a given
capacity V have the smallest surface .e. . if which the to say that the product will be greatest..
. the squares of the distances of which from three straight lines Q. From among all
rectangular parallelepipeds with a given volume V. Of all triangles of a given perimeter p. m
m M t . this is maximum will be the greatest value. the sum m. . y will the quadratic moment
the moment of inertia of the given system of points relative to the point P P P P P be the
least i. Represent a positive number a in the form of a product of four positive factors which
have the least possible sum. and Note that x The ploblem can also be solved of the
extremum. upon being revolved about one of its sides. And the function reaches a
maximum.P .
The velocity of Fiilt. . . heat Determine how to divide the current released in unit time is / into
. according to which a light ray is propagated along a line AMD which requires the least time
to cover. a f xy tf . resistance /.Sec. proportional to / then the quantity of /. Inscribe in surface.
v Applying the Fermat principle. a given sphere cylinder having the greatest total . Using the
Fermat principle. . y mately represent a parabola y x and a straight line x It is required to
connect these rivers by a straight canal of least length. If a current / Hows in an electric
circuit containing a . The beds of two rivers in a certain region approxi . The points A and B
are situated in different optical media separated by a straight line Fig. . . derive the law of
reflection of a light ray from a plane in a homogeneous medium Fig. Fig. At what point of the
ellipse r Ta TT b w z does the tangent line to angle of smallest area . light in the first medium
is v l in the second. Through what points will it pass . Find the axes it form with the
coordinate axes a the ellipse tri of x . derive the law of refraction of light rays. Finding the
Greatest and Smallest Values of Functions . Find the shortest distance from to the straight
line x the point Ml. y .
/. Agt. . When solving the problems of this section it is always necessary to draw the curves.
then then then M M M is is an isolated point Fig. . Definition of a singular point. b A . at once
A its is called a singular point if point Af x c y Q of a plane curve coordinates satisfy three
equations . . . t/ . yzzax Let us find the partial derivatives and equate them to zero . z x has a
node if a ax . if if a node double point Fig. by means the whose resistances are least i Rgt
possible a so that generation of heat would be the Sec. Basic types of singular points.
currents / t Functions of Several Variables of three wires. . or a tacnode Fig. Here. a cusp of
the first kind if a z Solution. the Fig. Singular Points of Plane Curves . lt gt xy f x x. let the
second derivatives be not then a all equal to zero and A /CB . fx. Show that the curve y
isolated point if a . Alt. Ch. . an Example . . or an isolated point. . Fig. f x. . . or of c if second
kind Fig. At a singular point M . is either a cusp of the first kind Fig.
. y which is a node Fig. if if if agt. a fagt Pig. . Let us find the second derivatives and their
values at the point . a. Fig. when JcO. . a lt . Singular Points of Plane Curves This system
has two solutions . Hence. Fig. there is a unique singular point Fig. first kind Fig.Sec. AjO.
the point M symmetric is a cusp of the . of the given curve. Hence. the curve is Hence. and
the point is an isolated point Fig. . and The equation of the curve in this case will be exists
only is a tangent. Fig. . and the .btit the coordinates of the point N do not satisfy equation .
Fig. or then then then Alt A gt / x Y about the xaxis. N a t Oj.
Functions of Several Variables Ch. y . . Determine the character of the singular points of the
follo wing curves . x . and at each point is tangent to some line of the given family. . ya x .
Find the envelope of the family of curves cosaf/sina p p const. cissoid. x y . Envelope The
envelope of a family of plane curves curve or a set of several curves which is tangent to all
lines of the given family. . of the a. x x x ayaVx. ftV agt. a xy . a . It should be pointed out that
the formally obtained curve the gtocalled quotdiscriminant curvequot may contain. which
locus ts not part of the envelope of this family. then the parafrom the system of equations
dependent on metric equations a single variable parameter of the latter are found t fx.
Equations of an envelope. . Consider three cases agt. Eliminating the parameter a from the
system the form we get an equation of Dx. y. jtyjt . of Descartes. alt. Example. . Sec. a xx
strophoid. . When solving the problems of this section it is advisable to make drawings. gt x
yx a jc y axy Q folium z f/O. pgt. y. . a a has an envelope. . If a family of curves is t. Definition
of an envelope. a locus of singular points of the given f amily. a fx t y. a o. conchoid. a amp.
Determine the change in character of the singular point curve y x b x ax c depending on the
values of caltftltc are real. y . ax y lemniscate. . in addition to an envelope if there is one. .
Envelope Solution. a Squaring both equations and adding. .Sec. Find the envelope of a
family of straight lines that form with the coordinate axes a triangle of constant area S. Find
the envelope of ellipses of constant area S whose axes of symmetry coincide. . the system of
equations Form J cosa y sin a x sin p . ay t cos a . Find a curve which forms an envelope of a
section of length / when its endpoints slide along the coordinate axes. . . The given family of
curves depends on the parameter a. we eliminate the parameter frig. . Thus. This particular
family of straight lines is a family of tangent lines to this circle Fig. . Find the envelope of the
family of straight lines k is a variable parameter. Find the envelope of the family of circles .
the envelope of this family of straight lines is a circle of radius p with centre at the origin. Find
the envelope of a family of circles of the same radius R whose centres lie on the jcaxis. of the
Solving the envelope system for x and y we obtain parametric equations r/ x pcosa. .
Sec. are parametric equations of the space curve. Investigate the character of of families of
the following lines C is the quotdiscriminant curvesquot a constant parameter cubic
parabolas y x C b semicubical parabolas t/ x C x c Neile parabolas y a . df . find the envelope
of all trajectories bf the shell located in one and the same vertical plane quotsafety parabola
Fig. then the arc length of a section of it from t t l to t t is /a f j / ltM dy dz y dr . y. C. Functions
of Several Variables Ch. Arc Length of a Space Curve The dinates is differential of equal to
an arc of a space curve in rectangular Cartesian coor where If x. . . . M . The equation of the
with initial velocity point resistance disregarded is air trajectory of a shell fired from a v at an
angle a to the horizon Taking the angle a as the parameter. z are the current coordinates of a
point of the curve. . d strophoids a x y C a x. Fig. .
/. f . of motion between times f l and . . to M . The vector a may be defined by specifying
three scalar functions ax t a t and a z t which are its projections on the coordinate axes
function a y t The derivative of the vector function aat with respect to the scalar argument t is
a new vector function defined by the equality da The modulus at Matda x t . z . j/ . The
derivative of the vector function of a scalar argument. . . here. . from / to t .Sec. it. x . aarcsin.
. f f/ to the point M . Up . day . f . The position of a point for any time by the equations fgt is
defined Find the mean velocity Sec. x t. which is called the hodograph of the vector r.from to
cos y e sin z e from to arbitrary t from JC to x y /. . z ln j from the point . cos y sin z . jcy from
the point . da f t of the derivative of the vector function is da dt The endpoint the curve of the
variable of the radius vector rr/ describes in space rxtlytJt. The Vector Function of a Scalar
Argument . y/ /.. . The derivative is a vector. tangent to the hodograph at the corre sponding
point. The Vector Function of a Scalar Argument In Problems find the arc length of the curve
. For example. . dr ds dt dt where s is the arc length of the hodograph reckoned from some
initial point. t. / t t A /. .
then jT /. w ma m. the velocity and acceleration. cpaa p . tf is the velocity vector of the
extremity is ne acceleration vector of the and JTS quotT of the vector r. of Example .. is Ch..
extremity . where qgt is a scalar function of . of the vector r.. we find the velocity and the
acceleration The magnitude of the velocity is We note that the acceleration is constant and is
dt L / iO. motion. where m is a constant scalar. differentiating.. lt. If Functions of Several
Variables the parameter t is the time. of tjtk. we find that the trajectory of motion is a straight
x u z quotquot From equation . The radius vector time defined by the equation a moving point
at any instant of ri Determine the trajectory Solution. . From we have Eliminating the line time
t. a. Basic rules for differentiating the vector function of a scalar argument. if a const.
the vectors a and b are perpendicular to each other. is the time. and c are constant vectors.
with respect to the parameter t. . the velocity and the accelWhat are the magnitudes of
velocity and acceleration and / and what directions have they for time y . /. ft. where a and o
are constants and / is the time. The equation of motion is eration. of the volume of a
parallelepiped constructed on three vectors where a i tjt z k. The equation of motion is r i cos
a cos at jsmacostot k sincof. r. d a cos f b sin r a cosh f sinh t t / a. Find the derivative
vectorfunction of the function ata /. where at is a scalar function. The equation r of motion is
/cos/j/sinf.Sec. and c. the and the acceleration. the equation of a straight line. . Determine the
trajectory of motion and the magnitudes and directions o the velocity and the acceleration.
Determine which lines are hodographs of the following vector functions a b at f c a/ r r c r ft/.
while a/ at is a unit vector. . in length and in direction general case. is /. Interpret
geometrically the results obtained. Using the rules of differentiating a vector functisn with
respect to a scalar argument. rr. derive a formula for differentiating a mixed product of three
vector functions a. / . l The Vector Function of a Scalar Argument . . Construct the trajectory
of motion velocity and the vectors of velocity and acceleration for times. in direction only.
Determine the trajectory of motion. . Show that the vector where r and r are radius vectors of
/ equation two given points. where / . Determine the trajectory of motion. for cases when the
vector at varies in length only. . . Find the derivative. .
which perpendicular to the first . . M . of then the accelerascrew being screwed . a is the
radius of the screw. Determine the velocity of the point. . Find the velocity of a point on the
circumference of a wheel of radius a rotating with constant angular velocity co so that its
centre moves in a straight line with constant velocity V Q . . The equation of motion is of a
shell neglecting of rt it Ch. sistance where V Q V OX v oy v oz is the initial velocity. . Prove
that if a point is in motion along the parabola . z in such a manner that the projection velocity
on the xaxis remains constant tion remains constant as well. all At the intersection we obtain
three straight lines. Functions of Several Variables . air re z/ . normal plane MMM which
perpendicular to is the vector at and rectifying plane MMjAIj.c. A point lying on the thread into
a beam describes the spiral a where is the turning angle of the screw. Find the velocity and
the acceleration at any instant of time. const. two planes. . and h is the height of rise in a
rotation of one radian. vector of the principal normal The corresponding unit vectors T o JV .
containing the vectors is TT r . The Natural Trihedron of a Space Curve At any nonsingular
point dicular planes Fig. Sec. z of a space curve r of is pos sible to construct a natural
trihedron consisting three mutually perpenand osculating plane M AjA S. the tangent the
principal normal MAf of which are defined by the appropriate vectors MM the binomial MM t
Trr the vector of the tangent lme NBXT the gfxp the vect r f tne binormal. //.
Sec.
The Natural Trihedron
of
a Space Curve
may
be computed from the formulas
t dF
f
dr
v
ds
dt
ds
If X K, Z are the current coordinates of the point of the tangent, then the equations of the
tangent have the form
Xx Y
C
y
if
Z
z
I
Normal
Rectifying
plane
plane
Osculating plane
Fig.
where
of
Tx
Tv
,
T
from the condition
of the
of
perpendicularity
the line and the plane
we
get an equation
normal plane
Z
If
.
,
Wy,
and , we replace get the equations of the binomial and the principal normal and,
respectively, the osculating plane and the rectifying plane. Example t. Find the basic unit
vectors T, v and p of the curve
in
equations
T x Ty by B x
,
,
Bv
,
B z and A.
V,
we
at the point
i
.
of
Write the equations
mal
at this point.
the tangent, the principal normal and the binor
Solution.
We
have
and
Functions of Several Variables
Ch.
Whence, when
,
we
get
dt
dt
X
dt
i
J
gt
Consequently,
T
lt/A
P
/ay
it
V
U/
Since for
we have , yl, ,
quotquot
follows that
quot
are the equations of the tangent,
xy
are the equations of the binomial and
lz
l
y
z
are the equations of the principal normal. If a space curve is represented as an intersection
of
FJC, y, z
two surfaces
,
and
Gx,
TTZ
y,
,
the vectors drdx, dy, dz
then in place
of
the vectors
we can take
and d r dx, dy, d z z and one of the variables x, y, z may be considered independent and we
can put its second differential equal to zero. Example . Write the equation of the osculating
plane of the circle
at its point
z,
J/
x
y zQ
Ml,
,
.
variable,
Solution. Differentiating the system and considering x an independent we will have
x dx
y dy z dz
,
and
dx
dy y dy dz z dz . ,
d/
Putting
xl, ygt
z,
we
get
Sec.
The Natural Trihedron of a Space Curve
is
Hence, the osculating plane
dx,
defined by the vectors
dx
t
and
o,
,
yd
,
,
jdx
or
,
,
and
.
is
Whence the normal vector of the osculating plane
B
and, therefore,
its
J
equation
is
lj
k
lxl
that
is,
as
it
should be, since our curve
is
located in this plane.
T, v,
. Find the basic unit vectors
p of the curve
z
xl
at the point
f
cosf,
ysin/,
t
g
normal
. Find the unit vectors of the tangent and the of the conic spiral
principal
at an arbitrary point. Determine the angles that these lines with the zaxis. . Find the basic unit
vectors r, v, p of the curve
. at the point x . For the screw line
make
y
x,
z
x
z
y
asmt,
bt
write the equations of the straight lines that form a natural trihedron at an arbitrary point of
the line. Determine the direction cosines of the tangent line and the principal normal. . Write
the equations of the planes that form the natural trihedron of the curve
x
/
,
x
if
z
at
points Ml, , . the equations ot the tangent line, the normal z t and the osculating plane of the
curve /, y t, plane at the point , , .
of
its
one
.
Form
M
Functions of Several Variables
Ch.
.
Form
and binormal
at
the equations of the tangent, principal normal, an arbitrary point of the curve
Find the points at which the tangent to to the plane x . y z
this curve is parallel
. Form equations of the tangent, the osculating the principal normal and the binormal of the
curve
plane,
at the point / at this point.
.
Compute the
direction cosines of the binormal
. Write the equations of the tangent and plane to the following curves
a
the
normal
x R cos
/,
y R sin /cos/,
z
Rsmt
for
/
b
C
zxy, x y
yz,
xz
at the point ,, at the point ,
/,
.
zx
normal Find the equation y x at the coordinate origin. . Find the equation of the osculating
plane
of
the
plane to the curve
to the
z
if,
curve
,
/
ltgt,
. Find
a
. at the point / the equations of the osculating plane to the curves
ty
b
JC
c
y , x y at the point , , y, x z at the point , , z a y fz at any point of the curve x
,
QJ
y ot z
.
the equations of the osculating plane, the principal normal and the binormal to the curve
.
Form
z
y
.
x,
z
at the point ,
,
.
normal and the binormal to the conical screwline A /COS/, j//sin/, zbt at the origin. Find the
unit vectors of the tangent, the principal normal, and the binormal at the origin.
pal
Sec. . Curvature and Torsion of a Space Curve
Form
the equations of the osculating plane, the princi
. Curvature.
By
the curvature of a curve at a point
M
we mean the
number
/
R
lim ASO
JL, As
f
Sec.
Curvature and Torsion
is
of
a Space Curve
where
the angle of turn of the tangent line angle of continence on a curve MN, As is the arc length
of this segment of the curve. segment R is called the radius of curvature. If a curve is defined
by the equation rrs, where s is the arc length, then
p
of the
For the case
of a general
parametric representation of the curve
we have
. Torsion. By mean the number
torsion second curvature of a curve
at
a
point
M
we
rlihn Aso As
Q
the angle of turn of the binormal angle of contingence of the where second kind on the
segment of the curve MN. The quantity Q is called the radius of torsion or the radius of
second curvature. If rrs, then
is
drd rd sr
dp
ds ds ds
ds
where the minus sign
is
taken when the vectors
sign,
is
and v have
the
same
direction, and the plus If ///, where t
when
not the same.
an arbitrary parameter, then
dr d r dV d/dquot d
Q
dt
d,
Example
.
Find the curvature and the torsion of the screwline
r
i a cos
t
j
a sin
/
k bt
a
gt .
t
Solution.
We
have
/a
sin
t
ja cos t kb
/
dr
/
a cos
/a
sin
/,
/ a sin/
Whence
J
a sin a cos
t
Ja
k
cos/.
a cost b
a sin
/
i
ab sin
/
jab cos
t
ak
/
Hence, on the basis
of
formulas
and
,
we
a
get
JDL
fl
R
and
VaTb
i
i
ab
a
/.
quotfl
a
Q
aa
bab
b
Thus, for a screwline, the curvature and torsion are constants. Frenet formulas
dr
v
v
s
T
iP
P
v
squotR
RTQ
dsoquot
zero, then the line
the curvature at all points of a line is a straight line. . Prove that if the torsion at all points of a
curve is zero, then the curve is a plane curve. . Prove that the curve
that
if
. Prove
is
xt t
is
,
y
t t zlt
a
.
a
plane curve find the plane in which it lies. Compute the curvature of the following curves
bx
x cost, y smt, z l, y //
z
xzQ
cosh
/
at the point / at the point , ,
.
. curves
a
je
Compute the curvature and
i
torsion at any point of the
b
y e sint, z xacosht, y asiuht.
ecos/,
e
z
at
hyperbolic screwline.
. Find the radii of curvature and torsion at an arbitrary point x, y, z of the curves
a
xbx
. Prove that the tangential and normal are expressed by the formulas acceleration v dv VOif
T ltMJ V V,
w
components
of
WTT,
Wv
R
where v is the velocity, R is the radius of curvature of the T and v are unit vectors of the
tangent and principal trajectory, to the curve. normal
Sec.
A
Curvature and Torsion of a Space Curve
is
t
.
point
sin
in
la cost
ja
btk
of
uniform motion along a screwline r with velocity v. Compute its accelera
S
tion w. .
The equation
motion
is
Determine,
at
times
/
trajectory and acceleration.
the
and / the curvature of the tangential and normal components of the
Chapter VII
MULTIPLE AND LINE INTEGRALS
Sec. . The Double Integral
in
Rectangular Coordinates
. 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,
ydx dy
lim max Ai max Ar// c
where A
values of
ampy k ykl yk and the sum is extended over those xg, and k for which the points /, y k belong
to S. . Setting up the limits of integration in a double integral. We distinguish two basic types
of region of integration.
t
Xf l
i
x,
x
Fig.
o
Fig.
The region
of integration
Fig.
is
bounded on the
left
and
right
while the vaFig. . In the region S, the variable x varies from x l to x riable y for x constant
varies from , x to j/ qgt x. The integral magt
Sec.
The Double Integral
in Rectangular Coordinates
be computed by reducing to an iterated integral by the formula
a
ltPa X
J
S
fx, ydy,
ltPaUgt
where x
is
held constant
when
calculating
/x, y dy.
The region of integration S is bounded from below and from above by the straight lines y y
and y yty z gtyi, and from the left and the y, right by the continuous curves x gt y AB and x y
CD t y each of which intersects the parallel y Y y Y y t at only one point
l
l
lt
i
Fig. . As before,
we have
Vl
U
JJ/.
S
ydxdydy
j/i
i,
J
/
f x.
ydx
t
here, in the integral
fx, y dx
we
consider y constant.
If the region of integration does not belong to any of the abovediscussed types, then an
attempt is made to break it up into parts, each of which does belong to one of these two
types. Example . Evaluate the integral
/
D
Solution.
Example
.
Determine the limits
of integration of the integral
x,
S
ydxdy
if
the region of integration Fig. is bounded by the hyperbola y and x we have in view the region
constraight lines taining the coordinate origin. Solution. The region of integration ABCD Fig.
is bounded by the and x and by two branches of the hyperbola straight lines
and by two
Multiple and Line Integrals
Ch.
x
yyT
that
is,
it
and
/
have
belongs to the
first
type.
We
J
dx
J
fx, ydy.
Evaluate the following iterated integrals
J
dcp
rrfr.
.
dyjf ydx.
.
Jdy
pTv
X dU
.
o
j
a
sin
ltp
Jt
COS
p
Write the equations of curves bounding regions over which the following dduble integrals are
extended, and draw these regions
t/
X
.
Jrf/
J
fx, ydx.
.
JdxJ/. ydy.
X
KJC
.
JdxJ IX
fx, ydy.
.
y
dx
fx, ydy.
X
.
o
dy
y
fx, dx.
.
d
ix
fx, ydy.
Set up the limits of integration in one order and then in the other in the double integral
JJ/. ydxdy
S
for the indicated regions S.
S is a x the region containing the origin is meant. with centre at the point and f . . . . S is a
whose circular sector arc endpoints are A . S is a circular ring bounded by circles with radii rl
and / and with common centre . . Set up the limits of integration in the double // integral x. . .
Fig. . . . A . The Double Integral in Rectangular Coordinates . .. JdJ/x. . Dl. . .. . . . . C. . is is a
triangle a with vertices . trapezoid with vertices . . a. Fig. e //ltxlty lt/ a. . . Change the order X
of integration in the following double integrals ydy.Sec. S S S is a rectangle with vertices . . .
S ydxdy if the region S is defined by the inequalities b c h// lta . . C. . . . . . . . parallelogram
with vertices . . fx. . ydy. S is a . Fig right parabolic segment bounded by the and a segment
of the straight line connecting parabola the points l. C. and . x // . . S is bounded by the
hyperbola if x and the circle .
. S A. dy /. ydy. J o dJ/x. .X Fig. . yi . S where the region of integration Sis bounded and by
the arc radius by the straight line passing through the points A . o sin t/dy fx. xdxdy. . A. ydy. .
fi. VTZxtf V . dx a fx. dy fx t ydx. Jdx J fx ydy. and Fig. Fig. Multiple and Line Integrals Ch. . .
x . . . ydy. of a circle with centre at the point C. . . Evaluate the following double integrals
where S is a triangle with vertices . RVT A V . xdxdy t . and B. dx J/jc. /d/. d J fx.
JJ S r Va . / bounded by Fig. Compute the ifcosjc integrals and draw the regions over which
they rt a j dx j tfsmxdy. . . ffTi. extended over the region S. and fll. . . .. S lying in the t
quadrant. parabola x is and the straight lines x Q. where S a parabolic segment the parabola
. Evaluate the double integral xydxdy. which and an upper semicircle x is bounded . where S
is a triangle with vertices . . is a . a region bounded by the parabola y px and the straight .
Evaluate the double integral it is abvisable to make if S is line x p. .Sec. triangle with vertices
V y dx dy where S . by the axis . . j /quot S ydxdy. rr ey J J S . A I. dxdy. . quot b COS When
solving Problems to the drawings first. The Double Integral in Rectangular Coordinates . and
fll. extend at yf and the straight line y x. c ltty / / x sin . where S xy is a part of a circle of
radius r first a with centre at . . where S y is a curvilinear triangle OAB bound ed by the .
. Find the mean value of the square of the distance of from the coordia point x. Evaluate the
double integral S in which the region of integration nate axes and an arc of the astroid S is
bounded by the coordi . . fx. r sin cp r dr e/cp. y to polar coordinates r. The mean value of a
function Hint. S where the region S of the cycloid is bounded by the axis x of abscissas and
an Rt y R sin/. In a double integral. we have the formula f gt S ydxdy S cos qgt. yxy in in the
region SJOJtl. ltyltl. M afy R Sec. Change of Variables in a Double Integral . y the region is
the number . cp. Evaluate the double integral dxdy / S f where S the area of a circle of radius
a. Double integral in polar coordinates. . Find the mean value of the function fx. ydxdy.
Evaluate the double integral is which first circle is tan quadrant. when passing from
rectangular coordinates x. y of the circle x nate origin. cos/. which are connected with
rectangular coordinates by the relations /Cosp. Multiple and Line Integrals Ch. . to the
coordinate axes and lies in the gent . r y r sin ip.
v//plane if the Jacobian and the points of some w V . Double if integral in curvilinear
coordinates. continuous correspondence between the points region of the S f of the UV.
ydxdy from the variables x. v. the region of integration does not belong to one of the kinds
that has been examined. cos we x if Yquot r cp r sin cp r . In evaluating the integral F p. in
both directions.plane. I/ tK. y to the variables u. in the double integral x. rrrfr we hold If the
quantity p constant. P /J ltltp C f S F q. where r l qgt r. cp F cp. y dx dy du da dx dy dv dv Du
retains a constant t sign in the region . which it is required to pass are connected with x. r r dr
dcp dcp f r. v that establish a onetoone and. evaluate where the region is a circle of radius // /
with centre obtain at the coordinate origin Fig . In the more general case. v / du du holds true
The limits of the new integral are determined from general rules on the basis of the type of
region S Example . rsincp. r/rcosp. i. and region S of the . w. r sin p. Putting x J/l rcoscp. it is
broken up into parts. then the formula f y dx dy I cp u. . a where F cp.Sec. each of which is a
region of a given type. In passing to polar coordinates. r r rfr. v. rpaltP a integration S is
bounded by the halflines r the curves r and r r p. Solution. y by the continuous and
differentiate relationships Acpw.cp r z fltP r i fP r Pl are singlevalued functions on the interval
then the double integral may be evaluated by the formula If region of Change of Variables in
a Double Integral the r cp i ltp and and and arpp.
gral S calculate the double inte where S is a semicircle of diameter a with centre at the poin
Cf. it follows that varies from S Jt varies from any qgt. . . . Jd//J . y x. Passing to polar
coordinates. i is . to for Since the coordinate r in the region to jt. ydy. JJ/x. ydxdy. . ydxdy.
Fig. Multiple and Line Integrals Ch. . is a triangle bounded by the straight lines / . where S S
bounded by the lemniscate Fig. and qgt ff yx coordinates r and cp and set up the limits of to
the new variables in the following Pass to integration integrals polar with respect . fx. K. /
where S lt/.
r . evaluate where the region of a V ax J Jdx Vx tfdy. Transform c p t dxf x ydy o ax ltaltp and
cgt by introducing new variables u . . Evaluate the double integral ll S T/R. ltp over a region
bounded by the cardioid r a coscp and the a. Evaluate the double integral of a function fr.
Passing to polar coordinates. circle r . x x j extended over the region S bounded by the
ellipse passing to generalized polar coordinates X u r wo Y . Passing to polar
coordinates.Sec. evaluate S where the region S is a loop of the lemniscate . y . . Passing to
polar coordinates. Passing to polar coordinates. This is a region that does not contain a pole.
evaluate the double integral S extended over a region bounded by the circle jc y ax. Change
of Variables in a Double Integral . evaluate the double in tegral S integration S is a semicircle
of radius a with centre at the coordinate origin and lying above the axis.
/ cp/ qgt. Area in polar coordinates.ydy. Sec. b j dy d. . qgt x y x . Evaluate the double
integral S where S a region bounded by the curve b fi k Hint. Construct regions whose areas
are expressed by the integrals x J ji a dx J j d. Area in rectangular coordinates. If a region S
in polar coordinates r and defined by the inequalities acpp. . Construct regions whose areas
are expressed dy the tegrals arc tan in sec ltp aicoscp a df J o rdr b J T Compute n n a
quotTquot these areas. v xy Ch. then ltF qgt P F / P S ffrdcpdr S C C /dr. . Make the
substitution gt. Evaluate these areas and change the order of integration. S dx a cp J x is . y
br sin cp. . in the integral i dxf x. dy. then X dy. The area of a plane region S is S If the region
S is defined by the inequalities b op axb. is Multiple and Line Integrals Change the variables
u i x y. . a /q . Computing Areas .
. Find the area bounded by the curves r acoscp and r acosqgtagtQ. x. . . Computing Areas
Compute the area bounded by the straight lines x y. V /V // quot quot J quot . Find the area
bounded by the ellipse xy y . and the straight line xy by this axis. x x y x. ltaltp. Compute the
area bounded by the ellipse . Find the area of a curvilinear quadrangle bounded by the arcs
of the curves ifax. Find the area bounded by the .alt. y y new variables u and x u. xy bx. Find
the area bounded by the straight line r cosql circle r. Hint. Passing to polar coordinates. Find
the area bounded by the parabolas f z y x.x. y y y and the . v.Sec . . Compute the area lying
above the xaxis and bounded a. y by y xQlt.b. Introduce the new variables u and u. and put
uy. Hint. and put yvx. ax. and y x . . xy Qlt. find the area bounded by the lines z Q. The area
is not to contain a pole. ltaltb. line . Introduce the . if a. Find the area of a curvilinear
quadrangle bounded by z the arcs of the parabolas x ay x a. the parabola y . if vx. ltaltp. x.
fll. . . is whose volume expressed by the in tfy dy reason geometrically to find the of this
integral. The volume V y. Find the volume of a solid bounded by the elliptical x y. Compute its
volume.. . In Problems to sketch the solid expressed by the given double integral . . . ZX .
and on the sides by a right cylindrical surface. the plane paraboloid z nate planes. . C. .
Multiple and Line Integrals Ch. Set up the limits ol integration. xdy. . value . Sketch the solid
a V a . . xl. and C. quot. Sec. Fig. which cuts out of the ju/plane a region S Fig. is equal to /.
Use a double integral to express the volume of a pyravertices . . Ccmputing Volumes a
cyltndroid bounded above by a continuous surface of low by the phne . be mid wiih . and the
coordix f f/ . A solid is bounded by a hyperbolic paraboloid z x tf and the planes / . e .x tegral f
dx YC . Fig. f J whose volume is dx f J x ydy. . Fig. A.
. In what ratio does the hyperboloid divide the volume of the sphere x . yquotx. . Find the
volumes bounded by the following surfaces . is bounded by the cylinder x y x. the surface ae
ltJ quot gt. Computing the Areas of Surfaces The area o of a smooth singlevalued surface z
on the jci/plane is the region S.if if x / f . . . h In Problems to use polar and generalized polar
coordinates. Find the entire volume enclosed between the cylinder a x y a and the
hyperboloid x f. z a and the Compute its volume. A solid . . y . . xtja.// and the a f. the
paraboloid z . Compute the volume of a solid bounded by the jq/plane t the cyl inder x . . . .
ax. Compute the volume of a solid bounded by the f/plane. . // a . Compute the volume of a
solid bounded by the jq/plane.y a. a. Find the volume bounded by the surfaces a . whose
projection JJ . / a. x if f/ . p a gt p. The volume lying sphere inside the paraboloid is meant. / .
. . .if .t . . y z x y y x //. Find the volume of a solid bounded jc h// a a by the surfaces Sec. Find
the entire volume contained between the cone z a and the hyperboloid x f.Sec. and the
cylinder h . x H a. az . and the cylinder x y / . . Find the volume of a solid bounded by the
paraboloid x H. . J . J/T. . Determine the volume of the ellipsoid . . y x y . . is equal to fx t y. . .
and the cone x f if . . / Computing th Area of Surfaces planes .
Find the area of that part of the surface of the cylin der x y R zO which lies between the
planes z mxand z nxmgtngtQ. Compute the area of that part of the surface ol the ax. Multiple
and Line Integrals Ch. of . If S is a region in an jq/plane occupied by a lamina. the area of
that of the y of the sphere x y z a cut out by the surface .. Find the area of the surface of the
sphere cut out by the opening. Compute ax which cut out part surface of the by the sphere
surface . y. Compute the area of that part of the surface of the ax which lies between the
xyplane and the cylinder x y cone x y z . Find the volume and the area of the surface of the
remaining part of the sphere. . A sphere of radius a is cut by two circular cylinders whose
base diameters are equal to the radius of the sphere and which are tangent to each other
along one of the diameters of the sphere. . cone x y which lies inside the cylinder x z f/ . Find
the area of that part of the plane j which lies between the coordinate planes. Prove that the
areas of the parts of the surfaces of the z az arid x az cut out by the cylinparaboloids x y y
der x tf are of equivalent size. The mass and static moments ot a lamina. An opening with
square base whose side is equal to a is cut out of a sphere of radius a. R . Compute z the
area of that part of the first in the surface of the octant and is Compute x the area of that is
part of of the it y cylinder x y z a . . cone x which is situated y bounded by the plane y za. . y
is the surface density of the lamina at the point x. Compute the area of that part of the
surface of the ax paraboloid y z ax which lies between the cylinder tf and the plane x a. . .
The axis of the opening coincides with the diameter of the sphere. and Q x. . . Applications of
the Double Integral in Mechanics . . Compute the area that part of the helicoid e carctan x X
which lies in the first octant between the cylin ders iy a and Sec. then the mass of the lamina
and its static M .
y dx dy. If the lamina is homogeneous. are is its static moments rela tive to the formulas . y
dx dy. y . The moments lamina relative coordinate axes see . Compute the coordinates of the
centre of gravity of the sin area OmAnO Fig. the centre of gravity of a lamina. y expressed by
the H Q jj x. The Q moments x dy. The coordinates of the centre of gravity of a lamina.
density at any point is equal to the distance of the point from the leg /. /r . y inertia of a lamina
relative to the origin Putting QX. Compute the coordinates of the centre of gravity of an area
bounded by the parabolas // . . y is relative to the x and t/axes are yQ x. and t/axes are. // in
formulas inertia of a plane iigure. equal to J J S is /X The moment of S S S yQ x. M x d
where the mass of the lamina and M x My . by the straight lines x y bounded by the cardioid r
a cosij. S S M Y If Jex. of gravity of a circular sector of radius a with angle at the vertex a
Fig.x f and if x. . then y My i M M J // . we get the geometric moments of . . and . Find the
static moments of the lamina relative to the legs / and OB. . const. . M M is Applications of
the Double Integral in Mechanics moments x and double integrals A MY C x. y . ydxdy.Sec. .
Mx J J dx dy. then Q x. of inertia of to the x If the lamina homogeneous. we can put Ql. . Ij of
a sine curve. Find the coordinates of the centre of gravity of an area . which is bounded by
the curve // and the straight line OA that passes through the coordinate origin and A lamina
has the OA b. then in of inertia a a lamina. respectively. Find the coordinates of the centre .
Find the is mass of a circular lamina of radius R if the proportional to the distance of a point
density and is equal to at the edge of the lamina. from the centre with legs OB a . Compute
the moment of inertia of a triangle bounded relative to the axis. and its shape of a right
triangle and the vertex A .
relative to the xaxis. y. . and b relative to diameter. Compute the moment of inertia of an
area bounded by and the straight line xy the hyperbola xy relative to the straight line x y. Find
the moment of inrtia of the cardioid r al cosltp relative to the pole. The triple integral of the /. .
Multiple and Line Integrals Ch. Compute the moment of inertia of a homogeneous lamina
bounded by one arc of the cycloid xat sin/. . plane . . . Compute the moment of inertia of the
area of the lem niscate r a coscp relative to the axis perpendicular to its in the pole.o gt o f i
fa tgt amp/gt z d Az/c gt o . the density is proportional to the distance from one of its vertices.
Cgtmpute the moment of inertia of the lamina relative to the side that passes through this
vertex. Sec. . Find the ters its moment a d and DdltD of inertia of an annulus with
diamerelative to its centre. y a cos/ and the xaxis. Triple Integrals . Fig. . Compute the
moment of inertia of a relative to the axis passing through its vertex the plane of the square. .
Cmpute the moment of inertia of the parabola if ax by the straight line x straight line // square
with side a perpendicularly to a a segment cut relative to oil the a. lt extended over the region
V is the limit of the corresponding threefold iterated sum function lim x A/j max max max .
Triple integrals in rectangular coordinates. In a square lamina with side a.
w.w. u i where the functions cp. which are connected with .Sec. i. y tyu. V where the region
V is defined by the inequalities Solution. in tha triple integral . ay. //. continuju
correspondence beV oi xf/zspace and the points of of l/l/lspace. z bv the relations x u v. a J V
x dxdy dz J x dA J J . //. z to the variables it is required to pass u.n r a We therefore finally
get V a If . JJy Z dd/dz. a t . .v. We have X lt/ T dy Example . Change of variables in a triple
integral. t in onetoone ind. in tween the points some region V of the region of integration both
directions. . dydz j xS yz dx t a M a JC here S yz is the area of the ellipse cons an S v z Jib I/
vz V lc a V . z Xquot. Triple Integrals o of the Evaluation of a triple integral reduces to th
successive computation ordinary onefold iterated integrals or to the computation double and
one single integral. x are continuous together with their paitial first derivatives. . three of one
Example . dxdydz from the variables AT. Compute /. Evaluate AT // Z j extended over the
volume of the ellipsoid Solution.
the ranges of the spherical tude. z / sin v we have / r cos Example . Fig. w dy dy dy da dv dw
dz dz dz da dv retains a constant sign in the dw use of region V. Fig. we r get / r for spherical
coordinates ap.y. For a sphere.dxdydz f IT w L gt w gt ty u gt y w lt du dv dw. determinant
Jacobian of these functions dx dx dii dx dv dw D u. . latitude. where X rcosrp. // h Fig. gt the
latitude. cp. Solution. cp. rsinrp. v. the radius vector Fig. for cylindrical coordinates r. and r
radius vector will be coordinates fp longi . compute JSJ V where V is a sphere of radius R.
the longitude. /cosi sin q. i cos . where f/ x r cos i. Passing to spherical coordinates. . then we
can make the for mula J V fx. In particular. the functional Multiple and Line Integrals Ch. r ltp
is z//.
. The moments of inertia relative to the coordinate axes are in the for T J J V y Y . . A. we
get the geometric moments the body.y. V of the centre of gravity are f M J M Al I then we can
put Y . U.y. dx dy dz..Sec Triple Integrals We therefore have f f f Vx z y zdxdydz dcp f dty f r
ACOS dr Ji . The volume of a region of threedimen ltn The mass of a solid occupying the
region V is M C f f V Y v z d where x.//. ygt z If the solid is homogeneous.z is the density of
the body at the point . zdxdydz for the indicated regions V. Evaluating triple integrals Set up
the limits of integration in the triple integral J J V fx. mulas for the coordinates of the centre of
gravity. The slal/c moments of the body relative to the coordinate planes are M yy quotV Myz
I quot MZX The coordinates Y A f/ f/ dx dy dz. Applications sional A//zspace is of triple
integrals. V l J Putting of inertia of J J V Y i in nese formulas.
a djt J dy J . fd I rfy JCt/ . . J X J o . V is a volume bounded by the surfaces Compute the
following . . . integrals V . . . . Evaluate d dy dz where V is the region ol integration planes
and the plane xyz. V is a cylinder JC bounded by the surfaces . Evaluate V where spheres V
region x of integration is the y z R and x f z Rz common part of the . p dx dy J o xyzdz. fl . J/ .
/f. V is a cone bounded by the surfaces . Jr yJrZ . Multifile and Lire Integrals Ch. Evaluate r r
r V bounded by the coordinate where V the region paiaboloid of cux y integration is the and
the sphere X common part . . of the y a . V is a tetrahedron bounded by the planes X X .
.. Evaluate where V the region of integration z is bounded by the cone. . first transforming it
to cylindrical coordinates.gee. transforming it to cylindrical coordinates. Evaluate V where V
Xz is the interior of the ellipsoid r quotr IJ . . z and the upper J. R. V where V is a half of the
ellipsoid volume bounded by the plane j. Evaluate di/ J dz. .l. Evaluate Triple Integrals
zdxdydz. // . jc hi/ and the plane h. Evaluate J y z Rz jc j a first . Evaluate R VRX Vflajira J dx
/ first dy J o A /fTJa it transforming to spherical coordinates. evaluate where V jc the surfaces
x is a region bounded by a z and containing the point . . Passing to cylindrical coordinates.
. In a solid which has the shape of a hemisphere . evaluate the integral V where V B. Find
the mass M if of a rec tangular parallelepiped Qxa. the deny. . is the interior of the sphere x y
z Computing volumes by means to of triple integrals of a solid . Compute the volume of that
part of the cylinder x f tf ax which is contained between the paraboloid y az and the xyplane.
Applications of triple integrals mechanics and physics . point x. . y. bounded by the
coordinate planes zO Fig. . Compute the volume of a solid bounded by the xyplane. . a c. z is
equal to the coordinate of the point. . x y sphere cut a solid OABC /. ltzltc. is Fig. z lt/lt. Use a
triple integral bounded by the surfaces compute the volume . and the plane . the density
varies in proportion to the . Passing to spherical coordinates. Compute the volume of a solid
bounded by the sphere x y z a and the cone z x /yquot external to the cone. Multiple and Line
Integrals Ch. Out of an octant of the c z x .gt . C. z z a internal ax and the sphere x the
cylinder x f. the z xyz. sity lx. lt .z to the cylinder. . Find the mass of this body if the density at
each point x. ampltc . at y. Compute the volume of a solid bounded by the paraboloid y y i to
and the plane xa. Compute the volume of a solid bounded by the sphere z x y z and the
paraboloid x ifz internal to the paraboloid.
Sec. a. whose altitude is h and the radius of the base is a. Show that the force of attraction
exerted by a homogeneous sphere on an external material point does not change if the
entire mass of the sphere is concentrated at its centre. In the case of certain imposed on the
functions / . . . f x. Solution. relative to the axis which serves as the diameter of the base of
the cylinder. Improper Integrals Dependent on a Parameter distance of the point from the
centre. and altitude. a. Hence. fa A.v. /i. To Cp.v. . Whence Cp lnp. p gt . Find the moment of
inertia of a circular con base. radius of density Q relative to the diameter of the base.
evaluate gt dx a gt . a and on the corresponda ing improper integrals vc have the Leibniz rule
a dx restrictions . a dx. By differentiating with respect to a parameter. Improper Multiple
Integrals . . Find the force of attraction exerted by a homogeneous cone of altitude h and
vertex angle a in axial crosssection on a material point containing unit mas and located at its
vertex. we put a We have In P CP. Let Then da a find quota in the latter Whence F equation.
Sec. paraboloid // z . . Find the centre of gravity of the solid. a i Example . Differentiation with
respect to a parameter. . a. Improper Integrals Dependent on a Parameter. Find the moment
of inertia of a circular cylinder. Find the centre of gravity of a solid bounded by the x and the
plane x. p Ina Cp.
then the limit on the ripht of is not dependent on the type of regions eliminated from S. For pl
we have . But if pgt . . If region . then we put fx. Let a be a circle of radius origin. then for the
con is necejsary on the right of lo exist at least for the region . otherwise it is divergent. such
regions may be circles of radius with centre at P. If has a limit that does not depend on the
tyre of small regions eliminated from . for instance. where a S signifies that we expand the
region o by an arbitrary law so that any point of should enter it and remain in it. then lim a O
S r lim Q /a the and integral the integral diverges. b. If a function / x. y is continuous in Ch.
vergence of an inirioper integral otherwise it is divergent. and sufficient for the limit one
system of regions o that exhaust is everywhere contina. a function f x. If the inUgrand / . U an
unbounded y dx dy lim quotS f x. /A. If there is a limit on the right and if it does not depend on
the choice of the region o then the corresponding improper inte. with centre at the coordinate
p we have If plt o . the improper integral under consideration If is called convergent. f QC
then lim a o P and converges. y dx dy. except the point . b A discontinuous function. //gt. P
then we put ydxdym fx. where S is a region obtained from S by eliminating a small region of
dia meter e that contains the roint P. Test for convergence dxdy is . where S is the entire
j/plane. // it yQ. is nonnegative f x. of The concept of improper double integrals is readily
extended to the case triple integrals. y uous in a bounded closed region S.ydxdy. Multiple
and Line Integrals triple integrals. Improper double and a An infinite region.v. Solution.
Example . Passing to polar coordinates for Q . W where a is a finite region lying entirely
within S. gral is called convergent .
the integral converges if for pgt . . that is. dz j l/ satisfies the Laplace equation u dx d Q U cy
is . f w . // Fp. . if a /Ol. integral diverges. c // sinp/.Sec. JT Improper Integrals Dependent on
a Parameter Q the / a Cdcpf fljilnlQ r J J r o o . Find / . . b //e. Taking advantage of the
ormula compute the integral xn l nxdx. The Laplace transformation defined by the formula F
p for the function // Find d cos p/. Thus. lim/a Q oo oo. Prove that the function u r J QC .
Using the formula evaluate the integral . .
Applying differentiation with respect quot to a parameter. . quotquotquot f dx a gt . GO
Multiple and Line Integrals Ch. evaluate the following integrals . Evaluate CO these formulas
and then passing to polar GO dz . where S is a region defined by the inequali . ties r . e X
smrnxdx agt. . e . pgt. . The EulerPoisson integral defined by the formula Eval may also be
written in the form Icdy. uate / by multiplying coordinates. x . Evaluate the following improper
integrals GO QC . eltquotd oO. . gt . rno arc tan ax dx . / . dyev y c c/ dx.
M As and let us form the and I t t integral sum S n is AV / s r ne im it of this sum. These
integrals are evaluated in like fashion A line integral of the lirsl type does not depend on the
direction of the path of integration. r/. y z . A. where V is a region defined inequality x if z
quotexteriorquot of a sphere. y be a continuous function p A alt lt. y. J . . . CCr JJJ V d xd t x
T . I .. ..v y qgt of A c/. Also considered are line integrals of the first type of functions of three
variables f x. when n oo and max As/ . . where S is a square A. if the integrand / is
interpreted as a linear density of Hie curve of integration C. called a line integral of the first
type n lim quotgtgtac cfs / i /xf. case of rt quot parametric representation the curve C .. n that
break up the curve C into elementary arcs Al . Line Integrals Test for convergence the
improper double integrals . /.Y . Line integrals of the Irst type. . / Let us construct a system of
points M. lt . . yl. Let / x. //.amp be the equation of some smooth curve C. nVxy dxdy. we
have qgt/. Line Integrals . .. z taken along a space curve. i/ ds c is is the arc differential and b
evaluated from the formula J f A / rfs C In the J a fx t cp . . Sec.vr. As/ / x. then this integral
represents the mass of the curve C. . by the .See. S where S is is a circle f/ ltK U TTiriva
where S a region defined by the ine quality x tf I quotexteriorquot of the circle.
Line uous functions integrals of the second type. ydy a P x. B is . Example . ydx Q x. y along
the as the work of an appropriate variable force curve of integration C Example . the equation
and the equation OA is / . cp ltp x Q x. Qx. y. If P x. X x varies. OB x . Solution. qgt x dx. AB
BO OA . In the more general case when the curve C is represented parametrically y tyt. OJ
Similar formulas hold space curve. for Une integral of the second type taken over a A line
integral of the second type changes sign when the direction of the path of integration is
reversed. ltP. Multiple and Line Integrals Ch. where / vanes from a to . and . This integral
may he interpreted mechanically P x. Here. Fig . y are continand f/ ltpA is a smooth curve C
that runs from a to b as B A Fig. then the corresponding as follows line integral of the second
type is expressed P C x. y and Q x. Evaluate the line integral Cyi . . Evaluate the line integral
where C is the contour of the triangle ABO with vertices A . the equation AB is /l x. . We
therefore have . we have y dy a P ltp .
/ and Q x. u from its total diflerential based on the use of line integrals which is yet another
method of integrating a total differential. y and dy x. . where P . We have y dx a cost. we
have y We get we have dx and along P t M II M J . ydx Qx. Solution. point then of the path
simnlvconnectH of integration C is contained entirely within some S and the functions Pr. P x.
integrating with respect to u P C P M. Then along P P.. we have x . of Line Inlegtah the
ellipse y where C is the upper half clockwise.i U dX dy d If conditions one and two are not
fulsee integration of total differentials does not guarantee the existence of a jillcd. then this
line integral is not dependent on the path of integration and we have the NewtonLeibniz
formula x yj the . ydy. b sin traversed x dy n b sin ta sin / a cos . yUv / Q ydyPx. The case of
ab f sin jt oo / b cos / dt dt ab C cos / dt t aamp. x . ydydUx. the initial and . where x lt y is if In
particular. y together with tlvir partial dcnviitives of the first order are continuous in . If the
integrand of a line integral second type is a total differential of some singlevalued function
UUx. Similarly. /. Vo ydx. y dx Q x. y P x.Sec. For the contour of integration C let us take a
broken line P P Fig . Ux. Px. ydy Ux . y is a variable point. M U x t yU x .v y is a fixed oint and
. is the the contour of integration C terminal is closed. t of a total differential. y dy y x. and
formulas and may prove wrong see Problem We give a method of finding a function U x.
ydxQx. y. that is. yJ U x l lt/. the presence of condition singlevalued lunction U. then a is the
necessary and sufficient coiditioi foi th existence of the function identicalquot fulJilment in S
of the equality If the contour reaio.
y Q. y. y are continuous together with their firstorder partial derivatives in the closed region
SjC. the work of a force field. Xx. is an arbitrary constant. . x Solution. Then Find U. Example
. y dx C where C // ic. Let AO Ux. Greens formula for a plane. U . or y y dx x y dy dU. YY x.
then Greens formula holds S here te circulation about the contour of line integrals. y and Q x.
y. Y Uo /I XQ Fig. X . of a force. Applications An y area bounded by the closed contour C is p
dx p xdy the direction of circulation of the contour is chosen counterclockwise. O. Z y t z or. If
C is the boundary of a region S and the functions P x. along a path C z. The fo lowing
formula for area is more convenient for application The work x. C is chosen so that the region
S should remain to the left. Multiple and Line Integrals Ch. having projections accordingly. is .
X.
A J . c K r . //. gt where C is an arc of the curve JK c .e. j J. where C the first arc of the
cycloid x a t sin /gt cos /.. q/ J A yds.Sec. Z. irrespective of the shape of the path C. where C
is the contour of the square x y a flgt. ral y d. from the point A to the point oo. J where v l f/ Zj
is the initial and x . where C an arc . . y. dU j . i. lt/. /. A. and A xyds.. is . z a If the force has a
potential. /J. s . dU J l/ lt Ux tli t y tt z . if there exists a function potential function or a force
function such that dV dx A dU . niscate r JU y c a cosltp. circle x c y / ds. equal to z t/ x lt y. of
the logarithmic spi . TT dy dz is then the work. rfs where C is the righthand loop of the lem . .
A if . a c rae m vmgtQ . Line Integrals of the First Type Evaluate the following . where C is an
arc of the involute of the x a cos x /sin/. . A Xdx Ydy Zdz. . . J line integrals xyds.. where is C
is a segment of the straight line I. expressed by the integral c Line Integrals UU L .
connecting the points . y asmt is tcost I/ji. Xdx Ydy Zdz Zi i. a ifds. where c C a quarter of the
ellipse i figt l lying in the c first quadrant. z is terminal point of the path. x..
act on a mass m located at the point A . z ae from the point . the straight line Om A .
Determine the coordinates of the centre of gravity of a halfarc of the cycloid x at first sin /.
zbt. Find the mass of the first turn of the screwline A. amp . AB parabola y x from the point .
to the point A a. . y asmt. . Find the arc length of the conic screwline C x sin /. z . the a ydx
xdy. Determine the mass of the contour of the ellipse yi T l. about the zaxis. . . x. i where z C
is the first turn of the screwline acos/ y asnt. and terminating at the point . y is equal to y.
where C an arc c arch of the cycloid x at al sin/. / bt. f bt. J t/quot I z ds y C is the circle x y z
a . J x xydx. a cos/. of the turn of the screwline x a cos/. first to point of B. y ae ae cos. ia Fig.
. where . T y i . if the density at each point is equal to the radius vector of this point. . where .
cos/ which arc runs in the direction of increasing parameter /. is AB the is an arc of the . . if
the linear density ot it at each point M x. With what force will a mass distributed with uni form
density over the circle x f y a z . taken along different paths emanating J xydxx dy OA from
the coordinate origin . Find the moment of inertia. y a cost lt/ltji. Line Integrals of the Second
Type Evaluate the following line integrals . x. c xy.xytfdy. . . . Multiple and Line Integrals Ch.
M B. a. . Find the area of the lateral surface z of cylinder y o the parabolic // x bounded by
the planes l . A . . . asin/.
. Integrals of b the /axis. b J O r xdx ydy. if AB of the bisector of the second quadrantal point
A is and the ordinate of the abscissa oi the B is is . . d the e broken line the broken line OBA
OCA. tfdx xdy. a. f along a path that does not intersect the . the axis of of is xaxis. . ydx d .
the axis oi symmetry symmetry which which is the the c the parabola OpA. cosydxsmxdy
taken along the segment angle. xdy . bsrt traced clockwise. i a xdy ydx. where C c is the
upper half of the ellipse xacost. Fig. Evaluate the line integrals with respect to
expressionswhich are total differentials rt . where C the righthand loop ol the lemmscate cop
traced counterclockwise. c . OA J xydx x dy xudx x x Problem . xfl d a dll f r . y
counterclockwise. . i. Lim parabola OnA. y . as in i/dy .Sec.
then lies in ellipse C W that function and is a c . Show that if fu is a continuous closed
piecewisesmooth contour. e f J dx dy x ry a i ong a p a th at joes intersect the straight line
A.fy . I J . . C xWx ydy j a. . the integration x. Find the antiderivative function a U if b c du x
du du ydx xydy xy y dx x xy y dy d dtt . i path does not intersect the straight line y A . .
Compute /. J xlt xy dx x y x V dy .. J/ f J . integrands and a i. Find the antiderivative functions
evaluate the integrals . lt/ qgt . f xdx yfy X taken clockwise along the quarter of the the first
quadrant. y Multiple and Line Integrals not of the C/t. j x the integration path does not
intersect the straight line y . x. V f X x y yA J fix I / y z I fat xZ y J . x. o J y b .
rA.Sec Line Integrals Evaluate the curves . of corresponding to the variation . . Evaluate the
line integrals of the total . . /. transform the line integral / c f if dx y xy In jc KT dy. zdxzxdy x i
where C is a turn c of the screw. b i. the integration path is situated D C.line / asin/. . y line
integrals taken along the following space tjdz. yte mv y xyz f J . where the contour C bounds
the region S. Using Greens formula. traced in the direction of increasing parameter. Formula
. . /cosa sin sina a const. Greens in the first octant. . i r xd idijzdz j J . xdxydy z zdz. x // i.
circle xydx yzdy zxdz.yquot M v d i. differentials a .. . where I . . where OA is an arc of the OA
situated on the side of the AZplane where //gt.t C is the circle c /y /cosacos/. p the parameter
/ from to ydx zdy xdz. dx f x dy xy dz . .
b . fl. is equal to S. evaluate the integral xy dx c xif dy. where ds is the differential of the arc
the contour C.. Evaluate Consider two cases ifcf. then when when where s is the arc length
and n is the outer normal. . Find lie y ydxx ydy directly and by applying Greens e y y dx f
xtjdy. . x Find AmBnA formula. Applying Greens formula. . a the origin is outside the contour
C. . if the points A and B AmB on the axis. c nds. Applying Greens formula. square with
vertices at the points . is drawn through the points A . Multiple and Line Integrals C/t. whose
axis is the axis and whose AnB. provided the contour traced counterclockwise. Applying
Greens formula. Evaluate the integral and n is the outer normal to taken along the contour of
a A is . tion Verify the result obiained by computing the integral directly. . evaluate / x t if dx x
y dy. andgt. Show that if C is a closed curve. Cl. traced counterclockwise. where C is the
contour of a triangle traced in the positive direcwith verlices at the points A I. bounded by the
integration path AmB and the segment AB. find the value of the integral I jxcosX. . . and Cl. .
and . fl f . . where C . the contour encircles the origin n times. nysmX. . is the circle x if R
chord is A parabola AmB. while the area.
The ellipse x acos /. function and determine the work done by the . Z where const and Yx if f
Newton attractive force and the material point moves from position A a. Y. Z force over a
given path if of gravity and the mate point is i moved . . . y bsnt. The cardioid y a sin/ sin . A
circle of rolling it. c to infinity. from position A f x lt y l zj jx to position B b r Uv x K. A loop of
the folium of Descartes x if zxy Q agt. K. R and outside sliding along a n is an Assuming that
integer. . cardioid. The curve x y . . r is rolling without it. asin /. . . Find the work done by an
elastic force directed towards the coordinate origin if the magnitude of the force is
proportional to the distance of the point fiom the origin and if the point of application of the
force traces counterclockwise a quarter of the ellipse . without sliding along D a fixed circle of
radius R and inside Assuming that is an integer. Line Integrals Samp D. sil Find the lying in
the potential first quadrant. b. wards.Sec. The astroid jt x a cos/ cos /. . Applications of the
Line Integral Evaluate the areas of figures bounded by the following curves a cos/. find the
area by some point of the bounded by the curve epicycloid described moving circle. find the
area bounded by the curve hypocycloid described by some point of the moving circle. Find
the work done by the force of gravity when a material point of mass m is moved ironi position
A JC P // l zjz the zaxis is directed vertically upto position B x // . A circle of radius fixed circle
of radius axy. Analyze the particular case radius r is of r R . . Analyze the particular case
when r j rgt astroid. A field is generated by a force of constant magnitude F Find the work
that the field does in the positive jtdireelion when a material point traces clockwise a quarter
of the circle x y R lying in the first quadrant. Zrng force a X rial of a force R X. .
cos Y. Let function and zcp. Q Q . z. continuous functions and S is a side of the smooth
surface S characterized by the direction of the normal n cos a. . If a projection a of the
surface S on the jo/plane is singlevalued. that to the zaxis intersects the surface S at only is.
Z the initial point of the . z are P dy dz Q dz dx R dx dy f f P cos a f Q cos p R cos Y dS. .
zdS n lim gt fas I is the area of the /th element of the surface S. the point /. //. where k const
elastic path is located on the sphere while the terminal point is located on the sphere Sec.
Surface integral of the first type. . Example Compute the surface integral where S is the surf
ace of the cube ltJtltl. lt//lt. . be a continuous The surface integral of the first type is the limit of
the integral sum x. then ihe corresponding surface integral of the second type is expressed
as follows R Rx. every straight line parallel one point. The value of this integral is not
dependent on the choice of side of the surface S over which the integration is performed. z.
Surface integral of the second type. //. Surface Integrals f x. Multiple and Line Integrals c Ch.
X force. Y ky. kz. then the appropriate surface integral of the first type may be calculated
from the formula where AS/ z/ y dx dl J S a . and the maximum diameter of elements of
partition tends to zero. upper edge of Let us compute the sum of the surface integrals over
the the cube zl and over the lower edge of the cube z integral is obviously three The desired
surface times greater and equal to If P Px. y. y a smooth surface S. cos p. y. //. . y lt belongs
to this element. x z / and kx.
//.xydxdy. the second type is yz dydz xzdz dx. then the direcIf the surface is represented
implicitly. .Sec. Surface Integrals When we pass to the other side. Find the mass ot the
surface of the cube Osz . z lt Oxl. where S is the external side of the ellipsoid . we then have
the Stokes formula R Rx. S. Stokes formula. rr. and the direction of the normal is defined so
that on the side of the normal the contour S is traced counterclockwise in a rigiithanded
coordinate system. where the external A s side of the surface of a tetrahedron y Q t . xy z a.
M x. where cos a. y. z. first type tfdS. Nzdxdy. z. .v.T D dz where and the choice with the side
of sign before the radical of the surface S. Evaluate the following surface integrals of the . a z
side of the surface of the hemisphere // . .. Q Q x. z are C rrfdR dQ . of s. o x dydzydzdx
zdxdy. should be brought into agreement P . FT D dy COS V T OF . If the functions P
continuously differentiable and C is a closed contour bounding a twosided surface S. cos p.
tion cosines of the normal of this surface are determined from the formulas OF Q COSapr.
Vx tfdS where the lateral surface the s cone g i Ozbl integrals of Evaluate the following
surface . where S is the sphere x // is z z a. fdQ dx dP T j dy J cos v y dS. if the surface
density at the point Is equal to xyz. COSB D dx dF . this integral reverses sign. z . y. y. t
where S is the external Oyl. cosy are the direction cosines of the normal to the surface S.
bounded by the planes . . j . of the surface. //. dz JJ ldy c s a J dt Id dR Tdx fi cos a l . F x.
surface S cosy are the direction cosines of the outer normal to the Applying the
OstrogradskyGauss formula. /. transform the integrals V lt lt a x . . P x. first partial R smooth
surface bounding the volume V. is B . zdxz where C is the circle . . z. are functions that are
continuous together with derivatives in the closed region V then we have the Ostroe t
gradsky. Find the moment of inertia of a part the lateral x f y z about the zaxis. where ABC A
is the contour of ABCA / z /IflC with vertices A a. c f y zdx z xdy x ydz. where C is the ellipse
JC // . . . find the given integrals and verify the results by direct calculations .yz dx y zx dy z
xy dz c b j ydxzdy xdz. c Applying Stokes formula. . homogeneous parabolic envelope azje
Multiple and Line Integrals C/i. z. transform the following surface integrals over the closed
surfaces S bounding the . a. X l. surface of the cone z ft . Applying Stokes formula. z a sin
cos/ lt/nj. . sin/. C equal to zero . y dx z dy x dz. cos p. and P / v. y.Gauss formula where
crsa. xdx c // xydy x y zdz. where C is the curve acos/. . a. In what case / the line integral Pdx
Qdy Rdz c over any closed contour Sec. . Determine the coordinates of the centre of gravity
of a ltzlta. The OstrogradskyGauss Formula If Q Q is a closed thMr A y. C . of . y xdy x ydz.
cosp. Oza. Prove that direction. /olume Vtosa. cosy are the direction cosines normal to the
surface S. where S is the exter nal total surface of the cone . the external side . dx z d dy. . .
cos y d. where a. Or/n. lormal to the surface . compute the following surface integrals o.
Using the OstrogradskyGauss formula. where S cube the external s side of the surface of the
Oxa. external z. the outer .y Q. xdydz ydzdx zdxdy. dydzif dzdx z dxdy. Prove that the
volume of the solid surface S is equal to V bounded by the M of where cose. then if S is a
closed surface and / is any fixed where n is the outer normal to the surface S.gtec. of a
pyramid bounded by the surfaces x yz x V xdydz ydzdx zdxdy. cosp. JJ s J J J The
OstrogradskyGauss Formula cosy are direction cosines of the outer S. is is xydxdyyzdydz
zxdzdx. . x dy dz y dz . where the side of the sphere x f // jc cosat y o a cos p z . is x Q.
cosy U in the direction The divergence of a vector field a P ai ay ja z k is the scalar diva The
rotation curl of a vector field Va. . y. cos Y. a a v ja z k. In coordinate form. Divergence and
rotation. Fundamentals of Field Theory . //. The vector lines force lines. The const. cos a .
The flux of a vector field aP in a direction defined by the unit vector of the normal to the
surface S is the integral through a surfaces ujcosa. z are projections of the vector a on the y
ayx. flow lines of a vector field are found from the following system of differential equations
dxdy dz does it A scalar or vector if field that stationary. are called level surfaces of the scalar
scalar field is where P x. y. z. cos p. VI. then the Ostrogradsky Gauss formula holds.cos p /.
t/. depends on t t. Sec. is called the gradient of the field U f P at the given point P ci. Sec.
where a x a x x. and n is a unit vector of the outer normal to the surface S. Th Gradient. y is
the Hamiltonian operator del. then the derivative of the function . where P is a point of space
and rxiyjzk is the radius vector of the point P. COSY an dS an dS ax cos a . //gt /. x. z A
vector field is defined by the vector function of the point a aP ar. A C z. a P ax ay j az k i is
the vector da z da y . cos p. defined by the scalar function is a point of space. . t is called
where V y. Ch. S s S If S is a closed surface bounding a volume V. . The gradient is in the
direction of the normal n to the level surface at the point P and in the direction of increasing
function U. the time. C.a y cos p az cos Y dS. z. y. where y. or nabla. Flux of a vector. ry. The
vector quot it not depend on the time is called nonstationary. ax i a and a z a z x. z fields.
Multiple and Line Integrals Ch. and has length equal to dn If dx the direction is given by the
unit vector / cos a. coordinate axes. Scalar and vector of the point surfaces f field.
curve C a s is the adr f f /i rotadS. that is. In it is necessary and sufficient that that case there
exists a potential U defined by the equation dU a x dx. then Stokes formula holds. then a dr
U B U A. Laplacian operator or AU where A . rota . What will the level surfaces be of a field
where . the circulation of the vector a is equal to zero m adrQ. it satisfies the Laplace
quotErltgt. given in a simplyconnected domain. If the potential U is a singlevalued function.
tial if Potential and solenoidal fields. If the closed curve C bounds a twosided surfaces.
Determine the level surfaces of the scalar field where r U FQ. AB in particular. that is.Q and
the potential function U is harmonic. where Ufr is a scalar function the potential of the field.
then the line integral is called the circulation of the vector field a around the contour C.
.Sec.gtgt isthe . fxyz. then div grad U. it be non rotational. Fundamentals of Field Theory form
is ff r r which in vector div a dx dy dz. For the potentiality of a field a. A If vector field a r is
called solenoidal if at each point of the field div in this case the flux of the vector through any
closed surface is zero. vector .a dz. the direction of the vector should be chosen so that for
an observer looking in the direction of n the circulation of the contour C should be
counterclockwise in a righthanded coordinate system. The line integral of the the formula a
dy f a z dz y dr a s ds V a x dx f C C C the If C is closed. a . The vector iield ar is called poten
U.a v dy f. the field is at the same time potential and solenoidal. where n is the vector of the
normal to the surface S. Circulation of a vector. which in vector form has the form and
represents the work done by the field a along projection of the vector a on the tangent to C.
the work a along the curve C is defined by f a of a Held.
cosp. b div i/c c grad /. y. Multiple and Line Integrals Ch. respectively. . In z what case in the
direction of the radius vector r of will this derivative be equal to the of magnitude of the
gradient . . Evaluate . Show that straight lines parallel to a vector c are the vector lines of a
vector field aP c where c is a constant vector. where C. . C K C gTadUV Ugrad grad t/ / grad
gradC l i/ gradf/ C gradV. . b r c j . if U is equal. . . In what case will this derivative be equal to
zero . Find div a for the central vector field where r aP /r . Evaluate grad f/. . and C d e V
grad/ U AfU grad J grad ltp / cp t grad U. and what will their position be relative to the vector
c . Determine the level surfaces ot the scalar field U arc sin . Find the derivative the function
U in the di rection of /cosa. cosy. Deterthe gradient of the field is perpendicular to mine at
what points the zaxis and at what points it is equal to zero. where C and C l are constants.
where c is What will the level surfaces be of this field. Derive the formulas a divaj CCjdivaj
Cjjdivajj. . where c is div/a grad Ua /diva. this point. . to a r. di a constant vector. d /rr /qr.
where CD is a constant. a constant vector. Find the gradient of the scalar field U cr. . Find
the vector lines of the field a CD// CDJC/. Derive the formulas t a b c j are constants. Find the
derivative of the function x U y ala given point Px. Find the magnitude and the direction of the
gradient x z of the field U if xyz at the point A .
it is directed to j and depends only on the distance r from where fr is a singlevalued
continuous point function. jt taking the hemisphere z J/quot/ if for the surface. Using the
OstrogradskyQauss theorem. Find the potential by a material point of mass nates U m of a
gravitational field generated located at the origin of coordi a r. prove that the flux of the vector
a r through a closed surface bounding an arbitrary volume v is equal to three times the
volume. Using Stokes theorem. Fundamentals of Field Theory . located at the coordinate
through an arbitrary closed surface surrounding this point. Evaluate the divergence and the
flux of an attractive jltf. Derive the formulas a b c constants. Evaluate the divergence and
rotation of the gradient of the scalar field U. force origin. . Evaluate the line integral of a
vector r around one turn of the screwline /cos/. of the field. . equal to a r b re and c frc. . rotC
a l l f . .Sec. Evaluate the divergence and the rotation of the vector respectively. Prove that
divrota . t C are . . . where c a constant vector. // . Find the potential U this towards a fixed
point F frr. then the field is a potential field. that is. where c is a constant vector. . Find the
flux of the vector r through the total surface of the cylinder lt/ ltelt//. Evaluate the rotation of a
field of linear velocities co r of the points of a body rotating with constant angular velocity lto
about some axis passing through the coordinate origin. Find the flux of the vector a xi yjzk
through . . b the total surface of the cone. rot Ua grad U a U rot a. a is if a is. . C a C rota C
rota where C and rot/c grad Uc. evaluate the circulation of the vector a zk along the
circumference x if xtfi R z. . Show that if a force F is central. Show that the potential U
satisfies the Laplace equation . y Rsmt z lif from / / n. Find the divergence and rotation of the
field of linear velocities of the points of a solid rotating counterclockwise with constant
angular velocity ogt about the zaxis. F of a point of mass w. a the lateral surface of the cone
ltelt//. .
fr where k is constant. Find out whether the given vector field has a potential find if the
potential exists U U. Will the vector field a constant vector a rcxr be solenoidal where c . and
. . Prove that the central space field lenoidal only is /rrwill be so when . a b c a a a . Multiple
and Line Integrals Ch.
. tt a n l tt is called convergent if its partial sum has a finite limit as n of the series. to take a
geometric progression aq no n a . It is convenient..Chapter VIII SERIES Sec. lta.. the series
is then called divergent. If the limit lim n Sn QO does not exist or is infinite. and the series .
The convergence or divergence subtract a finite of a series is not violated if we add or
number test of its terms... a Tests of convergence and divergence If Comparison I. If the
series J diverges. If a series converges. for purposes of comparing series. . then diverges as
well. .. . Number Series . A number series a.a t . For convergence of the series it is
necessary and sufficient that such that for n any positive number e it be possible to choose
an and for any positive p the following inequality is fulfilled is for N gtN Cauchifs test.. after a
certain n n . noo The converse not true. while the gt oo. . The quantity S lim S n n oo is then
called the sum number is called the remainder of the series.lt n of positive series.a . then lim
a n Q necessary condition for convergence. . then the series also converges. Fundamental
concepts. . converges..
converges. Example . since of its general term is greater than the corresponding term the
harmonic series which diverges. The diverges. . Example The series converges. . The series
J converges. which converges for ltland diverges for q. b n at and converge or diverge the
same time. since J while a series with general term converges. b Comparison if test II.
Example . since here a J quotn n while the geometric progression ni whose ratio is lt The . a
n b n . Example . since quot n J whereas a series with general term n diverges. then the
series lim y. series ln ln In /i diverges. and the harmonic series which is a divergent series. If
there exists a finite series and nonzero limit n in particular. Series Ch.
Let ofO after a certain n and lim n gt OD let there be a limif V n/ n if Then converges if qlt. d
Cauchys test. By means of the integral test it may be proved that the Dirichlet series
converges of if pgt . is q. Example Test the convergence of the series s Solution. and
diverges qgt. Then the it is series converges not known whether . if q the series lt and
diverges if q gt . Let an gt after a certain n and let there be lim n gt GO l an . quot i r quot i
We have J /i . e Cauchys integral test. c a limit Number Series DAlemberts test. and the
integral / x dx converge or diverge at the same time.Sec. If a n fn. then convergent or not.
comparing with the corresponding Dirichlet series Example Test the following series for
convergence UL L Solution. and lim H Hence. When ql t the question is of the convergence
of the series remains open. where the function f x rnonotomcally decreasing and continuous
for jcal. Here. If lt . and diverges tested series may be by The convergence of a large number
if plt. the given series converges. the series quot positive. .
the series converges absolutely. . Test for convergence the series Solution. we can make
use for the series of the familiar convergence tests of positive series. The series lim I n oo . .
For instance. But if lim n J I I gt or GO n f lim /a n n gt oo K gt . the divergence of does follow
from the diver gence of . Tests for convergence of alternating series. converges.. But if con lt
n or lim n n gt oo /KIlt not . then not only does diverge but the series does also. converges
absolutely if lim converges. Leibniz test If for the alternating series the following conditions
are . it follows that on the basis of test II we can say that the given series likewise converges.
. For investigating the absolute convergence of the series . verges and diverges.. for the
remainder of the series Rn the evaluation holds. Example . comparison Since the Dirichlet
series converges for p . Example . This series diverges harmonic series. since the series test
are fulfilled. In this case. since the conditions of the Leibniz converges conditionally.. lim n oc
bn then converges.. values of the then also converges and is called absolutely convergent.
Let us this series form a series of the absolute values o the terms of Since lim . . fulfilled bl s
b b . In the general case. terms of the series . . If a series Series Ch. then the series is called
conditionally not absolutely convergent. composed of the absolute fll. l.
S k S Squot k k .fl n .. where and limit lim k S k ccS k Sk exists is a partial is sum of the is
harmonic series. a . the Leibniz test is not necessary for the convergence series an
alternating series may converge if the absolute value of its general term tends to zero in
nonmonotonic fashion Thus. of course. the series its Number Series diverges despite the fact
that its general term tends to zero here. of the other hand. Series with complex terms A
series with the general term c n a n z ib n i converges if. .... the series with real terms ni an
and amp. For the convergence of an alternating series it is not sufficient that general term
should tend to zero. converges. if the whose terms are the moduli . k lim gt S fe oo. here.
The Leibniz test only states that an alternating series converges if the absolute value of its
general term tends to zero monotonically.. whereas the lim fe and finite S k a partial sum o f
the convergent geo metric progression. Indeed. for example. A convergent series if may be
termwise by t any number fc. a. although the Leibniz test is not fulfilled though the absolute
value of the general term of the series tends to zero. the series On an alternating .S then .
Operations on series. in this case ni SC B S n i Xn i The series definitely converges series
and is called absolutely convergent. Thus. that is. a quotquotn converges and it converges
absolutely.Sec / Note. it does not do so monotonically. hence. conver g e n a the same time.
and only if. a of the terms of the series multiplied . the monotonic variation of the absolute
value of the general term has been violated.
Ch.. S c The product of the series cl the series where If cn a. a n and bn is t . . .bn a AI
verges d terms verges the series and converge absolutely. b z we mean a series a. f . . /ith
term of the series using J I. .. . . . i j. . . then the series also conabsolutely and has a sum
equal to S S If a series converges absolutely. quot . t.. its sum remains unchanged when the
of the series are rearranged. . . .. S. flrt a din. . ...... .. By the sum difference of fli two
convergent series If flt.. .. . Series b m . cgt. Test the following series for convergence by
applying the comparison tests or the necessary condition . a n q.. .. . I .. a n . . . M gt Write
the simplest formula of the the indicated terms .. ... In Problems it is required to write the first
or terms of the series on the basis of the known general term a n . .fl n . This property is
absent if the series conconditionally.... . ....cn . . bl a l . H ..
. i .. .... f y. J .. lest.. i l j i . / OJ ... rT no nMnl . ....if . ..Number Series .. ... fn n . H ... V J/ n K
quot Using dAlemberts gence . i . ... sSFI .. . .. i I I J i X. TtjO r/lo/ . ... . test the following
bones for conver l . quot i j .. .. . using Cauchys test Test for convergence the positive series
... I i ../i Test for convergence.. Oquot .
. . n J .. . . . iJ .... Ulnfll.. . . gt .. ..... . .. .... n quotquot . .n n . .. quotquot . Series Ch. .... . yini.n
. .. . . .. rilnnln In n . .. I .. . . /J E STTOTJ .. .. sin.. I. ... quotquot .. . .. n . arcsin.. . ../ jT . .. g
quot Hn . Mquot quotquot .. ..
. / n n Q quot lkJlL i / .. / ...tquot. For convergent series.Sec.. . . and for gt. S ./ . Number
Series .. if pl.h. / K l . quotTV ii U . pgtl. i ni .... . . cos .uif.. Prove that the series converges for
arbitrary q t if and for lt/lt. pltl... n . if Test for convergence the following alternating series.. .
.... l i .. I .. ... .. . ....... .. quot . test for absolute and conditional convergence. . . if diverges for
arbitrary q. a . l l . pl.
Series Ch. Test the following series with complex terms for convergence CO .. sin na . Find
out which of these series diverge. r y . Convince yourself that the dAlembert test for
converdecide the quesl ide question of the convergence of the a gt where test it is possible to
establish that this series converges. Convince yourself that the Leibniz test cannot be applied
to the alternating series a to d. Xquot . v i quotL h u quot d T l T T TT T . . .. F . V. J i . . which
converge conditionally and which con whereas by means of the Cauchy verge absolutely a J
l quot j .j. ni lytan n Vn gence does not series . . lquotl. . i . i . . .
Form the product of the series V nVn . Estimate the error due to replacing the sum of the
series y by the sum quot J quot Hquot n terms. X Form the sum of the series and Does this
sum converge Form anc the difference of . . . the sum of the first five terms. Form the series l
. Between the curves of their point of to the t/axis at / nij to X and y and X the right of
intersection are constructed segments parallel an equal distance from each other. Number
Series V rQ quot l fa . .. the divergent series T r and rt es for convergence. Estimate the error
committed when replacing the sum of this series with the sum of the first four terms. Will the
sum of the lengths of the segments mentioned in Problem be finite if the curve x is replaced
by the curve y . and V. sum converges while CC QC . What can you say about the signs of
these errors . n i . Does this series converge . V.. Jf .Sec. Will the sum the lengths of these
segments be finite . Choose two series such that their their difference diverges..OTTM n tt
Does this product converge . Given the series .. quot of its first .. QC Does the series formed
by subtracting the series TT f sef i es n i converge n i .
n lim QO Sn x. where S n vergence. Functional Series . estimate the accuan approximation
for n . .... Estimate the error due to replacing the sum of the of its first sum iV.. . The function
S converges is called the region of convergence of this series. .f n x. The the functional
series /. R n x SxS n the x is region of conthe remainder of the series. by the sum n terms..
its sum to two decimal places three How many terms of its the series / MIS does one to three
have to take to compute to four sum to two decimal places series .lt set of values of the
argument x for which /. Estimate the error due to replacing the sum of the i JL LLLl. In
particular. Zl ni Iquotquot does one have to to take to compute decimals . Region of
convergence. . Estimate the error due to replacing series sum of the i by the racy of such
series n terms. estimate the accu racy of such an approximation for n. MM quot/iquot the by
the sum of its first n terms. .. . series Series Ch. Find the sum of the series Sec.M.. of its first
.. In particular. and x belongs to sum of the series. . x is fl x f z called the x . . Find the sum of
the L gL jL .
l. . . In special cases. Applying to the series of absolute values gt the convergence tests of
dAlembert and Cauchy. with in which the series a R the series diverges. to is sufficient. c.
Thus. fl . . For any power c lt. when determining the region of apply to this series certain
convergence tests. for x and oo. the power series may either converge or diverge.Sec..
When xl we get the harmonic the series series TTQ. the scries converges when . the
formulas lim n rt and Hm quot/c.quot t Solution. . we can that assert that the series
converges and converges is. . if an infinitude of coefficients cn . the terms of which are the
absolute values of the terms of the given series . or the series diverges. n Diverges Fit. the
converges absolutely. o series which diverges. In convergence of a series holding x
constant. which accord with the Leibniz converges conditionally. too However. At the
endpoints of radius of convergence R may also be equal to a the interval of convergence x
R.. Denoting by u n the general term of the series. if lt is. we get. Determine the region of
convergence t of the series x quot xy XY quot iy TTquot quotTr tquotquot n. For example. if
that ooltlt lt x lt t oo Fig.f Diverges X Example . for the radius of convergence of the power
series . c n and a are real gence x a lt numbers there exists an interval the interval of
convercentre at the point x a.. we will have lim I M U n lim IH Jxl n n n jc l Using dAlemberts
absolutely. The interval of convergence is ordinarily determined with the help of the
dAlembert or Cauchy tests. by applying them to a series. o T .. c n x B fl . and when in x we
have test .. Aa c R with . it . respectively. . one must be very careful in using them because
the limits on the right frequently do not exist. if test. Power series. Functional Series the
simplest cases. if ltxltl.
. the series may both converge and diverge. The power series may be termwise
differentiated and integrated within its interval of convergence for fl ltRY.. . . Uniform
convergence. no matter what e . then for the power n series zZo cn a n ib n Z . when axb I
and the number series n absolutely and uniformly Weierstrass test. by means of the
dAlembert test it is easy to see that the circle of convergence of the series quotquot quotquot
. . z Jt f/f/ with centre at the point . If Series z Ch. s quot n . whose terms are absolute values
of the terms of the given series. It is then advisable. The functional series converges
uniformly on some interval if. one cannot use these formulas.. If fn x f c n rtl.ncn xa. for Iz
cumference of tne circle of convergence. At points lying on the cirabsolutely.. . . this occurs if
the series contains terms with only even or only odd powers of x a. . series converges
absolutely and uniformly on any interval lying within its interval of convergence. that is. gt lt
ngtN given series. quotquot is determined by the inequality zf lt it is sufficient to repeat the
calculations carried out on page which served to determine the interval of convergence of the
series . the . cn xa n dx. cQ f x.. only here x is replaced by z. The centre of the circle of
convergence lies at the point z . then for any x of the interval of convergence of the series we
have a. ... without resorting to general formulas for the radius of convergence. if x a. Thus. . it
is possible to find an N such that does not depend on x and that when for all x of the given
interval we have the inequality R n x where R n x is the remainder of the e. then functional
series converges on the interval V The power . . to apply the dAlembert or Cauchy tests
directly. x ty is a complex variable. . when determining the interval of convergence. cn
converges. vanishes as a particular instance. while the radius R of this circle the radius of
convergence is equal to . dx d xa dx c xa dx .C n ZZ Q . . It is customary to determine the
circle of convergence by means of the dAlembert or Cauchy tests applied to the scries z z ltR
.. there exists a certain circle circle of convergence z z inside which the series converges z
gtfl the series diverges.. /. for example. as was done when we investigated the series ... c n
xa l .
nMn . Find the interval of convergence of the power series and test the convergence at the
endpoints of the interval of convergence . . the series J. D. convergence of the series . . p . . .
. Here. I . of n Functional Series n series Find the region . the series and have the same
interval of convergence as the . Xi. quot. . . n yin I . the number also belongs to the interval of
convergence of. . .. X /l sin quot J i .. lquot fI .Sec. ni . . M I lquot jbn Mx lA ni D jzrkv gt / I OT
O v sin n oMIfl OvQI ni ri o nr . X rr. . . /Il V X .
CO V x rr n n l ni n . rtfly Vquotquot r .. V . quot. n . gt . OCQA ltJO. /I l lquot gt l ATquot
/jquot .n . n ft . ni . ni nl xnl . CO . quot . V gt. l no XA. Y. CO Series Ch. . . ai . S V nquot . .
CO f n s . . gtquot nl quot l . T lquotquot r. Y . . . V quot X t . quot. .
it must be shown that for any e gt an N dependent only on e such that for any n gt N we will
have the ine quality I Rn x I lte for all let Taking any egt. To prove the uniform convergence of
the given series over the subintervai it is possible to choose a. . llHI . . Solution.. l r. n gt
since lnl alt and ln Thusgt is puttin g Nss when ngtN a t R n x indeed lnl less than a e . let us
take a submterval ct.l/ii .. . where a is an arbitrarily small positive number.gt . .. Proceeding
from the definition of uniform prove that the series . xla and. .. Within the interval .quot. Hm R
n x im X X J/I . Using the formula for the get. convergence. In this subinterval jcsl a. it
contains points that are arbitrarily and since x of the subinterval the given series on any sub
close to Jti. we all are convinced that for a. X no mattei how large n is. I lt /I atquot .
Functional Series Determine the circle . Aarf V W. but converges uniformly on any
subinterval within this interval. consequently. a. we jclt . . does not converge uniformly in the
interval . . . . us require that e. i h l . whence i a ea f lt nllnl lnea In altlnea. is thus proved. that
j is. aquot wl ltea. J . x of the subinterval under consideration. As for the entire interval . .. I .
arf no ni V ltgC fl. a.Sec. M quot. aj and the uniform convergence of interval within the
interval . of convergence .. . for sum of a geometric progression.
Using the definition of uniform convergence... Prove the uniform convergence of the
functional series in the indicated intervals on the inlerval on Iquot . . y y v y. ... converges
uniformly positive number. converges not only within the interval . x x . . . points x will be
found for which R n x is greater than any arbitrarily large N we number Hence.. . i.. . . find the
sums of the series . however the convergence of the series in . it is impossible to choose an
N such that for n e at all points of the interval would have the inequality R n x . . t y yLfl l .. x
quot x n h . is not uniform... . c the series F F.. .. is nonuniform.. converges uniformly
throughout the interval of convergence . lt gt . . I over he en tire number scale.. prove that a
the series converges uniformly in any b the series x finite interval. but at the extremities of
this interval. Applying termwise Y V differentiation n and integration. and this means that the
convergence of the series in the interval . . x .. co where is any x jc x.lL . I quot T on the
interval . jc jc f . . Series C/i. . d the series in the interval .. .
. / gt quot if n is odd. we have II n even. . sinh x... then this series called Taylors series is of
the form powers of A xaltR I When holds a if when the Taylor scries is also called a
Maclaurins series. f n x cosh x. . We Expand f if X .Sec..quotx x . fquot x /. and H / the
function f x cosh x in a series of powers of x. We oo convergence of the series we apply the
have lini ngt / .L i.. To evaluate fin x the remainder.v cosh x. in some neighbourhood a.. the
derivatives of the given function f x cosh x. if n is even. A . . and To determine dAlembert the
interval of test.... is smh x. . Equation I the remainder term or simply remainder of the xaltR
Taylor series as n oo. Taylors Series .. from . Taylors Series . .. . Putting a . Sec. generally. .
sinn A. if n is odd. . Expanding a function in a power series. in a series of expanded. /quot .
Whence. l. . generally.A. . fquot . Tl. v s r quotquot l x ... one can make use of the formula l
find fn l l fl lt fl where ltlt Lagrantes form./ quotquot Find the sums of the /r series ... If a
function f x can be of the point a... we get / A . Example Solution.
. this is therefore. made immediately evident with in Y n the help of for accord nl with
necessary condition lim and consequently lim /gt for any x. A series converges for
dAlemberts convergence. Techniques employed for expanding in power Making use of the
principal expansions I. if if n is is odd. gtgt . IV. Hence. has the form R n x . series. III. for
diverges when for it l.fj. i sinh . ooltltoo.. This signifies that the sum of the series for any x is
indeed equal to cosh.cosh . simply to obtain the expansion of a given function in a power
series. . it follows that e. Series Ch. II. sinh.. the boundaries of the interval of convergence
i.e. with the general term the X Iquot and therefore R n x I e . in for any. diverges on both
bounconditionally converges when x On and xl the expansion IV behaves gt gt mO x m x
daries.. when it converges absoas follows for m it and lutely on both boundaries. and also the
formula for the sum of a geometric progression. The . in accord with formula the interval
QOltJCltOO. y . and R n x Since n A n even.. . it is possible. It is sometimes advisable to
make use of termwise differentiation or integration when expanding a function in a series.
without having to investigate the remainder term. in many cases. When expanding rational
functions in power series it is advisable to decompose these functions into partial fractions.
the series converges remainder term. any x test.
Sec.. holds respectively. X xa ltgt dfx. x lt lty. Example Expand in powers of x the function
Solution. is the Taylor series is then called a Maclaunns y series. fc If a notation . a. for a
function of two variables..y dx a f dy W and so forth. Here and henceforward we mean quotin
positive integral powersquot. V n o i v j nlt n x. Decomposing the function into partial
fractions. we will have Since and . Taylors Series .e. for when i.. Expanding a function into a
Taylors series in the neighbourhood of a // point b has the form . n l fi Agt t it follows that we
finally get /i o nQ n v nQ The geometric progressions and and converge. . Here the as
follows .. Taylors series two variables /x. formula xltj when of .. hence.
of the following progression. .gt a.. cinriM Making use . Series Ch. . expand the given
functions in Applying powers of and indicate the intervals in which these expansions occur .
The expansion occurs if the remainder term of the series fel as n oo. O cosh . In x V . .
various techniques. xe . . cos . Expand the indicated functions in positive integral powers of .
Applying expand following functions in powers of . arc sin. rJT . . . J o . . e x . sin cos . . in
powers of . In . a x agt. cos. . X X x . occur . The remainder term may be represented ni in
the form where ltlt . cos . sinh. and indicate the intervals of convergence of the scries . arc
tan . / of the principal expansions IV and a geometric write the expansion.v . In f the . and
indicate the intervals in which these expansions differentiation.. . sin cos . functions. . find the
intervals of convergence of the resulting series and investigate the behaviour of their
remainders . . llnl. sin . a sin rr Jooo. . . . . .
Expand In of TJ . powers of l. quot z quot L V Write the first three nonzero terms of the
expansion of the following functions in powers of x . . in a series of Expand cos Expand cos
in a series of powers powers powers of of j . in/iL. . dx. . e in a .Sec Taylors Series ./v rMl fJl
o . Show of it may be taken that the . . . . . powers of x series of powers of . tan. Expand
Expand . . a series of powers of . Show that for computing the length ot an ellipse it is
possible to make use of the approximate formula . of . A tenary line J heavy string hangs. . . .
q tension of the string. while q that for small . e x . V f . Expand / in . to the order string hangs
in a is the weight of unit length. . tanh. Expand In Expand Expand Xz X Y in a series of in a in
a j powers of series of powers of series of . parabola the s ya . . sec. in a caand H is the
horizontal where a // J acosh a . X . under its own weight. J. . . Expand oX jc in a series of JL
powers f. in a series of in a series of . lnx x . where ellipse. ltsin. . . e is the eccentricity and a
is the major axis of the . . In cos. Expand function in a series of powers powers of . of f h
Expand fxh in a series of . .
How many terms do to three decimals series of powers x Exam we have to take of the
series cosArl in order to calculate cos j. ber j .. . . .. find the number e to four How many
decimal places terms of the series In do we have cimals to take to calculate In to two
decimals to de . . by taking the sum of its first five terms when xl by expandsee of ... . and
estimate the error.. evaluate it by means of Y t putting a .. tion l/S .. the in a series of powers
error of if What is magnitude of the we put appro ximately . . Calculate / to two decimals by
expanding the funcx in a series of powers of x. To what degree of accuracy if we make use
of the series will we calculate the num u O .. Calculate the number ing the function arc sin A
in a ple . Calculate J/I to three decimals. Series Ch. Find out the origin of the approximate
formula r / fxampa agt. Expand . How many terms of the series have to be taken to . . How
many to three decimal places terms do we have to take of the series . to calculate sin to four
decimal places .
For what values of x does the approximate formula sin XampX yield an error that does not
exceed . find the region of convergence of the resulting series and investigate the remainder.
. Evaluate yxedx to three decimals. Taylors Series . .Sec. . x y Expand fxh. JC yield an error
not exceeding . Wiite the expansions. . . . arctan l y j/ . Expand the function cosx y in a series
of powers x and y. to four decimals. Evaluate ydx to four decimals. Evaluate J . Evaluate e dx
to four decimals.ykin poFind the wers of . Evaluate l/ccosxdx to three decimals. . function ft y
xy. . Evaluate / jldx dx to three decimals. of this to the values ft. . For what values of x does
the approximate formula . Expand Expand . y k. y axbxy cy. in powers of x and y. . k. .
increment f/ when passing from the values . of the following functions and indicate the
regions of convergence of the series . nlxyxy. / . . fx. of . the function e xy in powers of x and
y the function sinxy in powers of x and . . . sin y. . xy and /. . sin xgt sin y.
belonging to the interval of a jt.v.. .e. .. satisfies the the function Dirich uniformly bounded.
which in the interval of this interval at ji. . in Write the first three or four terms of a
powerseries expansion x and y of the functions x . If a function / is even i. .. e cosy.. Ji
satisfies the Dirichlet conditions at any point x which /x is continuous. conditions in an interval
is We a. n.. where M is constant. Incomplete Fourier s/jc... Fourier Series let . . Dirichlets
theorem. a n cos nx b n sinnx. / x formula rt w l f . . and ji a r / /quot . Dirichlets theorem
asserts that a function /. say that a function b if. Series C/i. a of ltxlt b. function f . where the
Fourier ji coefficients a n and b n are calculated from the formulas ji JT n J fxcosnxdxn Q. ...
may be expanded in a trigonometric Fourier series fx a. of has no more than number kind the
points at all them are of each the function finite limit f x has a finite limit on left f g
discontinuity and g discontinuity Urn f I e and a quot on the right / lim / e gto e egtJ. e. then in
series. . then the sum of the Fourier series S x is equal to the arithmetical mean of the left
and right limits of the function SM At the endpoints of the interval n and X K. . a the that finite
first is fxM when of i. . has no more than a finite number of points of strict extrenium.. H Sec.
cos x b v sin x a cos xb sin ... J .amp f JI J jt If x is a point of discontinuity. f x in this interval..
..
e. . ji as well Q x when n Expand the following ji. in Expand . /i . M n e / v flllX f x cos . i . the
sum of the Fourier series is defined in a manner similar to that which we have in the
expansion in the interval Jt. nx cos j b sm. nnx . function specified in an interval . In the case
of an expansion of the function / x in a Fourier series in an arbitrary interval formulas a. and
bn JT J / x sin nx dx n . either as an even it may be expanded in the interval . .. nnx where . .
If a function f x satisfies the Dirichlet the discontinuities conditions in some interval /.
Consider c fl . .Sec. . smhax. d al. . the limits of integration in should be replaced respectively
by a and a/ in the the points of discontinuity and at the endpoints of the interval x ic. . Q t/i ..
cosh ax. fx . . . x eax slnax. fx .. bx when special cases .. construct the graph of the of the
corresponding series outside the interval JT.. . r x ct . functions in a Fourier determine the
sum of the series series at lt gt Consider the special case when fi . . the function fx jr a
Fourier series in the interval ji. fx .. Fourier series of a period . at our discretion.. f x cos ax.. .
/v t r v . . . . L fxslndxn. .dx . . then at of the function belonging to this interval the following
expansion holds A nued r.. a a b l b a . i. / x . xl At the points of discontinuity of the function f
x and at the endpoints of the interval. . . n. Ji in an incomplete Fourier series of sines or of
cosines of multiple arcs. X function itself and of the sum JT. nx . . x / a y . juc . n may. a
function Fourier Series then an If / x is odd i. . fx . IT / x / . / of length /. hence. the . be contior
an odd function. in the interval Jt. . interval ji. af/ of length /. ju fli cos J amp stay a .
Expand the following functions. jt. /A. x.. fx . Use the expansion obtained sum the number
series Take the functions indicated below and expand them. when when / x when ji lt.c
interval x when Expand the following sines of multiple arcs functions. fx . in when lt. Sketch
graphs of the functions and graphs of the sums of the corresponding seiies in their domains
of definition. in the . in the interval sines of multiple arcs u when in co . . Find the sums of the
following number series . b of cosines of multiple arcs. fx x by means of the expansion
obtained . into incomplete Fourier series a of sines of multiple arcs. g Ch. n lt x lt n. Find the
sum of the following series by means . . Ji. . in the interval . eax .. sin. K. Series Expand the
function fx n. in sines of multiple arcs to . fx of the expansion obtained IPP. i quot quotquot . x
xn when x. in the interval .
Ji expanded sines odd mul tiple arcs. n represents an expansion the function fx odd multiple
and in is it odd is arid /fyf / in y of then the interval Ji. n Expand the following function in
cosines of multiple T . ltJClt fjc Expand the following functions. ltlt . ltjclt. Expand the following
functions in cated intervals . x sin x. . in incomplete Fourier series a in sines of multiple arcs.
fx Fourier series in the indi x ltlt. cos x Fourier Series . and b in cosines of multiple arcs .
Qltxltl. . Using the expansions oi the functions x and x in the in cosines of multiple arcs see
Problems and it . Prove that L if the function /x . .Sec. . Jt fxx fx e /ltlt/. /M . . . is even and we
have the interval arcs. prove the equality interval . if x f x then its Fourier series in in cosines
of Ji. in the indicated intervals. when . / . . j when ltjclt. . / . . J when ltjrlt. / x when lt x Y cos x
when .
sin x f sin x . . get a differential equation of type whose general integral in the corresponding
region is the relation . /quot y /. while the graph of this function is called an integral curve. .
Find the differential equation of the family of parabolas yCxCJ.. The integral Cp . f/.. eters Cj
if we have a family of curves C n from the system of equations and eliminate the param . By
assigning definite values to the constants C. integrals. we. which contains n independent
arbitrary constants C t ... and . The function y yx..Chapter IX DIFFERENTIAL EQUATIONS
Sec... Conversely.. dx .... Cw of the differential equation . We have and. Forming Curves. we
get particular . C n in . Verifying Solutions. dx n . . is called a differential equation of order n.
Differentiating equation / twice. and the desired differential equation C from equations we
obtain . generally speaking... C n and is equivalent in the given region to equation . An
equation Fx. then it is usually called an integral Example . y of the type n t/ . Example . OA..
Solution. /quot we get C l C and C. Initial Conditions Differential Equations of Families of .
consequently. If the solution is represented implicitly. Eliminating the parameters C. is called
the general integral of this equation in the respective region. y. Check that the function t/ sinjt
is a solution of the equation where y Solution. yx is the soughtfor function. which converts
equation into an identity. Basic concepts. is called the solution of the equation.
c . . Cn are determined this is possible from the system equations lt/ cpo.. lt/o C lt . i/ ...
whence and. t C . // . t t/ x. jc // djc h dy . We in y have .. y y n l are given and we know the
general solution of equation y ltVx.. Initial conditions. cos lto/ . C B . Putting formulas and we
obtain ... . y Show . x C.. Determine whether the indicated functions are solutions of the given
differential equations . ltltgt. a y xe b y x quot // / . j/ . differential equation y the initial
conditions for the desired particular solution y yx of a n fx..C sin CD/.. / a jc that for the given
relations are integrals X. C n . differential f equations the indicated . for which y Solution.. i
Example . c . sinjc cos. . y . l. then the arbitrary constants of C.. Find the curve of the family .
JK x h // . . . x f X/ t / e.Sec Verifying Solutions to this It is easy verify that tha If function
converts equation into an identity. l y e. hence. // . if C lt .. . y... . C C. . . .
i where fx. . . yC. and or . xyxyquot Differential Equations Ch. y i ay . gt. .. Types of firstorder
differential equations. of the given families of curves C. differential . FirstOrder Differential
Equations P. a parameter. The general integral of equations x. . A differential equation the
first order in an unknown function yt solved for the derivative y is the form t of of y f. . y lt/ .
//plane. . . xy . lt. . y P and Q x. or xtyy I. In certain cases it is convenient to variable x as the
soughtfor function. Cequot. Form the differential xyplane.C. and to write in the is form where
gfr . j xe. Form the differential vertical axis in the yplane. . y y nxy. equation of equation of all
parabolas circles in with the all For the given families of curves find the given initial
conditions .ygt i Taking into account that and j the differential equations and may be written in
the symmetric form Px. yr Form thej . y C sin . lt/ all straight lines in equation of the . y Cx. l i
n . Sec.y . .p fo re P arameters gt amp M. Q . y C lt . By solutions to we mean functions of the
form t/ cpjc ihat satisfy this equation. C. lt/. . . . C. t/ltty. iiamp. are arbitrary constants Cx. x .
where y are knowri functions. xy yy equations t Form differential Q.ex the lines that satisfy yC
l . C. ydxQx.. y consider the the given function.
are called isoclines. By constructing the isoclines and direction field. of the form where C is
an arbitrary constant. . . y . rection field. field of By constructing the isoclines xk straight lines
and the diwe obtain approximately the field of integral curves Fig. y . Solution. y k. .Sec. lt/ x.
is y called a direction field of the differential equation and is ordinarily depicted by means of
short lines or arrows inclined at an angle a. FirstOrder Differential Equations is equation .
construct the curves of the equation yx. Using the method of isoclines. Example . Direction
The set of directions tana /x. y . regarding the latter as curves integral which at each point
have the given direction of the field. in the simplest cases. make approximate constructions
of fields of integral curves for the indicated differential equations is . it is possible. f . equal to
k. field. The family of parabolas the general solution. Curves fx. ylftf . Using the method of
isoclines. at the points of which the inclination of the field has a constant value. to give a Fig
rough sketch of the field of intagral curves.
find y A. t/ . we f /i one step of the process. y is continuous in some region in this region has
a bounded derivative .. . . Using Eulers method for the equation find /I. Differential Equations
Ch. / i. . .. .. yQl..x/. UaltxltA. / . ygt tnen through each point y Q that belongs to U there
passes one and only one integral curve y yx of the equation cp ol. i / . For the sake of
comparison.. the exact value is e T ss . Cauchys theorem. A .. /t . A . For an approximate
construction of the integral curve of equation passing through a given point replace the curve
by a broken line with vertices M. yx y. . yl. find y. t/ . where M . y y.. f y . Example .. .. Eulers
brokenline method. /. if We yl /i . b lt y lt B If and a function / . find lt/ find yl . construct the
table Thus. Using Eulers method. find the particular solutions to the given differential
equations for the indicated values of x . X.
In y is In x ln C in t where the arbitrary constant taken antilogarithms we get the general
solution In C. Solve the equation If for some value y is yb gt In particular. respectively. also
as lines x then the function y is ti Q r/ . we get fxdx Whence. After taking f where C C.
Differential Equations with Variables Separable Sec. by integrating. dividi dividing by y ila
contained in the formula When we could lose .. FirstOrder Differential Equations with
Variables Separable. yQ we have directly evident a solution of equation Similarly. the straight
a and will be the integral curves of equation . by the ieft sides of which we had to divide the
initial equation. consequently. Orthogonal Trajectories separable . Firstorder equations with
variables separable. dividing both sides of equation by X. Example . for C the solution . and
Y / . we the general integral of equa tion in the form Similarly. y dy Q by gy and multiplying
get i X x Y y dx Dividing both sides of equation by dx. Equation may be written in the torm dx
x Whence. is a firstorder equation of the type y An equation with variables fxgy X. find the
solution that satisfies the initial conditions Solution. x Y. we have and. . the roots of the
equations X. logarithmic form. x we get the general integral of in the form Y y and integrating.
but the latter is . separating variables. if a and b are.Sec.
Differential equations of the form with variables where u reduce to equations of the form by
means of the substitution u is the new sough tfor function Orthogonal trajectories are curves
that intersect the lines of the given a right angle. Utilizing the given initial sired particular
solution is conditions. Whence. we get C . ygt ort ia is a parameter at family is the difierential
equation of the family. the de y J x equations Certain differential equations that reduce to
separable. x. Fig. we get the differential equation of the orthogonal trajectories it x of
integrating. If F x. then O is the differential equation of the orthogonal trajectories. Differential
Equations Ch. and. Example . replacing if by . we have i x family parabolas Fig. tion of the
family / we find the duerential equa . Find the orthogonal trajectories of the family of ellipses
Solution Differentiating the equation . . hence. y. .
. When forming differential equations in geometrical problems. y xy . curve is the hyperbola
xy . Solution. we can readily pass to a differential equation by differentiating both sides. // .
for which the segment of any tangent line contained between the coordinate axes is divided
in half at the point of tangency. we get dy Ty In and. the desired Solve the differential
equations . use is made of the geometrical significance of the definite integral as the area of
a curvilinear trapezoid or the length of an arc. Q. Differential Equations with Variables
Separable . we can frequently make use of the geometrical meaning of the derivative as the
tangent of an angle formed by the tangent line to the curve in the postive xdirection. al . Find
a curve passing through the point . .y be the midpoint of the tangent line AB. satisfy Find the
particular solutions of equations that indicated initial conditions x . y is M M dy dx This is OA
OB d x y x the differential equation of the soughtfor curve. // jq/ x fan ydx . In many cases this
makes it possible straightway to establish a relationship between the ordinate y of the
desired curve. Solve the differential equations by changing the variables . xy // /. Example .
its abscissa x. x e sec ydy Q. r/sin x the yny yl when . tan A sin y d cos x cot ydy . e y y e //
when jt when x .Sec. y tan //. Forming differential equations. Let x. by hypothesis we have a
simple integral equation since the desired function is under the sign of the integral. x ydx x y
fo/ dij . nxny Utilizing the initial condition. In other instances see Problems . c .and axes. x /
ldx x . x l V. Cor xy C. The slope of the tangent to the curve at x. It is given that OA y and OB
Zx. which by hypothesis is the point of tangency the points A and B are points of intersection
of the tangent line with the y. . . xy x dxx yydy . that is to say. and the tangent of the angle of
the tangent line /. . . to obtain the difleiential equation. consequently. Transforming. . i/ //l.
however. In this case. xyy . we determine C Hence. Q.
. construct the families and their orthogonal trajectories. x . y are homogeneous functions of
the same degree. . . y and Q x. . . Find the equation of a curve which passes through the
point . Find the equation of a curve that passes through the point . by this curve and the
ordinate of any of its points is equal to / the abscissa of this point. Sec. We can also apply
the substitution xyu. Differential Equations C/i. Equation may be reduced to the form and by
means it is of the substitution y xu. if P AT. . a tf/ a. transformed to an equation with variables
separable.. Find a curve which has a subtangent twice the abscissa of the point of tangency.
t/ y a ffx. where u is a new unknown function. Find the orthogonal trajectories of the given
families of curves a is a parameter. xy . Find a curve whose abscissa of the centre of gravity
of an area bounded by the coordinate axes. In Examples and . Find a curve whose
subtangent is of constant length a. Find a curve whose segment of the tangent is equal to the
distance of the point of tangency from the origin. . t A differential equation Px ydxQx. . 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. pass to polar coordinates . . Find the general solution to
the equation Example . . . if the segment of the tangent to the curve between the point of
tangency and the t/axis is of constant length . .ydy is called homogeneous. Homogeneous
equations. . x a. . FirstOrder Homogeneous Differential Equations . for which the segment of
the tangent between the point of tangency and the axis is divided in half at the point of
intersection with the yaxis.
x ydx f/ /. and also indicate the curves that pass through the points .x . we FirstOrder
Homogeneous Differential Equations Solution. For the equation x y dx xydy of integral
curves.by separable. then./ . . . Equations If that reduce to homogeneous equations. then f
xu eu u t. What shape should the reflector of a search light have so that the rays from a point
source of light are reflected as a parallel beam . Find the equation of a curve that passes
through the point . .Sec. Integrate the differential equations . y x . . c. x xy jf dy f/ dx y jvy .
Find the particular solution of the equation x Q. and I Ue I . we get an equation with variables
. putting into equation x w a. andjl. Put y ux. j/ where the constants a and P are found from
the following system of equations. then. or Integrating.l. . provided that r/l when x . whence x
In In . respectively. xydy Solve the equations . .fp. and has the property that the segment cut
off b the tangent line on the r/axis is equal to the radius vector of the point of tangency. get w
Q In In . xdy ydx Vx ifdx. xyydxxdy Q. y. t get a homogeneous differential equation in the
variables u and v. . If u. putting in a. we a a b c . . find the family . .
Find a curve for which the segment on the yaxis cut by any tangent line is equal to the
abscissa of the point of tangency. . we apply a method that called variation of parameters.
Find the equation of a curve for which the segment cut off on the yaxls by the normal at any
point of the curve is equal to the distance of this point from the origin. off Differential
Equations Ch. . To solve the inhomogeneous linear equation . Find the equation of a curve
for which the area contained between the axis. we seek the solution of the inhomogeneous
equation in the form of .e f Example Solution.r I . we put into y and y which are found from .
one of which is a constant and the other a variable. the curve and two ordinates. then
equation . relationship . I. and we get the general solution of in the form y Ce is J P PX dx . . .
Find the equation of a curve whose subtangent is equal mean of the coordinates of the point
of tangency. A differential equation of the form yQ of x degree one in y and y is called linear.
To do this. We thus get the general solution of the inhomogeneous equation in the form /x J
Cx. is equal to the ratio of the cube of the variable ordinate to the appropriate to the
arithmetic abscissa. . If a function Qjt. Sec. the variables may be separated. is The
corresponding homogeneous equation get Solving it we C . and then from the differential
equation thus obtained we determine the function Cx. which consists in first finding the
general solution of the respective homogeneous linear equation. . In this case. Solve the
equation y tan /.cos x. FirstOrder Linear Differential Bernoullis Equation Equations. assuming
here that C is a function of x. that is. Then.y f takes the form Q and is called a homogeneous
linear differential equation. Linear equations.
and from we find u. we find. from Bernoullis equation. Solve the equation y y yugtv. Example
. VH Solution. A first order equation of the form y find y. sin cos C cos . the general solution
equation has the form COS In solving the linear equation tion we can uv t also make use of
the substitu y will where u and v are functions of f x. whence we have Putting this expression
into . we get .Sec. This is Bernoullis equation. we then from we find M. we require that PjcM .
Bernoullis Equation Considering C as a function of x. near equation by means of the
substitution z the or substitution the y uv apply directly reduced to a li r/ It t also possible to
method of varia tion of parameters. and differentiating. cos . Then equation have the form u If
Pxuv vuQx. be fulfilled. Putting we ijet uv vu uv x yquotuv or v f u u j f vu x VtaT. dC cos x A
dx I sin x /. we get dC djc . is called Bernoullis equation quotquot. To determine the function
u we require that the relation u w x u x.xgto cos x Putting y and y into . hence. or dC r dx
whence Ccos di of j Hence. P lt y Q x y It is is where a and a .
any x. . y . cut off on the yaxis by a tangent line. Find the equation of a curve. xdyyl sin A y
smxdx. . x x. . X y y j e Q y b when x a. xy x . . . l dx JL x Xy y m . . a segment of which. a
segment of which. y . is proportional to the square of the ordinate of the point of tangency.
yianx cos x / when jt . . the general solution is obtained in the form Find the general integrals
of the equations . ax . Find the equation of a curve. cut off on the yaxis by a tangent line.
Find the general solutions of the equations . find v and. a tangent line and the radius vector
of the point of tangency is constant. ydxxy Find the particular solutions that satisfy ditions the
indicated con . Prove that the expression y lt y of a linear remains unchanged for the
geometrical significance of this result . y when x. cul off on the xaxis by a tangent line.
whence we Differential Equations Cfi. . is equal to the subnormal. x equation. consequently.
Find the equation of a curve. . is equal to the square of the ordinate of the point of tangency.
What is . dx . Find the curves for which the area of a triangle formed by the axis. Given three
particular solutions y. a segment of which. .
then there exists a function U. is readily found in two cases . . or from the formula form x.y
the equality dxQx t ydy Q p t/ is fulfilled. The gen C. from this we get y f/r. . Exact Differential
Equations. is Solution.Sec. x fV / C is the soughtfor Differentiating s with respect we find l y
cp qgt / // qgt jt s . y integrating factor such that iPdx Qdy dU. Sec. Exact differential
equations. C . dU see Ch. If for the differential equation Px.V . VII. y mined by the technique
given in Ch. If the left side of equation lis not a total exact differential and the conditions of
the Cauchy theorem are fulfilled. This is an exact differential since the form J K/ i X y and.
Whence it is found that the function u. Exact Differential Equations. VI. x y y by y f y y C.
general integral of the equation. t/ fxV C consequently. Here. Integrating factor. satisfies the
equation The integrating factor u. Example . equation. qgt xgt V jc f qgt y. Sec. hence. Find
the general integral of the differential equation x Af/ dx xy y dy . y is detereral integral of
equation is U x. then equation may be written in the and is then called an exact differential
equation. U.a. Integrating Factor . We finally get and hypothesis. . . Sec. The function U x.
Integrating Factor . Find the equation of the curve for which of the tangent is equal to the
distance of the point tion of this tangent with the xaxis from the point the of segment intersec
M .A. and y whence U U xy dx to y. the equation of .
. Qxy quot r . r . y y . Differential Equations C/t. . V / x dx xydy Q. hencegt ft dx or r cty r dx Q
jB. . x . . Integrating it. Solve the equations that admit of an integrating factor of the form fi M
or M Hy . Example Solution. . jc S dx xy y d/ . . cos t/ . . Solve the equation f xy xy J and
Since quot d Here P y rVf . x dx it follows that rffi fdP dQ dxdx and JS Q. . jc y sin r/ rfi/ x sin /
// cos y dx . Find the particular integral of the equation which satisfies the initial condition . .
Multiplying the equation by i e t we obtain which is an exact differential equation. ydxxydy .dy
dxj . . we get the general integral Find the general integrals of the equations x t/d z/dy . yl . .
xydxxdy Q. In . .
Fx. has the form M AM/.y.ygt lt// lt. or a we get the general integral of the given equation y C
A xy to family of parabolas. C gt x. Differentiating the general find the singular integral
integral with respect C and eliminating C. . the. generally speaking. Solving for y we have two
homogeneous equations do fined in the region x the general integrals of which are or
Multiplying. through each point xQt / of some region of a plane there pass two integral
curves. y. generally speaking. FirstOrder Differential Equation not Solved for Derivative Sec.
there may be a singular integral for equation . in each case.i which for example two
equations is of degree two in y. a singular integral is the envelope of a family of curves and
may be obtained by eliminating C from the system of equations ltIgt and OJx. are Besides. /.
Example . C . note that the curves defined by the equations or are not always solutions of
equation . C . the general integrals of equations . gtc . we JtO. where tlgt COM. or . y.lt/
Thus. F p x. The general integral of equation then. / If an equation I . for .p. C by eliminating
p t/ from the system t of equations Fx y. by solving for y we get ytix. Firstorder differential
equations of higher powers. y.Sec. p. therefore. is It may be verified that yjc the solution of
this equation. FirstOrder Differential Equations not Solved the Derivative . a check is
necessary. Geometrically. Find the general and singular integrals of the equation A// We f Aj/
y Q. Solution.
Solution. putting p yj which is the t pay . Find the In Problems curves. Solving a differential
equation by introducing a parameter. It may be verified that / r. Find the general and singular
integrals equation yy xy. ft dp dp or pp xp x. Integrating we get p x C. we x x obtain the
singular solution / r. If a firstorder differential equation is of the form It is xp of the Ch. . then
the variables y and x may dqgt be determined from the system of equations where p t/ plays
the part of a parameter. we have the general solution or lt/ y Differentiating the general
solution with respect to C and eliminating C. we have x pppf . general and singular integrals
and construct the of the equations of integral field . Substituting into the original equation. .
Similarly. y JC. we get x x into the given equation. .is the solution of the given equation. If we
equate to zero the factor p x which was cancelled out. we get and. /MO. same singular
solution. Making the substitution t/p. orj l.Differential also Equations by differentiating
possible to find the singular integral with respect to p and eliminating p. we rewrite the
equation in the form Diffeientiating with respect to x and considering p a function of x. if y ty x
tnen x and y are determined gt from the system of equations Example .
Sec Q. If pqgtp. solve the equations . . Solve the equation lt/t/A i. y .v dp Solving this lineai
equation. a singular solution that is found in the usual way. Putting ing yp we have //pv
different a Ting and replac dy by pdx. where p f called Lagranges equation Equation is
reduced to a linear equation in x by differentiation and taking into consideration that dypdx p
dp. Solution. . An equation of the form ltPP gtP. The Lagrange and Clairaut Equations .
xsmylny. yij xyy Q. equation y that pass through the point M . general solution is of the form
yCx is also a particular solution envelope that results by eliminating the parameter p from the
system of equations Its I x lJ Example. then we get a may be raut s equation There C a
family of . If in equation lpltMp. then from and we get the general solution in parametric form
where p there parameter and fp gp are certain known functions. . Besides. . yyxyly x .straight
lines. we get p dx p dx or . y yegtquot. . y . we will have l . y xy . / Sec. . Introducing the
parameter y p. Clairauts equation. Find the integral curves tf is is The Lagrange and Clairaut
Equations y of the . Lagranges equation.
Hence. r . consequently. rFT . equation does not have a singular integral. y xy y. y y Clairaut
equa . f/ U d y . Miscellaneous Exercises on FirstOrder Differential Equations . . . Firid the
curve for which the area of a triangle formed a tangent at any point and by the coordinate
axes is con stant. by . Solve the Lagrange equations . Find the general and singular integrals
of the tions and construct the field of integral curves . therefore. Find the curve it the distance
of a given point to any tangent to this curve is constant. and. Find the curve for which the
segment of any of its tangents lying between the coordinate axes has constant length /. the
general integral will be ll To find the singular integral. . we form the system f/px in the usual l.
Whence Differential Equations C/t. Putting y into we are convinced that the function obtained
Is not a solution and. Sec. . . Determine the types of differential equations cate methods for
their solution and indi b /// c ju/ .. y lt/. y xyy. . way. Xy V y y xy j. y ij . .
yyl. . . . y d. X . . Miscellaneous Exercises on FirstOrder Differential Equations i j lt/ x cos f/ k
y sin x xy y . xy . . . . x l . xdy ydxifdx. / . / f . dydx v . . y xy . . . . H P P J S . cosA. / // x Hl/l j
lny t y . yl dx xy xyQdy Q. xrfx . J / ff . Ht ./quot xy f lt///. f/ or/ jc . . ln. a x b jcln ycos dy ydy. T
. y xcosyr a sin lt/ . xydx y y dy . ////cosArsin.y xtf nx. . y/l xegty.sin // . x n Solve the equations
.x /quot h f y . dx . xdy.Sec. xy . . i . axt . lt/. .tcosx. . dx .v . . y J . . // j . . f/yhf/ . . xyy dxxdyQ. .
x x d. m x xy dx dy . .y dx y dx./ Zxyy / . y dy J dJf . e y dy x ltje .
b y z for x. y for . Ya . t/ . cut off by the tangent on the xaxis. Find a curve whose angle
formed by a tangent and the radius vector of the point of tangency is constant. .q/ . the
segment of which. Find the equation of the curve if it is known that the curve passes through
the coordinate origin. x . y . xy . for the indicated initial con ditions . . dy x xy V dx . lt/p Find
solutions to the equations . yy f y y A. x y y y x y for .y . . Find the curve passing through the
point . Differential Equations Ch. yJdx f xy ydyQ. this curve and the ordinate of any point on it
is equal to the cube of the ordinate. is equal to the length of the tangent. e . . for . c y . . . Find
a curve if we know that the sum of the segments cut off on the coordinate axes by a tangent
to it is constant and y for . The is sum of the lengths of the normal and subnormal equal to
unity. . . cos for x . . . byl forjc. . xy y a yl for jcl. Find a curve.. xyy a // for jcl. / At. for jcl. a Q
for je . / . . . y . J/oM dy x y . Find a curve knowing that the area contained between the
coordinate axes.q/ p . xyy x yy x dx Q. cot . by this curve and by the radius vector of any
point of it is proportional to the cube of this radius vector. for x . for . . b y for x. e. . . xyyx .
Find a curve knowing that the area of a sector bounded by the polar axis. jt / y equal to a. for
which the subtangent is equal to the sum of the cooidinates of the point of tangency.
fact that one hour. xvhn f . . kilograms of u . and the radius vector of the point of tangency is
constant and equal to a . Miscellaneous Exercises on FirstOrder Differential Equations . will
contain . it When forming is and. Water is entering the tank at the rate of litres per minute.
problems. If the concentration is uniform. The quantity of solution in the tank at that instant
will be / litres.. The area bounded by a curve. Find the curve. Therefore.. . Find the equation
of this curve if it is known that the latter passes through the point .crements of the desired
quantities these relationships are accurate to infinitesimals of higher order are replaced by
the corresponding relationships between their differentials. and the mixture is flowing out at
litres per minute. the coordinate axes. Let the quantity of salt in the tank at the end of t
minutes be x kg.Sec. ft f . . found from the expiration of salt. is At . particularly in phvsical
frequently advisable to apply the socalled method of differenwhich consists in the fact that
approximate relationships between tials. This does not affect the result. Find the curve if we
know that the midpoint of the segment cut off on the xaxis by a tangent and a normal to the
curve is a constant point a. a tangent. firstorder differential equations. . Separating variables
and integrat we obtain ln or n lnC C M C x The constant C . How much salt will the tank
contain at the end of one hour Solution. then the quantity of substance in volume V is cV. .
the concentration c the QQ kg per litre. During time dt litres of This ing. Problem. of which the
segment axes is of the tangent contained between the coordinate divided into half by the
parabola if x. Find the curve for which the area of a triangle formed by the xaxis. Find the
curve whose normal at any point of it is equal to the distance of this point from the origin. The
concentration c of a substance is the quantity of it in unit volume. the tank . that is. .
consequently. and the ordinate of some point of the curve is equal to the length of the
corresponding arc of the curve. solution flows out of the tank the solution contains cdt kg of
salt. The concentration is maintained uniform by stirring. is the sought for differential
equation. a change of dx in the quantity of salt in the tank is given by the relationship dt. the .
A tank contains litres of an aqueous solution containing kg of salt. infinitesimal h.
and g is the acceleration of gravity. Find the relationship between the air pressure and the
altitude if it is known that the pressure is kgf on cm at at an altitude of metres. what part of
this quantity will reach a depth of metres . The rate of outflow of water from an aperture at a
vertical distance h from the free surface is defined by the formula where c. . The quantity of
light absorbed in passing through a thin layer of water is proportional to the quantity of
incident light and to the thickness of the layer. The retarding action of friction on a disk
rotating in a liquid is proportional to the angular velocity of rotation. W at constant
temperature a degrees. . Prove that for a heavy liquid rotating about a vertical axis the free
surface has the form of a paraboloid of revolution. it m a hemithrough a cir . If one half of the
original quantity of light is absorbed in passing through a threemetrethick layer of water. kgf
on cm . By how much will the band increase in length due to its weight if the band is
suspended at one end The initial length of the band is /. During what time will a body heated
to cool off and during the first to if the temperature of the room is minutes the body cooled to
. in the bottom. Radium disintegrates by one half in years. Find the relationship between the
temperature T and the time f if a body. is brought into a room Differential Equations Ch. Find
the percentage of radium that has disintegrated after years. sea level and . When solving
Problems and . . Find the relationship between the angular velocity and time if it is known
that the disk began rotating at rpm and after one minute was rotating at rpm. make use of
Newtons law. . . The rate of disintegration of radium is proportional to the quantity of radium
present. Solve the same problem for a weight P suspended from the end of the band. During
what period of time will the water filling spherical boiler of diameter metres flow out of cular
opening of radius . . According to Hookes law an elastic band of length / increases in length
klFk const due to a tensile force F. . by which the rate of cooling of a body is proportional to
the difference of temperatures of the body and the ambient medium. heated to T degrees.
. Putting . Rl and the electromotive force of selfinduction the current / L. eEsmat E and are
constants .Sec. for instance. assuming y p. we get . Assuming that the rate at which the salt
dissolves is proportional to Ihe difference between the concentration at the given time and
the concentration of a saturated solution kg of salt per litres of water and that the given
quantity of pure water dissolves / kg of salt in minute. we have / p. p t p. f/ when x . I a
differential equation does not then. a Solving the latter equation as linear equation in the
function p. resistance / and selfinduction L is made up of the voltage drop proportional ing
velocity of . contain y explicitly. we get an equation ot an order one unit lower. The is covered
with is resistance to a body falling with a parachute to the square of the rate of fall. The air
bottom of a tank with a capacity of litres a mixture of salt and some insoluble substance.
HigherOrder Differentia Equations of direct integration. HigherOrder Differential Equations .
whence p. The electromotive force e in a circuit with current i. Find the particular solution the
equation that satisfies the conditions Solution. If . ogt Determine at time / if and Sec. i when .
find the quantity of salt in solution at the expiration of one hour. The case then n i Miles If .
Cases of reduction of order. of Example I. Fx. Find the limit descent.
e. we have . Py Vy or Integrating. we have C . or dx whence. . then. we obtain C . whence
cosx or t/secx. for instance.and We have obtained an equation it. Put y tfp. we find C . i.
Putting yl and . yquot pgt we get an equation of an order one unit Example . Hence.
integrating once again. when then . w obtain Putting solution y is when x . Differential
Equations Ch From the fact that yp when x . putting lower yp. our equation becomes
Solution. Hence. of the Bernoulli type in p y is considered the argument. we have arc cos x C
. Solving we find From the fact that f/p when r/l. Cj. Hence.. Find the particular solution p t of
the equation provided that /. the desired particular y If a differential equation does not contain
x explicitly.
lt/quotT. f/quot . . Find the general integrals . . / . . . HigherOrder Differential Equations I
Solve the following equations / . solutions for the indicated initial con x Mz/quot l / .f/quot s . .
yl y Jt . yyquot . yquotnx . x x. yy yquot yquotyyquotify. l. and for j/ / yyquot . xyquot /quot . jfy
yquot. . . of the following equations yyVyifyyif. yl for jc . lt . y for xl. . for for A e s . . p lt/quot .
t/ t/ y. I. . for x . . y tor i. . j/l for / lt/ l J/ / jc / . t yTquot quot . i/Vl. i. xyquot i/ . for . Find the
particular . forjt . // /quot. f/i/quothf/ ylnj/ satisfy the indicated conditions l. . ytf i/quotyquot yl.
/i/quot F . . . JC forxl. y for . y . Find solutions that . . . lt//. yyquot . /nln. . t/l . yify /quot f yquoty
. y nx. lt/quot .y . r/l. yquot xifV y . j/ lt/ j/t/quot. tyquot// . y x . ditions xy x t/quota/quot. .Sec. /.
yy yyQ. y . /Y foi x . xyquot . yquotquotf yquotl.X lt/quot . J . . . y y j/ for . for x . . r/ln. f/ l.
. yl. at it. Find the curves of constant radius of curvature. . . / jc / l / egty . yyquoty I/quot/ /.
the point . . yl. forxl. . The weight of the cable is neglected. Find the position of equilibrium of
a flexible nontensile thread.. forl. is We may consider that is the air resistance in free fall of
motion proportional square if the initial velocity to the of the zero. . where V is the force of
reaction of the . Find a curve whose radius of curvature is double the normal. Find a curve
whose radius of curvature is equal to the .. H//yquotf y ff Solve the equations . for . yyyif .
plane. y for . 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. . . 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. velocity. / yquot y // . y e for . . and
tangent. for x H/ lquot x . j . . y y . Find the law . . . Q y . y . The frictional force is ji/V. . . Find
the law of motion if the angle of inclination is a. yy yquot y yquot Choose the integral curve
passing through x Q. y for . lquot. for . y l for xl. . . . z Gxyquot / A i I . Differential Equations
Ch. and the coefficient of friction is p. to the straight line y . . . . . Find a curve whose radius of
curvature is proportional to the cube of the normal. A heavy body with no initial velocity is
sliding along an inclined plane. Hint. Find the curves whose projection of the radius of
curvature on the //axis is a constant. jr/ t/ . i i //I. normal.
/ /. y .... . C n x ya . The resistance of the water is proportional to the velocity and is kgf at
metre/sec. Solve the equation . C . . . these functions are called linearly independent. t x y
..... where the functions Cjx ing system of equations lt/i . If the fundamental system of
solutions /.. / ..... How long will it be before the velocity becomes m/sec Sec. The general
solution of a homogeneous linear differential equation w . Example. . The general solution of
an inhomogeneous linear differential equation with continuous coefficients P. . P n x y .. y tt
fundamental system of solutions. coefficients P. .. Inhomogcneous equations.o. C lf .
Homogeneous / equations. x e/ltquot. Cn otherwise. x and the right side f x has the form
where and Y the general solution of the corresponding homogeneous equation particular
solution of the given inhomogeneous equation ... then the general solution of the
corresponding inhomogeneous equation may be found from the formula f/ is a y C l x y. n are
determined from the follow C t y Cn x /. . A motorboat weighing kgf is in rectilinear motion
with initial velocity m/sec.. n is of the form where are linearly .. . .Sec.c. th continuous P. such
that functions i qgtix.wlt/... . Linear Differential Equations //ltgt is Linear Differential Equations
f . qgt A t . y n of the homogeneous equation is known. .x gt . f/ if there are constants C.o.
The P/iW are called linearly dependent not alt equal to zero.. the method of variation of
parameters.. independent solutions of equation .
Test the following systems tionships a x. x g d x. e c . b of functions for linear rela x . e . yx y
sin x. . . into account that the reduced form of t/quot we obtain Whence and. x x x x . cos A.
ing its homogeneous equation. Solution. consequently. sin x. know fundamental system of
equations l a b c y cos x y xe ye. . . a linear h sin . Knowing the fundamental system
homogeneous differential equation of solutions of a linear find its particular solution y that
satisfies the initial conditions . differential . y x sin . x lt d f/. f e e . . Differential Equations C/i.
f/ cos A. it may of be taken that yl nx may and and the solution equation be sought in the
form Forming the system the equation is and taking jt. Solving the homogeneous equation
irifo f we get Hence. Form x . cos . gtl and y where A and B are arbitrary constants. . x.
A secondorder linear equation with constant and q without the right side is of the form U If k l
and fc are roots of the characteristic equation . p a . Sec. . xnxlyquotxy y . . arl or r . xyquot
xy yx . that is. particular solution y x. a . and . . . solve the following inhomogeneous linear
equations. Linear Differential Equations of Second Order with Constant Coefficients
coeflicients p t.. If a is a root of the characteristic equation . / x t here r is the multiplicity of the
root P n cos bx Q m x sin bx. that is. yquot The general solution linear inhomoge neous
differential equation py wfW may be written in the form of a sum of the corresponding
equation I without determined from formulas to . Homogeneous equations. . Solve the
equation knowing its By the method of variation of parameters. . k. Solve the equation
knowing its particular solution y l . then r ax is where y ritfht the general solution side and Yx
e Qn . then we put Y eQ n x where Q n x is a polynomial of degree n with undetermined
coeflicients. Inhomogeneous equations.Q t then the general solution three ways of equation
kz if is vritten in one of the following /CVV C.t xy . CV .Sec. cnspxHCmpA if are real and /.
yquot f /sec A.e . if / . Linear Differential Equations with Constant Coefficients . yeVC. and Y
is a particular solution of the given equation . The function Y may be found by the method of
undetermined coefficients in the following simple cases a e P n x.a p and a a of a pi p . .
where P n x is a polynomial of degree n. /quot . f x If a is not a root of the characteristic
equation . ltpa .
The characteristic equation amp fc . K oH ne gei eral solution of the given equation is
solution of the equation y xe y y characteristic equation k has a double root /ff. form Hence.
whence f o g and anc . Example . and nl. the double root k and. we is f/ C . since a coincides
nHth . A X e i Ax A zx e Ax H.l and Solution. The coefficients. and equating the Example .
amp lu has roots fc. Jo A Thus. r Diilerentiating Y twice. . the general solution of the given
equation will be written in the l fe / Example fc a Solution. The right side is of the form . and
equating the coefficients of identical powers of x arid Cancelling out e the absolute terms on
the left and right of the equality. then are polynomials of degree Stf x cos of m. T N x sin bx.
since nl and /. will see Find the general solution of the equation /f y The characteristic
equation and jlr has roots /. Hence. we obtain /l . If Differential Equations cpa i bi . Diflerenx.
we have bA and fl . is T N x sin to bi r is the multiplicity the roots a for secondorder
equations. substituting into the equation. fl . n. Solution. x ftl The riht side of the equation is o
the form fxxe here. The general solution of the corresponding homogeneous equation first xe
zx t ax P n tiating obtain type Y e C e The right side of the given equation is/ x Y e zx Ax B.
The general solution ot the corresponding homogeneous equation . then we put S N x cos bx
Y e ax Nmax . Find the general solution of the equation yquot y y xe . the method of variation
of parameters see Sec. where Sx and Tux But if cpa amp/ . K x re ax where rl. In the general
case.B xe zx . where a and P l be . Find the general f . consequently. i. twice and putting the
derivatives into the given equation. Ch. used to solve equation . The particular solution is Y
xe Ax B.
for x . We then get four . y for x . . . for a. quot . . To this side there corresponds the fcl. .
yquot y . . .sin . .. particular solution Y. n are the corresponding solutions of the equations lt
ypywfiW sum y is l i. b quot cos x . sin. . . n. C . . yquottjy . D Therefore... y for for jc.Sec. .. tf
yy Q. . Differentiating . a . for type of particulai solutions inhomogeneous equations Indicate
the a quot AV x . t y . . . . the given xl. Y Y . and xsmx. iT y . If the right side of equa and K//
then the l. ji for x for jc and and for for x .j.. . i/ j/ . . y . . P n jc. f / y Find the particular
solutions that satisfy the indicated conditions . xcosx. twice and substituting into the equation.
cos C jc x sin x j cos A sin . from which we deter. f Q .. /.. . . here. equations A D X X mine A
cos f. yquot tf . The tion is principle of superposition of the sum of several functions solutions.
Q OT . . if y y . C . rl.. Linear Differential Equations with Constant Coefficients where a . . K x
The general solution is C. f .Y n n the solution of equation . quot . yquot ky . . / yyQ. . we
equate the coefficients of both sides in cos. Find the general solutions of the equations . .
quot .
lt/ x le xe e yquoty xe cos x V sin x. . /g y xsmx. . e x . yquot y y . . x sinA. . Consider the
cases p ogt. y. . Differential Equations Ch.y x sin . j/quot . . f . Solve the equations . / H. . Find
the solution of the equation yquot . pco. t/ . . . . /quot . yquot jc i/ / . / yquoty y e x e x / cos. f/
/l for r/ y x x y lQysrxie /f/ f/ if y . . . yquotye. .cl /quot f/quot f . satisfies the conditions yl. . .
Find the solution to the equation that satisfies the conditions /. y sin x that Solve the
equations Wy . . . . / t/quot yquot yquot yquot y x . x/l o sinp/. / ye tx . / . y f y lt/quot f Find the
general solutions of the equations . . . / for x . . . / t / / . tf . . c yquot sin x e x d yquot y y e sirx
..
Then of one load where. . x xe. Linear Differential Equations with Constant Coefficients . .
yquot y ianx. Solution. V b yquot x y xe . k tion is . Find the equation of motion that will be
performed by one of these loads if the other falls. A load weighing kgf is suspended from the
spring. lt/quot y tanh A. Find the law of motion of the load if the upper end of the spring
performs a vertical harmonic oscillation / sin/ cm and if at the initial instant the load was at
rest resistance of the medium is neglected. obviously. Find the period of oscillatory motion of
the load if it is pulled downwards slightly and then released. yquot y x . tfy y xle . Let the
increase in the length of the spring under the action in a state of rest be a and the mass of
the load. yquot y x cos x cos x. /quot yy xe xe . of variation of parameters. and so . The force
stretching a spring is proportional to the increase in its length and is equal to kgf when the
length increases by cm. . and. . x . Denote by x the coordinate of the load reckoned vertically
from the position of equilibrium in the case of a single load. the . a . whence C a and C a .
yquot y cot x. A load weighing P kgf is suspended from a spring and increases the length of
the spring by cm. . consequently. . yquot y y x le. m. . quot initial The general solu xC l cos I/
/ C sin I/ t t. . Applying the method following equations . . yquot y yquot y y / . t/ y xecosx
sinx.Sec. solve . Two identical loads are suspended from the end of a spring. lt . L . The
conditions yield Jt u and d Tt when .
characteristic equation Namely. A ball inside the tube is sliding along it without friction. If the
motion begins when m of the chain is hanging from the support. A y xe x sin PX. . if k is a
real root of the equation of multiplicity m. A material point of mass centres with a force
proportional to the distance the constant of proportionality is k. A long narrow tube is
revolving with constant angular velocity ogt about a vertical axis perpendicular to it. .
Inhotnogeneous equations. . . b at the initial instant the ball was located on the axis of
rotation and had an initial velocity v . yt xe x cos PX. how long will it take for the entire chain
to slide down .. . at the initial instant the point was located on the line connecting the centres
at a distance c from its midpoint and had a velocity Differential Equations Ch. . considering
that a at the initial instant the ball was at a distance a from the axis of rotation. . A chain of
length metres is sliding from a support without friction. Homogeneous equations. Find the law
of motion of the point knowing that the distance between the centres is . equation particular
solution of the inhomogeneous is sought on the basis of rules and of Sec. .. m of zero. then
to this root there correspond linearly independent solutions of equation m if a i then to the p/
is a pair of complex roots of equation of multiplicity latter there correspond m linearly
independent solutions m. The fundamental system of solutions y lt Un f a homogeneous
linear equation with constant coefficients n ylt a iy n an . Sec. ex sin px. of equation x y l e
cos PX. is attracted by each of two . is constructed on the basis of the character of the roots
of the Q. Linear Differential Equations of Order Higher than Two with Constant Coefficients t .
Find the law of motion of the ball relative to the tube. the initial velocity of the ball was zero.y
a n y gt lt . .
yquot Find the general solutions of the equations . Then and Eulers equation is transformed
z into a linear equation . Solve the equation x yquot Solution. Eulers Equations A where
linear equation of the form n l yfx t I a. .. A . Eulers Equations n . . . . y / lt/quotlt/... ai/ b. y v
J/quot tquot . yquotyquotyy x . cos In C a sin In x y t t .y v yquotiJ x sss SB T MS... yquot ... is
called Enters equation. b.. lt. w . y yquot . with constant coefficients. Fxample . Find the
particular solution of the equation that satisfies the initial conditions y Sec. cos C. . . putting
ax be . . the given equatioi takes on the form whence or C. yv y nnl . Putting xe..Sec. Let us
introduce a new independent variable /. A n are constants. gt. wv O.. sin y C. we get e Sx xy
/ z Tr ze dx didi Consequently.
. Solve the equation X y Xy l/. In . . ln. we get the .sinplnx . then to it there a In x cos P In . .
miquot Inquot cos p I Example Solution. . y m x nx . Differential Equations Ch. Solving it we
find Hence. .. a p is correspond m pair of complex roots of multiplicity linearly independent
solutions / m. For the homogeneous Euler equation Q the solution may be sought in the form
lt/ . If k is a real root of the characteristic equation of multiplicity m then to it correspond m
linearly independent solutions m. k Putting into tion from which /. lt. a . . y . .. . the general
solution will be yC Solve the equations . tT . . y . found from . We put Substituting into the
given characteristic equation equation.. /.. . we get a characteristic equaexponent k. xYx// . .
y n we can find the t i If . y Mnx.. after cancelling out x k . g .. V xy / .... .
and C are arbitrary constants. xy Equations / when . into equation . for example. we arrive at
a secondorder equation in one unknown y . dx dx dx From the first equation will we
determine / T lx dy y and then j from the second we have ax amp . Substituting function into
formula . we find secondorder equation with one unknown func where C. Solve the system z
T Solution. first equation of the system / and substituting the A.into the equation obtained
after differentiation. we determine the function z without new integrations. a normal of of a
system the solved for the derivatives of them with respect to x. We the desired functions.
system of two form To find the solution. . Example. The set of formulas and . where y is
replaced by gt. amp y TTquotT ax Put ting z and j. . Sec. f We differentiate the first equation
with respect to x . that instance. of is. for firstorder differential equations. we get a tion u.
Solving it. yields the general solution of the system . Find the particular solution of the
equation Systems of Differential Equations V that satisfies the initial y x conditions y .
Systems of Differential Method of elimination. ..Sec. have. we differentiate one of
Determining z from the value found.
da yz zx xif isolate the integral curve passing through the point . . dz . Solve the systems dy
z . Solving it we find and then We can do likewise in the case of a system with a larger
number of equations. dy . . dz . dx xy c dy dz xy z dy t dz . . . . z when je . y when / . . . .
Differential Equations Ch. . dT dx dy dz /// . dx r/ . . I JJ .
Integration of Differential Equations If it by Means of Power Series not possible to integrate a
differential equation with the help of elementary functions. A material point is attracted by a
centre with a force proportional to the distance.c x.nncn x n nl s In rttl lt. Find the trajectory. ...
.. A shell leaves a gun with initial velocity u at an angle a to the horizon. . Substituting y and
yquot into the given equation. .. putting X Q in place of x Example . Sec. Find the equation of
motion if we take the air resistance as proportional to the velocity.. we arrive at the identity
Collecting together.. then in some cases its solution may be sought in the is form of a power
series y n o n o The undetermined series into the of the binomial x are found by putting the
coefficients c n n .Ct . We put y whence. quot y of the equation and the subsequent
derivatives y n x where yx Q y Q t/ y x f are successively found by differentiating equation
and by n.. Integration of Differential Equations by Power Series .. The motion begins from
point A at a distance a from the centre with initial velocity perpendicular to OA. equation and
equating the coefficients of identical powers x on the lefthand and righthand sides of the
resulting equation.. . We can also seek the solution of the equation in the form of the Taylors
series y . differentiating. on the left of this powers of x and equating to zero the coefiicients
equation. . the terms with identical of these powers. Find the solution Solution. .Sec.. we will
.quot. we get yquot .
In Examples .. k .. y for xl. l l... have cs and so forth. .. . find the solutions of the equations for
the indicated initial conditions. quot . etc.. y for xa / . y y . it is readily seen that series
converges . yquoty /oquot .. generally speaking.. Find the solution y of the equation We put
We have y . . y x y for xy lx y y f/ .. we succes/ . for . similar for differential equations of higher
orders. . Generally.. . Example Solution. . . Testing the resulting series for convergence is.
this solution may be written in final form as l The procedure is x or e x. /. . With the help of
power series. test the solutions . complicated and is not obligatory when solving the
problems of this section. i sively find Differentiating equation y x y.. X X . for . For the
example at hand. X where for l T . Differential Equations Ch. .. y x . y y x y . Consequently. .
Consequently. fc quot c f/ and c y Q Applying dAlemberts oo lt x lt oo . obtained for
convergence. test. .
y / Jt / Sec.. for x . Cn u n . which are linearly independent among themselves in any finite
number and which satisfy the given boundary conditions. . a. t of equation that satisfies the
boundary conditions a. . the equation t of the points of a string a dt dx where a string. there is
an infinite set of such solutions M ttl. for x . The desired solution u is represented in the form
of a series arranged according rt to these particular solutions u The coefficients conditions. .
each of which represents the product of functions that are dependent on one argument only.
first seek the particular solutions of this specialtype equation. T Q is the tensile force at and
Q time t is if the its linear density of the Find the form of the string ends x and / are U Fig..
required to find the solution u ux. A transversal displacement u ux with abscissa x satisfies.
the string had fixed and at the initial x Fig. Solution.. . Problems on Fouriers Method . yl. . In
the simplest case. . It is instant. . . for . y .Sec. the form of a parabola u / and points had zero
velocity. / for . . / . . its f . . C n which remain undetermined are found t from the initial
Problem.. at time . Problems on Fouriers Method To find the solutions of a linear
homogeneous partial differential equation by Fouriers method. /. for .
C. u . Tt Q and Xquot x KX x . will readily be seen that we do To not lose generality by taking
for k only positive values every value h k there corresponds a particular solution k . this
expression into equation and separating the variables. . .. p where k an integer. constants.
equation is possible only the general quantity of relation is constant Denoting this constant X
we find two ordinary differential equations . Ak . Ihat satisfies the boundary conditions We
construct the series f I .. . of the series should Since du amp it y CD A / kan sm . C cos Kx D
sin X. knx tj siny. . A cos at B sin aKt. /. B. follows that. We Putting seek the nonzero
solutions of equation of the special form X xTt. we obtain and V kan . kant cos . and X/ . we
get T t X where A. cosj kan kan . D D XA are arbitrary From sinX/ condition since we have X
is Let and at cannot be equal to zero determine the constants. we get ZLW a tO l x when by
Since the variables x and t are independent. by putting . Solving these equations. hence.
kant p cosyJ kant ..lt. C the same time as C is zero. and the Differential Equations initial
conditions Ch. us It For this reason. kant knx whose sum obviously satisfies choose the
constants satisfy the initial conditions We equation A k and B k . and the so that the
boundary conditions sum .. . knx sin . smyJ smj.
is pulled out to a distance A cm at point x . Find the form l are and x string at time t if the
ends of the string fixed see Problem . / kast fA Osin . The soughtfor solution will . VIII. . Find
the the points of a straight of the form of the string . at n From familiar formulas Ch. Sec. the
of a crosssection of the rod with t displacement u ux abscissa x satisfies. . and the points at
of had zero velocity. Determine the shape of the string at any time /. and then released
without impulse. Hence. / ft . sin /if jtx V . . jix . . the equation at the initial time. at time /.
Determine the longitudinal vibrations of an elastic horiand pulled zontal rod of length / cm
fixed at the end back at the end by A/ l cm. in Problems on Fouriers Method coefficients A k
and B it is x necessary to expand in sines only. du function x. T. cm then released without
any impulse. to determine the a Fourier series. initial time / . and is d/ f where a cu a ltPu dx r
E is Youngs modulus and Q is the density of the rod. ltjtlt/ receive a velocity . the function ux.
At the string time t. fl A . and Ak Q if amp is even. we have h knx . had / . a the it form string.J
be cos / d A . .. its ends. attached at of the sine curve u A sin y . whose axis coincides with
the jraxis. . In longitudinal vibrations of a thin homogeneous and rectilinear rod. and /. At the
initial instant x and /.Sec. A string of length / cm and attached at its ends. if amp is odd. lx
and the .
A. . where a is a constant. for any time t in a rod temperature distribution Determine the
temperature distribution of length cm if we know the initial ux. the temperature u ux. t in a
crosssection with abscissa x at time /. . . x. in the absence of sources of heat. For a
rectilinear homogeneous rod whose axis coincides with the Jtaxis. satisfies the equation of
heat conduction di Differential Equations C/i.
in the narrow meaning / w an . are numbers resulting place such. Aa is located within . Since
in limiting actual practice relative error. known in that the i . Absolute is called the limiting the
limits a absolute error. if the number a definitely If has n correct decimals of unit the absolute
error of the last decimal from measurements made to a definite accuracy. for the limiting
relative error. . Number of correct decimals. replaces the exact The absolute error of an
approximate number a which number A is the absolute value of the difference between them.
more The exact number A A briefly. Operations on Approximate Numbers error. if it . digit i .
then the number a has n correct decimal the k places narrow / meaning of word. Aa. If there
is a larger number of significant digits in the approximate number. A is called the limiting
relative error of the approximate number for a. which satisfies the inequality . then it is said
that all decimal places of this approximate number are correct in a broad sense. we often
take the number the . . By the relative error of an approximate number a we understand the
ratio of the absolute replacing an exact number A A error of the number a to the exact
number A. We say that a positive approximate number a written in the form of a decimal
expansion has n correct decimal places in a narrow sense if the absolute error of this number
does not exceed one half unit of the nth decimal place. In particular. Relative error. take. .
A/la A or. The number A. And conversely. The number . In this case. quotif I To approximate
number a does not exceed a for example.Chapter X APPROXIMATE CALCULATIONS Sec.
the number when n gt we can / where k is is the first significant / I . which satisfies gt the
inequality l . of the number a. 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.
is always advisable to avoid subtracting close approximate numbers and to transform the
given expression. If Aa lf . not one of the decimals it difference is correct. Therefore. /XM ..
Whence result.. f . the results of final calculations.. In subtracting the approximate numbers
and . Multiplication and division of approximate numbers.. and in each summand a reserve
digit should be retained. f. ... if the it number is of correct final. j Hence. correct all digits in the
initial data are the narrow sense. The limiting absolute error ot an algebraic sum of several
numbers is equal to the sum of the limiting absolute errors of these numbers. Aa. if need be.
. we find that the limiting relative error of the product is the . and round off the sum by one
decimal place If we have to add approximate numbers that have not been rounded off. so
that this undesirable operation is omitted. to four correct decimal places.. .. . The product of
the approximate numbers .. Assuming that all decimals of the factors are correct. of . .. or
more . . are the limiting absolute errors of the appro. decimals of the product is three and the
be written as follows . we get the difference .J.. We note that the examples of this section
are. and for this reason the answers to them are given as approximate numbers with only
correct decimals. Addition and subtraction of approximate numbers.. . as a rule. The results
of intermediate calculations may contain one or two reserve digits.t . we retain in the answer
only a definite number of decimals Example . . . The limiting relative . Therefore. The limiting
relative error of a product and a quotient of approximate numbers is equal lo the sum of the
limiting relative errors of these numbers Proceeding from Ihis and applying the rule for the
number of correct decimals . in order to have. The relative error of a difference is not
amenable to simple counting. in the sum of a small number of approximate numbers all
decimal places of which are correct. .. . Powers and mth power roots of approximate
numbers. Example . . they should be rounded off and one or two reserve digits should be
retained. Example . The limiting relative error equal to the mfold limiting an approximate
number a the is relative error of this number The limiting relative error of Is mth root of an
approximate number a the th part of the limiting relative error of the number a. The relative
error of a sum of positive terms lies between the least and greatest relative errors of these
terms. should correctly. error of is . only correct digits at least in the broad sense. Then add
the resulting numbers as exact numbers. all summands should be put into the form of that
summand which has the smallest number of decimal places. . Henceforth. we shall not
otherwise if Approximate Calculations assume that stated in Ch. Then be guided by the
foregoing rule of addition while retaining the appropriate extra digits in the sum up to the end
of the calculations. of the .. Calculating the error of the result of various operations on
approximate numbers. Particularly unfavourable in this sense is the difference of two close
numbers.
.a n . . U . . two decimal places will be correct. Operations on Approximate Numbers xitnate
numbers a t ... V... that enter into the operations the principle of equal effects.Sec. Evaluate
S ln ... quot.. AS Aa We have general form S TT. . log . In V .. the approximate is . an . It
should be pointed out that sometimes when calculating the admissible errors of the
arguments of a function it is not advantageous to use the principle of equal effects. / log .
Thus. AaAiQ. admissible .. Aa. . Solution..UUigt Thus. are correct to one decimal place..
Applying the formulas of for the quantities AS or S given us and considering all particular
differentials UM M Ac or the quantities equal. since the relative error of number y is equal to r
the absolute error then equal to SOTTJQ quotquot we can be sure f tne first decimal place. ..
strictly speaking. Establishing admissible errors of approximate numbers for a given error in
the result of operations on them. . Hence.. since the latter may make demands that are
practically unfulfilable In these cases it is advisable to make a reasonable redistribution of
errors if this is possible so that the overall total error does not exceed a specified quantity. we
leave . the approximate numbers .. then the limiting absolute error AS of the resuU S /a t. a
solid cut passing through the diameter of the base to the base. . a n may be evaluated
approximately from the formula The limiting relative error S is then equal to Ji/r .. . . Example
.. o the approximate numbers a t . . Let us first compute the limiting absolute error AS in the r
Aamp ln a V b. . . And we pet the answer . we calculate the dak da k f absolute errors Aa lt .
Now let us do the calculations with one reserve decimal . Example equal to The volume by a
at of a off a circular cylinder plane R tan a. an angle a segmentquot. the problem thus posed
is indeterminate. and . . . is computed from the formula quotcylindrical of V R To what degree
accuracy should we measure the radius .. that is. .
.. sin a Thus. for an accuracy of .. . Find the product of the approximate numbers. cm
accurate to .. which are correct to the indicated decimals a . kg.. c . Subtract the approximate
numbers. . . a . their absolute and relative errors. c ... an accuracy of l c . .. mm... Determine
the number of correct in the narrow sense decimals and write the approximate numbers a .
for an accuracy of . quot Whence cm . b .. Indicate the possible limits of the results.. b . c . . .
s cm and the angle of inclination a so that the volume of the cylindrical segment is found to
an accuracy up to Solution. for. ... b .. c . . If AV. Ihe radius to mm if we measure . We
assume A lt m and . which are correct to the indicated decimals . ltAa R quotNna . c .. b .
Measurements yielded the following approximate numbers that are correct in the broad
meaning to the number of decimal places indicated Compute quot. cm and . f mm. A/ and Aa
are the limiting absolute errors of the quantities V. Find the difference of the areas of two
squares whose measured sides are . .. we are calculating A/ Act . . cm. b . .. R and a. we
ensure the desired accuracy in the answer to and the angle of inclination a to . Add the
approximate numbers. b ... then the limiting relative error of the volume V that Approximate
Calculations is Ch. Compute the absolute and relative errors of the following approximate
numbers which are correct in the narrow sense to the decimal places indicated a a . which
are correct to the indicated decimals a . ..
b . to within .. The sides of a rectangle are . Operations on Approximate Numbers m a .
accurate cm. to Compute the volts volt . . . . Compute the values . / cone are . The side of a
square is . cm accurate to . .. The legs of a right triangle are . whose period is close to sec.
cm. Compute the tangent of the angle opposite the first leg. cm and . of radicands are correct
to the indicated decimals a /.. cm . r . b . . in order to obtain its oscillation period with a
relative error of . cm. cm. / to . The relative error in measuring the lengths is . and the
generatrix is m approximately. Use these data to compute the . . c . To what degree of . cm.
To what degree of accuracy one of the legs is . the cm roots accurate to the mm. ji . Find the
area. The hypotenuse of a right triangle is . It is required to measure. . Find the quotient of
the approximate numbers. cm can we determine the second leg and the adjacent acute
angle Find their values. the lateral surface of a truncated cone whose base radii are m and
m. b j/ c KsTT. . The period equal to current if the electromotive force is equal and the
resistance is ohms ohm. Calculate the specific weight of aluminium if an aluminium cylinder
of diameter cm and altitude cm weighs .. cm. which are correct to the indicated decimals .
cm. Evaluate the absolute and relative errors of the result. cm .. How accurate must the
numbers n and g be taken . cm . c .. . Compute the area of the rectangle.Sec. gm. while the
relative error in weighing is . To what degree of accuracy do we have to measure the length
of the pendulum. and . . Compute the indicated powers of the approximate numbers the
bases are correct to the indicated decimals a . . cm total surface of the truncated cone. of
oscillation of a pendulum of length / is where g is the acceleration of gravity.. . The radii of
the bases and the generatrix of a truncated .
.. it is advisabie first to set up a table of finite differences. which includes the points . then the
error interval a. hen x . x n and x. For . accordingly. hence. Q P flL dbs where / is the rod
length. then .. quadratic interpolation. . to accuracy do we have to measure the radii and the
generatrix and how many decimal places do we have to take the number n . t . the difference
of which h . if / x has a continuous derivative f ln l x on the x lt .. A . the following approximate
formula is .. b.. where o quot and A . b and d are the basis and altitude ol the crosssection of
the rod. n. To determine Youngs modulus for the bending of a rod of rectangular
crosssection we use the formula Approximate Calculations Ch. I. If y f x is a polynomial of
degree n.. To simplify the use of Newtons formula.. formula is exact In the general case. amp
n y. y n are the correspond. are successive finite the polynomial diilerences of the function y.
Let x lt . . linear inter potation. const and Aquot f// and. cm. Ax/ .. Interpolation of Functions . /
. for /i . A t/ A. and the quantities d and cm are known to an accuracy of .. x/. . s is the sag. of
formula is where is some intermediate value between .... .. i .. Sec.l Ax. xn be the tabular
values of an argument. x/ U .../ practical use. the tabular values y / particular cases of
Newtons formula we obtain for nl. To what degree of accuracy do we have to measure the
length / and the sag s so that the error E should not exceed .. s. n. . Newtons interpolation
formula. ... As takes on. and P the load. n is constant table interval and y lt ing values of the
function y Then the value of the function y for an intermediate value of the argument x is
approximately given by Newtons interpolation formula . . provided that the load P is known to
.. . n and more convenient x.
. l . Solution. . /i .. Here.. for q we take the common value to the given accuracy m gt. to find
the corresponding value of the argument x inverse interpolation. Example . To do this. Using
the table of two sue approximate the root of the equation smhx . Find sin using the tabular
data sin . will . .. q Applying formula have and using the first horizontal line of the table. . sin .
then / /l l. we Using formula have and taking into account Thus. Whence x jt cessive
approximations q m lt..lt lt//.. all the decimals of sin are correct. it is alsc possible. putting and
CH n iO. Interpolation of Functions If the number n may be any number. . then it is best to
choose it so that within the limits of the given accuracy. from a given intermediate value of
the function //. first determine the corresponding value q by the method of successive
approximation. Using Newtons formula. that Let us evaluate the error R if / sun. we sin . the
differences A should be constant to within the given places of decimals Example sin .. . We
set up the table Here. in other the difference Aquot / w words.Sec.
. which for / takes on given values yf / by the Lagrange interpolation formula x gt x x . a .
Taking y Approximate Calculations .. . .. In the general case.. . . .. Lagranges interpolation
formula. Make sure that JC jc all the finite differences of order are equal.oog . Given the table
log . .. Given a table of the values of x and y Set up a table of the finite differences of the
function y.. We can thus take . set jt up a table of differences of the function y x for integral
values of x lying in the range ljtlO.. . .. ... . . . Iog. log compute the numbers log . Use log . . n.
. linear interpolation to and log . . Utilizing the constancy of fourthorder differences. i.. . .
Solution.. . x xn o . . .. log log . . . log . . Set lip a table of differences of the function y x f for
the values . . .. .. Ch we have . a polynomial of . is given degree n. . . . .X Xk X quot . . . . .. . .
... x xn x XQ x .
A function For what x will y is given by the table Form Lagranges y for JC . . sin sin sin . ..
Experiment has yielded the contraction of a spring x as a function of the load P kg carried by
the spring mm Find the load that yields a contraction of the spring by mm.Sec. for the values
of the sine every half degree. Form Newtons interpolation polynomial for a function
represented by the table .. in the table by computing with Newtons formula.. interpolation
polynomial and find the value of . .. Interpolation of Functions . .. Form Newtons interpolation
polynomial for a function represented by the table Find y for x .. Given the table sin sin sin Fill
.....
b of the equation f A there is a unique root O. establishing the intervals as small as possible
within which lies one and only one root of equation . . Establishing initial approximations of
roots. /a. . . the intervals or c. then to evaluate the absolute error of the approximate root . If
on an interval a. Sec. a To obtain second approximation c a. Generally speaking./lt. . c.. .
Given a table of the quantities x and y Compute . take one or two reserve decimals This is a
general remark. b. The abscissas of the points of intersection of the graph with the xaxis are
the rools of the equation /x . Approximate Calculations Ch. b by Lagranges formula. and for x
a the values of y for x of linear interpolation. It is sometimes convenient to replace the given
equation with an equivalent as the absequation p if A. Computing the Real Roots of
Equations by means . that is. . If the function / x has a nonzero continuous derivative / x on
the interval a. then by replacing the curve y fx by a chord a. we should continue to calculate
the approximations c. gt lt it is advisable to use millimetre In approximating the root paper
and construct a graph of the function y fx. The sequence of numbers cn nl. where the
function f x is continuous on b. b and If a function /A is defined and continuous of equation .
computing the roots to a given degree of accuracy on an interval a. c . b at the . root . that is.
we obtain the first approximation of the root t / we apply formula to that one of ends of which
the function f x has values of opposite sign. passing through the points a.. roots of a given
equation The approximation of the / i consists of two stages separating the roots. The rule of
proportionate parts chord method. with the specified degree of accuracy.. until the decimals
retained in the answer cease to change in accord . . . Then the roots of the equation are
found cissas of points of intersection of the graohs y qx and y ty x. converges to the . The
succeeding approximations are constructed in the same manner. f b. then on a. b there is at
least one root or / x This root will definitely be the only one if / x when altxltb. f a and b. for
intermediate calculations.
. . we build a sequence of tial value numbers x lt x . To evaluate the errors we can use the
formula where u. . . we can make u a Computing the Real Roots of Equations use of the
formula . . is chosen so that the function x Kf x f Kf jc .. if xn and x n l coincide r within E. . crt
. it For practical purposes xQ is more convenient to use the simpler formulas a.. If / x axb.
Newtons method method of tangents. to the root rt ltxlt / x b . Proceeding from the iniwinch
belongs o the interval a. Let the yiven equation be reduced to the form ltPM...Sec. fafagtQ.
fbfquotbgtQ. according to the following law . then the limit is the on/r/ roo/ of equation on the
interval a. . b. . then in formulas and we should put x . Iterative method. The evaluation of the
absolute error of the nth approximation to x n is given by the formula IS it xn y I s to xn l j r xn
I Therefore. . where fafblt. i/i l.. amp. which yield same accuracy as formulas .. If altjc rt lt n l.
min a lt x ltb fx. xn are successive approximations to the root . In order to transform equation
fx Q to . then the successive approximaof an equation fxQ are computed tions rt . from the
formulas Under the given assumptions. . where If a .. that is. where min and fquot x for ... .
tonic and the sequence x n n .. .. we replace the latter with an equivalent equation where the
number A... then the limiting absolute error for x n will be . the a/. where y xrlt r is constant for
a xb. is mono lim n gt oo . .
Example . In . to two decimal between and . .. in the approximation to x is. .. for for the x In .
. we of the Hence. /x lent x Approximate Calculations Cft. . /jc . with one reserve decimal. we
can put X/x . of this we can take g. We write the equiva equation values of A. in the
Considering that get all approximations lie interval . Solution. .. the third decimal place has
Here. the limiting absolute error remark made above. the exact root g of the equation within
the limits lt g lt ... places. putting x .. the root of the preceeding Computing the root by th
Example . here since And so we can Become fixed Let us npw evaluate the p stop error.
Reduce the equation x Inx to the form initial approximation to the root . y In. by virtue A .
approximate number will . and all the decimals be correct in the narrow sense. lies Thus.
equation is close to rO..... .. lies Compute. . . and this number that close take . x I kx nx is is.
reduced to the form x x . should be small in absolute value in the neighbourhood of the point
example. We make use of the result We carry out the calculations using formulas f .. iterative
method. ln. lnx to and xn ltP. Here. as one of the suitable of the to root the equation l A. The
or X initial U . x Inx Example equation that of .ss.
system of linear equations Similarly we obtain the third and succeeding approximations. . At
this stage we stop the calculations. are We e We carry out the calculations using formulas
with two reserve decimals jtjS . Let it be required to calculate the real roots of a system of
two equations in two unknowns to a given degree of accuracy f. and there be an initial
approximation to one of the solutions . . . . . On the interval we have /f gt. In. . xx . In In . . a
Newtons method. . . take Computing the Real Roots of Equations Here. a x f/ x.y Then by
Newdoes not vanish near the initial approximation y Q y tons method the first approximate
solution to the system has the form where a P O are the solution of the system of two. . In . .
P . . Calculating the root by Newtons method. . and by determining the coordinates of the
points of interand tp x. . .jp dx. let this . evaluation o the error. for example. by plotting in the
same Cartesian coordinate system the curves fx. Hence. since the third decimal place .
graphically. where a p v are the solution of the Pi. linear equations The second
approximation is obtained in the very same way lt.Sec. We omit the does not change any
more.. the conditions ltxlt / x of for gt and gt . section of these curves. r of system Jt x ot y y Q
This initial approximation may be obtained. . . The answer is the root. j/. /flt. . . Let us
suppose that the functional determinant dtf. . The case of a system of two equations. f
xfulfilled.
y. Y. Condition will be observed in such a choice of parameters a. which converges to the
solution ol the system or.v . F x. given the initial approximation to the root .. . a. The
sequence of approximations x n y n . is constructed according to the following law Fx If t. y of
the jnd a. to the solution . system of equations o as derivatives of the zero in the initial
approximate solutions . y y ltSgt v x. y do not vary very rapidly in the neighbourhood of the
initial approximation X Q // . Y on the assumption that the partial derivatives of the functions /
x. partial equal or close to fi.. xgtf .. . y n belong to U then n lim oo . we find approximation. p.
yz t /. P. oo The following technique is advised for transforming the system of equa tions to
with condition observed. can also apply the iterative method to solving by transforming this
system to an equivalent one Fx. We consider the system of equations a/ x..ltx ltrltl. Example .
it in the form yFx the Y. lim n y n i. . ..c all . O x. ltrlt in some twodimensional neighbourhood
U of the initial approximation y Q which neighbourhood also contains the exact solution . t/ .
Rewrite y. f y Fgt. the system of equations Approximate Calculations b Iterative method.
which is equivalent to provided that . t Choose the parameters functions Fv. and assuming
thai t y D. Reduce to the form the system of equations . rj of the system. Ch We . y. y and
ltpA. t .y . . what is the same thing. in other words. amp such y will be that a.
la . . y Computing f x t the Real Roots of Equations t Solution. . p r . . Find graphically the
initial approximations and compute the real roots of the equations and systems to two
decimals . . x . p. . bQA. p. . e. Then the system of equations i y . x . t x yy x . . x.. x. cos. of
the rule of proportional parts compute them to and two decimal places. has the form .In ... .
Here.. . .. A . use Newtons method to compute the real roots of the equations to two decimal
places . use the iterative method to compute the real roots of the equations to two decimal
places . . . Yl ialt/ form / For suitable numerical values of system of equations a. . ltP ltgt. JC
jc In . yxy. sin . l a // l pi/.fx x .Ay.. . x Proceeding from the graphically found initial
approximations. .. . l. s x . . f x v . Y . . .. r/ . and in a sufficiently small neighbourhood of the
point x Qt y Q condition will be fulfilled.M . i. o Write down the system that J in the ltfy is y
equivalent to the initial one /la. . is by means Isolate the real roots of the equations by trial
and error. jc x .Sec. yx y . . logjcy. i/ c . Utilizing the graphically found initial approximations.
pi VI Y. x x. we put a . lY . t qgt x. xy which equivalent to the initial system. . . Y anc choose
the solution of the . X . .
we compute the integral by taking the values of the integrand with one or two reserve
decimal places.. n a with an absolute error of where M max f x when To attain the specified
accuracy e when evaluating the integral. . . this is what gives us the number of partitions n.
The value of h obtained is rounded b a n h should be an integer. is a function continuous on
a. Trapezoidal formula. . Approximate Calculations Ch. Simpsons formula parabolic formula.
x y . be the abscissas of the partition points. off to the smaller h must be of the order value so
that of Ve . let xn b. b we divide the of interval of integration calculations b into n equal parts
Q and choose the interval h . and // // be the corresponding values of the integrand fx. For
the approximate evaluation b of the integral If x a. . . e x e. / . the in terval h is found from the
inequality That is. Compute jc to three decimals the smallest positive root of the equation tan
jc. Then the trapezoidal formula yields y Let x ih Xi x a.. . yn nl . Having established h and n
from . .. If n is an even number. y Qgxl . x . Sec. Compute the roots of the equation xtanh xl
to four decimal places. . quot quot Lff . Numerical Integration of Functions . . . then in the
notation of b Simpsons formula h Q Ko n y ..
a material point is made to move along the xaxis . utilizing the error formula given in the text.
generally speaking. that is. val h in Remark. etc. h is halved. Approximate the work A of a
force F if a to x from x table is given of the values of its modulus F Carry out the calculations
the Simpson formula. otherwise the pro cedure is repeated. the number n is doubled. For an
approximate calculation of the absolute error R of Simpsons quadrature formula . according
to which where and S are the results h and // /i. . Under the action of a variable force F
directed along the xaxis.Sec. Numerical Integration of Functions holds with an absolute error
of where M max / x when integral. Since. after the result is obtained. practical work h is
determined in the form of a rough estimate. Approximate J xdx by the trapezoidal formula
putting rt. Evaluate this integral exactly and find the absolute and relative errors of the result.
the specified accuracy e when evaluating the interval of calculations h is determined from the
inequality To ensure the That is. it is difficult to determine the interand the number n
associated with it from the inequalities and . use can also be made of the Range principle. is
The number h is rounded off to the smaller value so that n an even integer. the interval h is of
the order J/JF. i by the trapezoidal formula and by . Establish the upper limit A of absolute
error in calculating for n. respectively. one to the number of decimal If the new result
coincides with the earlier places that we retain. Then. then the calculations are stopped. of
calculations from formula with interval .
. Using the Simpson formula. Equations method.. to four r decimal places. FT X Jcos . .
using the error Formula given in the text.. . . Q i fcos dX. . Numerical Integration of Ordinary
Differential of successive approximation be given a firstorder differential equation . C . gt
places the . lt. where b is chosen . Evaluate to two decimal a PPlyi n S the substitution
improper integral Verify the calculations ft Jy by applying Simpsons formula to the integral
QO J rr x r . . Let there Sec. Ife. Approximate Calculations i Ch.so that lt plane figure
bounded by a halfwave of the sine curve axis is in rotation about the xaxis. calculate to two
decimal A f/sin and the places the length ot an arc of the ellipse in the first y situated
quadrant. . Using Simpsons formula. gt . A method Picards subject to the initial condition /
yfx.y when x lt . dr. calculate J o V . taking n . xogxdx. calculate the volume ot the solid of
rotation to two decimal places. Establish the upper limit A of absolute error. . Using the
Simpson formula. Calculate the following definite integrals to two decimals . Jl ..
then the process converges in the interval successive approximation definitely I y.Sec.
where ft mina. / And the error here is . y is defined and continuous in the neighbourhood R
and satisfies. we choose the interval of calculations /i by dividing the interval X into n equal
parts so that h Xf are determined from the formula X lt e. to normal systems of differential
equations. .. Numerical Integration of Ordinary Differential Equations The solution y x of . The
partition points x M i. n. . x x in this neighbourhood.. The RungeKutta method. with slight
modifications. the Lipschitz condition fx. Differential equations of higher orders may be
written in the form of systems of differential equations. be represented in the form yx where
the successive approximations y .. yti dx If the right side fx. i/JKJltfitfil of L is constant. Let .
Vifx. the corresponding values function are successively computed from the formulas /. on a
given interval to find the solution y x of to a specified degree of accuracy e. / x/ of the desired
. To do this. which satisfies the given initial condition. t lira yi x // are determined from the
formulas X ltgt / x. .y n W ltML The method of successive approximation PicarcTs method is
also applicable. By the RungeKutta method. generally speaking. . it be required. . can. M/ /
and Af max/U.
or find these values by the method of successive approximationor by using the RungeKutta
method.. Milnes method.. Sec. / / of the desired function y x for instance. The RungeKutta
method is accurate to the order of h A rough estimate RungeKutta method on the given
interval x X may be obtained by proceeding from the Runge principle . ygt z. To solve initial
conditions yy when X values initial z when x x the Milne method. y. n and iquot // A. and so
forth. where f . To check we calculate the quantity . ltpx.. y m and y m are the results of
calculations using the scheme with interval h and interval /i. . yi. by in some way find X we
Q.. . . it is To check the the quantity correct choice of the interval h advisable to verify e The
fraction should amount reduced. of the error of the to a few hundredths.. The RungeKutta
method is also applicable for solving systems of differential equations y fx. Approximate
Calculations Ch. n R I ym Urn I where /i m. subject to the the successive yiyi. otherwise h
has to be .. . n are quot Hi vhere fi fx it y andi /i. The ap proximations y t and y for the
following values of successively found from the formulas r/ t i. z . t/ . with given conditions / .
one can expand the solution y x in a series Ch. IX. ..
. lt . . . . . does not exceed the unit of the last decimal y m retained in the y x. Example . Carry
out the calculations answer for t gt . Then the entire interval of x to jc. integration from ... /
choose the initial interval h from the condition /rlt.Sec.. . . . . Table . Solution.. /i / i s n oy.
then for f/ we take I// and calculate the next value y/ w then one has to start from the
berepeating the process. the Milne formulas are written separately for the functions y x and z
x. Calculating the value y r Here. D .. Calculate to two decimal places the value of the
solution of this equation when the argument is x . // . . .. . We To avoid involved writing. . But
if e/ ginning and reduce the interval of calculations. ... gt Xo / y. / / o h fc . .Q. the remaining
three values we calculate by the Milne method from formulas /. .quotquot By in this criterion..
fl /i .j .. The order of calculations remains the same. . . . amp . We carry out the calculations
with two reserve decimals according to a definite scheme consisting of two sequential Tables
and . Given a differential equation // x with the initial condition y. . .. h o . . the interval h that
Similarly we calculate the we chose was rather rough... let us take h . / o. . . A i . lt by a
combined RungeKutta and Milne method.. values y t and y The results are tabulated . is
divided into six equal parts of length . The magnitude of the initial interval is determined
approximately from the inequality h m For the case of a solution of the system . If Numerical
Integration of Ordinary Differential Equations e/ . we denote by y f and y the corresponding
values of the solution y and the derivative y We calculate the first three values of y not
counting the initial one by the RungeKutta method from formulas .. x . . At the end of Table
we obtain the answer. /s yamp The value of // will obviously be the answer to the problem. .
Let us check the first three decimals in tins approximate number are guaranteed. / x. by
means of points x f i . .... . . .
. . Applying formulas . Ch.. . . ... . .. Calculating.. y y... . . y by the RungeKutta Method. fx.. y
jr . Table Approximate Calculations . . /z .. . . We have fx. ..... . . Calculating the value of j/ jf .
we find . y t .. r/. . . .. A OK . no need to reconsider the interval of calculations. there is . . .
hence. h y t .
// . Sec. The results are given in Table . we finally have y . for instance. y I/O f/ i/ . . . With the
help of the numbers X Q x it x t and / y lt y zt ys we calcu. Thus. We then form a diagonal
table of the finite differences of q Vfx. . late lt/ . . Adams method. where h y hf x . Numerical
Integration of Ordinary Differential Equations We obtain i/ y . . or they may be found by the
method of successive approximation . or by applying the RungeKutta method and so
forth.Sec. To solve by the Adams method on the basis of the initial data /x we in some way
find the following three values /o g of the desired function y x these three values may be
obtained.. by expanding y x in a power series Ch IX. q lt q . Similarly we calculate the values
of y and y. qs hy hf x. . in this approximate number the first three decimals are guaranteed..
. yl. we use formula in it to calculate A// / R and q t and obtain the next diagonal putting n A lt.
A desired solution yx. Alt Using this diagonal we calculate the value of / of the ft. W m .. lt. we
calculate we calculate qhfx f/ . A . differ by not more than one or two units of the given
decimal place not counting reserve desimals. Then. A and into the . it. is equivalent to the
formulas of for the Adams method is complicated and for practical purposes is useless.. /quot
. . . utilizing the numbers of the new diagonal. . y . . of the . calcuwhen . since in the general
case it yields results with considerable excess. r After . and so forth. ces with the aid of the
Approximate Calculations Ch. IZ A lt. which are situated together with along finite differences
Alt/. that we obtained in the solution Their calculation is given in Table . O finding A//.
Example . the quantity h of the initial interval of calculations is determined from the inequality
/t lt m if we wish to obtain the value of y x to an accuracy of / A/ And when we know and lt/
introduce / . z . z yz f/l. the Adams formula and the calculation scheme shown in Table are
applied separately for both functions yx and zx. Alt . In this sense the Adams formula . late to
two decimal places differential equation Using the combined RungeKutta and Adams
method. in utilizing i the numbers e lt/ . We . The Adams formula for calculating by proceeds
from the assumption that the third finite differences A lt/ are constant. . in which case there is
an increase in the number of first values of the function y that are needed when we first fill in
the table. y x y . . difference table and then fill into it the A lt.. To increase the accuracy of the
result. .. We calculate the subsequent values // f/ by the Adams method see Tables and . We
shall not here give the Adams formula for higher accuracy. we A calculate. Accordingly. In
actual practice. by o means Alt . A. choosing the interval h so small that the adjacent
differences A ltf and Alt/. Adams formula maybe extended by terms containing fourth and
higher differences of q. differential equations Find three successive approximations to the
solutions and systems indicated below. use the values y lt y z //. we follow the course of the
third finite differences. formula and putting n. Milne and RungeKutta Evaluation of the error
Example . A of lt/ situated diagonally in Hhe difference table. A a new diagonal parallel to the
first one. The Adams method consists in continuing the diagonal table of differen Adams
formula Thus.. see yy Example of . For solving system . yjfy yQ Q. The answer to the
problem is . . Solution. the value of the solution of the x with the initial condition f/ .
.
lt/ y y . fc y . Italicised figures are input data Answer . g . /.Table . Table Auxiliary Table for
Calculating by the Adams Method . Basic Table for Calculating /. by the Adams Method.
use the RungeKutta method to approximately the solutions of the given differential equations
and systems for the indicated intervals . . Kxlt. Approximating Fourier Coefficients
Twelveordinate scheme. Let y n of fx n n .. J z.. and y Q We up the tables y y yz y y yu y ygt
y Sums j Differences A Q u l uz Iu U M U Sums Diilerences s t s l s s. z A when x . . . x. z
Compute y and . Jc cos/ AJI x Q. when x .. tv Sums Differences t . Putting the interval
Approximating Fourier Coefficients l . . calculate to two decimal places the solutions to the
differential equations and systems indicated below for the indicated values of the argument.
..ji. Compute when x . Find and x n. when jcl.. x . . .Sec. A .. z when . Compute y when x . . .
z when x .. when x . I z x z. Compute when xl. calculate /i . when x x . . . . l when f. combined
RungeKutta and Milne method or Applying RungeKutta and Adams method. . .. Sec. . x. f
when x . . . z ltxltl. y z. a .. Compute y when . of be the values the function y f/ fx at equidist
equidistant points set xn the interval lU. z. yz z x. ry and Compute . Compute y when xl. . .
when x. V .
s. . a . s. . Example.a a .rc by the . . Approximate Calculations Ch The Fourier may
coefficients a n b n n .. . . / x . . . . of the function . Other schemes are also Calculations are
simplified by the use of patterns. s . Find the Fourier polynomial for the function y
represented by the table fx From formulas we have . . .. sin x Consequently.... where . We
have fx zz used. . .. amp . an cos nx b n sin nx. ordinate scheme.. Using the sin x . fx t a s
amp. .. a. sin x. a. y be determined approximately from the formulas . . . a a . . cos x . cosx
cos for . find the Fourier polynomials the following functions defined in the interval . i.
/. first . . . t/ . .. y. . lt/ . evaluate the Fourier coefficients for the following functions a several
fx xgt . . y. . . y. .Sec. y . y. . Approximating Fourier Coefficients tables of their values that
correspond to the of the . . y. t/. . y. / n . s . lt/ . / . . Using the ordinate scheme. y. J/ / y . y. . n
. y y . . y. y. . equidistant values argument. y. . . y. lt/. . t f/ t/ s yg . . .
b xV y . . e nonperiodic. is. . .. a . e.y oo.oo. / gt when ltxlt . . . . . . . It lt. . . x y gt when gt . qgt
. . e.px . b lt when oo lt x lt and lt x lt f oo. ltxlt. . b x lt then a ltab b. y x. a amp a a/ . . / . A x /
xf x.. x ltlt Jt lt lt ltlt. . t/ if Clt when / ii when /ltlt/. ab a b lta lt x lt . .ANSWERS Chapter . i/ a y
Q when x l l Jf . . /quot quot./. . ji. ArjiltAlt/ Jt / .. . that AlO. this i. impossible. . . u n c periodic..
. . . Jt. Since b gta Besides. boo be ltlt oo. . d xgt . . Hint. I Solution. ltxlt . . T JI. d periodic .
lltxlt.. x aa t and I . . . e odd. lt.. Klt. c even. Thus. oo. llt. . i / m q when m qil q q /A. Utilize the
identity if x / x r r. . pxjc . T n.. . ylt whenOltlt and x . .. Even. when x when jclt and . Whence a
b b ab amp a a Hence. lt. Whence A or KO. x /and A. dodd. /ltlt oo b x. y ltQ . bodd. .tlt. x gt c
. a Periodic. . if ltxltc. xltl. b periodic. a Klt. r/gt when Vlt xlt and d j/ i/ and lltxltf oo. a ltxlt. . .
Solution. / . c z/gt when ooltxlt . igtx x. but or is . . . . . a oo ltlt should either i. when x .. jc K.
x.. /. s x l l l .
lt/ . Fig. . a c / VlogT. Fig. See Appendix VI. . . ax where b. . The desired graph the sum of
the graphs sinjc. . . . points . Hint. Hint. See Appendix VI. Hint. . See Appendix VI. Hint. Hint.
. ltylt a . . . The function even For xgt we determine and y . The period of the function sin x
with amis T njn. Fig. Hint.V sin cp. Appendix VI. . VI. . . . . See . Hint. See Appendix VI. . Hint.
a . Hint. Fig. Appendix VI. e x y lanyf . See Appendix VI. Hint. Hint. Hint. . jrlcosjc. Hint. . Fig.
. . Taking the integral part. See Appen dix VI. y is is y. . . Fig. . Hint. See Appendix VI.
Appendix VI. Hint. Appendix VI. Fig. The desired graph the product of the graphs sinx. Hint. .
. See Appendix VI. Hint. Hint. . The graph a hyperbola . Hint. . b sinw. when ooltlt. . Hint.
Hint. Hint. . .Hint. Hint. . See Appendix VI. . . . The desired graph is the sine curve y plitude
and period n displaced rightwards along the xaxis by the quantity Fig. Fig. Hint. Hint. Fig.
cosjc when /quotquotlt JC j ooltxlt and b y log x when ltxlt oo. Fig.r. Hint. See Fig. See
Appendix VI. . Fig. . a sin x. we will have yA sin x cp where A Vamp b i and arctan In our
case. See Appendix VI. Fig. . . . . c u u x . . See Appendix VI. is See Appendix . Fig. . See
Appendix VI. Hint. Hint. . See Appendix VI. Fig. Hint. . . Appendix VI. Hint See Appendix VI.
and . . . Hint. Hint. . . x/l/ when x . Fig. Fig. . ooltxltl. . Fig. Appendix VI. See Appendix VI. .
See Appendix VI. y shifted along the axis by X Q and along / the t/axis by yQ . Fig. Cf. Fig. .
Completing the square in the quadratic trinomial we will have y y and yQ ac b ja. See
Appendix VI. .. See See See See . . Fig. . Hint. Fig. . Hint. d c . . Fig.Appendix VI. . .
Appendix VI. is . . arc tan . See Appendix VI. d x gt f ooltlt oo. . Hint. . Hint. When x at which
gtlt. Hint. . b a cos. VI.Answers c if ooltlt yu arc l t/ oo. Fig. Fig. Putting aA cos p cp and b A .
See Appendix VI. See . .a Whence the desired graph is a parabola y ax displaced along the
axisby See . Hint. See Appendix VI. . . Hint. y lt /S. . we have y o / x . cos x yl yl x x and and
t/ i/ the Fig. if gt. Hint. cp. Fig. y x if xltl. Fig. . . See Appendix . Fig. . See Appendix VI. . Fig . .
. Hint. Fig. XQ and the t/axis along yQ by Hint.
. . we . the parameter t cannot be laid off geometrically . a c A when . . Fig. . . See Appendix
VI. oo.s Xl . Hint. I. C . a . f b . yi A . Fig. Sit i d . rquot x . Fig. . .. . Appendix VI. . we get now
easy to construct the desired curve from the points.. /i T JV. cfxgtN when xgtXN. . eltl. ..
Here. . . . Fig. . we get the desired curve see Appendix VI. oo. .. Fig. . See Appendix VI. y
corresponding to the abscissas x . d e x. . . . . . . . a n . b . Answers a table of values Form
Constructing the points x. c f/ . . y. Use the formula . . Hint. Solving the equation for y. oo. x It
is Fig. Hint. . l /i . r YX f. .. . . b . b . .. p . . c . . . haver . . . j. Passing to polar coordinates mind
that log y . . a. . Solving the equation for x. y obtained. x Fig. a a . f . e y. we will have x . . . ..
f x . Passing to sin cp cos cp cos qgt gt polar coordinates r e see rcosq. . . It is sufficient to
construct y . we have ogy Whence we get the y points x. . . . See Appendix VI.. . . . W . b /i gt
n gt a a c A. . . c .. . . .. . y . oo. i g . gtN when xgtXN. Fig. . Fig. the points x. . See Appendix
VI. d oU . . Appendix KK VI. n gt V e b . .. See Appendix VI. Hint. . . . a . . . . .y and tancp .. . .
See Appendix VI. when rsincp.. . Hint. . . ft / . . assigning to the ordinate y arbitrary values /gt
and calculating the abscissa x from the formula Bear in oo as y . F // i l. . See Appendix VI. . .
. . J . y z x. / . . y of the soughtfor curve. .. . . Fig and will sin M. . . logArlt N ltltAO.
ji. . . . A. . . . . . In . cos a. . x . . . . o . . sn z sin a. b . I. . Jt . it follows that lim cos gto e . a . . .
. follows that lim cos . . . . lim cos x X e it . a .o lt y e . . e Solution. a b . . b . . . . . . . . I Z b . . j.
. . . . y . j/jt b . Since lim lim J. a . b . M i a Ina. i . . . . . . . where a see Example . equot . . . . . .
. b . a . . . . lOloge. . . a oo. . A . . . . . . As in the preceding lim case see a. . . i quot. . . . . a . .
b V i quot/. . . quot. .i . b . . b . . . . . a . Solution. e . .lim v n . . . . where a. . ji. . . o . Jt. cos A.
z K Hint Put . . a . Put a. Utilize the identity a e na a . a . . . . . In a Hint. . . Hint. . . JL
.Answers . equot .. e. lim cos X lim X cos x sin a quotm Si nee lim V lim XQ . . Irt i m . . .. b . .
. jx a lt . is a . / when xgt x. are infinite discontinuities. . . . . ... . are infinite
discontinuities.AJC lt Ax I Ax x kn. .. . Q t Qltfkt Ugt gt. .. .i/ y when xgt. Take advantage of
the l. x is a removable discontinuity. Hint. x is a discontinuity of the second is a removable dis
is a discontinuity of the second kind. x . y . .. No . b . discontinuity of the first kind. . .. x kn k. .
. . g. .. where k kn. . b . c / x A b x lt. a have J/xfAx. a . a .. si. x is a point of discontiauity of
the first kind. . . Take the is inequality cos integer. j/ whenx . .. . . . e . . x is a remo vable
discontinuity... when xl. . when xlt. . . .. b x isa removable discontinuity. d . . . then cosxltl and
y. . a . ... . . Hint. a / n. when but x /Swi... . x . . .... .. . x is is a discontinuity of the first kind.
Second. . . kind. where k is an integer . x oo. if If x amp kn fe.. yx j/ ltx lt . .. b a c . b third. / f /
.. .. a lteltl. b d . the i . . x xkn . Hint. . v nj . fL e .. . c . .. c c x lt x lt . . No gtl.. x is a discon . y ..
b x b lt. . . gt . are infinite discontinuities . c . . x when lltxlt oo.. . xjifcy fc . Hint. x gt . /quotlO
l/quotTl If y . then cos xl and . . i . f d f e . when ltxltl. ./quotil advantage of an Ax/V A Ax
lquotxlt Ajc/Vx. /. . . . e t ... . J/ /. x lt when . x is a removable continuity. a x discontinuity a
discontinuity of the second kind.. a b Ugt. o lim ltC n nMo quot quot of is . . c . the straight
line yx the asymptote is the curve yzT Q/Qo H where k . x . proportionality factor law of a
compound c interest. . . r then when cos ampxltx A we . . . d j . . Answer Solution. b x
inequality xf.
. m/sec. of the first kind. Use the results of Example . x . a . . . lim T is temperature of at time
t. . . sin x Ax hm Ax AAO Ax J/ Ax / Ax j/ Hm . c quotquot h. a is a discontinuity of the first
kind. a k k is integral are discontinuities of the first kind. a abx.. AJCOJ/ . x is a discontinuity
of the first is a discontinuity is a kind. we have P x P x lt.. Axo / Ax / Ax . A . discontinuity of
the first kind. a . since k the function y Ex is discontinuous at xl. . . AX b o Ax . No. l. . xl
tinuity of the second kind. is continuous. . . Ax a L . a . sec x. a fQ lim Ll Ax lim AX . . The
function is continuous. c ft A . Hint. . L tanqgt. . Solution. c lim . . where where is cp is the the
angle turn time t. b . b xampx c e . . . b x is integral are points of discontinuity of the first kind.
. . rl . sec x. r . E AJO X Ax .. c . lim Axgto cos x cos quot d .. A AX gt IT a l. II b . jtAxfJx l b gt
.. . .oo. nx . quot . . x x x xk Chapter . b . For the given function the equation / x . . . gi. . . /Ax
/x Ax l f x im fx AO of Ax Ax at / x Ar . Show that when is sufficiently large. Hint. b JP dt lim
AfMgt Ar . ... . Ax cos x cos xf Ax . cm/sec. . gt . . // x has the form x . quot J. . rX dt a lim A/H
A At b where Q the quantity substance b at time t. a x x cgt V v quot n lim AMgt A Ax . a a .. .
cos x x Ax Solution. . b the function kn k is integral are discontinuities of the first kind. . c . Ax
lim AJCM . . Hint. . x.. . c x . b .Answers is a removable discontinuity. xo and Problem . ..
Solution.
. . a faxb fax b c I I . . Answers b / lim r Ax lim I . . I /I tan x B . X f cos x . n In . e cos x sin x.
x cos x sin x V cot x sin sin. x . I Y . r. . x e x . x arc tan x. A lim . . xsmh x x . f arc sin y. mat
m l bmnt n ax . y y A x x x z i JL . . on x . . y x x arc coshx Vx . xe x O . c l . x x lnx.. . . . y
arcsinxfj/i quot X v . . x sinx. H/A. / . c //fl x . wy X .. . UU.. . . .. cot x sin x . x / . cos Ssin /quotl
x cosx VlSsinx gt x gt. sinh x x coshx. xVxf. y . . sinx lOcosx cosx cos x . x x x. . . oA OIS. . T
. x . JL . x x ./j/l. lim l sinAxl Agto Ax x .Hint. l/x dx x x . l x . x arc tanhx . e x gty f. x sinx cos x
sin/. r / . OOU. rn . . . toco. tan x COS X quot . . sin x x . . . . e x . x x x x In x . fWft VJA . . x .
x . . c . gt singt gt . l x . x .zo .hdnh cosh x . ji . V . . L lJt Varctanjc jaril x arc tan x . /arc sin x .
tanhx xln sinhcosM xln arc sinh x . JL x lim sin Ax LiJLJl. K . . .arc sinx . Hint.. gtX . fl fex //. . .
cos x sinx. . fel .sin / ..
cos In . . x . . x lA f x x/V f pX . . . X cosx xl X COS . . x p. . . x sin . . . x xa V x arc sin x .
Solution. n a quot x cos x sin x . sin x cos x . t / xe . . OB . . . cos x . . tan. . . . L In x x jx J .. / I
arc . . . tanxsec . Solution. . X asinax . T Jjc . xcosxcos x . x K a sin x p . . arc sin . . xi . x . cot
x log e. .sin sin x cos tanx x sin x cos x In xjx sin J x cos x x f f sin x sin . sin x cos x. xlnx x
sinx arc sin x VI x . . x x pjf sin t . x a x a x . In . . quot . . . x . . . . x . .Answers . xcosx . x fx . s
l .. x In . . e . coscos. . f sin . . arc sin x arc cos x r /quotl x x. x X sin/qgt. Vl . . V . . x cos x cos
xcos x r xx sin . i sin quotquot . . AQC Vx x . . / sin /. y H. . x j . . x arc sin x . .
.
,
.
Answers
.
.
.
xcosa xt
.
sin
sin x
.
cos a w lna. .
f
.
..
/
sin xe
eacos p/
.
x p sin pO . equot sin p.
.
ecosx. .
afn
xMna.
l/tanx
l/Tosxln a.
quotquotquot
/.
Psinhpjf. .
tanh x
.
tanh x
.
.
cothx. .
.
L
xarctanhx
y
.
.
sinx
.
.
y
.
x y In x
. xarcsinhx.
cosx
.
a
yl when xgt
.
x
when x
lt
does
not
.
exist
b
,,.
.
,
/
.
.
fW/
I
b
y
I.
c
n. .
a
/
,
/.
,
/
V
do not have
/I
,
/
.
d/
/
.
it
,
.
e /.
and
/
exist. .
x.
.
.
. Solu
tion.
We
y
equot x
.
or
xyyx
lx xl x lx x x l x.
x.
Since e x
t
follows that y
Answers
S
gt
.
,
.
.
.
n.
.
.
/
li
cos x Inx
.
SIn
J
cos x
cos In cos
sin tan x
arc tan xX
r
xarctanx
.
l
a
..
quot
.
fir ra.
tan
f.
lt quotquot ltt
.
.
.
a
.
.
.
W.
.
tan..
oo.
,
.
.
J
.
an iden
. tan. .
tity.
.
.
. No. .
.
.
.
Yes, since the equality
is
.
j/
.
.
lfju/ f/
.
.
r
cosy
x
xy
.
y
.
y,
ex
arc tan
yVxy
.
y
nxxx
.
.
ab
c
.
.
.
.
.
.
arc tan
r
.
.
,
,
,.
.
bd
.,
.
lt/
.. k ..
fi o
lo
y
x
.
x
x
x
y.
.
a
y
x
y
yx
,
l
yl
.
ycexy
for
x Jt // x f/fl
.
for the point ,
the
point
,
.
.
GA
.
.
.
yf .
x
t/
.
.
At
the
point
,
/
Ilf
at the point , /
xf y
x
at the point
,
t/
x
.
c
.
Answers
.
Hint.
The equation
of the
tangent
,
is
crosses the xaxis at the point A x and the yaxis at B , f/ the midpoint of AB, we get the point y
Q . . . . at the at bolas are and intersect point , tangent
,
ZXQ
.yQ
Hence, the tangent
.
Finding
angle
The paraan
arctani
at
the
point
,.
.
S,
T
arc tan
.
.
.
yp
.
Sf
S,
jia
/JT
gt
rt
a
fl
t
n
/lji
tani
.
jt.
a
,,
I
p.
tani
.
qgt
.
. cm/sec. .
g
in/sec.
The equation
is
x
.
The
range
u
cm /sec
cm /sec
of the trajectory sin
syx tan a
velocity,
a
.
The
K
uj
u g/sina
g
/
the
slope
of
the
velocity
vector
is
v
cosa
.
To determine system. The range
Hint.
of
the trajectory, eliminate the parameter t from the given is the abscissa of the point A Fig. .
The projections
velocity
on
the
axes
are
and
jis
The magnitude
of
the velocity
is
the velocit y vector
directed along the tangent to the
CO
.
of
Q
,
he
cm
. cm/sec, the area, at a rate diagonal increases at a rale ol . The surface area increases at a
rate of ji m /sec, /sec
f
the volume, at a rate of .
is
JT
m/sec. .
is
g,
is
the
density
at
M
.
y cm/sec
.
.
The mass
of the rod
g/crn,
,Y
.
at
B
.
g/cm. quot
.
.
x
the density at x . e
A
is ,
.
.
the density cos x .
X
.
arc tan x
xv
x
.
cosh
a
i
. yquot
.
a
/quot
.
is
y
The acceleration, a .. . The law . the is x acoscof the velocity at time / is point M, acD cosof.
Initial velocity, awsincof the acceleration at time / is aw velocity when initial acceleration is
aco acceleration when The maximum absolute value of velocity is aagt the maximum is . n nl
value of acceleration is aw . y n absolute na . a n x
of
u
.
vl
//
sinx
.
.
.
The velocity
motion of
..
.
b
ir
.
i
.
.
Answers
d
n jc
OJtn
.
a/
b/
c
/quotT.
.a
a
sin
a gt b egtal a
XU
b
lt
.
.
.
.
l
quot
.
.
.
.
a
b
AJC
o
j
ft.
cf/
//
.
quotJ
.
,.
.
v
y
f .. Ar/
.
d/..
. cfl
x
when
x
.
and No.
o
d//
Ax, A Ax
.
AA
.
. For
.
.
..
.
.
d/
i ..
.
dy
..
.
A
M
I
.
.
.
.
xedx. . nxdx. .
x
yxe
.
IJ
dx.
.a.b.c.d
.
.e
.
b
.
.
.
.
. cm . . /. J/ . . . a /.
quotquotquot
quot .
d ..
.
..
. ..
. ..
.
.
d
iy
.
.
l.lc. dx
.
/i
.
sin,ln,
fc. .
X
/a
.
dx. .
e
..
.quot
sin
.
.
Answers
.
e
xcosa
sinsina nad
xx
is
/l
.
No,
since
of
/
does
not
exist.
No. The point
,
.
a
discontinuity
the
function.
.
amp
.
.
.
a
bg .
.
In
xx li l
f
U
gt
where
x,
ltlt
.
. sin
x
where where
sin
J,,
where
,, lt lt
.
sin
xx
sin g ,
,
Clt,lt.
c
s
ix JJ. ... lL.jeS
,
ltBlt.
. Error a
b
A
.
L
i
n
both cases g
jc
ltlt
.
.
The
x
J
error
is
less
i
.
than
.
Solution.
We
have
j
/
we
a
H
i
x
l
x
I
Expanding both
factors in
powers
x
of
,
L
.
L
xx
.
/
.
,
x
f
,
get
f
I
,
x
J
y
I
.
y
H //Tquot
x
y
x
a
Tne n
expanding
.
e
in
powers
.
for
oo
of
,
we
get the
same polynomial
.
.
e
x x rlHrn
.
oo.
IT
.
.
.
...
.
..
...
.
.
.
. a.
.
oo
for
n
gt
.
a
for
n
l
nltl.
..
.
.
.
.
o
.
.
e.
.
.
.
.
e
.
.
e
.
.
e
is
.
.
.
.
.
.
for
Hint.
area
Find
S
lini
,
where S
R
a
sin a
the exact
u gt
of
expression
the
quot
the segment R
is
hh
the radius of the corresponding circle.
Chapter
.
III
. increases , oo, decreases. oo, , decreases . . increases. , oo, increases. oo, and , oo,
increases , , decreases . and , oo, decreases. . , . increases , oo, decreases. , and , oo,
decreases. . , , decreases , oo, increases. . and , oo, in,
,
,
,,,
.
,
,
,
creases
,
,
decreases
,
oo,
increases
.
,
oo,,
,
de
creases
f
,
oo
j,
increases. .
,
,
increases. .
decreases
Answers
increases.
decreases.
,,
oo,
,
.
,a
and
a,
oo,
.
,
.
and
decreases ,
oo, increases
. ymax
j
when
.
No
when x . extremum. . t/m in when xQ t/ mln when x t/max when x t/ mfn . when . when x. i/
max . t/min . max . No extremum. x.. No extremum when x .
when
/
m
i
n
when
x
f/
.
max
yg
when x ..
.
t/
max
/ when x
.
i/
when
.
x
j/
max
mm when
Vquot when x ma x when x
/
L yO
min
/quot
l
when
r
.
r/
max
/
r
quot.
y min
.
Kquot
when x K
y mta
s/quot when
.
/
m ax when
rr
n /maxcos
Ji
when
x
x
Scoswhen
,
fjkJ
y minl
x
when
fe
JT
fe,
,
....
.
ymln when
x
.
. y mln
.
when x
f/
.
.
t/max.
wn
// min
when x
.
when
.
.
f/
x
mln
mn
l when
x
y mln
f/
when x
.
.
when x
value
is
c
//max
,
.
min
m
for
g when xl. x greatest
No extremum.
Smallest
value,
.
M rwhen
x
.
m
when x
Af
and x
M
,
for x
.
....
m
when x fc
x
f
j
l
for x
fcjT
j fcO,
,
.
m when
l MJI
.
when
x l.
l when . /ns when xl Mo when x
x
b
M when x . m when x
M
a
m
when
x.
.
.
p
,
lt
.
.
a
Each
of the
terms must
gt
be equal to
The rectangle must be
square with side
Isosceles.
. The
of
tlit
side adjoining the wall
must be twice the other
.
side
.
The
side
cutout square must be equal to
base. .
is
.
The altitude must be
to
of
half
the
g
. That
whose
altitude
is
equal
the
diameter
of the base
,
Altitude of the cylinder,
Ly
radius
.
its
base
of
R/
ro
where
ilt
the radius of the given sphere.
Altitude
the
cylinder,
RV
where
R
is
the radius of the given sphere.
. Altitude of the cone,
where
where
cone
Answers
R
is
is
the radius of the given sphere.
. Altitude of the cone,
.
R,
R
the radius of the given
r
is
sphere.
Radius
of the base of the
jrr,
where
the radius of the base of the given cylinder. . That
whose altitude is twice the diameter of the sphere. . qgt section of the channel is a
semicircle. . The central
jr, that angle of
is,
the crossthe sector
zero
/
is,
.
.
The altitude
of the cylindrical
part
must be
that
the vessel should be in the shape of a hemisphere. . h
l
d
.
yo . XQ
.
The
sides of the rectangle are
of the ellipse.
lie
aTand
The
a
J/,
where
of
a and b are the respective semiaxes
the vertices of the rectangle which
.
coordinates
on the parabola
is
I/ M
/
.
Y.,
j J.
.
The angle
.
equal to the greatest of the numbers
arc cos
and
arc tan
b,
AMa
If
P
L
.
.
.
r
m
.
a
ty.
.
.
.
,
l
P min yaqQ.
spheres,
the
velocity
a
impact
/HI
with
.
mV w
the
fMm. Hint. For a completely elastic impact of two imparted to the stationary sphere of mass
m after sphereof mass m moving with velocity v is equal to
quotI/
v
.
n
is
r
,
if
this
number
is
not an integer or
is
not a divisor of
V,we take the closest integer which
of
is
a divisor of TV. Since the internal resistance
battery
the physical
meaning
of the solution
obtained
is
as
follows the internal resistance of the battery external resistance. . y h. . o
must be
oo,
as close as possible to the
,
concave
down
,
oo,
concave up
..
M,
,
oo,
oo,
inflection. , oo, concave up. , point concave down, , oo, concave up no points of inflection.
and , , concave up , and , oo, concave down
of
.
points
quotquot
of
inflection
,
M,
, Vo,
,
Af ,
V
oo,
.
,
and
of
, concave up
/,
and ,
,
,
concave down
points
inflection
M lj
/
/e
and
.
.
jr
J,
concave up
,
concave
down
J
amp
,
,
...
points of inflection,
lit,
concave up k
ampJi,
ffely, concave downfe,
oV
,
. /m,
i,
,
...
the abscis,
sas of the points of inflection are equal to
xkn.
.
VampJ
concave
Answers
down
,
oo
Y
J
concave
up
V/gt
oo,
up
,
M
oo,
f
Y
oo,
JL
W
is
a point of inflection.
is
.
,
concave
inflection.
.
,
and
,
l
concave down , concave up ,
a point
,
of
concave
A
. of inflection are M , T and M x. . , . . , x, x , y . . right . jc x, right. . left, left, left, , left, yx ,
right. . yx,jt. , right . . . , . . i/O. . . , left, , right. x n, left y right. . y a. . ma x when x . when x
point of inflection, Af,l, /max when min r points of inflection Ai lf when jfV m when x . y min Q
when jel, point of inflection, M, , . m ax f/max wnen , m n when point of inflection M , . when A
of inflection . M lt Y, and points //min
down points
,
j
J.
iy
.
,
l,
,
y
JC
JT,
in
r
jc
f
,
//
i
Jt
,
max
Points
when xO
of
f/
/y
min
when x asympT asymptote O. asymptote, AT
totes, x
x
.
,
.
xl.
max
.
when x
point
mm
inflection M, lt when il
l,
. farm
when jcl
of
inflection,
M
/,
asymptote,
J
x
.
.
maxQ wnen o
,
points
of
inflection,
asymptote,
.ymn
. ma x
when
x
J
m jx l whenxl
asymptote,
when x asymptotes,
M lt x and
l when
.
rJ.y
points of inflection, ,
.
and
Ai
lf
J/quot,
inflection,
.
Ai,
and
max
x
Point
,
point
of
of
y
/
asymptotes, x
x
and
.
inflection,
,
asymptotes,
x
,
.
.
when
jc
asymptotes,
x
and
x
. nnx when max
f/
A
,
min
vvnen
x
asymptotes, r
and /,
B
,
,
V when are endpoints ma x are endpoints. Point of inflection,
Q and
c
v
.
.
/I
.
,
and
,
.
Endpoint,
A
m in
when x
.
.
Endpoints,
point of
and
BY
Q
max
V
when jg
AY , . inflection, M f fJ,
points
of
V M, tl,
x.
V
s
l
YH
max
f
nax
when
x
t
,
inflection,
.
. Points of inflection,
.
when x
when
.
mn
x
points
when x
x
of
inflection,
asymptote,
.
max j
when
.
jc
.
ym n
y
K asymptote, when x point of inflection, , / when x when x , mi n
i
m M gt ,
in
M
l
Q
and M f , when x
asymptote, . max
.
.
m
n
when
.
x
K
asymptote,
x.
.
Asymptotes,
and
,
min
T
/
when
Answers
f
J/max
TT when V
/JT
Q
,
I
points of inflection,
Af,
nl,
J
,
,
of inflection,
of
o
J
asymptotes,
xl
.
/
m
i
D
M
/,
jrV, asymptote, x
.
/
/ max
.
when x
point
when xl
of
point
V
inflection,
/iooy
/
I
M
,
j
asymptote, y
a,
Q.
.
Points
inflection,
M,
x
a,
and
of
M
asymptote, y
/ i/quot
,
. f/max
wn en
y
//
points
inflection,
MJ
g
,
asymptote,
.
J.
t/
.
.
rnax
when
w min
gt
points of inflection,
.
M
f lf
max
when
point
.
jcl
of
when .
t/max
when
e
inflection,
M
equot
.,
.
asymptotes,
inflection,
and
,
.
/min
f/
j
wnen
.,
point
f/
point of
inflection,
M
rr
quotquotquot
.
min
g
when x g
x.
of
M
,
e
V,
jc
asymptote, Ay
yQ
/min
when
.
max
F
Ai
when
.
as
when
point
of
inflection,
fe
.
e
.J
y
of
limiting endpoint. .
y mn
when
x
.
V
points
y
inflection, gt ., . Asymptotes, r/ Q.
M
,
.
asymptotes, xl. when x oo and y
not
x as
Asymptote, x oo.
the
.
Asymptotes,
x
.
,
the function
is
defined on
interval
,
jn
Jfeji
. Periodic
function
with
AJi
period
fc
n. y m n
,
when
jc
t/
max quot when A
j
e
,
,
...
points
of inflection,
Mk
o
j
n kn,
Oj.
.
Periodic
function
with
period
ji.
y mln
fe
,
when
,
j Jif/Ji
of
/
max
q
V
when
and
xc.
x
the
fen
,
...
points
inflection,
M k kn,
Nk
On
fare cos
f
Jfejx,
yg lj.
ji,
t/
. Periodic function with period
interval
i/
ji,
max
n
M,
nal
m in
i/
when points
..
j
when
x
/
m in
i.
when
inter
of
inflection,
M
l
,
it,
.
Odd
.,
. and
periodic function with period
/
On
,
max
when x
ml n.
f
when
.
i/maxl when
Answers
T
Ji
nin
When
ymax
points
l
wnen
f
min
when
l
when
n
of
inflection,
AiO.,
.
M
.,
.
.
M,.,
Periodic
M .,
with
ji
.
M
,
.,
.
A e ., .
function
period
n.
V ym n L
,
when
...
when
foi
fc
,
asymptotes, xrJi
inflection,
Jt
kn,
. Periodic function with
period
n
points
of
Mk
.
Even
yJ
fc
,
t/
,
,
...
asymptotes,
the
j JI/JJT.
,
periodic
function
with
period
n
On
interval
c
rrs
when
arccos
min
max
whenxnt/ mln
of
when
arccos
wnen points
inflection,
M
l
,
Oj
Af t
skill,
function
arc periodic
lO with
m
i
M
jtarcsin
it.
J
,
period
On
,
.
t/
Even
the
interval
n
ma x
when
vvhen
F
A
when
x
arccos
.
of
minFM
f
arccos
zr
/
n
when xJi
points
inflection,
y
,
Oj
..
./Si/S Odd
Limiting
points
of
function. Points of inflection, M Even function. Endpoints, , , Af J. cusp points of inflection, lf
/I
fcjt,
i/g,
fejt
/j
,
,
....
A
t/max
.
when
t
..
,
Odd
function.
graph
,
oo
and
,
asymptotes, x.
/
.
Odd
function.
O
JI
oo. Point of inflection, JT when max t/
x
j
/en
M k kn,
function,
fcrc
J asymptotes, xy , asymptotes, t/mln when x
m in n n when
k
q
A
/JJT
points
,
of
inflection,
,
....
.
Even
and
/
as
x
oo
//
.
as
oo.
.
/
m
n
when x
quot
node asymptote, y
l.
fr
when
JC
max
Whe quot
JC
XS quot I
left
P oint
J
right.
inflection centre of . Odd
l
. when M , asymptotes, when of inflection, point y and In. as . . Asymptotes, oo and y
function.
t/
symmetry
, Ji
asymptotes,
yxn
t/
and
yx
m in. when
f
max
t/
ji
j
of fe Periodic function . Mk x fcji. a cusp when min awhenji// . min when e. of when Q.. j XJ..
.. monotonic x function. . x. asymptotes.. . Periodic . . . The domain of definition of the
function is the set of intervals kn t where fc Periodic function with period ji. Limiting endpoint.
i/ . feln.. fl cusp when lt y U . Domain definition. as x endpoint.. period ji.ri period jt. lf . y mln
when . x of gt .e. jT j Jf m in r wnen f I/quot .. . function with x period jr y mm l when x /jJi . B.
asymptote. / max e wnen .. . . as x . . TT. /e .. e fc . . r/. max awhen / . r t . definition. domain
where of definition is the set of intervals fcjrjJi Points of k is an integer. M Urcsin I i/ /m. Jc
inln when / x ... inflection.. .. /jJtfji. p. Asymptote. t/ . points of inflection when . with .
asymptotes.. . ... . min . t/ . . points of inflection when i. jc and y i/ xmin l i/ andymln i when
cusp. . . and when t i. . .. . Endpoints. to l max wn n / j ljK e. . y i as . asymptote.. and M . x
kn. Domain . . .. Points of inflection.. asymptotes. . lt fc. To obtain the graph it is sufficient to
vary / from to jt. . f /quotquott . Periodic function with . Answers j/ m n . y mm M. . M.
asymptote. . and M A and arc sin t/ max when The ampJi amp . m in . increasing /jn x . . x
when t oo.e... . / . when i . ... asymptotes.. x . . . t/max e when / .. when asymptote. VSe . jT ..
. points of inflection. UlA g I . . point inflection. Point of inflection. . ..e .. oo and r/ . Endpoints.
i/ . limiting kn. .. when V asymptotes. fejt.. as oo. t/ min g xji ampJT. jT / ma x jl .
cosp . . where a Chapter IV In the answers of this section the arbitrary additive constant C is
omit ted for the sake of brevity. y aV. l cos. . . quotquot at quot amp . . sin p cos . c . KB . x .
R . ds a Vl p . . ds dp. a pK c l . n az .. . . ltgt y where c Ka . . . . . ds asinrff. . a sin t . / f k . . .
ds cos dcp. . . dltp. cosa tp cos/. f / . / . . ds cosa u . . zzz . ds x sin I/ a tanh X . K . quot. Z sin
a cos . ilfl. K.. ds Z asin cos/ d. . Z tit . .. . ds cosh dx cos a cos a sin. . . .. . . sin a sin/. p
semicubical parabola. . cosp quotp. . and f . . . . R a/.. X .Answers d dx. a cos . both vertices.
.
. /x l. JC .. xl . In / /quotlO . a i a J Solution. . In . arc sin x . . x the . L arc tanx a ln / . cl. . x x
// . Hint. f Ili . Dividing numerator by the denominator. . Solution. . lnx / x a . / . . a in a c . . y. .
arc sin ln x /x . arc sin / cotx . . j/ x J. Jiya. quot xW Solution. . . . j . rj . . . tanx x. a xax fi / lt .
Put sec x. b x tanhx. tan . . I j . In b x V ab / . . s x . Hint. . a b xcothx. Put tanh z . Answers . .
Inlx fll r . we get . . . . . I T . . . . x /quot arc tan In j V x . arc . .
. e x . . . lJL. V ft lnjt . In tan . . au. cosl r. a . . tan U .. . lnsinx. sin . arcsmx . . . lln tanf . . ./ r .
. ncos/quotx. Jr r . . . ftX Xln sin a . See . arc sin e HOI. lncosx. ft. sin lOx . x cot x . tan . hint
lilt. . . sin . . . Put in jc cos . Ina i . y .Answers . . In Hint. l e V a . . l n . In I . . are tan . U. nquot
. Hint. H . j. V sni r. Hint. . LJ. tanx . . . jJ o . . a K . sin /quotx. . . i arc sin . . M. x cosax. . . sin
l o . a . ax a In tan ft . . In XcoslogA . In a arc tan . tan fljc . In e . arc tan a. l nx . i In .
lntanhx. . . iln tanl. In x r V f sin x sinx . . In sinhxl . lr lnx . Latanh. In cos Vx K. . .r . . .
Incoshx. . . Vex . . a f lntanox. x . . fls n . . x . arc cos . . . in a amp arc sin x . . ne . cotx. .
lnsecx l/sec xi. . x Xarc tan .. .. .In a I ocotx r. s inhx . /quotl lt . . . . . In tanh . . . . . n lnx tan p.
. arc sin . . quot. . arc tan /tanx I v Hint. a taniwx. arctane. b In . a arccos when x gt . arctanx.
J . n J/ arc tan / . . . / . . . /x. . . . larcsinA . lnxx . . Answers D . . . In sec .. tan jr . . . . e sln . jc .
In tan x .
In In n Inxf . . In V x . x. . and arccos J if xlt Hint. . sinx . x arc sin x . x arc tan . in similar
cases we shall sometimes give an answer thai for only a part of the domain of the integrand.
Solution. r . V x . . arctan e .J/quotl sin . c f the eneral C orrn p n to e ax dx Q a xe where P rt
x polynomial See of degree is the given polynomial of degree n and Q n x is n with
undetermined coefficients . See is good Henceforward. . and Cancelling oute equating the
coefficients of identical powers of whence ax . . . . X e x Ax Bx f C e Ax f BeX Jf . x. . we get
C In . . TT arc sin x. B . xlnx . x st In x . o igf y . Hint. e x f X Hint. after differentiation.
Problem . Hint. xcosx. TYf x xf. . a. d e lnsinx. may be used in place of . Putx . The
substitution x the trigonometric substitution.. e T quotxhx x . . arccos if x gt .Answers c . Vx
Vx . aarccos Put x . Note. In place of undeter of repeated integration mined coefficients by
parts we can use the following method xVdx x Bx C ex or. .
Infx. Utilize the identity a x . x . . is sin x . sinx . dx J J a r . x . X arc tan x Xarcsinx Xarcsin . .
Ka x and xdx x a xa r whence f / Ka . Answers Problem . I . X . rt x.. arc tan x o Inx l. x .
xcotxf cos x In . x /m . x arc sinx . . y o xln x xlnx x. V . O ij. x x. Solution. arc sin x . x /TIxx V
x V . arc tan x . xtanx In cos x x ltrt X / cosx sinx V V du dx quotquot x . cos In xJ. . is x cos Q
n x cos PX R n x sin of the given polynomial of degree rc. x . arc sin . Qn x see and R n
Problem ln x are . . . x xy a x s j. . and degree n with undetermined coefficients . j X X arc
sinx sin l/T x In x Vlx . x x .. . . X In x In x X X lnlnxl. x x . Ay Ar x i .f x dxa l . Put a x. ft . we
have f /aa a . It also advisable to apply the method of undeter mined coefficients in the form
px dx Pn where P n x polynomials . In x arCMn . . xcos In x xsin In x x dx d . . .lnx. . . . . ax
and we get and . . fy Inx . f . . /quotxlnx x. cos x Hint. x . x arc tan x x. .. l Inx. x r Whence C
xdx x l K Hint. tan e sin x quot .
xfnx . . . In x . In .. y jc In x In x l arc sin lL. x l . LarcsinP. .Answers dx x Consequently. . In . /
.. . arcsinjc . Hint.b x a . l . A. arc sin x o arc tan o . . yax . ln x In Ul In X . arc sin o . . o . arc
sin . In . . See Problem . . arctan A arc tan a t . .. Vamp / f ltT Hint. r . . r In /quot . quot . . V xx
arc sin . See Problem . . T ln . . n jc z arc tan T arc tan y y . . . ln x arc tan x . .lncosx . Va x fa
arc sin . x xf U . . . U x x . nx nA. I . arc x sin y . In jcf. l Fn TTT . j . arc sin .
arc tan x. arc tanx. . Answers X arctan .. x In x tan x. . jii . . where . jclgt . arctanl. x s J J . arc
tan X l/l. . lnx y ln H .arc o SxSx . . In / /quotx . arc tan o A . r . i . . lquot jc I FT A x X /a In . . .
. . arc tan Kquot . In arc tan . jfI In / il.. arctanx. . n x . In . x x . .TTjT f l . arc tan Xarctan/. Put
J x . . J/l arc tan /l. In arc tan . . In . Hint. xx V . arc tan x .. tan x l . . .
Answers
l.
.
.
li
.
.
i
.
arc sin
l.
.
.
/
retan
.
.
.
where
.
,.
.
.
.
sin
x
gsinx.
.
cosxjtosgt
.
x
cos.,.
.
J. .eo..
gt..
,.
i
. ..
.
isinx.
cot
jc
,.
.
cot
Jt
.
.
tan x
tan o
.
x
T tan o
x
.
.
In
tan x
lnt.n.
COS
.
.
quot
ln
t,n
i
.
cos
.
.
sin A
In
siux
,.
Slfinsinx.
cot
x
cot x x.
.
.
tan f
.
JJ
.
J/x J/cSS J/cw
. . Jtwix. .
xlB
SrRjquot
.
arc tan
.
where I
quot iain
cosH
,.
.
.
..
cosx.
..
.
,.
lEMltP.
,.
Answers
sin
,
sin
Jo
.
.
cos cos cos . z o lo
tan
tan
.
f In
.
tan
.
.
In
tan
. arc tan l
tan V
/
.
In lo id
a
,
sin
a
r
cos x Solution. Whence P sin cos .,, Q
We
put u
,
sin x
P
f
We
lO
have
ln
p psinx crs
,
cos x
sin x
a
and,
cos x consequently,
J
rfx
.
In cosx
situc.
J
.
Si
I
iiC
f
j
.
gt
lt.
arc tan
Hint. Divide the numerator and denominator of the
fraction by ccs x
.
arctan
V
.
.
Hint.
See
V
See
Problem
.
.
Hint.
Problem
.
In x
.
/quot f
sin
V
quot
Vi
r
t.
sni
ln.
Hint.
.
arc tan
x
/
/quot
arc ian
Use
the
identity
.
.
sit
sin
sin
sin
n
tan
.
Hint.
sin
Use the identity
..
xcos
jsmx
cos
.
jsin
,
Sm
.
cos
.
JT
gt
In
tanh
r cosh
.
.
coth.
.
In cosh
cnth
.
.
coth
arc tan
,
/
. arc tan tanh. ..arctan
V
Hint. Use the identity
,
or
If
j. O
r
fV
iAl
I
.
m.
sinh
.
sinh x r
A
.,
..
..
...
J
sinh
cosh
,
sinh
J
cosh.
In /quot cosh
cosh .
Answers
.
.
.
x In x
.
H.
.
.
J/quotIT
.
In
x
yV
x
I
Kx
In
x
.
VV
r
ln x
x
.
.
F
JT
.
.
I
.
Uarctan.
.
x
Lln
.
Y
.
.
sin
.
TT
sinx
sin xclt sx c
.
..
e
.x.
oZ
lnklO
.
xlnx
Vx V
.
arCCOS x quot
X
.v.
.
X
a
, arcta quot
.
,.
l
M
n cosxsin x
n
tanx.
.
/.x..
...
x...
.
.
arc tan
. In
Vgt
arc tan x
..
.
.,
.
In
x
arctanxl.
.
j In
.
.
Answers
lx
.
In
quot
x
.
l
.
x
arc tan
r
.
lquot
.
.
/
.
/X J/jc V
.
.
.
J/x
J/x
I
In
J/JTt.
.
X arc sin
x T,
.
.
.
lVrgt
arc sin
.
xl
.
.
hi
arc tan
,r
,
where
.
In
tan x
cot
.
cottcosx
T In
.
cos
,.
x
cos
coslt
tan
.
t
tanS
.
sinx.
.
tan
In
X arc tan
.
.
arc tan
.
. arc tan tan
x.
.
ln tanjc
secjcicosecjc.
.
In
.
xarc
j/
an
tan
V
A
.
T rX
/
/
y
Xarctanl
A.n.
gt
tan
y
quottan
tan
In cos x
.Jx
a
.
X In sin
ax
ax.
.
x tan
.
r.
.
uu.
le.
.
fin O
Answers
. . .n
.
lcotco tje .
J.
.
.
r
x.
.
quot cosh x. .
coth
ln
.
sinhxl
.
.
gL lln e
.
.
iarctan
/
xfir
.
In
.
coslnx
sinlnx.
.
.
f
x
tan
y x sin x x cos x
.
In
,
lcos
sinV
arc
t
r
Chapter V
.
fc
a.
.
vjg.
the
L
t
.
.
.
.
.
Hint. Divide the interval from tervals so that the abscissas of
x
x
to x
.
points of
x b q,
.
the xaxis into subindivision should form a geo.,
on
metric
Hint.
progression
x l,
.
See
Problem
sina
sino...sin/io
a
at
sin
xl
xq
xn
xn
.
.
In.
formula
.
cos.
Hint.
Utilize
.
the
cos
l
cos V n a J
.
da
r
in
in o
.
.
In.
.
.
.
.
f cos
x
x
.x
.
/
,,, .... . In
.
,
.
o
.
.
sinh.
/i
.
sin x.
.
Solution.
.
The
sum
as
,,
gral
i
sum
of the function /A
/,, rV T
Tquot J
sn quot
n
may
,.
be
Therefore,
mtc sn
X on the interval
lim n
oo
rH M a
.
. In . Solution.
The sum
iH
n
/i
o.quotH may
iJi n
...
H
/I
M
sum
be regarded as the integral
of
n/
J
the function /x
Answers
on the interval
,
where the division points have
lim
n gt
oo
the
form
xk
ln
.
amp
,
..... n.
Therefore,
sn
ro
fJ
o
In .
H
t.
p
.
j
.
o
.
.
oo
.
quotT
o
.
jln
.
.
In
j
.
.
.
.
n.
.
. arc tan
arc tan
.
arc tan
.
.
ln
.
.
.
.
.
.
.
In
O
V
.o
. arc
.
o
In .
cosh
eJ
..
.
ir.
tanh
if
tang
.
. sinh
i
ji.
.
.
tanhln
.
In
.
.
sinh
.
..
. Di.
.
verges.
,
p
.
pltl
.
diverges,
if
if
pl.
pgtl
Diverges.
if
.
n.
. Diverges.
.
r
.
,
diverges,
.
T
pltl.
.
.
.
.
Diverges.
Diverges.
.
Diverges.
. . Converges. . . Diverges . Converges . Diverges verges. . Converges. . Diverges .
Converges. . Hint. Bp
.
t. ,. . Diverges. . i k Ina
.
l ln
Con
q
fxdx fxdx where
t
/
jc
x
xquot
since
lim
/xx
l
qltl,
where
is
o
j
and lim
gt i
quot/x
,
both integrals converge when
plt and
f
that
is,
when pgt and
l
e. The
/gt.
.
Hint.
pf x dx
x dx,
xp
first
integral converges
when pgt,
IL
the second
when p
In
arbitrary. . No.
./
quot
/rff.
r
.
Cf f quot. J
.x
a/a. .
.
n. .
JT.
/quotquotS
.
n.
,
.
/
.
.
.
.
quot
o
H.
. In
y
.
.
.
.
T
nr o
Answer
.
.
.
Z
I
.
rr o
gt.
l.
xPe
dx.
.
.
.
z
.
.
.
z
bz
I
Solution.
rplC
U,
Applying
the formula of integration by parts,
we put P
dx,
e x dx
dv.
Whence
du px. p l
v
e
and
If p is a natural number, then, applying formula account that
p times and taking into
we
get
gt
k
quot
,
fe
is
ii
an
even
number
n
/
k
is
an odd number
/
... k
quot
l
.
jt
quot
Hr
Hint.
t
.
V
W
...
.
B
.
,
.
.
Put
. a Plus b minus c plus Hint. Sketch the graph of the integrand for values of the argument on
the interval of integration . a First b second
c
first.
.
.a.
.
.
.
.
arc sin i.
.
lt/lt/quot.
The integrand
.
lt/ltl
.
jtlt/ltyJi.
.
lt/lt
.
Hint.
.
increases monotonically. .
lt l lt.
s
J
.
Hint.
Take account
of the sign of the function.
.
.
i
.
. . . In . .
.
m
In .
.
naa
.
.
.p
.
.
O
..
In
.
.
O
.
.
O
n
t
coshl .
.
. ab .
Vquotl
.
/.
na
g
.
Hint. See Appendix VI, Fig. .
. . jra .
altr
.
a
.
.
n.
.
yln.
Hint.
See Appendix VI, Fig. . . a f
.
V
Hint. See
Appendix VI, Fig. .
.
n
and
n
.
.
jii
and
np
.
Answers
. na
.
. nb
t
ab.
.
na
.
.
fl
Hint
For
the lo P
the
parameter
.
varies within the limits
.
ltoo
Fig. .
.
See Appendix VI, Fig. .
na
VI,
Hint. See
Appendix VI,
.
. jtV. Hint.
See Appen
dix
Fig. .
.
.a
.
.
Hint.
See Appendix VI,
..
.
.
Hint.
Pass
to
polar
coordinates.
.
VhTtf.
Hint. Utilize the formula
.
cosh q
sinh
/. a .
l/Tlnl/n
lflln.
.
.
. In e
Vquot. .
lnVT
fcln
. Ia. .
.e
.
l.
aln.
.a
/Tquot.
.
ln
Hint. See
Appendix VI,
Fig. . .
.
.
.
. a. . a
Vquot ln quot
In .
..
.
.
,.
JQ.
.
.
Ji
.
.
.
no.
.
av
c.
je
.
.
.
.
t
amp
n. .. o
.
.
Jia
.
A quotP o
.
.
U
c
In .
l
n a.
.
.
.z
O
.
nha.
.a
b
jia
n.
na.
.
.
na.
..
.
,.
.
x
fl .
,
.
.
J
.
lUi
U
nabc.
JW amp
..
ao
s
.
.
na /quotp.
.
nabh
/ oC
.
.
DOT
iT.
.
.
nT
nln.
V
l.eff
sinh. z
.rta .
.
. nab Hint. Here. y b external surface of a torus taking the
tfx.
Taking the plus sign, we get the minus sign, we get the internal sur
Do
a
.
e
face of a
torus.
.
nf
of
arc sine na
.
In
,
where
eccentricity
ellipse.
o
bn
ac
a
Answers
.
jia
.
.
ita
/.
.
na
.
.
.
M
M amp .
x
.
M x V M X M K na
.
Jy y.
.
MxMKa
.
y a.
lt
.
.x
a
a sinh
asina
.
,
sstion.
It
SBa
Jia lt
a
gt
Divide the hemisphere into elementary spherical slices of area da by horizontal planes. We
have da jiadz, where dz is the altitude of a slice.
ft f
az dz
z
Whence
tance of
z
j
nfl
o
Ct
r.
Due
the
to
symmetry, x y Q.
of
.
At
a
dis
altitude
from
vertex
the
cone.
Solution.
Partition the
cone into elements by planes parallel to the base. The mass of an elemen tary layer slice is
dm,YftQ dz, wnere Y s ne density, z is the distance
of
the
h
Jt
cutting
plane
from
the
vertex
of
the
cone,
Q
JZ.
Whence
n
z
h.
/
.
f
g
a
Solution.
Due
to
symmetry.
T
we partition the hemisphere into elementary . To determine layers slices by planes parallel to
the horizontal plane. The mass of such an elementary layer dm nrdz, where Y s tne density,
a is the distance is the of the cutting plane from the base of the hemisphere,
ra
o
a
a
z
zdz
Qa. . /jta
radius of a crosssection.
.Ia
quot
zab
J
circles.
We
s
have
z
quotquot
b.
I b za
/J.
.
/amp
.
.
/
quot JWamp
a
.
r
Ib W
T nab.
.
/yjt
Solution.
We
each
partition
the
ring
into
elementary
concentric
The
mass
of
such
element
dm yn.rdr
/
and
the
moment
of inertia
nCrdr
y n/
JY..
nRHy.
to
Solution. We partition the cone into elementary cylindrical tubes parallel the axis of the cone.
The volume of each such elementary tube is dV nrhdr, whe r e r is the radius of the tube the
distance to the axis of
the
cone,
h
H
s
is
the altitude of the tube then the
moment
of
Answers
inertia ia
/Y
/
jttf
r
dr
lr
where
Y
is
the
densit y
of the
cone. .
cylindrical
tubes,
Ma
.
Solution.
axis of
r
We
which
partition
is
the
sphere
into
elementary
the
the given diameter.
a tube,
a
volume dV
is its altitude.
nrhdr, where
is
the
radius of
Then the moment
of inertia I
nay
An elementary
I/ f
ha
/
r
quot
a
I
I/
L./dr
A
nav,
where y
is
t
e density of the sphere,
.
and since the mass
Sn ab.
Hint.
M
a
o
itay, it
fol
lows that y
b x/
Ma
o
the
.
Vn a b
f/rr
r
.
// o
a
Tn.
U
. a x,
jcaxis
o
The coordinate axes are choand the origin
of the solid a
is
sen
so
that
coincides
with
the
diameter
the
centre of
the
circle b
o
is
Solution.
The volume
its
double
cone obtained from rotating
a triangle
about
of
base,
is
equal to V
o
rft/i
,
where b
is
the base, h
the altitude
Jtx
zbi.
the
triangle.
is
By the Guldm
of
tlugt
theo
rem, the same volume
of
V
the
where x
hence
the
distance
centre
vt
.
gravity
trom
base.
x
H
.
v
t
i
.
lnl.
.
xltMuagtt
D flt,
f
.
/
weight
//
Hint.
The elementary
force
force
of
gravity
is
equal to
the
the volume of a layer cf thickness dx, that is, dF where y is the weight of unit volume of
water. Hence, the elex dx, where x is the water level. mentary work of a force dA ynR H
ynR
of
water in
dx,
.
AryR
A ynRH.
mass
lamp
H
.
.
.
.
tnh
A RTM
Q
Solution.
kgm.
acting
/lo.mg/.
The
force
on
a
m
is
equal to
Fk
j
,
where
R
r is
f
the distance from the centre
of the earth. Since for r
R
we have
Fmg
h.
it
follows that
kMgR
.
The
soughtfor work will have the form
A
m
J
krdr
.
kmM
.
.
KKnJ
ergs.
,
When/i oo we have A
mgR.
Solution.
uP rdr.. pw p ergs. f a the gravitational constant. . The is frictional force a ring of width dr. Q
vnrdr rrdr Q n a UJ A Hlnt . .. Solution. Thus. ltD . kgm. of work required fe equat . . the work
JS Jd. The amount p . . M . the Poisson igti law pv pjgt t where . .x oi/ . . to the reserve of
kinetic energy. . the complete work Ax of dr JifiPa. cal. holds true. . Consequently.Ms k k v I
jiu. The work perrevolution r formed by frictional forces on a ring in one complete f a is . P
abyxh. j M da where dKL. Igm . them of the pressure on unit area of the support p p ta a z . o
. nln kgm. pfo is fr b n gtr . cm. Solution. X MW. R Whence . . to volume v an is A J UQ p
digtp igt In . Q j. Q the surface density. Solution. Therefore. X X / isothermal process. KJ. k a
is . . at a distance r from the centre. For an expansion of a gas tV The work performed v l in
the from volume v Solution. . r the distance of from the axis of rotation.Pa.The Answers force
of interaction of charges is / i dynes. . . Hence A J P k dv. kgm. If a is the radius of the base
of a shaft. Solution. A k . The da is kinetic energy of a particle is of the disk it t an element is
area. For adiabatic process. Hint. P the vertical component gf a directed upwards. is K / co .
Represent the given function replace in the form /M I/ f J and by . on the other hand.
excluding the coordinate origin . a and. V i/ x . ua it in the middle.. Kxl. and consequently.
/ww a.. . TRI cal. . Use the JouleLenz law. . Chapter VI . Solution. at ta whence dt dv. r/ and
n Jiltxltn /i it. x xx . /lt/ Solution. . then In the hence. where n is an integer. y o Solution.
identity /VT put . /lt. . lti/lt. . . f x jS. V l/ Hint. a. h the lti/lt and lt. f part of the plane adjoining
the ing its sides x including these lines and axis and contained between the straight lines
when x x when . S J vt a dv. . Hint. lt . . and y. and the yaxis. includsquare formed by the
segments of the straight lines lltt/ltl. b bisector of quadrantal angle x ygtQ d strip contained
plane located above the straight line lines between the straight lines / . Answers
perpendicular to along the large lower side of the rectangle. Oje// . Designate x yu. including
these e a and / . UV v. Q . j that part of the plane located above the t Klt. / a . . Then . f x. . T
Ri / . z xVquoty. x It remains to name the arguments u and Hint. xyx y lt gt xltt/lt . i strips /uK
jc n. When we have quot the identity Then / a z Single circle with centre ai origin. the
acceleration time is /l. M xtktdt Axjlx. S x t/ /quotz xy f. x. S. c halfincluding m the circle x y . g
two strips x. a and x ff/ a including the ring contained between the circles ri/ boundaries.
y gt V x . the level lines are equilateral hyperbolas. . y. . bolas xyC C .. u .c . . k limit does not
exist. xx parabola y with the exception of the coordinate origin. a Discontinuity at je . p. c a
cube bounded by the planes x . paraboloid. cylinder. when From . both families of surfaces
za are divided by the cone r/ . rc. m that part of the plane located above the parabola y x and
to the right of the /axis. we get the function . . whence it is passing to polar coordinates x gt if
x and y in such manner that p const Osqx. . dx dy x z l dy . the set of variables x. . xyzQ b
concentric spheres with centre at origin. r cos cp. consider the variation of x pass kx and
show that the given expression and y along the straight lines y may tend to different limits. X
evident that then z sin . the function z discontinuous at the point since there no limz.f/ r sin
ltp. . a . l. b . t/rf. Hint. Continuous. Sixth and Eighth octants excluding the boundary. qMO. f
limit does not exist. the level lines are at the origin. const. it follows that z does not have a
limit as x and y . b hyper C d straight lines y axC. is f/ .k the entire j/plane. a hyperbolic
hyperbolas. . b a paraboloid of revostraight lines parallel to the straight line f / lution. Indeed.
for u . is Similarly. lt gt is Hint. including the points of the t/axis and excluding the points of the
parabola . c for u onesheet hyperboloids of revolution about the zaxis. . the generatrices of
which are parallel to the straight line x the level lines are parallel lines. including its surface ..
depending an the choice of k. the entire t/plane. . which continuous everywhere.. Since these
limiting values of the function z depend on the direction of cp. In Itemb e d e to polar
coordinates In Items e and f. including its faces. family of concentric circles n k . the level
lines are concentric circles with centre at the origin. Answers y gt . . a Parabolas are the
circles C C y . e a parabolic . jc ay. a a plane. c circles jt f/ lines yCxxQ. . lt yCx gt . the level
lines are equilateral c d secondorder cone. f the lateral surface of a quadrangular pyramid. .
discontinuity the tines of the coordinate axes. o the n the entire place except points of the
straight lines and t/ x y n k . and function the v u y is everywhere continuous. y b all points of
the straight line of line of is the circle x c line y discontinuity. the level lines are the outlines of
squares. d discontinuity are . glevel lines are parabz C fx i the level Ifnes olas yCx h the level
lines are parabolas y . d a sphere of radius with centre y and z lt lt . we get z sin ltp. twosheet
hyperboloids of revolution about the same axis. Putting y y l const. Third. a First octant
including boundary. a Planes parallel to the plane c straight . b First. f/ since for yl the
denominator jt when ltp .
. tan p . . ox xy Vxy dz yx Vxy oz . . y zdxxzdy r/dx xdt/dz. m. dz .s. du dx /x dy xy dz. dz
xydx xy dy. fy . . tan Y . dz ydy. Accurate to metres more exactly. du r x dx y dy z dz.. . duxy
y. . . d x sin d . . . Ax A/ Ax x dx amp f AxAi/tAx Ar/. dzxydx yxdy.. . . n ag Ef. ri/ln. Hint. . . ..i/
ox x ln x dy x cos. a A/ b A/ df dy d/. d/l. dw. . tana . . dx x . Put the differential of the area o
the sector equal to zero and find the differential of the radius from that. of cm. . ... r. / xy dxx
y. . . b . A/ Be convinced that f . . r . de I v i. dz . . dz . da dy cos a dxsia a. dx di/. . t/z .. x dz .
x . . . d/ . z arc tan . dz sin xdx xdx sin /di/. xquotln. cm. . . d/ . . a . cm rela live to inner
dimensions.. xzxtfquot . /. A/.. . and take advantage of the definition of partial derivatives. r
cm. df . . . y IfT Check to see that the function is equal to zero over the entire xaxis and the
entire i/axs.. tanpco. c . cp x. dx . y dy. Answers . ..
tan p . mn. and consequently. y ty x.. . /f / O.Answers sin In sin. and any / .. y . x .... m m of f
x. o. cosasr. . y we y. . A . lfiol . v . l . . j l lt/ S . y xlt . irpyii yvj for .. d . ltfc . O . o cp i/. J/ MO.
M. when /. . grad a. . I / .. x/. dx W dx . r r y .. . a rate of m/sec.l. yi A f/ cos jry jc sin xy. j //.
quot gtlt perimeter increases at /. . c . x . /in of . y / w . // . . . J. . . cos QQOQr . y that For x
ticular. COSY . . Whence fquot. . xd x . /quot / Jll . v yltf x. . . . find that Q. .. r dy dz . . u. . / V
km/hr. A. . Similarly.. the area increases at a rate ofm/sec. . . cos p . . xly dx . The . quotii if
gr/iiquot. Using the rules verify that differen tiation and the definition /y a A partial derivative. .
. IJC f Q. zx. Hint. . in par fj..
d / . C. sin x C. Hint.ri dx a. xy . c. tan C. qgtW t. y. C. x dy dx dy. z t/z C l . arc . sinxdxdy
cosy dxdy xsiny dy. I axy or dx . x . dx ay lines. dzey x . d fl. . p C. JC // y C f . the zjC. . y .
d/l. Write condition of the total xy differential for the expression XdxYdy. z . a . / f a M a. . y ln
a . dzx ltjc yz z dzGy xz xt/ xf/ z a dzzsinx cosy dz d/y dx cosx t/sinz . . The equation
defining// M amp ln the equation of a pair of straight r . x y dx dx y dy . . Y X . afzd C. . dz dz
lttquottxdx ydy ltftdx dy. o x x dz yef eVfquot ltye yf ye w dx nv v uu ampfu efv xeyf e x y xy f
ye x fo dx dy uu . dz sin y dx dy cos y dx dy sin y dy cos x d . . . dz e cos y dx xeyfu xeyf xeyf
eQ dy uv uu . u. X Iy dx . x . or . / Answers . . is y a ITI. o . . f tlv . . / . x axy dx x dx . . /cp du
xdy dz ydzdx zdx dy. . xt/ . d z . . . . ff y . . gtlt/.
. .Answers dz dz dx abzgt dy . a dz c cos y . . a .. . d . qgt cp cos ltp . . dxdy . .. r . d Tc IO a
rf l/ l dz z dy. . pp. . . sina .r. i/ . . F. . tt . z . . . cos dz . dr/ f . r . . dw . . dx . O. Tquot quotquot
T dxfy u . .. b g. d dx dw. . . sin / q.
i . / k /t. a b ab points jc i/ when . . Hint. Projection on the . jcft/lf/ f /ifcK/z .. parallel to the
xzplane. y zm i n aquotd quotquotFf wne quot .Tnax quot nonrigorous maximum y . . / . zl.
Answers the tangent planes are the points .ax byh b cy k iah . . tions mentioned in the
answer are explicitly defined by the of Y eacn these .. X lyl y. . A/x.y y in the neighbourhood
of the point . a b the points at the XTTL. t/ and . Hint. . .lzl. x. z H zf /lt/. xH / /. z/ . to the
j/zplane. . . y. a . . The equation defines two functions.w on the /zplane lt y IQ Projection
xzplane x z A . f and . y cy. /.. .. . . of which one has a maximum z m ax when . / a f amp l n
the xplane. Zimn functions has a boundary extemum z . a . . zmax whenx . f/. . . The line of
tangency of the surface with the cylin der projecting this surface on some plane is a locus at
which the tangent plane to the given surface zis perpendicular to the plane of the projection
ax bxy . Projection on the xi/plane . . . . w min when JC J/ z . / x. Apply Taylors formula for . .
Ai . x y z . quot at points min of the circle y when x . x . i/ . n zMpj yy . There are no points on
the surface at which the tangent plane is parallel to c j/Hz . z min . at points of the circle x t/ f
.l. the functions a fx. z fx / /ixt/ z kyx x H. . z max whenxl. the other has a minimum jwhenx . y
when xl.. . rx in the neighbourhood of the point . j / . z xl b f x. . NocxVcmum..t/ . The func
equalities .. at the points .. bJ. i/ /i fcM /i/ef/i . T . when x Y y and when x no extremum for xj/ .
max at V . . z mm t/lxl . z min There is . jt y Dl r gt Un b Hint. I // I y k. Vf y F . .
and a ssi v ll m. J/V. Greatest value value for x y yQ boundary minimum. Greatest value for
x t/ in ternal maximum. The radius of the base of the cylinder t/ xy provided Bxy . .Mor x . x . r
r/ . I . . Isosceles Cube. . are the semi axes of the ellipsoid. . f T . . . . y for x dary minimum. r/
equation has a maximum maxquot for x minimum min for x . Cube. . parallelepiped are rp Y
V j/ quotT where V c b. for x V Y /quotquotquot j y z I/ IT. . .l/ gt functions assume the value .
gt/a./gt. V. smallest value z quot TFT . m f m. imta . triangle. quotmax . The dimensions and c
of the c a. . y. b Q. . . . for x . . . x f/f and consequently exist only inside and on the boundary
of thecircle x . both functions have a boundary extremum . . . w max . Hint. Ai . . . max j/ i. .
max for for xl. . . igt a Q b . smallest value for x . . . yl. f . One of the functions defined by the
the other has a . Jgaa f . . max j for . smallest value for x .y of the ellipse from its centre
coordinate origin The problem reduces to finding the extremum of the function is equal to x fr/
. for x . smallest bouninternal minimum and for x . . Sides of the triangle are fe p. y e Answers
at the points of the curve jt s i/ r/ min . y T Major axis. b greatest value for y. at the points of
which both I . . V . a . . a Greatest value for / quot T /quotT . u max . y boundary maximum. a
j/ f f T Greatest value for x . . a i . MMIX fl . f/ . . . at the points . b smallest value for x .J .
WminC for x t/ C. r j/ f f fl I/ / f . . The square of the distance of the point x. This value is the
least for the first function and is the greatest for the second. t/ . . a /a /a gt /a . j/V. . minor
axis. .
. are . the altitude R / . v . then B b. f / . . for A. . . is an isolated c. Cusp of that provided
second kind / / /. w i. a The discriminant curve the locus of points of inflection and of the
envelope of the given y of the envelope is the locus of cusps and family. of the ellipses taken
as coordinate axes. if yx. The duration of motion to finding the a y cosp of the function /a.
then A a. . //ji/ function . xS. Node . ycos/lAr. The isolated point .c the discriminant curve y
velope. . Find the minimum / the Kl /J/ t . UjCOS . . BM b tan p. none are equal. a b tanp c. . .
If among the quantities a. . /sinH any /cos ysin t wtor f. XQ Z Q. M. . any /. the point of the sin
straight line its cos . / l. sin / ellipse. . z . d hyperbola. cos A. b the discriminant curve y Q is
the locus of cusps and is not an enof the family . . whose equations. c ellipse. Node . .. i . .
and B t . at which the ray passes a frm one medium . must lie between A. u /.M a tan a. J
The problem reduces r minimum . . for f/ sin/. the parabola o quotT with . . if . Isolated point .
The channel must connect the point / . . cos. of ap. fo / . Answers is / . Tacnode . a is BM of
the ray J . e / / w for screwline. lt altb of the first kind. then the curve does not have any
singular points. a Straight line. the point into the other. and c. Obviously. . t/ provided that a
tan COS p . . . y . . . . w. cos . Node . is a cusp point. . yR.a. for t Q. y px. b. A pair of
conjugate equilateral hyperbolas. b parabola. . zyft. EE p Uj j length is VL ft Hint. is a node.
Origin is isolated point it if agt. vj. where of R is the radius of the sphere. / /J/. . . .. is a cusp
of the first kind if a and a node if a b. . /V . . ifafcc. . . Cusp of first kind . / AI t Hint. have the
form is xy g . . then A a. / the f / / . d the discriminant curve decomposes into the straight lines
locus of nodes and x a envelope. . If a bltc. . the axes of symmetry .
plane. quot. V . x tangent. / . y cos/. a sin y cos sin acoscof. . z b a . cos YI normal plane. . y
tgtr z z TT binormal. . j normal plane.. ing plane. cosv. z . T . . ejz a normal plane. t t quot . .
cos Y cosJ The direction cosines of the principal normal are cosa. b tangent. I pri quot t cipal
normal. tangent. osculating x z rectifying plane. . j principal normal. ilx jc quotquotquotquot i/
M . . quot/ a r /i . v k /.z. . . P. x x . / sin r r ampcos/ y . z CD co/ sin of uy cofc . p y fc. cosp . .
uK . a cost b gt tangent. sincof w i circle. . / . sin/.. . quotT T r/fz TT tangent.zbt asint . . . .r/z
osculat quotquotJ A f P rinci P al normal. r/ /z a . y .Answers .. x r/fz . . co / cos v a cos CD/
cof cos a sin art cousin a cos agt/ . where w aagtu sin o is the angular speed of rotation of
the screw . c Z f quot O gt /. / oscuating plane. . xcosacoso. x acost cos ysmt are cos a VI
normal / The a cos directi on the tangent a sin cos p f . . . x .. v co i . t cos . agt . a tangent.
./J. cosv. COST.l il / . / acos/ . p j tangent. a binormal cosines b of u . . y z . . T cos t sin sin t
cos tJ v / sin / cosO /sin . u osculat z i/ g . V aV u . . c bxx ayly a ing plane. y . x / z normal
plane. z x normal plane.
plane. . x . t/ x . TT/ . JL t ./T y T I . x . ln .. J dj/ J JC / x. r . L . J y o binormal TSBS / i M . x y
y . r/ JC . dy f x. A xl. . JJf. a Vi b . It yd/. w w . . O . When i t . .. f/ i/rfx ix JdxJ/. oy x . . . . O . .
x . t/y . . .. . u gt principal normal. . Af J/ . osculating .ydy. y . / gt J v / jc z . . . . a K K . . /C/T /
r Chapter VII w. . flgt i when . . . . . quot j wn /TQ I/ . Answers binormal. . bx . / . t X fx..
/quotTo /. t/ . .
quotquot j dx j x f x. . / ./ quot dlt/ J dx j . i/dlt/ j f d J r /. . ydx r Vtp j dy y c/ vry yrp Kl JC V X .
y dx d dx j f x.Answers . i ydy K a J dy J fx. y f x. dx. dy JL y dx d/ / x t y dx Ji Vu l y a Va y l
dlj I fgty y J fx. y dy i/ dy f x t w quot a a r e r J dy J ax oo ao K dt/ f x.v o Vx o jc a Kl f/ a a f /a
V iKr Jf x. .ydy vihT a rn i rfy o i i IA dx / . fx. yd dy j i fx. dx /Jt. f f a U.x a Vci . o y dy
/AT.ydx.ydy J V n . . t Vl fx. t/ d r.
y fx. . Chapter VII Q . y . y u z x.n Ke When / . . oo t J ZX i y ydy. . . . r . a fl . .. .ydx.
binormal. .y.ydx dx fx t ydy i /. /. a K . y oo X .c . lt/dx d/x. . x / y t/ lt/ . a gt. i/ fx. dx JC Ja f x.
Q gt principal normal. . OC . K . jf TT/. . . t/ . M Pnb RQK n flfj Q ii when . x Ing. f dx X / x. . o /
jc Vltte /U. . X ydlt/. . . . y jcl.. . b . v. y x r. . . x . . y . ydydy f x t ydx. y / . . . Answers bx x
osculating plane. . r dy /. /. t/ x . . jc . . dr x u y dy. . . O .. .
ffMir. . y dx j dy J / x. y dx. y dx. . ydx. y j EHi f x. y dy dx f x. SI y dx . y dy J quot dy dx j fx. f
dy / x. y dx. d Ki a ta a / u ax oo S a a . y f dx c/ C i/ /. . y dx J J rr V X dx Vl .x b dx y dy a f Xt
y dx c fxgt P dy a / x.x f xgt a Va L. y dy i Jdy J f x. t/ J dx j f x. J quot dx j Vamp / x.Answers .
f x. ydy f t/ dy C j fx. y dXr JL Ji lt JL . ydx /x. j dy j f x. a o j a T dy f x. dy / x. Jltx i / j vrgt x. y
dy j J / x. y dxt Vx d L . y dx r r J dy J /x. a . y dy J dy dx J / x.
j J VT VT VTlfi /A. C j S // dx dr/ C dx o o Jl /COS/ / cosO by ydygt where the last integral is
obtained from D o o the preceding one the substitution xR sin t. J o . I dx / x. fdf/ o f arc in
f/dx. g . . . a. c . In . . / . dy o j /. . . . . .ydx J a a V J dy a c/ fx. VRy dy f. J o dyfx. quot y jt . i
dx o JT/ // xr/dt/ . . . . Hint. rsinpdr. .ydx. flf i o Jt T cos p Tquot p sin qgt no T J r/rcos. . d di/
.ydx. . /dx jd/ o P P J fx. i/ Va za a r a fl / G yl aa y p a V j o a . . Ji/. b . . Answers a a a . a . .
p . jt arc sin y /A. . t/efy.y jdxj T y.ydx aVtfH o yf dx fx. . . J J i fx. . .
/w wy. dltp r/rcosqgt. variation for since find wraxjt B follows that uvau . . quotquotTquot T a
cos ltp . . t. Hint. .. r sin qgt dr. r sin qgt dr dqgt r/ r cos ltp. . uvudu. y ftx we urr . / i fr TT dw i
y dv quot Of f k J quot u Hint ter chan S e w variables. We the q o have limits since x u u as v
and yuv of u c. f ltp cos /tanpdcp i sin T ltp sin p n r dr cs a f ltp / tan qgt dcp tan ltp dqgt r dr.
the equa tions of the sides of the square will be u v. The limits of integration are p icfu . The
Jacobian P i is I g abr. r/ r cos cp. Solution. f J dcp f J r sinpdr. . the Jacobian xr. Solution. is I
u. . We of define functions . u . u u .Answers jn i sin sin ltp . of whence v a v when x Limits .
nab. a /COS dltp ltp Jt a /COS J . of u the w. rsinqgtdr. The equation curve . ablf L/i
rarctanrbli Ahk . n Ji o oo t . when ulu. whence a .
. . .. . Since r must be real. .. /quotT o o . a . . a . Change . . T. .. ht upper limit. . Pass to . a.
jw LV I Hint. a l flZ w Hlnt . Hint. . . o ap IR o . J. . to symmetry of region of integration relative
to the we can compute the entire ak tan integral. dy b dy . . yT . . .. . Hint. . . o nafrc. . i O . .
Paas to polar coordinates. . lOjt. confining cos ltp ourselves arc dy y fa . the j/zplane. . Hint.. .
. sin qgt . a ln. d . . a g. L KT. r / Answers rjcosp.. variables u. Change the variables .
Integrate in .. . . it follows that p pDue sin whence of the for the first quadrantal angle we ak
have fanltpraxes. udi/lCrlxdy. jJia . quotquot V quot . . . . aarcstn Q . m ft/ . . L xt/ J ft . r y cos
qgt rr cos cp cp T . a . . the . A. na . a . xy u. . ..b sincp fo the first quadrant i dxdy S Vy j dx
Vy j jgt o aVax dx b abrdr. iiaa P. . a. IB . . . dx dy. /quota c c a . . . f.. IX . Hint.rrsin cp
whence the lower limit for r will be and J.
Hint. Hint. y from the straight lint / is equal d V JL and found by means In of the normal
equation of the straight line. Placing the coordinate origin at the vertex. we direct the
coordinate axes alonjj the sides of the square. For the variables of integration take i / and x y
see Problem . Hint. a . Dd . . T T r C a i ady .quotquot o . where k is the proportionality factor.
we have jt jt a sec dltp jp a cosec dqgt ltp /x oo I kr r sin qgt r dr kr r sin qgtV dr . A . C fl a arc
sin Hint. x In to . .a dx J J change ya the x r J Integrate by parts. /x a / Voj a . . n d . The
distance is of the point x. /. /o jia . quot m . / Jg Jifl . . . o L. Pass to polar coordinates . /t . lz J
. and then variable a In sin t Iransform Hint. a. xdi R . . fca /T quotl. Hint. T . a arc tan V L /. b
Vb ca /a c c .lro. the distance from which is proportional to the density of the lamina. . a and a
aa . Answers polar coordinates. the answer. The moment of inertia is determined relative to
the xaxis Passing to polar coordinates. Pass to polar coordinates.f f KI . . . J. JT . Hint. VR H
a . a b H. dx J dy J / . fl .
f. .. The component of attraction. jtfcQ/i the xaxis is equal to r sin cpf z proportionality factor
and Q is the density. jt . where Q the density. and the square of the distance of . . The
moment of inertia is computed about the xaxis. The moment of inertia is computed about the
xaxis. by this element tj the follows that the resulting stress is directed along the of unit mass
lying at the point is equal to j sin ip Q sin i cos ty dtp dqgt dr. the element cos a. c a . . The
base of the cone is taken for the xr/plane. Hint. .Pass coordinates. Hint. acosp . Solution.
rdydrdz from where k is . After the passing to cylindrical coordinates. . a b c. . iz to spherical
coordinates. v . a /i . y J J /. symmetry The mass of an element of volume dm p/ cos dcp
dtydr. . Introduce . Passing to cylindrical coordinates. jt. ip From is zaxis. o . . for the plane of
the base of the cylinder. u /.lln. jutfc. Hint. nnot h na . along the zaxis. Hint. . the cone is
taken for the coordinate origin and its axis is the zaxis. a. t/ . . n . /z a . Jia Selut ion. x D Hint.
i . nabc. square of the distance of an element r z sin qgt rdydrdz from the xaxis is equal z . C
dx Q C f L dz fdqgt f L COS ltp rdr facoscp to spherical quotquot TT . for the axis. . . we have
for points of the surface of the cone r j / . c r . the axis of the cone. . a . .lt/. Pass to cylindrical
coordinates. . dy . . i . fl. Answers i VTT /. . and the equation it of the plane of the base will be
rsin . . . the equation of the lateral surface of the cone will be of the The vertex rt a. If we
introduce spherical coordinates. For the axis of the cylinder the jq/plane we tke the zaxis.
. . arc cot p . . J yey dye x p . n . Solution. . / JJ oo dp r Inr dr JL In r leJJ rfcpjtjQlnen rrfr e
Whence lim egto . . a p . Converges for line / agt . . polar coordinates. S K K v I/ y/ . L.. pgt .
Sur round the straight with a narrow strip and put I i . . Let r gtQ h z tance from the element of
volume dv to the mass m. consider In V x ydxdy... it follows that F kMm . The attractive force
of the element of volume dv of the sphere and the material point m is directed oo dqgt Q sin f
cos gt dr. . . Hint.Answers La i h cosec d f The resulting attraction is equal to . . o . . cp.
Converges.a J. P . a Hint. c p gt Q. dQ dz. . Eliminate its from S the coordinate origin
together with eneighbourhood. Solution. a circle of S radius e with centre at Passing to polar
coordinates. we have T. . Converges. The pro dF . l n la. Differentiate p Cf/ J o twice. .
Whence R But since xynR M. arc tan AW arc tan AW . where the eliminated region the
origin. o this point along r and is numerically equal to kym is . b P w for p gt a. a n L. Pass to
.. . is that is. kmydv / c s rz f H z kmy Q dltP . . . V a . We introduce cylindrical coordinates Q.
We denote by the distance of be the disfrom the centre of the sphere. . P . d PP A gt ln. z
with origin at the centre of the sphere and with the zaxis passing through a material point
whose mass we assume equal to m. Hint. where y la is the density of the sphere and dv
qddQdz jection of this force on the zaxis is the element of volume.
. a . b C . raamp o . nab. xa sin . a V. where is S is the area bounded by the contour C.
Converges fot . a . Solution.e.o a V u ita f ji V r fl jTu quot rln .. d . with base. . z c Vlfft e y x . .
c T o . . y a x xy C. J . r bgt c d K . ltcos J X. . . V a amp . Greens formula . . . V r arc sin h .
jia a . .n cosy. rc. cases . n ds J . jta . . lOVTO ftl . fi jquotM //. . b . naa . . / Hint. x yx n/ cos b.
. f f jcrfjcf J //a J y ty y dy . then cosX. S. . ll . equations of a . b . . t IJi . fl / . a . . parabola t/x /
that connects the l/a a b b and . m . aTT . all . ltP ds J dy Q as . Greens formula is used in
the region between the b contour C and a circle of sufficiently small radius with centre at the
coordinate origin . not applicable. . M yds may be interpreted geometrically as the area of a
cylindrical sur c face with generatrix parallel to the zaxis. . o . . agt f/ C. d lnJt / . d A . and
with altitudes equal to the values quotof the integrand. . l arc flu tan d aV . Hint. c L ln. a . q V
. o b . . e nx y. Jia Hint. V . as hence. Put . Hint. a b a . . . Therefore. d . c . /. d Y x . o a Ol
Hint In Case b. / S o xds. / JJ S /iJi. jra . In circle./ quotquot fl . where C is the arc OA of the e
points . the contour of integra tion. . If we consider that the direction of the tangent coincides
with that of positive circulation of the contour. Answers i . . Use the parametric .
. nR r / r. div grad U grad U . f / r ot for IJr. the level surfaces are planes perpendicular to the
vector lgrady when a c. the hyposee z . . . /i. x /c xy c . b J j cos a cos p . na . a. a . The
equation of an epicycloid is of the form x R f r cos f r rcosJL. / COS gtr . b n/ // // . . r r . a .
circle drawn to the point of of tangency. . . The flux is equal to jtm. rotr. rot /r .. . a . Hint.
divFOat all points except the origin. Circles. rotv co. . mgz. . y c Jfe. Ljc f work. Cones. / r r. . .
gradrr . Ji/ r nR for of r Hint. . . . O Jia . . div a . . . n ytx where t is a parameter. . /i . . . . where
/ is the angle of turn of Answers the radius of a stationary r.for .i. r t divr. FR. b . a divr. V
/Jquot a. R r Hint. . .i/ //. c div LW Ct coco/f /rlt . cylinders. rot . . work. b ii. . / factor. Potential.
a a . where k is a proportionality potential. jta. a jJi/ /// // . epicycloid . . cxr. . potential. V . /
mgz l . unit vector parallel to the axis oi rotation. c. a/i where is a . . div tf. V c mgz. . grad / gt/
x y z. . The equation cycloid is Problem . d // c. Spheres. where . b r z . . work. y /frsin/ rsinJ/.
. obtained from the equation by replacing r by the corresponding . When calculating . . .
Di. Diverges. . . VIII L. Converges absolutely. Converges. Converges. . In examples a and d
CO consider the series fei a ki a zk anc in examples b and c investigate separately the series
a ki k and a ki lt Diver es Converges absolutely. . d converges conditionally. The remainder
of the series may be evaluated this by means of the sum . Converges absolutely. Converges
. Diverges. nh. Diverges. . . . /f rfr dr. Converges. . . Converges. . . . Diverges. Diverges. .
Converges. Converges. Converges. . Converges. Diverges. Diverges. . . Converges.
Converges. c diverges. . b Uxyz C. . . /Di. . Yes. Diverges. Diverges. Converges. Diverges.
No. Converges. converges. Converges conditionally. . . Converges. Diverges. . T nl / . . .
Converges. . . . . . Diverges. Converges absolutely. . . Converges. Converges. . . . . .
Converges. Converges conditionally. a Diverges. Converges. of a geometric progression
exceeding remainder Rn an quot . . Converges conditionally. . . . . . . . Converges. a No
potential. Converges absolutely. . Converges. . fii fniv converges. Hint. . Converges
conditionally. Diverges. . absolutely. b converges absolutely. . . . . . Converges. . Diverges. .
Diverges. Diverges. . . . . Converges. Converges. . Converges. . Converges. Converges
conditionally. Converges absolutely. Diverges. . . TtD . Converges. Converges. . Diverges.
Diverges. Hint. Converges absolutely. . . Converges. . Chapter . Converges. . Converges.
Diverges. verges. Diverges. Converges absolutely. . . Converges absolutely . . Diverges.gt . .
the Answers flux. . . . . . . ConCD verges absolutely. Converges. . . k. Diverges. . . .
Converges. Yes. . . Converges. Diverges. . . . . verges. . . Con verges conditionally. . use r
the OstrogradskyGauss theorem. Hint.
. Solution. the series SpY . i. I n From this / zn VTj / / T r we find the above value of Rn .
converges conditionally Converges or c. . xltl.. . . . x Diverges everywhere. . . Putting /i. xgtl..
We multiply by quot / M quot Whence we obtain is nv nv . and when x gt the series with
general term . diverges absolutely quot for absolutely forxgt. . it diverges. at the remaining
points ltfc ljifcsO . when x lt . xl. when x . . m A quot. an x Q. Hint. Solution.ltx lt oo. .
.lt.converges. gt xlt . . the necessary condition for converwould follow that cos n. diverges for
xlt. converges conditionally for ltxltl. xltl. f m quot For tne iven series it is easy to find the
exact value of the remainder Solution.. oo. we find the sum of . diverges for X lt. and cos
n.xltl.. x lt . . . . . Converges absolutely for it . . S when gt . . verges . . lt. Converges diverges
for xlt Converges absolutely for xgt for for xgtl. x. Kxlt x k and yltxltl. of x. For these values
both the series the series convergegt When l ..oo ltlt.Answers . . .. . xgt. Converges
absolutely for x .x does not tend to zero as n thus. Converges absolutely for x gt . . lt. since
from cosngt xsO. . . . / i v n . Hint. xj. S . I . f . .v when /m lt x lt gence is violated when xlt. .
. farctanxof sum . ltxlt diverges. lt lt lltxlt ltxlt ltx. lltxltK . llt. . ltxlt. . . it is possible not only to
find the interval of convergence. this equality lltxltl. to note that the divergence of the series at
the endpoints of the interval of converthe aid of the necessary condition is detected not only
with gence gence ltxlt . . . .. sin .t n ltlimAiljcr rc. o . ye ltlt . ltxlt. . . ltxlt . x see Problem for jc x
. it is possible not only to find the interval of convergence.ltxlt . lt lt x lt I Hint. lt x lt . . . Kxlt. x.
. . gt xllt Hint. . ltXlt. . but also by means lim of the dAlembert test. however. . . . Consider the
. . I ltxlt z . lltjclt. . lt x lt ltxlt. . . ltxlt. . ltxlt. x . lt x lt . Using the Cauchy test. x of the series for
is obvious it is interesting. . x e . Inlfx . ltlt . ltxlt. ltxlt. . Whenx lim n oo . z I lt lti lnlA llt x ltl . . .
K V ooltxltoo. . but also to investigate the convergence of the given series at the extremities
of the interval of convergence. . lt we have lim n x n is nIn rtl. ltlt . e ltxlte. z lt ao oo lim n j e .
zlt V . ltxltl. series since . . but also to investigate the convergence of the given series at the
extremities of the interval of convergence. x . . . ltxlt . . . Hint. ax l ltxlt . . Using the dAlembert
test. z lt z / lt . The diver x of convergence.xl Hint. the general term of the series does not
tend to zero . x the series x xlt . when Answers and . ooltltoo. . . . ltxlte . ltlt. . . . .. lt x lt
Solution. C. ll n lfxl In arc tan x x lt .co oo x readily obtained by means of Hospitals rule. For x
the G . .
. e V ni .. T quot H /S yn OOltACltCO. cos jcacos a / xsm a cos a r smarr cos a . quot rn ltxlt.
o. . on Hint.Answers x . . f . . . a When power investiseries remainder. o TTiquot Z Z gating
the quotquot use S A ltxlt. f s . oo oo oo ltxlt oo . quotquot W quot quotquot l Iquotquot W JC
ooltxltoo.. . . rz . the theorem integrating r lquot oni vn no oo lt lt ltxlt oo . j . oolt x lt oo . ..
. x jxgt. quot .l x o .. . V ni n l Ar quot x ltJClt.. . Answers ltlt ltX ltoo..... . x A z oo h oo. . V n n
lt x lt . lt lt . ltxlt . Hint. i ni JL ltxltoo. x x xH x of powers . . D nI IKD. /l ... . J c quot ... e.
Proceed ampsinltp. . r . .n .. g V / Ti orFT quot lt lt . / je x ooltAtltoo. . . quot quotTTI X U
quotquot quotquot .n L. V AIO quot Q lt x lt .lt.. ni JDquot ttll OK .. .. S x ing from the
parametric equations of i/ pute the length of the ellipse and expand the expression obtained
in a series of . . .iltjcltl. e . TS. comthe ellipse x acoltf. ln V f I l l. . V.nx .. .
. X TT. f VL jL n . x yW I Hinl. oo ltt/ltoo.. Z Z o o xquotlt . x lt . . . . .ly. . .e. /i.. . Make the
substitution .e... . . i ltxltoo. quot r /y . T. / Jt Hint. fh. . .Answers Jgt Hint. . ax bxy cy a fy bx cy
k bhck . Eight terms. i. . . . . lx . o ... R lt ... . that is. . A. Z . Jlt fi. . . x . V n i . R lt . .. lt . I Y / / .
f t /i /r f/i . RIltlt.. T p r ve that exceed . i a geometric progression . Ult . j . t and expand nx in
powers of t. ly J ry l. lJC i/ lt j Hint. V lJfZL. Two x terms.. . it is necessary to evaluate the
remainder by means of a geometric progression that exceeds this remainder. . . .. i . i. V JO //
xquot. . x ... . tan /for /i K//lt. arc tan i arc tan x arc Ay . Two terms.
. y lgtfl tlyl l a m nf a is nonintegral. ni J . S Jl cosan. n . if Ji I a jL i B y if a is or nonintegral.S
ni n n. X . La V tr lta rt fl cos n n sin n J . S Ji cosh an. cosa a is an tr L integer. sin ax if an
integer. . Answers l no n. Sln . lquot. V CO tin . a is gn n X x ni . . sin quot .
. / will t readily be seen that o n n amp.P / x cos nx dx. T Sin The same substitution Identity f
as in Case t . .. b il A V J S n mx . i nl . with account taken of the assumed . Solution. taking
advantage of the assumed lt identity n . /cos . /lt . rs mx no . / . then. . gt quot sin n cos where
. sinh / J t /IJTJC ...HlJix o nx quot ni ...Answers slnn/t . a f / cos /i dx jjJi / W x cos /ix d H .. .
n sin T / sin nx dx f . If we make the substitution t in the first Jl integral and t x z in the
second. y / / leads to the equalities II /il.
. . exact . . no. . x. .u/ xi/ / r/ f . y yquot f/V . . xyquot yQ. y . . . . xl C. tan yC . / . Fa . C sin i/
jc. . hypothesis Differentiating . xln. . . . x x yltanx C. . . . yV Family X n mily of similar ellipses
x . . . . Yes. y . . . . Straight line is f/Cx hyperbola . of circles x y C . xy Q. i/ . Hint. . a b . exact
value y. . if twice with respect to x. . f/ r/ . isrrfj. cot //tan x C. j/.rSg g g o e . r/ y xe X . t/ ax . y
C n lOx y . . . Yes. . lnjc xy . or J/ C. . . . exact value y . Yes. Yes. . . . . wxr/ln. . x C t/ C . yl. r/
Ce a . V if e lnCx . X cos. . No. xyy xy x y y. . ment . e le x . xyy. C. /quot i/ r/ I/ // t/ f/quot // . x
l/. Yes. The seg of the tangent equal to By I/ C. e. quot . we get a differential equation. y t/ . y
. . ln/C. x i/ jc. y a . y . x x amp z . . . x . . Yes. . . . . . . . Answers Chapter IX . t/ px. . exact
value value . y b b . // x . .. . . . quotO. . . / . .t C. Pencil of lines ykx. x . x . . . / . . x Ce y . Hint.
. x . . . y x yC. Family of hyperbolas x y C. / . .
If a tangent at any point x. . . . y The desired differential equation is CxC. / . In solution t/ . y x
y p p . . . Hint. The coordinate origin is located in the source of light. x n . . y forms an angle
a. . . Paraboloid of revolution. r/ xln. General integral C x x . X is equal to y dx. x . Solution.
no singular / integral. . In p x C ye . c. . f/ Cx x . forms with the xaxis an angle qgt. . . x y C/ .v
/ . . arc tan C. By virtue of symmetry tUe soughtfor mirror is a surface of revolution. // . . is
singular integral . y respect . y . the xaxis is the direction of the pencil of rays. . i/ xn Singular
solution . tan qgt . . then tan a ing the origin with the point Answers M M y // bola. f/cosxi / sin
x. y of the curve. t/ cos y C. But tana x . . . x . . arc sin x . there HO singular integral. x . . . V p
r arctan C. C/ xsiii/jf/cost/ sinc tC. . . C. The plane section is a parajq/ and its solution is y The
desired surface is a paraboloid of revolution. Use the fact that the area x Cx x C C. Q/ . tancp
j/. v C. and the segment connecttan qgt x. generated by the desired surface being cut by the
xi/plane. x Y / Cx x In x . C.C singular integral x . The equation ab is linear with tgtx and .
Singular . . A . . lnU . A C x y C . Hint. Cx. . e . General integral ry. .. General integral there is
quotoquotquotquot rquot x x C J.
y X f. differentiate with respect to x h Bernoullis equation. . /i. f Clairauts equation. // LJ C. d
Bernoullis with variables separable. Q/ / . . arc tan In Cx. x C. differentiate with respect to x k
Bernoullis equation in r. y uv n Bernoullis equation. . In x lnC. A circle and the family of
geneous. . y . . m linear. xi/C . reduce The y uv. nlPLzIlsin P C . l. a Homoyigt uv c linear in y
y uv. X / arc sin Cx. . /./ . lnx xy . l y.sin y cos .xl . . arc tan y . The differential equation from
which x is defined as a function of p j/ is homogeneous. / no singular solution . . Cy . . its . . .
f. r C. y uve to yxy Vyl g Lagranges equation. Cx g / . Ce y . . x yxu tangents. x Ce sIn alsini/.
x CeVlt/f. a sin In jc C. Answers . . equation. b xy. y i Cequot ix jc . . ln yCx VaC singular . b
linear in x x astroid xquot a z igt. . . Arlt/Cegt . o l. . Hint. . i Sx y ln x y C. y uv i leads to
equation with variable separable.lt L .. gC. uxy j Lagranges equation. y xlCe. solution. xuv
exact differential equation. . y. . py . . y . JO/C . . singular solution. . . Cx C. x Ce y sin .cos. .
i/ .
Hint. p e ltwi. sec. // . c that the area is equal to J ltgt y dx and the arc length.ev e. the point .
Hint. . Hint. . Hint.v a singular point. . . Equation ae ki . . Hint. Equation kQ. Q Qo lt u quot
yquot t K T . sklw. . f/ . kg. e x is arbitrary. a . s/gtja.x . to f / Hl/ o . t/i. . . r/x. where C . i/ x.
The soughtfor differential equation . . a y x h px. is . Hint. . one hour. . Use tha fact .
Hint.kwdx. Equation fV y / gtanh ft I/ . The prts sure at each level of a vertical column of air
may be considered as due solely to the pressure of the upperlying layers Use the law of
BoyleMar otte. In V olOOgJ rpm. x . e get for the plane axial crosssection of the desired
surface the differential equation go x. Hint. Hint. . ds . c x x.// . . is dp kpdh. C A . .C C j a x.
according to hich the density is proportional to the pressure. . . t/ singular point of the
differential equation. n QQ n . a x C / C b no solution. Hint.. Use the fact that tha resultant of
the force of sgtravit and the centrifugal force is normal to the surface. a y . // equot. the initial
quantity Q t will decay in years.x.v as quot is a tionality factor. . Ta T Q of . i/ /. Pass to polar
coordinates. lakmg the r/axih as the nxis of rotation and denoting by co the angular velocity
quotof rotation. y . . i/ e x f . . y x ij . wf . y sin x cos x. f/. . Equation m mgkv .. . / sin . . xf Cf/. .
. . // . f . . Hyperbola r/fq C or circle x ij/ . Equation n/i f fcd/i V udf. / . rCe. b Cx. yVa Kx . t/ is
Q/ . y i x x l. is a singular point. yCx . . . p . dQ kQ dfi. xfx / / C. .
lt/ singular solu tion. x z C. . y ln quot . . . x equot No solution. In . Hint. . y C xx l / . . ex .
CkC . / r/ x C z. . JtC. //. e . cosh .. /. Catenary. r/ C. Singular solution. . yx . C . . x at/ a
Cycloid. . . . x i/ . x. . Circles.y C. . . /y quot . . equotquot e quot . Answers . a . x x a t y a lt
cos . y Cx . i/ in C . x x z . // In A . . l. . In xC l y . x sin /. . . . . w . er . C y / singular solution.
//C. . x. . . lt. x . . . Parabola. . r/ . r/ a . y . singular solution. t/sec . y . . y . Equation Ri L E
sinco. x aln . Circle. t/ . . C. t/ . . y C ed l e . // . lt/ O. y e . . x n i C A C Singular solution.
Parabola. C. . w Cj D oCjl . . sinC. . . // C .
sin a mr at u. . j/ . cos x C. f/ C. c c Hint. e no. j/C. y sin x B HCjC .v. / C. a t/y Q b yquot y f f/.
In . .fiv C cos x gtxf cos h x . seconds. cosv .C Xcosh The C t Answers where H H is a
constant horizontal tension. // r/ sin x. Hint differential equation y T El uation of . Equation of
motion. g no. // . i/l. f dx AZ ylt . f/e. a cosh e e .xe yA BiC av sin /.C e. y .c I C v tgt . s e lv c .
.C. .equot C e V . . . C. // . . . . C sin /J nx v . cos a x I/ g . . cos x y C. //C. b fl sin x. a No.e r A
. s. . . tn/ mg k quot/ .equot if k lt .e.j C a x . f /Jsinx.sm xC cosx. CiCx . ax / x a x f .CV .lt H. f/
C. L O AT / Hint. Particular sothe homogeneous equation r/j. // C t . f C P X ltgt.C e x . b
yesgt yes. . Equation of motion. r C. a A.lt x Fx X sin xJ . and l a.//tf . //e CicosAlC Sin A . p o
sin xcos . A Bx x . C z l v y AX Ccos xHnsec xftan Ifsinx In cos v xcosx.x x r. C rl . i/. e CvV x
xe Ax Bx C. cos xfi/ I C sin x x sin x. . d yquoti/ // Use . //. the substitution // /.v. . . . C . /y
.cosx s smx. jt x f x . sin // . h yes yquot / x//f/. no.. C sin x X x e sin x. . C. A d yes. cos x v .
.e C e X x . Hint. y C. y C.C. z . . . / . C. . By x B method . y C ic iA .g C. . s gsina Xlncoshf/ r
jiicosa. // cos x f sin x sin x.Cquot xex . If ft gt . of the variation of parameters we find C A.e h
C sin x. y the Hint. .x. y e C t cos vC sin x . . Law of motion. /C. cos Jfex i/quotTT J C sin / x
C t e x C e / I .xf fC lnx. d g cos x xc Av j. f/ .
y sin C quot C IJC a . X lt C x x . y . iL dt kx y. . . . y C e C sin x x x lt x . therefore. sin f. C C
e x. cos x C e Cf e ex C sinx cos sinxlnjsin. . sec. f / l sin/ t/ gsin . x t/C C xe xe . C e cos sin
U o . k . hence. ex / C x sin cosx j . C. t/ . / .cosorf. a y V /t Xarctan. from the position of rest
of the load. C. then xy I is the length of the spring at rest. y o CD/ Jf . C c y jce x x . k x g is
the distance of l.gx C x V w . Answers A x C cosco/ C sincif j slnp/. y iex l j. cos x . .EquatIon
of motion. l Ti / . ff X rec koned t . . xslnx C. . x cosx. b C. m dx. y / . where x jjtquot the point
of rest of the load from the initial point of suspension of the spring. yAe cosx sinx . .f jx Ie IJC
. y C CjCosjc C slnA. cos x xcosx. cos x X . / yCfV CStx. . sin . t/ xx xe X . X e. C. . C ex
xesin x e cos x. / . t/ C e x l s / I . e. . . . cos x C tan . . cos. y Cj cos x C sin x sin x cos x In
cos x C. . .x J f xe. t/C. Cf . cosx. .cln cot x x x . x quot cm . . x Cj cos w/ C. cm/sec . Hint. . i/
C e x HC e. . y l Transform product cosines to the sum of cosines. where . kbxkb x . y Cj cos
x C the x f of cos x j sin f rsinSx. y C C xex xe nx.
I C In x C .H . ye x e /j cos V xH. y C. wC equot x C c. . f/Q cos C sin In x. C. b r e oV. y t is .
Cj Qx e . . y C. .x y psin r. .x x . j/ In .C. .e C sin x e x C cos x . etf or e wi Hint. . . . The
differential equation of motion equot.C. j/ C. . y C. C x C. . C equot x .sxl c. cos x C x sin x y
C. y C.ar In . C sin x I/quotQ t . . y C. . sin In x. y . .C. /. / . z C cos x C sinr x y e C cosx C / f/
t i l . llt gs. // ce C. r/QxH i/ . j.Answers and x ccos f t y J . i/ C. cos In x . rgs. .C jcHIn xf / . C.e
C e X .sinjcHg sin v . // Cj C cos xC / x sec x j cos x In i/j cos v tan x sin x x sin x. / / . lMC C
lnjcH f xl C x C x o xlnx ln x. .C x cos x / i/T V .JChC x C x . . C ltT C e. . cos x C sin x.
.. w ry v / ln C. C . .if dy ..t f// z Co z From the initial conditions. Answers . Hint. r r t whence
dx dyd x dx . using the properties . y C. In In x . A. C xltr jt. e C cos C sin I/quotQ tj. we have .
z // . x z C we have dx J/ . . y z quot t amp amp the amp / f x . y dy v . r hence. of derivative
proportions. . y . f y z x consequently. f / C f z . gt .. and. xCV C C sinx. y C C.A. Integrating
the In homof we find first integral / jc p arcxtan dz . properties of derivative proportions.Q
and. z C.. r C. e /l. Z f Vx y / Hint. A. In v .z dz n zx y C Thus. dz Applying the dxdt/dz x dx
Similarly. . b In x arc f/ geneous equation x tan C . . z . xdx r y . Jtl /. / Then. r i/ . Whence lnz
. / . xdx udy . C . C A sin x cos . we will have C l . .. the integral curves are the circles x // z C
. . . Q .. . C V Solution. cf/y x y .. c x /z . yzx . . e x . x e x x e. mjjkv x mkvy mg for the initial
conditions when . zQfCj.. x C L l f I/.
Answers .I j yy Q ... x. . Hint. //I Hint.. . SCneS Conver es . . Use the condi lions u.. ft Q
Integrating. r/ . . u t A n cos no quotquot sin n quot . the series converges for l l quotquot the
serics conver es for ltxlt joo. quot. . igt sina. for l v llt Hint. quot /. we obtain k . x I/Q . I I Hint.
vx quot f . the x r A quotquot of ig quot . x acos y sinTr Vm t. yl lx .. differential mu equations
of motion . Use the conditions w. / x . . where the coefficients / . . x a loT r t o . . . // JT .
ttltquot/l .. A sin . //m qi x v ie series converges for ooltxltoo. cos /m sin / sin / . . u n y cos
Tquot sin T quot Hint se llie conf tlons du r forOltxlt. e u JfH . . .. . /t . r Xo i. r/ x X z prr o Z r X
X . Oo. Use the method of undetermined coefficients. . sin ux. Use the method x of
undetermined coefficients. . Hint. f/x x .. Hint. . . J . . UV . . cosa.. Mcossin. . M x. kv v mg kv
a Q sna mge m The / . Use the t method t o undetermined coefficients. .
. . ... a ltquot... . . . cm..t and q to three decimals on the principle of equal effects. f quot JtJf
dx. . to one hundredth Hint. u x. Hint. ./ . .. . Take the number n to three decimal places on
the principle of equal effects. . to. c the quotient may vary between and . c b a .. there is
sense in writing the result in the form .. and . . . . a . Answers . or . s ... b .... . a . c the result
of c b subtraction does not have any correct decimals. Hence. or ./ c b a lt.. . ... The last digit
may vary by unity. . . b ... b ./ ./ . . since the number lies between ... a . Use thequot
conditions u /./ . For practice purposes . u . . ampere should be measured to within . The
length of the pendulum . . . lt./ gm. Chapter . o. . . cm s cm Absolute error.. a . Measure the
quantity / to within and s to within . b . not a single decimal place in the quotient may be
considered certain.. lt lt lt .. . b two decimals. c . . . lt. a two decimals. cm. take the numbers .
v . . lt. Hint. Use the formula for increase in area of a square. cos rui sin . c one decimal. . ...
cm .. lt. m . . Use the conditions sin . X . a . . . .. ../ on the principle of equal effects. The side
is equal to . .. . lt. . Measure the radii and the generatrix with relative error /./ . c . . since the
difference is equal with a possible absolute error of one hundredth..a ioo a gtu . lt... x x. c b .
lt . .. .. . cm Relative error../ mm. . . nnx e jrrr lt. .. . a . sina ..
. after obtaining A /o. is tilled in by the operation of addition moving . the Hint. repeat number
throughout the column of fourth diilerences. Compute the first live values of y and.Answers J.
After this the remaining part of the table from right to left.
and . . where x e is the eccentricity of the ellipse. . .. y l x x ... . . . . . ... . cosx . A . . . Make
use of the parametric equation of the ellipse x cost.. sin x . JT foim the formula of the arc
length to the form /l e cos /d/. . .. ... . .. .. a . J/t I . . dal formula. x ji .. . . . sin x. . cos x . . and .
. . . cos x . a .. .. . . .. cos x . . when x . . .. . . . . . ... x v xn o Qy o y xs x x x T o uo . . . . . .. . . b
.. . . . .. . . sin x cosx smx. cos x . sin/ltnd trans x . .. . . x f/ x x . .. cos x . cos xj.. S. . . ... b . .. .
x x. sin x. Hint.. J/ fr x . . . x x. . . . . W . y x . . .... . . . . . sin x. . The interpolating polynomial is
y A. . . sin x...... . z . .. x n . . xx .. . .. ... by Simpsons formula. . . Answers b . Hint. ... cos x sin
x cos x f. sin x sin x f. sin x . . ..s AO. .. . . kg approximately.. . . z . . . a .. . y x T o J X x z W T
x x X T T yl x Zl TTquotquot . . ... . ./ . i x. for x. . . . When computing . . . and .. .. and .. .. . z
.cos x. T T x take . . . . iW . . x oe QC JC x for x. . . ... . sinx . By the trapezoi. . . x . . cos x . i/
. . .. .. ... . . .
Some Constants .APPENDIX I. Greek Alphabet II.
. Logarithms III. Appendix Inverse Quantities. Powers. Roots.
Appendix Continued .
Appendix IV. Trigonometric Functions .
Hyperbolic and Trigonometric Functions . Exponential.Appendix V.
/ The witch of Agnesi. Appendix VI. Parabola upper branch. . Cubic parabola. . Rectangular
hyperbola. Graph function. . of a fractional . Some Curves for Reference f . / . Parabola. . .
Cubic parabola.
Appendix / Semicubical parabola. Sine curve and cosine curve.v and // . Neiles p arabola. y
sin . xt yc or . and / cot. Tangent curve and cotangent curve. . x or a.
Appendix . . Graphs of the functions f/ //secx and yarc sin x x yarc cos x Ji A arc cos x gt y
arcsinx Graphs of the inverse trigonometric functions arc sin x and y arc cos x.
Appendix arc cot x cot x T quotV arccot . Graphs of the inverse trigonometric functions // //arc
tan x and arccot. . Graphs f/ g of the exponential functions and y e. . .
Appendix . . . Bernoullis lemniscate. Cardioid. . Cycloid. Strophoid. Hypocycloid i astroid. .
Evolvent involute of the circle sin . x af a sin. y cos / alcoscp. . /. f/y a v . u y sM AT ax . x a
cos y or A S fl sin / .
spiral. a . . . . Logarithmic r . r a sinSip. . r asitip. Hyperbolic spiral. Spiral of Archimedes.
Fourleafed rose. Threeleafed rose.Appendix .
Comparison test . . Area in rectangular coordinates . Binormal Boundary conditions Branch
of a hyperbola . Catenoid Cauchys integral test Cauchys test . generalized Approximate
numbers addition of division of multiplication of powers of roots of subtraction of Cardioid .
Brokenline method Eulers Witch of . . . Area of a plane region Area of a surface . . Cauchys
theorem . Angle of contingence of second kind Antiderivative . . Arc length Arc length
Archimedes of a curve of a space curve Characteristic equation Characteristic points
Chebyshevs conditions spiral of . . Cavalieris quotlemonquot Centre of curvature Change of
variable in a definite integral in a double integral in an indefinite integral Approximation
successive . . Algebraic functions Angle between two surfaces. Catenary . . . Angle of
contingence . Agnesi Bernoullis equation Bernoullis lemniscate Betafunction . Acceleration
vector Adams formula Adams method .INDEX B Absolute error Absolute value of a real
Bending point number Absolutely convergent series . Chord method Circle . . Diodes . .
Asymptote horizontal inclined right horizontal right inclined vertical Clairauts equation Closed
interval Coefficients Fourier . . . Composite function . of of convergence curvature Argument
left left osculating Circulation of a vector Cissoid of Astroid . Area in polar coordinates .
Conditions boundary Chebyshevs Dirichlet . . initial . sine tangent Concave down Concave
up Concavity direction of Conchoid Condition Lipschitz Cusp Cycloid . of centre of gravity
generalized polar Correct decimal places in sense Correct decimal places in a narrow sense
Cosine curve Cotangent curve logarithmic nth righthand second Derivative of a function in a
given direction Derivative of functions represented parametrically Derivative of an implicit
function Derivative of an inverse function Derivative of the second order Derivatives of higher
orders onesided table of Descartes folium of . firstorder Curvature centre of circle of of a
curve . Curl of a vector field Critical series Differential of an arc . . a broad Determinant
functional Determining coefficients first method of second method of Diagonal table
Difference of two convergent Coupling equation Critical point of the second kind points Cubic
parabola . radius of integration of second Curve cosine cotangent discriminant . . Gaussian
integral logarithmic Differential equation homogeneous linear inhomogeneous linear
Differential equations firstorder forming higherorder linear . Conditional extremum
Conditionally not absolutely convergent series Contingence angle of . . Continuity of
functions Continuous function proper lies of DAlemberts test Decreasing function Definite
integral Del Dependent variable Derivative lefthand Convergence circle of interval of radius
of region of uniform Convergent improper Convergent series Coordinates integral .Index
probability . . . higherorder principal properties of second secondorder total.
of a tangent of a tangent plane with variables separable Equivalent functions Error absolute
limiting absolute limiting relative relative Euler integral EulerPoisson integral Eulers
brokenline method Eulers equation Even function Evolute of a curve Evolvent of a circle
Evolvent of a curve Exact differential equation Exponential functions . . function series .
Double point Factor Elimination method Energy of Ellipse . . . Divergence of a vector field
Divergent improper integral . linear of a normal . . Flux of Folium a vector field . Domain
Domain Double in . theorem Discontinuity of the first kind infinite removable of the second
kind Discontinuous function Discriminant Dicriminant curve . kinetic Envelope equations of of
a family of plane curves Epicycloid Equal effects principle of Equation Bernoullis
characteristic Clairauts integrating Field direction field nonstationary scalar potential vector
scalar solenoidal vector Field cont stationary scalar or vector Field theory Firstorder
differential Firstorder differential or vector vector equations Flow lines of Descartes . Index
Differential equations of higher powers firstorder Differentials and higher orders
Differentiating a composite function Differentiation of implicit functions of of third method
coupling differential Eulers exact differential firstorder differential homogeneous . .
homogeneous linear differential . Direction of concavity Direction field Dirichlet conditions . . .
of definition integral curvilinear coordinates Exkemal point Extremum in polar coordinates in
rectangular coordinates conditional of a function . . inhomogeneous . Divergent series .
tabular Diocles cissoid of . Lagranges Laplaces .
of a function Greatest value . equation . . Hypocycloid . Fundamental system of solutions
Gammafunction . . Increasing function Increment of an argument Increment of a function
Independent variable Indeterminate forms evaluating t Functions cont inverse circular inverse
hyperbolic inverse trigonometric . integration of trigonometric trigonometric. hyperbolic .
Greens formula . Incomplete Fourier series . Gradient of a field Gradient of a function . .
Lagranges Lagranges interpolation Leibniz Maclaurins. continuous continuous. hyperbolic. .
Formulas reduction . . . Higherorder differential Higherorder differential equations
Higherorder partial derivative Holograph of a vector Homogeneous equations . . Gaussian
curve General integral General solution General solution of an equation General term
Generalized antiderivative Generalized polar coordinates Geometric progression . Fourier.
integration of inverse Hyperbolic spiral . Fourier series . Taylors . properties decreasing
Dinchlet discontinuous of Harmonic even of a series . integration of Functional determinant
Functional series Functions algebraic equivalent exponential . multiplevalued periodic
singlevalued vector rectangular Hyperbolic functions . . NewtonLeibniz . trapezoidal
logarithmic transcendental. . . Guldins theorems Graph H Hamiltonian operator Function
composite . integrating . . . linearly dependent linearly independent . function Homogeneous
Hyperbola linear differential implicit increasing Lagrange . Four leafed rose Fraction proper
rational . Newtons interpolation OstrogradskyGauss parabolic Simpsons Stokes . I Implicit
function Improper integral convergent divergent Improper multiple integrals . .Index Force
lines Form Lagranges Formula Adams Greens .coefficients . Hyperbolic substitutions .
Level surfaces LHospitalBernoulli rule quadratic . . . Integration of functions Lagranges
equation Lagranges form Lagranges formula Lagranges function . mass and static moments
of a moments of inertia of a Least value Lefthand derivative Left horizontal asymptote Left
inclined asymptote Leibniz rule . region of by substitution hyperbolic functions interpolation
trigonometric functions . . Lagranges interpolation formula Lagranges theorem Laplace
equation . line particular probability singular surface triple Integral curve Integral sum
Integrating factor Integration basic rules of under the differential sign direct by parts .
Bernoullis . Laplace transformation Laplacian operator by means Lamina coordinates of the
centre of gravity of a. Infinite discontinuities Infinitely large quantities Infinitely small
quantities Infinites Index Interpolation formula Lagranges Newtons Interval of calculations
closed of of Infinitesimals of higher order of order n of the same order Inflection convergence
monotonicity Interval cont points of open table Inverse Inverse Inverse Inverse Inverse
Inhomogeneous equation . path of . Involute of a circle . Iterative method . . interval circular
functions functions Integral convergent improper definite divergent improper double Euler
EulerPoisson general improper multiple . Leibniz test . Inhomogeneous linear differential
equation Initial conditions . Integration of ordinary equation numerical differential Integration
of total differentials Integration of transcendental functions Interpolation of functions inverse
linear . Lemniscate . numerical . . Involute of a curve Isoclines Isolated point . Kinetic energy
Integration of differential equation of power series . . Jacobian .
. . Picards . . Milnes . reduction series Numerical integration of functions . . Maximum of a
function . . NewtonLeibniz formula . of variation of parameters . Maclaurins series . Newtons
serpentine Nieles parabola . . chord method of differentials of elimination Method cont Eulers
brokenline iterative . Node Nonstationary scalar or vector Normal to a curve equations of
principal Normal plane field Maximum Mean value point of a function Meanvalue theorems .
Linear differential equations . M Maclaurins formula . Newtons interpolation formula Newtons
method . .Index of successive Pascals Limit of a function Limit on the left Limit on the right
Limit of a sequence Limiting absolute error Limiting relative error Limits onesided
approximation . Linear equation Linear interpolation of a fa nation Linearly dependent
functions Linearly independent functions Lines flow force Multiplicities root N nth derivative
Nnbla Napiers number Natural trihedron Necessary condition for convergence an extremum
vector Lipschitz condition Necessary condition for Logarithmic Logarithmic Logarithmic
Logarithmic curve derivative functions spiral . Minimum Mixed point partial derivative Moment
of inertia static Line straight . . Mean rate of change Method Number Napiers real Adams .
Numerical integration of ordinary Number differential equations Onesided derivatives
Onesided limits RungeKutta . Line integral of the second type . Newtons . . Ostrogradsky .
Open interval . . . Line integral Monotonicity intervals of Multiplevalued function application of
. Minimum of a function . of the first type . Newton trident of . . of tangents of undetermined
coefficients .
. . . Quadratic interpolation Quadratic trinomial . . . . . of superposition of solutions Probability
curve . . cubic . Parabolic formula Product of two convergent series Progression geometric . .
Operator Index critical stationary . . . Parametric representation of a function Partial
derivative hirhegorder quotmixedquot second Partial sum Particular integral Particular
solution Pascals limaon Path of integration . Nieles . Remainder term Removable
discontinuity tangency Points characteristic Righthand derivative Right horizontal asymptote
Right inclined asymptote Rolles theorem Root multiplicities . Radius of second curvature
Radius of torsion Rate of change of a function mean Ratio of a geometric progression Real
numbers Rectangular hyperbola Rectifying plane Reduction formulas . Reduction method
Region of convergence Region of integration Relative error isolated singular stationary of
maximum minimum Remainder Remainder of a series . safety semicubical . Polar subnormal
Polar subtangent Potential of a field Potential vector field Hamiltonian Laplacian Order of
smallness Orthagonal surfaces Orthagonal trajectories Osculating circle Osculating plane
OstrogradskyGauss formula OstrogradskyGauss theorem Ostrogradsky method . Proper
rational fraction Proportionate parts rule of Parameters variation of . Quantity infinitely large
infinitely small Period of a function Periodic function Picards method . Plane normal
osculating rectifying tangent Point bending critical of the second of discontinuity double
extremal of inflection kind Radius of convergence Radius of curvature . Power series
Principal normal Principle of equal effects Runge . Probability integral Parabola . .
Subtangent polar Successive method of approximation .Index Rose fourleafed threeleafed .
. Safety parabola Scalar field hyperbolic . Subnormal Scheme twelveordinate polar
Substitutions Second curvature Second derivative Second deferential SecondordeP
differential Second partial derivative Segment of the normal Segment of the polar normal
Segment of the polar tangent Segment of a straight line Segment of the tangent Semicircle
Semicubical parabola Series . . Fourier . Straight line . Strophoid . . . Rotation of a vector field
Solenoidal vector field Solution of an equation general . hyperbolic . . trigonometric .
Serpentine Table diagonal table of standard integrals Table interval Tabular differentiation
Tacnode Tangency point quotof Tangent Tangent curve Tangent plane equation of Tangents
method of Taylors formula . . . of two convergent series Superposition of solutions principle
of Surface integral of the first type Surface integral of the second type Surface integrals
Surfaces convergent . HospitalBernoulli of proportionate parts Archimedes . . with complex
terms conditionally not absolutely convergent convergent absolutely Scries cont Dirichlet .
divergent . functional level orthogonal harmonic . . number series operations on power
Taylors . Maclaurins . Stationary scalar or vector field Stokes formula . Newtons Simpsons
formula Sine curve Singlevalued function Singular integral Singular point Slope of a tangent
Smallest value . . . . Sufficient conditions for an extremum Sum integral partial of a series . .
particular Spiral of Rule Leibniz . . RungeKutta Runge method . principle . Static moment
Stationary point . . . . logarithmic . . incomplete Fourier . M.
least cont Lagranges OstrogradskyGauss Rolles mean Theorems Guldins of a function .
dependent independent Variables separable an equation with . Uniform convergence
Weierstrass Theorem Cauchys . Leibniz . Variable meanvalue Theory field . . Torsion
Tractrix Trajectories orthogonal Transcendental functions integration of Transformation
Laplace Trapezoidal formula Trident of Newton Trigonometric functions integrating . .
Dirichlets Theorem Value greatest . . . Index Term general remainder Test d Alemberts
Cauchys . Trihedron natural Trinomial quadratic . . Trigonometric substitutions . Taylors
series . Cauchys integral comparison . . Work of a force . . computing volumes by means
evaluating a of in rectangular coordinates Trochoid Twelveordinate scheme U Undetermined
coefficients method of . Vector acceleration of binomial Threeleafed rose . smallest . Triple
integral applications of . . change of variables in normal tangent line velocity Vector field
Vector function Vector lines Velocity vector Vertex of a curve Vertical asymptote Vertices of a
curve of principal of Volume Volume of a cylindroid of solids W Weierstrass test Witch of
Agnesi . Variation of parameters .