q> ) sin (p. y
=
= =
622. At
N
2
the
sect
=
=
what angles do the sine and y= s'm2x inter-
axis
of
abscissas
at
the
origin? 623. At what angle does the tanianx intersect the gent curve y
Fig. 15
=
axis of abscissas at the origin? tx 624. At what angle does the curve y e*' intersect the line x 2? straight 625. Find the points at which the tangents to the curve 4 20 are parallel to the jc-axis. 12x* y =z 3* -f. 4x* 626. At what point is the tangent to the parabola
=
=
+
=
+ =
3 0? 5x y 627. Find the equation of the parabola y~x*-}-bx-\-c that is tangent to the straight line x y at the point (1,1). 628. Determine the slope of the tangent to the curve x*+y* Q at the point (1,2). 2 629. At what point of the curve y 2x* is the tangent perto 2 the line 0? straight pendicular 630. Write the equation of the tangent and the normal to the parallel to the straight line
xy7 =
= + 4x3y =
parabola
y=
at the point Solution.
We
= [y'] x= i = -T-
K
x
with abscissa x = 4. have y
f
2
1
k
,/-
whence
7=;
V
the
x
slope
the
of
Since the point of tangency has coordinates
*
#2 =
= 4,
follows that the equation of the tangent is 1/4 (* 4) or x Since the slope of the normal must be perpendicular, *,
whence the equation
of the
normal:
tangent
y = 2,
4 (x
4) or
4x
+y
It
4# + 4 = 0.
= -4; t/2 =
is
18
0.
_
Geometrical and Mechanical Applications of the Derivative
Sec. 4\
63
631. Write the equations of the tangent and the normal to the 4# 3 at the point (2,5). 2x* curve y = x' 632. Find the equations of the tangent and the normal to the curve
+
at the point (1,0).
633. Form the equations of the tangent and the normal to the curves at the indicated points: tan2x at the origin; a) y b)
= y = arc
sin
^^
at
the
point
of
intersection
with
the
A:-axis; c)
d)
y = arc cos 3x at the point of intersection with the y-axis; at the point of intersection with the #-axis; y = ln* ~ x* at the points of intersection with the straight y=e }
e)
line
y=
1.
634. Write the equations of the tangent point (2,2) to the curve t*
and the normal
at the
'
635. Write the equations of the tangent to the curve
at the origin
and
at the
point
^
= j-
636. Write the equations of the tangent and the normal to the x* 6=0 at the point with ordinate y 3. 2x y* 637. Write the equation of the tangent to the curve x* y*
curve
=
+ +
+
2xy = Q at the point (1,1). 638. Write the equations of the tangents and the normals to the curve y = (x 3) at the points of its intersection 2)(x l)(jt with the #-axis. 639. Write the equations of the tangent and the normal to the 4 curve y* = 4x 6xy at the point (1,2). 640*. Show that the segment of the tangent to the hyperbola xy = a* (the segment lies between the coordinate axes) is divided in two at the point of tangency. 641. Show that in the case of the astroid x 2 8 + y*t* = a*/ J the segment of the tangent between the coordinate axes has a constant value equal to a.
+
/
_
___
_
[Ch. 2
Differentiation of Functions
64
642.
Show
that the normals to the involute of the circle
x = a(cost
+
t
y = a(sinf
sin/),
t
cost)
are tangents to the circle 643. Find the angle at which the parabolas y (x 2 _. x i ntersect. 4 _|_ 6* y 2 x and y 644. At what angle do the parabolas y
=
=
Sect? 645.
=
Show
2
+
that the curves y 4x 2x 8 and y are tangent to each other at the point (3,34). Will
same thing 646.
at
Show
2
2)
and
= x*
inter-
= x*
x
we have
-\-
10
the
(2,4)? that the hyperbolas
intersect at a right angle. 647. Given a parabola y* 4x. At the point (1,2) evaluate the lengths of the segments of the subtangent, subnormal, tangent,
=
and normal. 648. Find the length of the segment 2* at any point of it. curve y 649.
Show
that in the
equilateral
of the
subtangent of the 2
2
=
2
a the y hyperbola x to the radius vector equal
length of the normal at any point is of this point. 650. Show that the length of the segment of the subnormal 2 2 2 a at any point is equal to the abscissa in the hyperbola x y
=
of this point.
651. x*
y
Show
that the segments of the subtangents of the ellipse
2
jjr+frl
an d the circle
2
x -+y
2
= a*
at
points
with the same
abscissas are equal. What procedure of construction of the tangent to the ellipse follows from this? 652. Find the length of the segment ol the tangent, the normal, the subtangent, and the subnormal of the cycloid ( \
x = a(t a(ts'mt), y = a(l
an arbitrary point t~t 653. Find the angle between the tangent and the radius vector of the point of tangency in the case of the logarithmic spiral at
.
Find the angle between the tangent and the radius vecthe point of tangency in the case of the lemniscate 1 a cos 2q>.
654. tor r*
=
of
Sec. 4]
Geometrical and Mechanical Applications of the Derivative
65
655. Find the lengths of the segments of the polar subtangent, subnormal, tangent and normal, and also the angle between the tangent and the radius vector of the point of tangency in the case of the spiral of Archimedes
at a
point with polar angle
= 2jt.
Find the lengths of the segments of the polar subtangent, subnormal, tangent, and normal, and also the angle between the tanat an gent and the radius vector in the hyperbolic spiral r= 656.
=
=
r rQ cp arbitrary point cp 657. The law of motion of a point ;
.
Find the velocity of the point at / in centimetres and / is in seconds).
on the *-axis
= 0, ^
and
is
t
2
=
(x
is
658. Moving along the #-axis are two points that have the 2 5t and #=l/2/ where t^O. following laws of motion: x=\00 With what speed are these points receding from each other at the time of encounter (x is in centimetres and / is in seconds)?
+
659. The end-points of a segment AB the coordinate axes OX and OY (Fig. 16).
,
^5 m A
is
are sliding along at 2 m/sec.
moving
A Fig.
17
is the rate of motion of B when A is at a distance OA = 3 m from the origin? 660*. The law of motion of a material point thrown up at an angle a to the horizon with initial velocity V Q (in the vertical plane OXY in Fig. 17) is given by the formulas (air resistance is
What
3-1900
66
_ _ Differentiation of Functions
[Ch. 2
disregarded):
#=i; /cosa, where
trajectory of motion speed of motion and
A
661.
i>
/sin a
^-,
point
g is the acceleration of gravity. Find the and the distance covered. Also determine the
is in
its
direction.
motion along a hyperbola
r/
=
so that its
x increases uniformly at a rate of 1 unit per second. the rate of change of its ordinate when the point passes
abscissa
What
=
the time and
is
/
*/
is
through (5,2)? 662. At what point of the parabola
y*=\8x does the ordinate increase at twice the rate of the abscissa? 663. One side of a rectangle, a 10 cm, is of constant length, while the other side, b, increases at a constant rate of 4 cm,'sec. At what rate are the diagonal of the rectangle and its area increas30 cm? ing when 6 664. The radius of a sphere is increasing at a uniform rate of 5 cm/sec. At what rate are the area of the surface of the sphere and the volume of the sphere increasing when the radius
=
=
becomes 50 cm? 665.
=10
A
point
is
in
motion along the spiral
of
Archimedes
so that the angular velocity of rotation of its radius constant and equal to 6 per second. Determine the rate of elongation of the radius vector r when r 25 cm. 666. A nonhomogeneous rod AB is 12 cm long. The mass of a part of it, AM, increases with the square of the distance of the from the end A and is 10 gm when 2 cm. moving point, Find the mass of the entire rod AB and the linear density at any point M. What is the linear density of the rod at A and S?
(a
vector
cm)
is
=
M
AM =
Sec. 5. Derivatives of Higher Orders
1. Definition of higher derivatives. A derivative of the second order, or Ihe second derivative, of the function y=f(x) is the derivative of its derivative; that is,
=
(
The second derivative may be denoted
If
* = /(/)
is
or
^.
as or
f"(x).
the law of rectilinear motion of a point, then
eration of this motion.
^
is
the accel-
Sec. 5]
_
Derivatives of Higher Orders
Generally, the ith
a derivative of order (n
derivative of a function y f(x) is the derivative 1). For the nth derivative we use the notation
(v
y
Example
t
or
~^,
or
f
Find the second derivative
t.
_
y
= \n(\
(n)
67 of
(x).
function
of the
x).
/.JZL; /
Solution.
=
2. Leibniz rule. If the functions u q>(x) and v=ty(x) have derivatives to the nth order inclusive, then to evaluate the nth derivative of a product of these functions we can use the Leibniz rule (or formula): up
(uv)
<">
= u<"
3. Higher-order derivatives
= -r
then the derivatives y'x
functions
of (
*
I
i^
== f/^jc
represented
can
^2
successively
xt
we have
For a second derivative
Example
Solution.
2.
F^nd
We
the formula
w /
if
,
have -
If
.
.
_ /
(a cos*),
& cos .
&
f
"~~ "~~~ .
asm*
a
LUl
I
and .
*
(acosO
If
= q>(0, = *(0,
by the formulas
xt
parametricaKy.
-asln<
osln
be
calculated
68
Differentiation of Functions
[Ch.
A. Higher-Order Derivatives of Explicit Functions In the examples that follow, find the second derivative of given function. 667. y 668. y
669.
= x* + 7x' = e*
5x
+ 4.
671.
672.
.
= \n t/\+x
2
v
Show
y= (arc sin x) = acosh u
673. 674.
.
V^
675.
=
2
y=sm*x.
670. j y
//
th<
OY
I
.
.
*/
y=
that the function
2
a
-\
satisfies the
2
differ
ential equation
676. tial
Show
that the function y
2y'+y = e
equation y" 677.
Show
equation y" 4-3y' 678. tion y" 679. 680. 681. 682. 683.
Show
= -|-2y
the
satisfies
differen
.
y=-C
l
e"
for all constants
the
that
2
x
function
the
that
= -^-x e x
function y
x
+ C 2 e' 2x C and C l
= e 2x s'm5x
satisfies th 2
.
satisfies the equa 4y s 2 5x + 7x 2. Find y"' if y = x 5 Find /'"(3) if /(*) - (2^: 3) Find y v of the function # = ln(l+x). Find t/ VI of the function y==sin2x. Show that the function y = e~ x cosx satisfies the differ ential equation y lv + 4y = Q. 684. Find /(O), f (0), T(0) and /'"(O; f
+29y = 0. ,
.
f
if
= e x sinx.
f(x) 685.
The equation of motion of a poin along the jc-axis is
O.OOU 8
X-100-H5/
.
Find the velocity and the acceleration the f
t
point
for
times
=10.
A
/
= 0,
t
M
l
=\
y
c
an
is in motion around with constant anguls Fig- 18 velocity CD. Find the law of motion of i1 projection M, on the x-axis if at time / = the point is at Q (a, 0) (Fig. 18). Find the velocity and the ac celeration of motion of M,. at the in What is the velocity and the acceleration of tial time and when it passes through the origin? What are the maximum values of the absolute velocity and th
686.
circle
x
2
point 2
+y = a
2
M
M
absolute acceleration of Ai,?
l
Sec. 5]
_
_
69
Derivatives of Higher Orders
687. Find
the nth derivative of the function 3 natural number. 688. Find the nth derivatives of the functions:
where n
y=
(
js
a)
and
y^T^x*
y^^**'
b)
689. Find the /zth derivative of the functions:
j/=sinx;
a)
e)
b)4, = cos2*; = e~ *; c) y
f)
y=^j\ = yJ; J/
9
g)
|/=ln(l+x);
d)
h)
y=sin*jr,
y=
690. Using the Leibniz rule, find y a)
y = x.f\
b)
x y = jc .e-*
{n
l
\
d)y =
2
c)
691. Find
//
= (!
(n)
/
(0),
if:
2 A:
)
e)
;
y
= x*
~
cos x\
if
B. Higher-Order Derivatives of Functions Represented Parametrically and of Implicit Functions d^u
In the following problems find
K
692. a)
693. _,
= \nt,
'
b)
,
:asin/; :
= 0cos'/,
.
= arc tan/, \0-l x
J
"'
iv
f
c)
* = arc sin/
= a(l -cos/); x = a (sin/- /cos/), = a(cos/-f-/ sin/).
\ y
f
696. Find
^
[Ch. 2
Differentiation of Functions
70
d*u
^
697. Find 698. tions x
ax 2
Show
== sin
if
y=
{ \
that y (as a function
^ + be~ iV y = ae' 2
t,
* = ln(l-f-/
I
/=0,
for
satisfies the differential
*
= sec, { = tan/. \ y = ^"'cos/, {xy-e-<*tnt. x
699.
703. jt'"
701.
Solution.
By
/,
= y'
/
r/
,
.
^
y = f(x), x = f~*(y).
/=-
and
if
'
find
y
~
a
composite
^^
+ **_ *
=
function
we have
-2^;^.
1
if
if
required to determine represented implicitly.
is
it
= f(x)
=
the
|y
l.
707.
# = *-}- arc tan
708.
Having the equation y = x + \ny
709. Find
/
|/.
at the point (1,1)
x
2
5xy +
710. Find y" at (0,1)
711. a)
*
2x
find
t
+
6
-j
= 0.
if
The function y
is
defined
implicitly
= 0.
at the point (1,1). b)
and
if
2
Find
the derivatives x",
finally get:
In the following examples derivative y" of the function y 705. y*
^+f=
^.
s
differentiating
-jwe ^
706.
the equaand 6
a
+ y*=*l.
Substituting the value of #',
y
=
x = e -1 = ^
function
the rule for
2* + 2i0'=0; whence
by
any constants
7 2 ' Find
x'
if
< 1
of the inverse function
704. Find
defined
x)
y'" f
the
.
of
for
'
Knowing
),
equation
In the following examples find (
t
2
Find
,
if
x
2
+
2
i/
==a 8 .
by the equation
Sec. 6]
Differentials of First
and Higher Orders
71
Sec. 6. Differentials of First and Higher Orders 1. First-order
differential. The differential (first-order) of a function the principal part of its increment, which part is linear relative to the increment Ax dx of the independent variable x. The differential of a
y
= f(x)
is
=
19
Fig.
function
is
equal to the
product of its
derivative
by the differential of the
independent variable dy--=y'dx,
whence u y
MN
If is an arc of the graph tangent at M. (x, y) and
dy dx
,
'
the function
of
= f(x)
y
(Fig.
MT
19),
is
the
PQ = Ax-=dx,
then the increment in the ordinate of the tangent
and the segment AN by. Example 1. Find the increment 2 = 3x x. y Solution. First method: A//
= 3 (x + Ax)
and
2
(x
the
differential
+ Ax)
3x 2
of
the
function
+x
or At/
Hence,
= (6*
dy = (6x
Ax + 3 (Ax) 2
1)
1)
Ax = (6x
1)
.
dx.
Second method: t/'
Example
2.
and Ax = 0.01. Solution.
= 6x
Calculate
A/ =
At/
1;
df/
= j/'dx = (6x
and dy
of
the
1)
dx.
function
y = 3x
l)-Ax
(6x
=
(6jt
x
+ 3 (Ax) = 5- 0.0 + 3- (0.01 = 0.0503 Ax = 5- 0.01 = 0.0500. 2
and
2
1
1)
2
)
for
x=l
72
Differentiation of Functions
[C/t.
2
2. Principal properties of differentials. = 0, where c = const. 1) dc 2) d*- Ax, where x is an independent variable. 3)
4) 5)
= cdu. dv. v) = du d (uv) udv + v du. d(cu)
d(u
7)
differential to approximate calculations. If the increment argument x is small in absolute value, then the differential dy of the function y = f(x) and the increment At/ of the function are approximately
3. Applying the
A*
of the
equal: that
A # =^ dy,
is,
whence 3. By how much (approximately) does the side of a square change 2 to 9.1 m 2 ? area increases from 9 Solution. If x is the area of the square and y is its side, then
Example
if
m
its
It
is
given that #
The increment mately as follows:
At/
=9 in
and A*
0.1.
the side of the square
ky^zdy--=y' Ax
4. Higher-order
differentials.
=
A
j=z
-0.1
may
be calculated approxi-
= 0. 016m.
second-order differential
is
the differential
of a first-order differential:
We
similarly define the differentials of the third and higher orders. If y f(x) and x is an independent variable, then
=
But
if
y
= /(),
where w
d*y
and so
forth.
= cp(x),
then
= y"' (du) + 3y" du 9
d*u
+ y' d'u
(Here the primes denote derivatives with respect to
M).
712. Find the increment Ay and the differentia! dy of the func2 # 5* -f x for x 2 and A# 0.001.
tion
=
=
=
_
Differentials of First
Sec. 6}
713.
_
and Higher Orders
Without calculating the derivative,
find
73
d(l-x') for
x=\ 714.
and Ax =
The area
.
square S with
of a
Find the increment and the
side x
differential of
is
this
=
x*. given by S function and ex-
plain the geometric significance of the latter. 715. Give a geometric interpretation of the increment and differential of the following functions: nx*\ a) the area of a circle, S b) the volume of a cube, v=^x\ 716. Show that when Ax *0, the increment in the function
=
= 2X
corresponding to an increment Ax in x, is, for any x, equivalent to the expression 2* In 2 A*. 717. For what value of x is the differential of the function 2 y = x not equivalent to the increment in this function as Ax >0? 718. Has the function y = \x\ a differential for x = 0? 719. Using the derivative, find the differential of the function //
y
,
cos x for x
=y
and Ax
--=
~
.
720. Find the differential of the
for
for
=
function
Ax-
0.01. 9 and x 721. Calculate the differential of the function
x-^-J
Ax^.
and
problems find the differentials of the given functions for arbitrary values of the argument and its increment. In the following
727. y
722.
y^'-m-
723.
<,=
724.
#= arc sin
= cot s = arc
729. r
.
730.
725. //--=arctan~.
=
= x\nx
q>
x.
-f
cosec
(p.
lane*.
x
726. y e~ \ 731 Find d//
if
x*
+ 2xy
y*
= a*.
Solution. Taking advantage of the invariancy of the form of a differential, x dy) 2y dy (y dx
we obtain 2x dx + 2 Whence
+
=
74
Differentiation of Functions
[Ch. 2
In the following examples find the differentials of the functions defined implicitly. 732.
=
733.
734. X
y'y = 6x*.
735. Find dy at the point (1,2), if 736. Find the approximate value of sin Solution. Putting
3) we have
(see
sin
*
=
arc30=-jr-
and Ax = arc
+ ^~ cos
31=^ sin 30
31.
1=^, from formula
(1)
30=0.500+0.017-J~^=0.515.
737. Replacing the increment of the function by the differencalculate approximately:
tial,
a) cos
b) tan c)
61; 44;
d) In 0.9; e) arc
tan 1.05.
e*\
738. What will be the approximate increase in the volume of a sphere if its radius 15 cm increases by 2 739. Derive the approximate formula (for \&x\ that are small compared to x)
#=
mm?
A*
Using
it,
V5
approximate
Y\7, /70, /640.
,
740. Derive the approximate formula
and find approximate values 741.
for
for
*=1.03; * = 0.2;
for
* = 0.1;
for __
c)
d)
j!/TO,
j/70, jI/200.
Approximate the functions:
/(x)- !/"}= y=e
l
~ x*
x =1.05.
for /
742. Approximate tan 453 20". 743. Find the approximate value of arc sin 0.54. 744.
Approximate
\
Sec. 7]
_ _ Mean-Value Theorems
745. Using
Ohm's
/
law,
= --,
show
the current, due to a small change found approximately by the formula
2/
the
in
resistance,
in
may be
A.
A/ 746. Show that, relative error of 1/
that a small change
75
of the radius, a error of approximately in calculating the area of a circle and the surface of a sphere. 747. Compute d*y, if y co$5x.
determining the
in
length
results in a relative
=
Solut ion.
748. u 749. 750.
//
d y = y" (dx = 2
2
25 cos 5* (dx) 2
)
= x\ = arccosx,
find d*u.
{/^sinxlnx,
find d*y. find d*y.
= ^, find d'z. z = **-*, find d'z. z = flnd d4 2=TJ, M = 3sin(2jt-f 5), find = e* cosa sin(;t sin u),
751.
.
e
752.
753.
*'
754.
755.
//
d"w. n
find d y.
Sec. 7. Mean-Value Theorems
1. Rolle's b,
then the
theorem. If has a derivative
argument
x
has
at
the interval is continuous on f (x) every interior point of this interval, and
function
a
at
/' (x)
least
one
value
,
where a
< 5<
b,
such that
2. Lagrange's theorem. If a function f (*) is continuous on the interval and has a derivative at every interior point of this interval, then where a
<5<
ft.
3. Cauchy's theorem.
If
the functions
a
a^x^b
and for simultaneously, and F(b)^F(a) interval
-f (a) 756.
Show
l
that
and
the g.
and F
(x) are
derivatives
continuous on the do not vanish
that
then
_ function
0
appropriate values of
t
f (x)
have
f(x)
= xx*
satisfies the
Rolle
on
the
intervals the
theorem. Find
_
76
_
Differentiation of Functions
The function / (x) is continuous and different! able for all values = /(0) = /(1)=0. Hence, the Rolle theorem is applicable on 1) intervals and 0<*<_1, To find form the equation _ we
Solution.
of x,
the
and /(
Kx<0
n*) = l-3* = 0. Whence 2
0
and
[Ch. 2
1
757.
The function
=
fc^-J/l;
2
=
J/l
,
-i< g,
where
1.
=
=
2
takes on f(x) 2) \/(x equal values the end-points of the interval [0.4]. Does
/(4) j/4 at the Rolle theorem hold for this function on [0.4]? 758. Does the Rolle theorem hold for the function /(0)
on the interval 759.
Show
[0,
JT]?
Let
that the equation /'(*)
=
has three real roots. 760.
The equation
obviously has a root x
any other
real
= 0.
Show
that this equation cannot have
root.
761. Test whether the Lagrange theorem holds for the function
on
the interval value of
[2,1] and
the
find
appropriate
intermediate
.
= xx*
all
Solution. The function f(x) values of A:, and /' (x)= 1 3x 2
h3ve
/(l)-/(-2y = 0-6
=
2 2 and 3^ 2 inequality 1
continuous and difTerentiable for by the Lagrange formula, we = [l-(-2)]/ (E), that is, Hence, the only suitable value is 1, for which the is
Whence, /
/'(E)2 =
g=l; < < holds 1
762.
Test the validity of the Lagrange theorem and find the intermediate point for the function f(x) x 4/s on the interval [ 1,1]. 2 763. Given a segment of the parabola y x lying between two points A (1,1) and 3(3,9), find a point the tangent to which is parallel to the chord AB. 764. Using the Lagrange theorem, prove the formula
=
appropriate
=
sin (x 4- h)
where
sin
x = h cos
,
Taylor's Formula
Sec. 8]
=
2
77
+2
* 765. a) For the functions /(x) whether the Cauchy theorem holds on find E;
b)
do
same with
the
on the interval
~1
To,
and F(x)
the
respect to /(*)
= x'
interval
= sin*
test
1
and
[1,2]
and F(x)
= cosx
.
Sec. 8. Taylor's Formula If
(n is
function
a
f (x)
is
continuous and has continuous derivatives up to the
l)th order inclusive on the a finite derivative f (n} (x) at
lor's
a
interval
each interior point
&<*
(or of the interval,
and there then Tay-
formula
fw = f (
-V(a)
+
-V
w+
.
.
.
.
nl
where
= +
a 0(jc In particular, f
W =/
(0)
a)
and
when
+xf
(0)
a
0<6<1,
=
holds true on the interval.
we have (Maclaurin's formula)
r (0) +
+
.
.
.
+
(
"-'>
/
(0)
+
/<>
(I),
(
where
= ?
0<9<1.
0jc,
=
766. Expand the polynomial /(A:) tive integral powers of the binomial x
n*)=3jt
Solution. for
2
4A-
= 6^
f
8
2;c
A:
+ 3;
/'
f'(2)
= 7;r(2) = 8;r(2)
7
(jc)
+ 3^ + 5
in posi-
2. 4;
/'"
(x)=6;
n^4. Whence H;
=6.
Therefore,
or
2x z
8 Jt
767.
+ 3x + 5 =
(n)
Solution.
=
+7
(.v
2)
+ 4 (x
x Expand the function f(x)=e
term containing (x-f-1)
where
1 1
l
1
(x)
= e*
in
2
2)
+
3
2)
(A-
powers
of
.
x
+
to
l
the
8 .
for all
n,
p)(
1)=JL. Hence,
+6(*+ 1); 0
768. Expand the function /(x) the term with (x 1)*.
=
lnjt in
powers
of
x
1
up
to
_
78
_
[Ch. 2
Differentiation of Functions
= sin x
in powers of x up to the term containing x e* in powers of x up to the term contain770. Expand f(x) n~ l ing x 771. Show that sin(a /i) differs from
769.
Expand 9
/ (x)
and to the term containing x*.
=
.
+
+ h cos a
sin a 2
by not more than l/2/i 772. Determine the origin .
of the
VT+x&l+x y* yi+i&l+x~x*, 2
a)
b)
,
and evaluate their errors. 773. Evaluate the error
Due
774.
to
line
a catenary
in
of the
shape
775*.
\x\
\x\<\
is
formula
in the
weight, a heavy
y = a cosh.
thread
Show
own
its
approximate formulas:
Show
approximately
suspended thread
that
the
by the parabola
expressed
that for \x\<^a, to within
small \x\
for
lies
,
(^-J
we
have
the
approximate equality
Sec. 9.
The L'Hospital-Bernoulli Rule
Evaluating Indeterminate Forms
for
oo
1. Evaluating the indeterminate forms
and
and u
.
Let
oo
the
single-valued
a (*) are both infinitesimals or both infinites as x functions
if
is,
/ (x)
the quotient
(p
(x)
^-4
.
\
at
x = a,
is
one
of the
indeterminate forms --
oo i
then
00
.
,
*-+<*
provided that the limit
.
lim /(*) (p (x)
^lim x-+a
f
(*)
q>' (
x)
of the ratio of derivatives exists.
or
_
V Hospital-Bernoulli
Sec. 9]
The
rule
is
when a =
also applicable
fix)
If
the quotient
/
,,
_
Rule for Indeterminate Forms 00.
yields an
again
indeterminate
at the point
form,
= a,
of one of the two above-mentioned types and /' (x) and the requirements that have been stated for f(x) and q? (x), pass to the ratio of second derivatives, etc.
x
79
q>'
However,
may
exist,
it
should
borne
be
whereas the ratios
mind that the
in
limit of the ratio -^-~
derivatives do not tend to any limit
the
of
(x) satisfy
we can then
all
Example 809). 2. Other indeterminate forms. To evaluate an indeterminate form like 0oo, transform the appropriate product fi(x)*f t (x), wne re lim/, (jt) = and (see
K+O.
/(*) lim/ 2 (*)
= oo,
^^
into thequetient
form
(the
-
U
*
*->a
(T^T\ /i (X)
( the
f
orm -). oo
M*) In the case
of
the indeterminate form oo
appropriate difference /,(*)
f 2 (x)
oo,
one should transform the
into the product / t (x)
first
evaluate the indeterminate form 7*7^; r
lim
if
(X)
i
and
l
L
/ 1
7^7^=1, \x
x-+a i\
(
x )j
then
we
re-
)
duce the expression to the form
form
(the
).
/Tw The indeterminate forms I, 0, 00 are evaluated by first faking loga rithms and then finding the limit of the logarithm of the power [f l (x)]^ (x} (which requires evaluating a form like 0oo). In certain cases it is useful to combine the L'Hospital rule with tht finding of limits by elementary techniques.
Example
1.
Compute lim
JL1
*->o cot
(form "). 7 oo
x
Solution. Applying the L'Hospital rule
we have
JEfL^llm pL*r x+ocotx jc-o(cot*)
lim
lim
We
get
the indeterminate form
-jp
jc-*o
however,
L'Hospital rule, since
Um C-frO
We
sint
X
*
Hm
sin
~~*-H)
X
*
thus finally get "
JC->0
COt X
we
.
x
do
not
need to
use the
[Ch. 2
Differentiation of Functions
80
Example
Compute
2.
limf-J- x
*-M)V sin
Reducing
xL^ J
(form
common denominator, we
to a
lim
oo
oo).
get
(4__J_Ulim*.!z^lf x x J x-o x sin x
x-+o \ sin
2
2
v (fonn
2
z
'
Before applying the L'Hospital rule, we replace the denominator of the 2 2 fraction by an equivalent infinitesimal (Ch. 1, Sec. 4) * sin A;~ x*. obtain ter
lat-
We
o
The L'Hospital
rule gives
lim
L)
(
Then,
in
A "~ S
= lim
m *=lim-
a
elementary fashion, we find
_ ___ 2
x-<> \ sin
Example
3.
cos 2x
1
x
x
2 sin 2 *
,.
1
2
Compute 8
*2
lim (cos 2x)
00
(form
I
)
X-M)
Taking logarithms and applying the L'Hospital
lim In (cos X-*0
lim 2*p = X-+Q
31ncos2* X
rule,
= _ 6 lim
we
%X
x-+Q
get
= _6
.
J^
x
Hence, lim
(cos 2x)
*^e-*.
Jf->0
:
F
ind the indicated limits of functions in the following
ples. x '-
77G.
lim ,_>,
*'
Solution. lim
" ; X->1
777.
lim
xcosx v
m
r
^r 2 "~~ 7/
OJi
~~
9
*
sinx ,
779.
lir
*7on
is
I-*
tj 11IU
'
ji
x
*->il_si n :r
780.
lim
tan*
sin*
exam-
sec
,.
781.
2tan* 1+COS4* 2
'
x
785.
lim-
*-cot^ 2
lim^. tan5x
782.
S
^Ji
786.
lim
y
s
*-"
787.
.
i .
1.
lim
[,
cos x) cot
(1
x
x->o
.
n x lim -r-?=r
,
)
lim 1
783.
81
n
A-
x +
1
lim
Solution,
(1
.
^cos
x) cot
^
= Hm
,
cos A
'^- lim X->0
Sill AT
;
Forms
L' Hospital-Bernoulli Rule for Indeterminate
Sec. 9]
Sill
A
linrZi!iJ^(
788.
lim(l
A*)
Ian
~
a
792.
.
^
,V-^l
X^X
lim arc sin x cot x.
789.
-, V
lim x" sin
793.
'
hnilnxln
(x
1).
X-+Q ri
790.
n>0.
lim(jc e?"*), v^o
791. lim x sin
794.
lim ^
*i
f^ \
A
n"^ lr %
]
.
Solution.
A
=
[-
A
n
1
A'
1
,
= lim
lim j
^^MiiA-l
A
795.
lim
796.
lim
(A
1)
x
*
11
A
= lim
j
\nx
[
-1
*^
A -7
T-
A
h~? 2 A'
]
L__l
limf-^^' *
1
X
y 797
~ >l
^Vc
_^/
A)
AO J
^} 2cosx/
2
798.
lim
A;*.
Solution.
s
lim
p
We
= lim j~
have
0,
**
=
r/;
whence lim//=l,
In
y=?x
that
In
is,
A".
ImiA^
lim In
l.
t/
= limjtln x
=
82
(Ch. 2
Differentiation of Functions
799. limx*.
804.
li
V-H tan
a
800. limx4 * "*.
\
805. Hmftan^f) 4 /
1
X-+l\
1
801. linue s/n
806. lim (cot x) ln
*.
*->0
*.
X-H) cos
802.
lim(l-*)
803.
lim(l+x
ta
807.
2
)*-
lta(I) x / x-*o \
".
808. lim (cot x)* in
*.
X-+0
809. Prove that the limits of
X a)
cannot
sin*
be found
by the L'Hospital-Bernoulli
rule.
Find these
limits directly.
810*. Show that the area of a circular segment with minor and central angle a, which has a chord (Fig. 20), is
AB=b
CD=A
approximately
with an arbitrarily small relative error when a
->0.
Chapter III
THE EXTREMA OF A FUNCTION AND THE GEOMETRIC APPLICATIONS OF A DERIVATIVE
Sec.
1.
The Extrema
1. Increase and
of
a Function of One Argument
of tunctions. Tlu Junction y f(x) is called on some interval if, fo. any points x and x 2 which belong to this interval, from the inequality A',
decrease
increasing (decreasing)
l
i
ffxj
xz
i,
X
(a) Fifi.
Fig. 22
21
In the simplest cases, the domain of definition of f (x) may be subdividincrease and decrease of the funca finite number of intervals of tion (intervals of monotonicity). These intervals are bounded by ciitic-' or f' (x) does not exist]. points x [where /'(jc) Example 1. Test the following function for increase and decrease:
ed into
=
Solution.
We
find the derivative t/'
Whence
y'
=
for
x=l. On
= 2x a
2
= 2(*
number
scale
1).
we
get
two intervals
of
monot-
oo
+
Extrema and the Geometric Applications
84
Example
Determine the intervals
2.
of
and decrease
of increase
[Ch. 3
a Derivative
of the func-
tion
Solution.
=
, i
o\\2
oo<*< Example
2
Here,
2
*^~
for
Hence, the
the
of
discontinuity
and
function
(/'
=
decreases in the intervals
function y
and Test the following function for increase or decrease:
3.
/i y
Solution
a
is
~
*v s
.
5
*y a
'
3
Here, (2)
=
2
find the points x l Q, we Solving the equation x* -x 0, x,= l 1, *2 at which the derivative y' vanishes. Since y' can change sign only when passing through points at which it vanishes or becomes discontinuous (in the given case, y' has no discontinuities), the derivative in each of the intervals 1, 0), (0,1) and (1, (00, 1), ( +00) retains its sign; for this reason, the function under investigation is monotonic in each of these intervals. To determine in which of the indicated intervals the function increases and in which it decreases, one has to determine the sign of the derivative in each of the intervals, To determine what the sign of y' is in the interval 00, inter1), it is sufficient to determine the sign of y' at some point of the val; for example, taking x= 2, we get from (2) f/' 12>0, hence, y'>Q in the interval (00, increases Similar1) and the function in this interval we find that y'
=
1
~
'
v
the
in
Y\
/'A\
I
interval
(0,1)
**
(here,
we can
interval
(1,
x=l/2) and y'>0
use
in
the
+00).
the function being tested inThus, creases in the interval ( oo, 1), decreases in the interval 1) and again increases in the interval (1, -f oo). 2. Extremum of a function. If there
(1,
exists a two-sided neighbourhood of a point X Q such that for every point X^X of this Q
neighbourhood we have the inequality f(x)>f(x Q ) then the point x is called the J
Fig
23
minimum
point
of
while the number
mum
the
/ (x
)
of the function y neighbourhood of the point
function is
y
f(x),
called the
mini-
f(x).
Similarly,
if
x lf the inequality any point xj^x l of some f(*)
The Extrema
Sec. 1]
The
of a
Function of One Argument
85
and absence of an are given by the following rules:
extremum
sufficient conditions for the existence
continuous function
/ (x)
+
of a
1. If there exists a neighbourhood (X Q 6, * 6) of a critical point * such that /'(x)>0 for X Q d
+
+
/(*)-
Finally,
function 2. if
f
there for
/ (x). r If f (XQ)
f' ( XQ )
(*
if
unchanged
sign
)
=Q
= 0,
is
some
0<|jc
positive
number 6 such
then x
X Q |<6,
is
that /' (x) retains its not an extremal point of the
then X Q is the maximum and /"(*<,)<(), )>0, then x is the minimum point; but if f )^0, then the point X Q is not an extremal point.
$
and
f"
and /'"
(*
(*
point; (*
)
= 0,
More generally:
let the first of the derivatives (not equal to zero at the function f (x) be of the order k. Then, if k is even, the is an extremal point, namely, the maximum point, if f (k) (* )<0; and it is the minimum point, if / (ft) (x )>0 But if k is odd, then A- O is not n extremal point. Example 4. Find the extrema of the function
point x point X Q
the
of
)
<
i/
==2*
+3
j
Solution. Find the derivative (3)
V
x
Equating the derivative
Whence,
x-
:
x-= function
if
we -/i,
find
then
is
we
get:
x l =-
a sufficiently small
Hence,
/'<0*).
and //max=-l. Equating the denominator r/
to zero,
the critical point
where h
\+h,
y'
v
*,
1.
From formula
(3)
we
have:
if
positive number, then /y'>0; but I is the maximum point of the
f
of
the expression of y' in (3) to zero,
we
get
=
\\e find the second critical point of the function A'2 there 0, where for*/! we have no derivative //' For *== /i, we obviously have //<0; is the minimum point of the function y, and //>0. Consequently, * 2 m jn (Fig. 24). It is also possible to test the behaviour of the function 1 at the point x by means of the second derivative
whence is
=
i/
/=--4^=
=
1 is the maximum I x, and, hence, *, point of the function. 3. Greatest and least values. The least (greatest) value of a continuous function f (x) on a given interval [a, b] is attained either at the critical points of the function or at the end-points of the interval [a, b].
Here, r/"<0 for
p
to determine the sign of the derivative y', one can *) If it is difficult calculate arithmetically by taking for h a sufficiently small positive number.
86
Extrema and the Geometric Applications
Example
of a
Find the greatest and least values
5.
on the interval
Derivative
[Cfi.
3
of the function
P/2
Solution. Since
it
follows that the critical points of the function y are *,
=
1
and
Y
Fig. 24 of the function at these points and the values function at the end-points of the given interval
Comparing the values
we conclude the
(Fig.
x=l (at *=2 /i
point
at the point
that the function attains
25)
J
the
minimum
point),
and
its least
the greatest value
Determine the
812. 813. 814.
y=l
m=l,
at
A4 = ll
o
(at the right-hand end-point of the interval).
intervals
of decrease
and increase
1ions: 811.
value,
of the
4*
jf.
= (* 2) y = (A:+4) = *'(*- 3).
*>-
i
2
.
{/
s
817.
=
818.
= (x
.
{/
of the func-
The Extrema
Sec. 1]
819.
of
a Function
y^ \-V~x.
822.
y = 2e*
823.
= x -f sin x. y = x\nx. = arcsin(l-f-x).
820. y
821.
One Argument
of
'
y
825
-
^T"
g
We
find
the
zero,
we
*
for extrema:
= +
Solution.
-'*.
_. 2~<*.
24
t/
Test the following functions x* 826. y 4*4-6.
z
derivative
of
the
given function, value of the argument x= 2. Since i/'<0 when x< 2 is 2, and y'>Q when *> 2, it follows that *= the minimum point of the function, and #min 2. We get the same result of the second derivative at the critical point y"~< by utilizing the ~ sign
Equating
to
y'
the
get
critical
=
827. y 828. {/ 829. (/
--
= =
Solution,
.
!
We
find the derivative
= 6*
y'
4-
6x
12
=6
+*
2 (jc
2).
2 Equating the derivative y' to zero, we get the critical points x,= and *,= !. To determine the nature of the extremum, we calculate the second derivative ^"^ 6 (2* 4-1). Since /( 2 2)<0, it follows that x,= is the maximum point of the function y, and #max 25. Similarly, we have the function y and t/*(l)>0; therefore, x 2 =l is the minimum point of
=
i=
2.
<-
12 )
2 '
840.
1
(*
I)
y-
2)'.
841.
836. V
= rr4=^.
837. t/=
w=J/(^
OQQ ooy.
it */ ===
Determine the indicated
843. y
=
844.
=
847.
citi ^. v -+ sin ^k*. I
least
intervals
=
/
,_
846. y
1)'.
O sin cin ZA Ov z
842. #
(if
f/
/1ft // M oto.
= x'e-*. = -. X -
ar/* fan ^ idii ^t. /t-~drc
interval
is
^
of the functions on the not given, determine the
and greatest values the
ln(l+*).
je
845>
^_.
838.
=
t/
_
88
greatest
Extrema and the Geometric Applications
and
[Ch. 8
of a Derivative
values of the function throughout the domain
least
of definition). 849. 850. 851.
= rih&. -= y x(lOx). y= sin + cos A;
imum
interval
= 2x* + 3*
A;.
12*
a)
b)
2
on the interval on the interval
that for positive values of
856. Determine
nomial
V = x* on the
854. y
= arc cos x.
Show
-
4
4
852. # 855.
853
!/
*we
+
[
1,3].
1
1,6];
f
[10,12],
have the inequality
coefficients p and q of the quadratic triso that this trinomial should have a minJt= 1. Explain the result in geometrical terms.
the
y*=x*+px + q
=
3 when t/ 857. Prove the inequality e*
> +x 1
when x
4* 0.
Solution. Consider the function
In the usual
way we
find
lhat
this function
has a single
minimum
/(0)
Hence,
/(*)>/ (0) and so e* as
we
set
> +x 1
when x when x
0,
^ 0,
out to prove.
Prove the inequalities: x sin x 858. x
<
^< o
when *>0.
^
when
~
when
859.
cos*>l
860.
A:
JL
861. Separate a given positive number a into two summands such that their product is the greatest possible. 862. Bend a piece of wire of length / into a rectangle so that the area of the latter is greatest. 863. What right triangle of given perimeter 2p has the greatest area? 864. It is required to build a rectangular playground so that should have a wire net on three sides and a long stone wall it on the fourth. What is the optimum (in the sense of area) shape of the playground if / metres of wire netting are available?
Sec.
1]
The Extrema
of
a Function
of
One Argument
89
865. It is required to make an open rectangular box of greatest capacity out of a square sheet of cardboard with side a by cutting squares at each of the angles and bending up the ends of the resulting cross-like figure. 866. An open tank with a square base must have a capacity of v litres. What size will it be if the least amount of tin is used? 867. Which cylinder of a given volume has the least overall surface? 868. In a given sphere inscribe a cylinder with the greatest volume.
869.
In a given sphere inscribe a cylinder having the greatest
lateral surface.
870. In a given sphere inscribe a cone with the greatest volume. 871. Inscribe in a given sphere a right circular cone with the greatest lateral surface. 872. About a given cylinder circumscribe a right cone of least volume (the planes and centres of their circular bases coincide). 873. Which of the cones circumscribed about a given sphere has the least volume? 874. A sheet of tin of width a has to be bent into an open cylindrical channel (Fig. 26). What should the central angle cp be so that the channel will have maximum capacity?
D N
I
M Fig. 27
875. Out of a circular sheet cut a sector such that when made into a funnel it will have the greatest possible capacity. 876. An open vessel consists of a cylinder with a hemisphere at the bottom; the walls are of constant thickness. What will the dimensions of the vessel be if a minimum of material is used for a
given capacity?
=
OB of the door of a ver877. Determine the least height h so that this door can pass a rigid rod tower of length /, the end of which, M, slides along a horizontal straight line AB. The width of the tower is (Fig. 27).
tical
MN
ABCD
d
90
Extrema and the Geometric Applications
M
of
a Derivative
[Ch. 3
878. A point (x # ) lies in the first quadrant of a coordinate plane. Draw a straight line through this point so that the triangle which it forms with the positive semi-axes is of least area. 879. Inscribe in a given ellipse a rectangle of largest area with sides parallel to the axes of the ellipse. 880. Inscribe a rectangle of maximum area in a segment of the parabola y* 2px cut off by the straight line x 2a. ,
=
881.
On
=
the curve y
=
find a point at
,
1
-\-
Xt
which the tangent
forms with the A>axis the greatest (in absolute value) angle. 882. A messenger leaving A on one side of a river has to get to B on the other side. Knowing that the velocity along the bank that on the water, determine the angle at which the is k times messenger has to cross the river so as to reach B in the shortest possible time. The width of the river is h and the distance between A and B along the bank is d. 883. On a straight line AB=a connecting two sources of light A that (of intensity p) and B (of intensity
M
square of the distance from the light source). suspended above the centre of a round table At what distance should the lamp be above the table
portional to the 884. A lamp of radius
r.
is
an object on the edge of the table will get the greatest illumination? (The intensity of illumination is directly proportional to the cosine of the angle of incidence of the light rays and is inversely proportional to the square of the distance from the so that
light source.)
885. It is required to cut a beam of rectangular cross-section ont of a round log of diameter d. What should the width x and the height y be of this cross-section so that the beam will offer maximum resistance a) to compression and b) to
bending?
I i/J
Note. The resistance of a beam to compresof its crossis proportional to the area to the product of the section, to bending width of the cross-section by the square of sion
its
height.
886. A homogeneous rod AB, which can rotate about a point A (Fig. 28), is carrying a load Q kilograms at a distance of a cm from A and is held in equilibrium by a vertical force P applied to the free end B of the rod. A linear centimetre of the rod weighs q kilograms. Determine the length of the rod x so that the force P should be least, and find P mln Fig.
2
.
Sec. 2]
_
The Direction
887*. The centres on a single straight locity v strikes strikes C of mass
fi,
of Concavity.
Points of Inflection
91
of three elastic spheres A, B\ C are situated Sphere A of mass moving with vewhich, having acquired a certain velocity,
M
line.
m. What mass should B have so that C will have the greatest possible velocity? 888. N identical electric cells can be formed into a battery in different ways by combining n cells in series and then combining the allel.
in pargroups (the number of groups is ] The current supplied by this battery is given by the formula
resulting
,
where
NnS
~~
the electromotive force of one cell, r is its internal and R is its external resistance. For what value of n will the battery produce the greatest <
is
resistance,
current? 889. Determine
the
diameter y of a circular opening in the which the discharge of water per second Q will be greatest, if Q = cy Vhtj, where h is the depth of the lowest point of the opening (h and the empirical coefficient c are
body
of a
dam
for
constant). 890. If x lf # 2 ..., x n are the results of measurements of equal precision of a quantity x, then its most probable value will be that for which the sum of the squares of the errors ,
(*-*,) 0=2 1=1 value (the principle of least squares).
is
of least
of
Prove that the most probable value of x the measurements.
is
the arithmetic
mean
Sec. 2. The Direction of Concavity. Points of Inflection
1. The concavity of the graph of a function. We say that the graph of a differentiable function y f(x) is concave down in the interval (a,b) [concave the arc of the curve is below (or up in the interval (a p 6,)] if for for the tangent drawn at any point of the interval (a, b) lt above) or of the interval (a,, &.)] (Fig. 29). A sufficient condition for the concavity downwards (upwards) of a graph y f(x) is that the following inequality befulfilled in the appropriate interval:
a.
a
rw
irw>oj.
2. Points of inflection. A point [* f (jc )] at which the direction of concavity of the graph of some function changes is called a point of inflection ,
(Fig. 29).
Extrema and the Geometric Applications
92
of
[Ch. 3
a Derivative
point of inflection x of the graph of a function derivative f (* ) or /" (x ). Points at which Q or f (x) does not exist are called critical points of the second kind. f'(x) The critical point of the second kind x is the abscissa of the point of inflec* an * the intervals x 6 tion if I" (x) retains constant signs in .d
For the abscissa
y
f (x)
there
is
of the
=
no second
< <
+ 6,
where 6 is some positive number; provided these signs are
x?
Jc
And it is not a point of inflection if the signs of f (x) are the same in the above-indicated intervals.
opposite.
y~f(x)
Example 1. Determine the intervals of concavity and convexity and also the points of inflection of the Gaussian curve
I
Solution.
We
have
i
I
bx
a,
X
b,
and Fig. 29
Equating the second
we
derivative y* to zero,
find the critical points of tHe
second kind
*
=
and
7=r
*o
=
T=-
+
These points divide the number scale OO
Fig. 31 -f(this is obvious if, for example, we take one point in each lively, +, of the intervals and substitute the corresponding values of x into y ) Therefore: ,
1)
the curve
curve
is
is
concave up when
concave down when
oo< -=^
F
2
x
<
7= and F2 ==
V
2
.
F=
< x <-f oo;
2) the
V 2
The points
(
-=^
\V2
,
r=]
are
VeJ
points of inflection (Fig. 30). It will be noted that due to the symmetry of the Gaussian curve about the #-axis, it would be sufficient to investigate the sign of the concavity of x this curve on the semiaxis +00 alone.
< <
__
Sec. 3]
Asymptotes
Example
Find the points
2.
of
93
inflection of the graph of the function
y=*/7+2.
Solution.
It
is
We
have:
obvious that y" does not vanish anywhere. Equating to zero the denominator of the fraction on the
right of (1), we find that y" does not exist for x 2. Since y" for 2 and f/"<0 for 2, it follows that ( 2,0) is the point of inflection (Fig. 31). The tangent at this point is parallel to the axis of ordinates, since the first derivative y' is infinite at x 2.
x<
>
*>
Find the
intervals of concavity and the points graphs of the following functions: 6x* 12x + 4. 891. y = x* 896. y cosx.
of
inflection
of the
+
892. 893.
y = (x + l)\ y = -4r o
897.
898.
.
X-\-
894. 895.
ff
=
X
9
i
899.
. ,
12
900.
\2x.
y=i/4x*
= y=x y=x
sin*. 2
In x.
= arc tanx y = (l+x*)e*. //
x.
Sec. 3. Asymptotes
1. Definition. If a point (#,/) is in continuous motion along a curve f(x) in such a way that at least one of its coordinates approaches infinity (and at the same time the distance of the point from some straight line tends to zero), then this straight line is called an asymptote of the curve. 2. Vertical asymptotes. If there is a number a such that y
Jim /(v)--=
then the straight line x 3
a Inclined asymptotes.
is
If
00,
an asymptote (vertical asymptote). there are limits llm X
->>
+ 00
K
and lim [/(*)X-++ 00
=
Ml = *i.
then the straight line y k l x+b l will be an asymptote 0, a right horizontal asymptote). asymptote or, when ^
=
If
there are limits llm
(a
right inclined
Extrema and the Geometric Applications
94
of a Derivative
[Ch. 3
and Urn
=
+
k zx y b^ is an asymptote (a left inclined asymptote horizontal asymptote). The graph of the function y f(x) (we assume the function is single-valued) cannot have more than one right (inclined or horizontal) and more than one left (inclined or horizontal) asymptote. Example 1. Find the asymptotes of the curve
then the straight or,
when
fe
2
= 0,
line
=
a left
Solution. Equating the denominator
to zero,
we
get
two
vertical asyinp-
lotos-
x= We
x=l.
and
1
seek the inclined asymptotes. For x
kl
=
lim
l
lim
x)
(// *
+ oo
X
= lim
we obtain
~l,
lim
*-+o> b =-
>
v
2
}^x z x
y^2
=0,
\ \
S
-/ Fig. 32
hence, the straight line y
=x
the right asymptote. Similarly,
is
when*
oo,
we have fc
fc
a
=
Hm
=
~=
1;
lim AC->~
Thus, the totes
is
Example
2.
y= -x
(Fig. 32). Testing a curve for asympconsideration the symmetry of the curve. Find the asymptotes of the curve
left
simplified
asymptote if
we take
Is
into
__
Sec. 3]
95
Asymptotes
Solution. Since
lim
t/
=
oo,
=
is a vertical the straight line x asymptote (lower). Let us now curve only for the inclined right asymptote (since x>0). We have:
k= b
lim (y *-*+ 00
lim
=
X++OD X x)
=
lim #->+<
test
the
1,
\nx
oo.
Hence, there is no inclined asymptote. If a curve is represented by the parametric equations x cp(0i */ ^(0 then we first test to find out whether there are any values of the parameter / for which one of the functions cp (t) or \|> (/) becomes infinite, while the other remains finite. When (p(/ )=oo and ty(t ) = c, the curve has a horizontal asymptote y c. When \j)(f ) = oo and (p(V ) = c, the curve has a vertical asymptote x = c. If
=
=
lim
then the curve has an inclined asymptote y kx+ b. If the curve is represented by a polar equation r /(cp), then we can find its asymptotes by the preceding rule after transforming the equation of the curve to the parametric form by the formulas x r cos cp /((p) cos q>; r sin
=
Find the asymptotes of the following curves: 901.
11
903. y
=
^rr.
-,
=
908. u
=x
909. y
= e-
2
910. i/=
.
911.
905.
y^Y^^l.
912.
906.
y==-
913
907.
.
r
* ~~"
-
914. x
=
/;
j/
=
*
915. Find the asymptote of the hyperbolic
spiral r
=
.
of a Derivative
Extrema and the Geometric Applications
96
[Ch. 3]
Sec. 4. Graphing Functions by Characteristic Points In constructing the graph of a function,
find its
first
domain
of definition
and then determine the behaviour of the function on the boundary of this domain. It is also useful to note any peculiarities of the function (if there are any), such as symmetry, periodicity, constancy of sign, monotonicity, etc. Then find any points of discontinuity, bending points, points of inflection, asymptotes, etc. These elements help to determine the general nature of the graph of the function and to obtain a mathematically correct outline of it. Example 1. Construct the graph of the function
The function exists everywhere except at the points x 1. and therefore the graph is symmetric about the point
Solution, a)
The function 0(0,
This simplifies construction of the graph
0).
The
b)
lim
odd,
is
discontinuities
are
x=
hence, the straight lines
t/=oo;
and
1
jc
and
1;
lim
and
oo
J/=
V-M + O are vertical asymptotes of the
#=1
X->--10 graph. c)
We
seek inclined asymptotes, and find
,=
lim X -> +
bl
oo
lim
-x
= 0,
y
oo,
#->-t-oo
thus, there is no right asymptote. From the symmetry of the curve it follows that there is no left-hand asymptote either. d) We find the critical points of the first and second kinds, that is, points at which the first (or, respectively, the second) derivative of the given function vanishes or does not exist.
We
have:
,
x=l,
The
derivatives y' and \f are nonexistent only at that is, only at points where the function y itself does not exist; and so the critical points are only those at which y' and y" vanish. From (1) and (2) it follows that
when x= when x =
y'=Q r/"
y' retains a
Thus,
(-V3, intervals
l), (
(1,
00,
V$\ and
x=
3.
in each of the intervals constant_ sign
1),
3),
=
(
(l, 3,
V$)
and (V~3
1),
(1,
t
0),
+00), (0,
1),
and (1,
/ 3)
(
00,
in
each
and
(3,
J/T), of
+00).
the
To determine the signs of y' (or, respectively, y") in each of the indicated intervals, it is sufficient to determine the sign of y' (or y") at some one point of each of these intervals.
Sec
Graphing Functions by Characteristic Points
4]
97
It is convenient to tabulate the results of such an investigation (Table I), calculating also the ordinates of the characteristic points of the graph of the function. It will be noted that due to the oddness of the function r/, it is enough to calculate only for Jc^O; the left-hand half of the graph is constructed by the principle of odd symmetry.
Table I
e)
Usin^ the results
function (Fig
of
the investigation,
33).
-/
Fig. 33
4-1900
we
construct
the
graph
of
the
Extrema and the Geometric Applications
Example
2.
of
a Derivative
[Ch. 3]
Graph the function In x
x Solution, a) The domain of definition of the function is 0
= limw = lim
JL? =
oo
X
X-*0
JC->
we
=
Hence, the straight line jc (ordinate axis) is a vertical asymptote. c) We seek the right asymptote (there is no left asymptote, since x cannot tend to oo ):
k=
lim X<-++
=
.
The
00
-^ X
lim # x->+<
=
0;
= 0.
right asymptote is the axis of abscissas: find the critical points; and have d)
j/
= 0.
We
Inx
1
y 3
y'
and
y" exist at all
of the
points
y' = Q
domain
when ln*=l,
that
definition of the function and
of is,
when x =
o
(/'=0
We
when
form a table,, including io the characteristic points
Inx^y,
the it
is
that
is,
when x~e*l*.
characteristic points (Table 11). In addition useful to find the points of intersection of
34
=
=
1 the curve with the coordinate axes. Putting / 0, we find * (the point of intersection of the curve with the axis of abscissas); the curve does not intersect the axis of ordinates the e) Utilizing the results of investigation, we construct the graph of lunction (Fig. 34).
-h
CM
CO
CM
O
'
" >
non
> C a c.u
-~
Extrema and the Geometric Applications
100
its
of
[Ch. 3
a Derivative
Graph the following functions and determine for each function domain of definition, discontinuities, extremal points, inter-
vals of increase and decrease, points of inflection of the direction of concavity, and also the asymptotes. 916. y
=x
9
3x*.
"
9
918.
u = (x
919.
y-
921. 922. 923.
924. 925.
\
= = y= y= = (/
(/
926. y==
928. 929.
930.
__
=,-
,6
3*'+! 932. 933. 934. 935.
936.
938.
_
y = 2x + 2-3'l/(xl-
963
z
l
.
'
#=
its
graph,
Sec.
an Arc. Curvature
Differential of
/>]
964. y
=
.
*
sin
+
-7-
y=
976.
.
4 /
\
978. ,,=
967.
979. ^
= y = *-l-sin*. y = arc sin (1 /F).
970. 971. 972.
and 973.
974.
953.
982 tan
.
983.
984
.
<>arcsin
f/
= lnA:-arc tan*. In cos x. y = cos^ = arc tan(ln = arc sin In (*' 4-1). ,/
,/
985.
je).
y==x
-^-
987. y
= lnsin*.
A good
exercise
ples 826-848. Construct the
is
to
>
j
Af
i/
K
= garcun* = n sin x
,,
A;.
= x arc tan - when when * = 0. y= = 2 arc cot*. = + arc tan*.
975. y
arc cosh
I
965. */= sin*- sin 2*. 966. (/ COS*-COS2*. 968.
101
*
=
graph the functions indicated following functions
the
graphs of
parainelrically. l 2/. 988. x=--t* //----/ 2t, sin/ (a>0). 989. x=--acob*/, te~ 990. jc /e', y fl / 4-g991. x i/=2/ + ea (cosh / a (sinh/ 992. x /), i/
in
Fxam-
represented
+
y^a =
= = =
l
.
1
.
,
=
I)
_
(a>0).
Sec. 5. Differential of an Arc. Curvature
1. Differential of an arc. The differential of an arc s of represented by an equation in Cartesian coordinates x and y the formula
dshere,
the equation of the curve
if
a)
2
J/~(d*)
//
= /(*),
b)* = /,Urt.
then ds
then
= 0,
ds^-
d)
^(*,
f/)
of the
',
form
ds
then
*
2
(dy)
-
= q>(0, y = +(0,
c)
is
+
ds-
V F* + F
then
'
.
2 /;
V F'
a is
plane curve expressed by
Extrema and the Geometric Applications
102
of
a Derivative
[Ch. 3
Denoting by a the angle formed by the tangent (in the direction increasing arc of the curve s) with the positive ^-direction, we get
of
dx
cos
a = -3-
,
ds
dy -r
sina
.
ds
In polar coordinates,
Denoting by p the angle between the radius vector of the point curve and the tangent to the curve at this point, we have a 008 P
sin
= dr
p
of the
'
/
2. Curvature of a curve. The curvature K of a curve at one of its is the limit of the ratio of the angle between the positive direcpoints and N of the curve (angle of contintions of the tangents at the points
M
gence) to the length of
the
M ^MN^\s
arc
K=
iim
As \\hore
a
point
M
is
the angle between
The radius i.
circle
line (/C
the
*
o
AS
when
M
.V
(Fig.
35),
that
is,
= ^,
positive
*
rfs
directions
of
the
tangent jt the
arid the .v-axis.
curvature,
The
Au
= 0)
f
of
curvature
R
is
the reciprocal
of
radius
the
the absolute value of the
e.,
K=
,
where a
is
the
are lines of constant curvature.
of
circle)
and the straight
Sec
an Arc. Curvature
Differential of
5]
103
We have the following formulas for computing the curvature in rectangular coordinates (accurate to within the sign): 1) if the curve is given by an equation explicitly, y f(x), then
2)
if
the curve
given by an equation implicitly, F(x,
is
0,
then
represented by equations in parametric form,
*=
Fxx.
Fx
Fxy '
Flx
y)
F
F'yy
'y
Fv
F'
x
3
/j
3) /
if
(/),
\j)
the curve
is
then f
y y
*,'"
x
^
,
where dx
dy -
^
'
~~dt* In polar coordinates,
when the curve
is
~
given by the
equation
/(q)),
we have r
z
+ 2r'
2
rr"
where ,
r
= dr
d"r
.
and
-
2
dcp
.
dtp
3. Circle of curvature. The circle of curvature (or osculating circle) of a is the curve at the point limiting position of a circle drawn through v and two other points of the curve, P and Q, as P and Q M. The radius of the circle of curvature is equal to the radius of curvature, and the centre of the circle of curvature (the centre of curvature) lies on the in the direction of concavity of normal to the curve drawn at the point
M
>
M
M
M
the curve.
The coordinates X and Y computed from the formulas
X=x-
the centre of
of
L,,
,
-
curvature
-jf-r
of
the
curve
are
.
{j
The evolute
of a
curve
is
the locus
of
the
centres
of
curvature
of
the
curve. If
ture
in the
formulas
for
we regard X and Y
determining the coordinates as the current coordinates
of
of the centre of
curvapoint of the evoof the evolute \vith
a
then these formulas yield parametric equations parameter x or y (or /, if the curve itself is represented by equations parametric form) z Example 1. Find the equation of the evolute of the parabola // x
lute,
.
in
Extrema and the Geometric Applications
104
Solution.
X=
4* 8
1
Y--
,
+ 6*
2
involute
curve
a
of
curve
a
is
the parameter x,
Eliminating
Y
the equation of the evolute in explicit form,
The
[Ch. 3
a Derivative
of
o'
we
+ ^lT")
which the given curve
for
find
an
is
evolute.
MC
The normal
CC
length of the arc
involute P 2
of the
of the evolute
l
is
a
is
the evolute
to
tangent
equal to the corresponding
P,;
the
increment
the radius of curvature CC, in AfC; M,C, that is why the involute P 2 is also called the evolvent of the curve P, obtained by unwinding a taut thread wound onto P, (Fig. 36). To each evolute there corresponds an infinitude of invowhich are related to different initial lutes, lengths of thread. 4. Vertices of a curve. The vertex of a curve of the curve at which the curvature is a point has a maximum or a minimum. To determine the vertices of a curve, we form the expression of the curvature K and find its extremal points. In place of the curvature K we can take the
radius
R
curvature
of
I
points
Since
// J
= sinh a
1
a cosh
, 2
and, hence,
x
J
*= of
is
2
X .
a
0).
=
cosh a
We
follows
it
that
tf
=
a
X
rf/?
have -j- = sinh2 dx a
.
Equating M 6
y
Q,
the point
we
the
A
we
get sinh 2
Computing
(0,
get
we
whence
0,
a
-r-y-
,=
2
A:
cosh2
catenary.
The vertex
=
a a curvature (or of
of
the
catenary
the
sole
putting
into
find
and
the second derivative
a point of the radius of
minimum
the
curvature) of
thus,
and J(/"
= acosh
to zero,
=Q
point *
value x
the
/?
>
(a
I")
ax
it
extremal
a
the derivative -j critical
its
I
the computations are simpler in this case. 2. Find the vertex of the catenary
a cosh
y
.
^
Example
36
Solution.
if
and seek
7-7^
> 0.
Therefore,
the
maximum
f/
= acosh
is,
a).
Find the differential of the arc, and also the cosine and sine of the angle formed, with the positive ^-direction, by the tangent to each of the following curves: 993. *
2
2 */
=a
2
(circle).
+ ^-=l (ellipse). y* = 2px (parabola).
994. ~2
995
+
Sec. 5]
_
996. x 2 / 8
997. 998. 999.
-f
2/
t/
Differential of an Arc. Curvature
=a
2 /'
(astroid).
_
105
y = acosh (catenary). x = a(ts\nt)\ y = a(lcost) (cycloid). x = acos*t, y = asm*t (astroid).
Find the differential of the arc, and also the cosine or sine of the angle formed by the radius vector and the tangent to each of the following curves: 1000. r^atp (spiral of Archimedes). 1001. 1002.
= (hyperbolic spiral). r = asec*-|- (parabola). r
1003. r
= acos*-
(cardioid).
1004. r=za.v (logarithmic spiral). 2 a 1005. r a cos2q) (lemniscate). Compute the curvature of the given
=
points: 1006. y 1007. x*
= x* 4x* ISA' + xy + y* = 3 at 2
+ = =
1008.
=1
at the
coordinate origin.
the point
A
the vertices
at
curves at the indicated
(1,
1).
and 5(0,
0)
(a,
b).
=
*' at the point (1, 1009. * /*, f/ 1). 2 2 2a eos2q> at the vertices cp 1010. r 2 1011. At what point of the parabola t/
=
and
= 8x
is
= n. the curvature
equal to 0.12S? 1
1012. Find the vertex of the curve y^-e* Find the radii of curvature (at any point) of the given x* (cubic parabola). 1013. y .
lines:
=
1014.
5 + S =1
(ellipse).
= -!^. * = acos y = as\n*t (astroid). = a(cosM sin 0; y = a(s\nt
1015. * 1016. 1017. circle).
8
/;
/
A:
/?osO
a
curvature of
the
y* = 2px.
1021. Prove
y = acosh
of
=
1018. r ae k v (logarithmic spiral). 1019. r- a(l -f-coscp) (cardioid). 1020. Find the least value of the radius of
parabola
involute
that
the
is equal to a
radius of
segment
curvature
coordinates of the centre of Compute at the indicated points: curves given the
of
the
catenary
of the normal.
curvature of
the
Extrema and the Geometric Applications
106
1022. xy=l at the point (1, 1). x* at the point (a, a). 1023. ay* Write the equations of the circles curves at the indicated points:
of
a Derivative
[Ch. 3
=
= =
x* 1024. y Gjc+10 at the point 1025. y e* at the point (0, 1). Find the evolutes of the curves: 1026. y* 2px (parabola).
1027.
= J + g=l
of
(3,
curvature of the given 1).
(ellipse).
1028. Prove that the evolute of the cycloid
x~a(t is
sin/),
y = a(l
cost)
a displaced cycloid.
1029. Prove that the evolute of the logarithmic spiral r is
also a
1030.
logarithmic spiral with the same pole. that the curve (the involute of a
Show x
is
= a (cos + /
/
sin
the involute of the circle
/),
;c
#=-a(sin
= acob/;
//
/
circle)
/cos
= asm/.
/)
Chapter IV
INDEFINITE INTEGRALS
Sec.
Direct Integration
1.
1. Basic rules of integration. 1) If F' (A-)--- MA), then
where C 2) 3)
4)
an arbitrary constant. f f (x) dx, where A
is
^Af(x)dx=-A [ft
\
U'H:
f f
If
f2
(
v )l
dv
-
( fj
(*)/*
i
and
(x)dx-~F(\-) -f-C
is
f f2
/-cf
a constant quantity.
O)
dx.
(v),
then
In particular,
r
i
ax
a
J
2. Table
II.
of standard
=
\
dA'
f*
III
.
,*r
IV
==
-r-:
\ 2 s J X -f- a dx C \ )
-=2
;5
a*
A'
dx r
V.
integrals.
dx
1
a
\:
arctan
a
I
r
~-,T-
2a
a+v
|-C
=
v:
1
arc cot
+C
(a
-f-C
(a*Q).
yS-
0).
(a^O).
VI.
VII.
a
(>0); fc x d* =
+C '
(a
^ 0). '
__
108
VIII.
IX.
\ f
X
cosx +
sinxdA;=
_
[Ch. 4
Indefinite Integrals
C.
cosxdx=sinx + C. COS 2
XI.
J XII
--= x sin
cotx+C.
2
'
tan
smx
XIII.
XIV.
x
H-C = In
|
cosec x
cot x |-f-C.
cosx f
sinhxdx=coshx-|-C.
XV. dx
XVL XVIL Example f (ax
2
1.
+ bx + c) dx=
f
Applying the basic rules tion,
1,
2,
3 and the formulas of integra-
find the following integrals:
5dVd*.
1031.
1040.
1032-5(6^ + 8^3)^. 1
x
033.
(x
r 1
034.
J (a
1035.
-i
a) (x 4 b) dx.
+ bx
J
)* clx.
yZpxdx.
-
(V
1042 ' U4 ^'
3
_
-- dxv
V x)
j/51
1043.
1036. '~ n
T
1037. J
(nx)
"
1045
^x '
dx.
}
1046
'
1047
'
1038.
1039.
^+ljc-/x-dx.
f
^4+^
'
,,
'
Sec
_
Direct Integration
1]
2
tan *dx;
1048*. a)
109
1049. a)
2
Jtanh *d*.
b)
b)
1050.
Jcoth'jtdx.
$3Vdx.
differential. Rule 4 considerably sign expands the table of standard integrals: by virtue of this rule the table of integrals holds true irrespective of whether the variable of integration is an independent variable or a differentiate function.
3. Integration under
Example
of the
the
2.
2
where we put u
Exa m p.e
We
5*
We
2
took advantage of Rule 4 and tabular integral
xdx
1
d(x
f
1.
2 1
)
3.
2
implied u
Example
2.
4.
jc
and use was made
,
( x 2 e xl}
dx-^~
( e*' d
3 (jc )
of
-
Rule
4
and tabular
i-e^ + C by
integral V.
virtue of Rule 4 and
tabular integral VII. In examples 2, 3, and form before making use of
of
we reduced the given
/ (
\
This type
4
a tabular
dx\
transformation
is
Some common transformations ples 2 and a)
3,
I (u)
the following
du
t
where a
=
(p (x).
called integration under the differential sign.
which were used
of differentials,
b)
(a
^
0);
b)
xdx =
Using the basic rules and formulas
,053.
to
in
Exam-
are:
dx=^d(ax-\
tegrals:
integral
integral:
of
2
^d(x
)
and so on.
integration,
find
the following in-
112
Indefinite Integrals
1140. f-^f-. sinh x
1143.
H41. f-*_. J cosh x
1144.
1142
/f
\
. '
Find the indefinite
< 1
Ad 46.
,,47.
integrals:
(x 1/5=1? dx. i
v*
(cothxdx. J
v
J sinh x cosh x
1145.
[tanhxdx.
J
J
r*
[C/i.
1163.
A \f
I
J?=CT1
f-^-. ]cos@
i
*J
" M -j
f^d,.
1165
1148. \xe-**dx.
1151.
,, 67 .
Jl_!d*.
--
i
1168.
i/^c' V e
J
,-
jcdx
JV^L l
\
J I
1169.
~ cc^s
sinx sin
A: -f-
1
Jx + cos*
1 i
ftan/JC
1166.
U50.
djt
x
'
-
-
A*
cos x
*
dx. 2
51
f v
"-^) v ^
sin SA:
1170.
J
^tan^-2 >
'
ff 2 -|--^- )-^-. 2x*+l/ 2x*+l ^
,)
1157.
^a
ilax
cosxdx.
7^Ti
1161.
"
J
M7.:.
!!!_!-.
+ !'
^
^
to
f
fl
_t ^
H76.
!
K4-tan'x
*
(<
jsin -Jdx. (
.rfy.
[
1174.
Jlan'aArdx
J
3<
5
1173.
dAt>
f-i^=. 1160.
I179 1172.
j -j^* .
ii77. f _
si J sinajtcosax'
4
Integration by Substitution
Sec. 2]
J
113
l/T=3
Sec. 2. Integration by Substitution
1. Change of variable in an indefinite integral. Putting
where
t
is
a
new variable and
cp
is
a
continuously differentiable function, we
will have:
The attempt is made to choose the function q> in such side of (1) becomes more convenient for integration. Example 1. Find
Solution.
It
is
natural to put
t
=
V~x
1,
whence
A-
a
way
=/ 2
-}- 1
that
the
right
and dx = 2tdt.
Hencu,
Sometimes substitutions
of the
form
are used.
Suppose we succeeded
in
transforming
the integrand f(x)dx to the form
f(x)dx=g(u)du where u = q>(jc). t
_
114
If
[g(u)du
known, that
is
_
Indefinite Integrals
is,
[Ch. 4
then
we have already made use oi this method in Sec. 1,3. Examples 2, 3, 4 (Sec. 1) may be solved as follows:
Actually,
Example
2.
u
= 5#
cfw
2;
1
Example
3.
u
= x*;
Example
4.
w
=
jc
s ;
du
du
= 5c(x;
_ -~
d
2
'
+c_
1
dx\
2
,
xdx = -
2xdx;
= 3x
-du.
dx
2
x dx
.
=
.
-
2. Trigonometric 1)
If
~a sin
2)
/;
If
_
substitutions.
an integral contains the radical ]fa z
x2
the usual thin^
,
is
to put
whence
an
integral
the
contains
radical
V*
a2
2
,
we
put
xawct,
whence /^x
3)
If
2
a2
= a tan
an integral contains the radical V'V
^.
+a
2 ,
we put* = atan/; whence
It should be noted that trigonometric substitutions do not always turn out to be advantageous. It is sometimes more convenient to make use of hyperbolic substitutions, which are similar to trigonometric substitutions (see Example 1209).
For more details about trigonometric Sec. 9.
Example
5.
Find
i
^
and hyperbolic substitutions,
dx.
see
Integration by Substitution
Sec. 2]
xlant.
Solution. Put
f
J
y~x*+l *
f V~ian 2
_
1?
dt
"~ f
J
= In
sin
|
/
tan 2
J 2 /
tan
/
+
/
+ sec
r-r
cos 2
cos'
t
,
/
i
dt
.
cos ! si "
2
sin
J /
f sec
dt
1
/
~~ f
cos
dx=
Therefore,
1J5
sin 2
J
*
4- cos2
2
/-cos/
-J 1
<
M-~
*
+ C = In
cos 2
/
_
/
1
^/
^ J cos
tan
f
/ -{-
|
_
cos/
~~
,
2 J sin
/
V\
/
-j-tan
2
/
1
tan< find the following
1191. Applying the indicated substitutions, integrals:
c)
f e)
COS
A'
d*
\
J
Applying suitable substitutions, 1192.
S*(2x+5)"djc. 1
1193.
('
J
1194.
+ *..d*.
the following integrals:
find
>97. n-
csinA )"
1198.
l+^A-
f-
J!
JxK2t+l
,, 99<
**
1195. r
.
Applying trigonometric substitutions,
find
the
tegrals:
,201.
('-=*. *' K
J
,202.
1203.
l
-=.
1204*.
f J
iZEl'dx. ^
following
in-
__
U6
1205.
\Ch. 4
Indefinite Integrals
1206*.
(f^+idx x
J
f ff 2 J x y 4
__ x2
1207. 1208. Evaluate the integral
by means
r
dx
J
/*(!-*)
x=sin
of the substitution
2
/.
1209. Find
by applying the hyperbolic substitution x = as\n\\t. Solution.
We have:
]Aa
2
+x = 2
]/~a
2
+a
suih 2
2
/=a cosh
^
and
dx=a cosh
/d/.
Whence \
ya
2
z
-\-x
dx=
^
a cosh t-a cosh
f ctf
=
~~~2~
Since __ x
~
a
and
we
finally get
where C 1=^C
In a
is
a
new
arbitrary constant.
1210. Find C
putting x
}
= a cosh/.
Sec. 3. Integration by Parts
A formula for integration on tiable functions, then \
by
parts.
udv = uv
H
If
(
=
vdu.
and u
= i|)(*)are
differen-
Sec. 3\
_
_
Integration by Parts
Example
Find
1.
\
In*, dv
Putting u
H7
x In xdx.
dx
xdx we have da
x*
2
v
,
%
Whence
~~9
x*
dx
Sometimes, to reduce a given integral to tabular form, one has to apply the fcrmula of integration by parts several times. In certain cases, integration by parts yields an equation from which the desired integral is determined. Example 2. Find \
We V
e* cos x dx.
have
=
e* cos x dx
\
e
xd
e* sin x
(sin x)
+
\
e
\
e
x sin
dx
AT
x d (cos x)
e
x
e* sin x
sin
-\-
x
e
+
x cos x
\
e* cos x dx.
Hence, e* cos x dx
\
e
x sm .v-j-e^cos x
V e
x cos x dx,
whence cos v dx --
~
(sin
x -f- cos
Applying the formula of integration by
.v)
-f C.
parts,
find
the following
integrals:
1211. ^
1212.
\nxdx.
Jarclanjcdx.
1213.
Jarcsin
1214.
A-rfjc.
Jjcsiiucr/.v.
1215.
Jjccos3A'Jx.
1216. Urfjc.
1219*.
1220*.
x sin x cos x d\ a
1222* $
1223.
(jt
+5x+6)cos
x*
\nxdx.
^
1224.
Jln'xd*.
1225.
(^djc.
f^d*. K A
J
Jx-2-*rfx.
1218**.
{
1226.
J
1217.
1221.
JjV'd*.
1227.
1228.
Jjcarctanjcdjc.
Jjcarcsmxdjc.
2
(x ^2A' + 5)^*dA:.
'-
1229.
1230
'
\n (x
+ V T~x*) dx.
118
Indefinite Integrals
1231
1234
-
1232. 1233.
[C/i.
eax *
'
\
JisFr**'235.
mb * dx
4
-
$sin(lnx)dx.
Je*sinxdx. $3*cosjtdx.
Applying various methods,
find the following integrals:
2
1236. (x*e~* dx.
1246.
V\-x
J
1237.
v
\e 2
1238. 1239.
(x
*dx.
1247.
-2x + 3)\nxdx.
1248.
2
jttan 2*d;t.
f
2
^x\n~dx.
1249.
1240.
f^dx.
1250**,
1241.
fllL^d*.
1251*.
cos (In x)dx.
J
1242.
1243. 1
244.
*n/ir
1245.
2
f
{ j
arctan3jcdA:.
jc
x
(arc tan
( (arc sin
fare
sin
\
^2
J
5
x
2 A:) 2
jc)
a2)2
2 :
-|
'
1252*.
dx.
1253*.
dx.
\
1//1
^ j
254*.
f
z
-4-
-4^-
x dx. .
,
^.
Sec. 4. Standard Integrals Containing a Quadratic Trinomial
1. Integrals of the form f
mx + n
\
2-7-7
.
7dx.
J
The
principal calculation procedure the form
ax z
is
to reduce
+ bx + c = a(x-}-k)
the
quadratic trinomial to
z
(1)
-{-l,
/ are constants. To perform the transformations in (1), it is best to take the perfect square out of the quadratic trinomial. The following substitution may also be used:
where k and
If m=0, then, reducing the quadratic trinomial to the form the tabular integrals III or IV (see Table).
(1),
we
get
Sec
Standard Integrals Containing a Quadratic Trinomial
4]
Example
119-
1.
_
dx
1
-(-*)
H-
g \2 5
-7
2
Qi
.
arc =-o*~7= -o 2
tan
(-TJ+S '*?
4
2
"S?+c. If m&Q, then from the numerator out of the quadratic trinomial
the derivative
-+.
f J
we can take
a.v
2
+^
and thus we arrive at the integral discussed above.
Example
f J
-> ;5=^n
2.
,v d -
i
f"
^J
-
A
'-x-i
IV 2.
-
Integrals of the
5
form
2x-l
are similar to those analyzed above. 0, and VI, if a lar integral V, if a
>
Example
r
J
The
The methods
integral
of calculation
finally reduced
is
< 0.
3.
dx
Example
d*.
I
+
1
f*
dx
1
4.
^+3 ^^2 f *" yV+2x + 2 J ^ ^ + 2x4-2
JC ^^
.
2
4jt
3
to
tabu-
J20
[Ch. 4
Indefinite Integrals
*
3.
of
Integrals
the
form
f J (mx
+ n)
V^ax'
=
+ bx + c
.
By means
of the in-
verse substitution i
mx+n
=
/
these integrals are reduced to integrals of the form
Example
5.
2.
Find dx
We
Solution.
put
whence
4.
Integrals of the form
\
]T ax*-\-bx
+ cdv.
By taking
the perfect square
out of the quadratic trinomial, the given integral is reduced following two basic integrals (sec examples 1252 and 1253): 1)
('
J
V
a*
A'
2
dx
= 4 fa^x* + ^ arc sin ^ 2 (a
2)
J
>
to
+ C; c/
0);
Vl?
Example
6.
sin
Find the following
1256.
integrals:
S' '',c 00
P
J
xdx
^7^+
one
of the
Integration of Rational Functions
Sec. 5]
oKn \o\J.
i
j
---
i
<
-
c r\
<
121
dx
I
\
1270. v2
dy
1
O *7 1
*~
J
1272.
'
2* 2
V~2 {-3* d* r I
J
X
'
f* t
1273.
X 1
1-7/1
J f
/"*l/"o
f~=.
,H4. 1265.
1
1266.
V
[
J
^
>1J -
T
1267.
xdx
V %
J
x
5
-
."~
-,-^ /5, -2,
dx.
1277 197 n 1278
dK.
2
f-1
JYKI
2
,
-v
279
x cJJt
sin
x
dA
(
J
1'-^=.
1268.
g
t'
T^PTTT^TlTInjfdi
i'
J x
^ _ ilnA._ !n 1
m
2
K
Sec. 5. Integration of Rational Functions
t. The method of undetermined coefficients. Integration of a rational function, after taking out the whole part, reduces to integration of the proper rational fraction
where P rator
P
and Q
(x)
is
(x)
(A-) are integral polynomials, and the degree lower than that of the denominator Q (A-).
If
Q(jr)
= (*
a)*.
.
of
the
nume-
.(A'-/)\
/ are real where a, distinct roots of the polynomial Q (x), and a, .... ., K are natural numbers (root multiplicities), then decomposition of (1) into .
.
partial fractions
is
justified:
^ calculate the undetermined coefficients A lt A 2t ..., both sides of the identity (2) are reduced to an integral form, and then the coefficients of like powers of the variable x are equated (llrst method). These coefficients may likewise be determined by putting [in equation (2) or in an equivalent equation] x equal to suitably chosen numbers (second method).
To
122
Indefinite Integrals
Example
4
Find
1.
xdx (*-!)(* +
We
Solution.
[C/t.
"
2
1)
have:
Whence t
a) F/rsf
method of determining the
We
coefficients.
the form x^(A-{- B^ x 2 -{-(2A-{- B 2 )x-\-(A ents of identical powers of x we get:
B
B2
l
)
(x\).
(3)
rewrite identity (3) in Equating the coeffici-
t
Whence
=i
,
,
= _i
Second method of determining the
b) <3),
B
;
we
,,4.
:
1=4-4, Putting x
we
1,
4
i.e.,
=
'/ 4
x=\
in identity
.
get: 1
Further, putting *
Putting
coefficients.
will have:
= 0,
we
=
2
B2 =
i.e.,
-2,
l
/2.
will have:
Hence,
TJ *
\
~~ Av
1
1
T t
4-1
f
\ J Ax
AT
Example
Solution.
2.
1
Find
We
have: "" 2
x(x tind 1
When
=A
2
(*
I)
x
x
I)
+ Bx (x
1)
1(JC
1)*
+ Cx.
(4)
advisable to combine the two methods of determining coefficients. Applying the second method, we put * in identity (4). We get 1=4. Then, putting jc=l, we get 1=C. Further, app2 lying the first method, we equate the coefficients of x in identity (4), and get-
Hence,
solving this example
it
is
= 4 + 0, 4=
1.
fl-=
=
i.e., l,
B= and
1.
C=l.
In t egration
Sec. 5]
Rational Functions
of
123
Consequently,
/=
f dx
dx
\
p
,
\
'
J
the polynomial
If
dx
r- = ln jc17+ J (*l)
f*
\ J X
Q
1
,
2
;
,
,
'
JC
,
,
'
A-
ib of
~
r-f C.
1
'
has complex roots a
(x)
,
,
In
JC '
1
multiplicity k, then
partial fractions of the form (5)
will enter into the
expansion
Here,
(2).
Bk are undetermined coeflicients which are determined and A lt B lt .., A k by the methods given above For k~\, the fraction (5) is integrated direct,
ly;
k>\,
for
use
is
made
of the reduction
~]
q
Example
and make the substitution 3.
method;
+
z sable to represent the quadratic trinomial x
here,
it
px~{-q in the form
A--J-
=
z.
j2
^
is
(
advi-
first
x-\-~
\
-f-
Find
Solution. Since A
then, putting x -\-2---z, r==
r
*\._ dz=z
jrc
wo
2 -|
4x
got 2
r
_j_^
tan ?
Hit i!ini
r
-
-
2. The Ostrogradsky method.
is
(A:)
derivative Q'
X
and
Y
the greatest
-- -- arc
-
Q
If
P(x)
where Q,
5-
i
z=
(A)
has multiple roots, then
"-
p
A'(v)
common
tan
r
(.Y)
A
divisor of the polynomial
(6)
Q
(x)
and
its
(A-);
undetermined coefficients, whose degrees by unity than those of Q, (A-) and Q 2 (x). The undetermined coeflicients of the polynomials X (x) and Y (x) are computed by differentiating the identity (6). Example 4. Find (A-)
(x)
are polynomials with
arc, respectively,
less
C }
dx
U'-
124
[Ch. 4
Indefinite Integrals
Solution.
Ax z + Bx + C
dx
we
Differentiating this identity,
?Dx
.
2
-\-Ex
+ F ax .
get
Dx z +Ex
\=(2Ax-\-B)(x*
3x*(Ax* + Bx
1)
+ C) + (Dx* + Ex + F)(x*
Equating the coefficients of the respective degrees of x,
4 = 0;
D = 0; E
F
D + 3C = 0;
2fl--=0;
we
+ 24 = 0;
B
i).
have:
will
+ F==
1;
whence
C = 0; D = 0; ~; O
5= and, consequently,
C 9
^ ~ _
To compute the r
-^
into partial
8
1)
on the
integral
__ 2
x
1
3^ -l
2
(x
= 0; dx
P
3 J x8
right of (7),
F= -4O 1
we decompose
the
fraction
fractions: 1
x8 lhat
is, 1
Putting
#=1, we
= L (x + + 1) + MX (x 2
A:
1)
+N
(A:
(8)
1).
L=-.
get
Equating the coefficients of identical degrees of x on the ,of
we
(8),
find
right
and
.
or
Therefore,
r
_
dx
\yS J *
i
p
^\V
o J x
1
i
dx
r \_
1
1
Q\
J j
~~ =ll
3
]X
11-1 '
6
and
^^^^^(J^\\ + ^ ln \ x Find the following
1280
'
^ +^^
r
l
integrals: '
1282
'
mian
^^+ C
'
left
Integrating Certain Irrational Functions
Sec. 6]
1284 ' **
C
J
1285
'
5 *' + 2 * - 5* + 4* dx*
12Q1 29J '
f j (Ii_4jt
-
125
dx
+ 3)
dx c J *(*
+ !)*
-~
d* C
1296.
-r-g^gdx. ^"* " I
R V2
C*
(
Av
t
C/
Q
rt
^4
.--.
,
*
I
I
J
'
.
1288.
-290.
.Tjpfc
l291 '
J^Sf*.
'292.
J^d*.
1299. .300.
Applying Ostrogradsky's method,
Applying
l306
different procedures, find
integrals:
the integrals:
'
I3 7 '
l309
find the following
l312
rf *-
'
-
Sec. 6. Intagrating Certain Irrational Functions
1.
Integrals of the
f.rm i
/?
a rational function and p lt q v p a
,
q z are whole numbers.
12J
[Ch. 4
Indefinite Integrals
Integrals of form (1) are found by the substitution
where n is the least Example 1. Find
common
multiple of the numbers
q2
,
...
* dx a I
Solution.
f
_
The substitution
____dx
2,x
f 2z'dz
J V"5jc=T-- J/2?=l""j
z
2
2
1
-^ =2
4
leads
to
an integral
of the
form
r^i " 2
1
J
=2 = (1+
2
/23T-i)
+
Find the integrals: 13 15.
f-^^djc. /*-! A
r
1316.
('J^Zdjc.
1321.
J .v-f-2
J
dx
c
1322.
7,7==.
J
1U7. .
^.---
-. ---('-. J
/v+l
.
I-
--^7= A (jv
(o
-A)
V
1-
x
1323.
.
v+1
)/(v+l) 1324.
1325.
,320.
2.
Integrals of the form "
r
where P n Put
(x)
is
ai*
+
- dx,
t,
x
(2)
+c
a polynomial of degree n
dx
a polynomial of degree (n with undetermined coeffi1) number. The coefficients of the polynomial Q n -i(x) and the number K are found
where Q n _i(x) cients and A, is
is
a
by differentiating identity
(3).
Integrating Certain Irrational Functions
Sec. 6]
Example
2.
Whence
Multiplying by V^* we obtain
2
+4
and equating the coefficients
of identical
degrees oi
A-,
=
-
;
D-0;
X=_
Hence, -
3. Integrals
of the
form dx
i;
a)
n
V ax
are reduced to integrals of the form (2) by the
They
A
substitution:
a
Find the integrals: 1326
.
1327.
j__^_. (-f^dx.
1329.
1330.
v
-^ fv
-^^rfjc.
1331. FA'
Vl-i
4.
2
X+l
Integrals of the binomial differentials
x
(5)
t
where m, n and p are rational numbers. Chebyshev's conditions. The integral (5) can be expressed in terms of a finite combination of elementary functions only in the following three cases: 1) if p is a whole number; "
2)
=z
s t
is
if
where
m 3)
a
n
if
s
is
whole number. Here, we make the substitution a + bx n =
the denominator of the fraction p;
+p
is
a
whole number. Here, use
is
made
of the substitution
128
(Ch. 4
Indefinite Integrals
Example
3.
Find
m=-^;n = -r ;p = -^;
x.
,
Solution. Here,
we have here Case 2 The substitution
yields *
= (z
4
3
I)
\Y dz
(z*
y
e
1
Find the
+ J/T
f (2
z
T
1335
e
),fe=J2ef_
dx
'
^-
T
133G 0. f 2
J
* f_-
,334.
Hence,
.
~
1333. r
=2.
integrals: *
1332.
:
Therefore,
= 12 where
2
integrability.
dx=\2z*
;
-1+1 -T
=
.v:
i'
337.{
.
J
(2
^
+ ,
3
.
3
3 JC
)
r*
y
1+
Sec. 7. Integrating Trigonometric Functions
1. Integrals of the form :
where 1)
m If
and n are
an odd positive number, then we put
is '
We
n
\
do the same Example 1.
if
0)
integers.
m^2/j+l ,
'm..
s l(]Zk
n
( sin 10
x cos " xd ( cos
is
^)
\
cos 2 x) k cos" xd (cos
( 1
an odd positive number.
x cos 8 * dx
=
f sin 10 x (1 sin
11
x
sin
2
sin x) 13 .v
d
(sin *)
=
A:).
Sec.
7]
_
__
Integrating Trigonometric Functions
the integrand 2) If m and n are even positive numbers, then formed by means of the formulas sin
*= y (1
2
sin JCGOS
Example
= =
2
If
2
3x
1
3-
4
cos 6* dx
cos 12* n ^
1
\
=
3* dx
--
f /
J \ x
m=
sin
-- -
sin 6x1 p \
^>
3)
f cos
2.
cos
cos 2*),
,
=
= sin .
, 2
x=
2
*= y (1 + cos 2*),
3* sin 3*) 2
sin
2
3x dx
=
'
f \
c
GA:
,
,
Zc
.
2
sin
(sin 6A:
c \ dx cos 6^ ,
2C 6x cos fi o^) ax v
2
.
=
=
/
sin 12AC
and
[i
n=
v are integral negative
numbers
of
parity, then
_
C
dx cos
J
v
In particular, the following integrals reduce to this case:
Example
3.
f
2
-^-= JC sec xd (tan *)= Jf (l+tan*x) = tan x-p--=-o tan '* + C.
J COS X
8
d (tan
x)
jc
_1 r ""-
tan 2
5-1900
(I) is trans-
sin 2*.
f (cos 1
129
-
~
identical
130
form
4) Integrals of the
(or, respectively, cot*
Example tan
1
x
=
f tan
5.
P.
x
= cosec* x
tan*xdx=
\ .
tv
(or
cot w xdx),
V
where
m
is
an
in-
1).
l)dx=
6.
=
tan*xdx =
2
V tan x (sec*x
tan* x
.
(sec*x
\
w xdx
number, are evaluated by the formula
tegral positive
=
[Ch. 4
Indefinite Integrals
-5
1)
\
dx=-^-^ ~
tanx + x+C. ,
,
,
3 J 5) In the general case, integrals 7 m>n of the form (1) are evaluated by means of reduction formulas that are usually derived by integration by parts. 5
Example
\
J cos* x
sin
=
\
cos
J
dx
A;
1
1
C cos x
2^^"j -^
f
. ,
cos x
c
sin
x
Find the integrals: 1338. 1339. 1
340.
1341.
1352.
Jcos'xdx.
J
C \
sin sin
2
sin*
x
8
-jr
*
cos
5
1346.
'T
^
-~-
*
dx.
x
J
sm*xcos*xdx. sin
4
1 A:
cos x dx.
Jcos'Sxdx.
1353.
x
1343.
345.
-2
8
x cos x dx.
-
1342.
1
COS
sin
Jsin'xdx.
J
1344.
{;
J
1354. J
1355.
JsecMxdx.
1356.
Jtan
1357.
!
5jcdA:.
jjcot'*d*.
1358. 1359.
1360.
' COS X
1362. 1363.
1364.
J
sin'xi/cosxdx.
f J
V sin x cos*
dx .
f-==. J l^tan*
*
_
_
Integrating Trigonometric Functions
Sec. 7]
2. V
cos
Integrals of the form
mx cos
1)
sin
2) sin
3)
mx
sin
V
cos
nxdx,
sin
\
131
rnx
sin
nx dx and
nx dx. In these cases the following formulas are used;
mx cos nx = -~ mx
sin
nx
cosm*cos ^^
Example
7.
I
= -^
= sin
-9-
(m + n) x + sin (m
[sin
(m
[cos
[
c
s
9* sin
x
n)
n) x] ;
cos(m
+ n) x]\
n)*-fcos (m-f/i)
(m
xdx=
8* -^ [cos
\
x].
cos
lOjt]
dx=*
Find the integrals: 1365.
^
1366.
J
sin
cosSxdx.
sin 10* sin 15
1367.
f cos
1368.
f sin
3.
3jc
~ cos |-
sin
A:
1370.
d*.
djc,
1371.
~ djc.
1372.
^-
J cos(aA:
sin
$
a>/
f sin
(co/ -f-cp)
cos x cos* 3x dx
^
sin x sin 2* sin
J
dt.
.
3* dx.
Integrals of the form f
where R 1)
1369.
is
/?
cos x) dx,
(sin *,
(2)
a rational function.
By means
of substitution
x
.
tan
t,
whence 2*
integrals
of
new variable Example
form
(2)
are
reduced
t.
8.
Find f J
Solution.
f*
J
Putting
to integrals
__ *E
1
+ sin x + cos
tan -pr^*'
we
wil1
x
have
of rational
/.
functions by the
132
we have
2) If
the identity
R( Ihen
we can
use
4
[Ch.
Indefinite Integrals
sin*,
the
cos*)
s/? *
tan
substitution
cos*),
(sin*,
=
to reduce the
/
integral (2) to a
rational form.
Here, sin
jg
=
COS*=
t
,-
and
*=arc Example
tan/, dx
<3 >
Jnn-'-
tan* = f,
sin
2
*
=
t
2
dt
dx =
^,
:j
-jz
have
will
dt
dt
p
*
note
denominator
that
r
1
the
integral
+ C = -7=2 arc tan F
(3)
it
is
(
/T tan *) +
evaluated faster divided by cos 2 *.
is
useful
to apply
C.
the numerator and
if
procedures (see, for
artificial
1379).
Find the 1373
2~)
of the fraction are first
In individual cases,
example,
/"
(/
2
K
=
i
= -^=arctan We
.
2
Find
9.
Solution. Putting
we
.
'
1374.
J 3
integrals:
+ 5*cos*
C^
1382 *'
'
*!.
J sin *
+ cos *
cos*
.
,
dx J 3
sin 2
2 J sin *
iooyi*
1384 *1
385.
-|-
P \
.
(1
^^.
T
I39
'
X r-. dx. 3 cos*) Sin2x <*-
P
COS
jj
,
^
-
p-j-^-rfx S
'
1389*.
cos 2 *
3sin *cos *
si n
T
J.
\
'
dx
P
1386.
1388
*
JiHiJT 5sln*cos* J
100T 1387.
2
**
1383*. f
10 or
mi-.
+ 5 cos
*
Is. n
*-6 sin* + 5
^
^ '
sin*) (3
sin*)
o-.j |;;;;;i::;>.
'
Integration of Hyperbolic Functions
Sec. 8]
133
Sec. 8. Integration of Hyperbolic Functions Integration of hyperbolic functions is completely analogous to the integration of trigonometric functions. The following basic formulas should be remembered: sinh 2
cosh 2 X
1)
2) sinh
2
3) cosh
2
*=l;
*=
~ (cosh 2*
x=
Y (cosh 2x+
4) sinh x cosh x
Example
1.
1);
1);
= ~ sinh 2x.
Find cosh 2 *djt.
V
We
Solution.
have
2
f cosh x djc-=r~(cosh 2x + \)dx
Example
2.
( cosh
8
We
xcU=
f
jsinh
2x +
^
Find \
Solution.
=
cosh 3 ****.
have
cosh 2 xd (sinh
x)=
J
(l
+ sinh
2
Jc)d(sinhA:)
=
= smh x-\ .
,
,
Find the 1391.
integrals:
1397.
Jsinh'jtd*.
1392.
1398.
^cosh'xdx.
1393.
sinh'* cosh*
1394.
1396
Sec.
9.
Integrals
C 2 J sinh
.
Using of the
^ A:
cosh 2
1402.
.
A;
Trigonometric
and
Hyperbolio
Substitutions
for
Finding
Form (1)
R
is
a rational function.
134
Indefinite Integrals
_
[CH. 4
the quadratic trinomial ax* + bx-\-c into a sum or difference Transfor Transforming integral (1) becomes reducible to one of the following types
t of squares, the
of integrals:
2)
R
(z,
3)
R
(z,
The 1) 2 2) 2 3) 2
m* +
Vz 2
z*) dz\
m
z
latter integrals
= msin/ = mtan/ = msec/
Example
1.
or z or 2 or z
)
dz.
are,
taken
respectively,
by
means
= m tanh = msinh/, = m cosh/.
/,
Find dx
==/.
(x+\Y Solution.
Putting
We
have
*+l = tan2, we
then have dx sec 2
dx
= f tan \
Example
Solution.
Putting
2.
Find
We
have
-
2
zdz
= sec*zdz
=
x
zsec2~"J
and
~ dz =
f cos 2
.
of substitutions:
Sec. 11]
we
finally
Using Reduction Formulas
135
have
Find the integrals: 1403.
S/3-2jt-x'd*.
1409.
J
W-6*-7d*.
"
1404.
1410.
1405.
1406.
Sec. 10. Integration of Various Transcendental Functions
Find the integrals: J (*
1416.
$*
1417. 1418.
1419.
1420.
+ I)
2
1415.
2
]
cos
2
e
2x
1421.
dx.
2
1422.
SJCC/A;.
x s\nxcos2xdx.
^
(*-/= J
1423.
f^lnj^djc. *
J 2
$e *sin x
\^e
jcdjt.
1424.
smxsmSxdx.
1425.
x ^ A'e
2
cos x dx.
1426.
I
1
J J ?
In (x jc
+ /FT?) djc.
arc cos
(5jc
5
smxsinhxdx.
Sec. 11. Using Reduction Formulas
Derive the reduction formulas for the following integrals: 1427.
/..Jj-^,,;
1428. /
=$
find /.
and
sln"xdje; find / 4 and
/,.
/,.
136
[Ch. 4
Indefinite Integrals
/w
=
1430. 7 n
=
1429.
Sec.
J^h;;
find
(x n e~*dx\
I*
find
and / 10
7 4-
.
12. Miscellaneous Examples on Integration
dx
1448.
xdx f
JO+jt x
5
dx
J *'-2*+2 1433.
P
T
J
)
xdx
1449.
C
..
2^ j
j y^i
1450.
dx.
KI-J
1
^
r
+1
x
dx
t
2 '
dv
dx
-
dx
1452.
dx
1453.
dx
P '
J (x* r
+ 2)
i
f-7^ A
'
t/
dx
y l/^v2 * r
-
1455.
Jx-2A' +
l
dx
C__^L_ 1457.
'
Jf *
VT^X*
,458. '
J 5jt
1442.
f
J 1460.
1443.
-1
1461
dx -
1444.
IHB^T^
1462.
+
J/ Jl a!+i)' (
1446.
ax
'-^-
1463.
r
1464. P
1447.
y2
j^===d.
f^ ^
^
dA;
1465.
sin' A;
\
COS '
cosec'SxdA;.
Miscellaneous Examples on Integration
Sec. 12]
1466.
1
M67.J.-(f,4 '
l469
-
dx,
fsinh^l-A
f
J
J 2sinx
X
+ 3cosx-5 "
J 2-+w-*-
A
f.
-
,)sinh'*" dx
1470.
A:
d)c
f*
!468
sinh ^ cosh
484.
1485
137
]>
1488.
C_^_ ,. 2e*
2* J e
d.v
.471.
1489.
"
j sin
2x
sin
jc
dx
1472.
rfx.
J
i
j1474.
1476. 1477. 1478.
1479. 1480.
JV? + A:
a*
J
sin x dx.
2
Jxe^djc.
J(x
1493.
J/?Tirfx.
J
Y
i
sin*
2>
J
.j-^z
y
\
(sin
C-
>
(
x
-s-
1)
src tdn x
*
dx.
1494.
\
1495.
(Varc
1496.
J cos(lnx)dA:.
1497.
J (x
1498.
Jxarctan(2x+3)dx.
1499.
fare sin
dv
>-
I
/\
\nYT=xdx.
x'
!
10~ *cfx.
1492.
/
^*V'djt.
r*
1483.
sin*
CJiLdx.
2
J
1481.
1491.
aX
<
f
1
1
sinyd.v.
3x) sin 5*
*^i*
cos
"2"^*.
+ cos x)*'
1500.
V^ dx.
rfjc.
Chapter
V
DEFINITE INTEGRALS
Sec. 1. The Definite Integral as the Limit of a
Sum
1. Integral sum. Let a function f (x) be defined on an interval and a=xc
.
.
into
(1)
where /
is is
= 0,
1,
2,
...
(n
1),
called the integral sum of the function f (x) on [a, b]. the algebraic area of a step-like figure (see Fig. 37).
Geometrically,
S,,
10
Fig. 37
Fig. 38
2. The definite integral. The number of subdivisions n tends to zero,
from
is
x=a
limit of the sum S nt provided that the to infinity, and the largest of them, Ax/, called the definite integral of the function f (x) within the limits to * &; that is,
=
max
(2)
A*j
-> o
^
The Definite Integral as the Limit
Sec. 1]
of
a
Sum
139
it is integrable on [a, b]\ If the function / (x) is continuous on [a, b], i.e., the limit of (2) exists and is independent of the mode of partition of the interval of integration [a, b] into subintervals and is independent of the choice of points / in these subintervals. Geometrically, the definite integral (2) is the algebraic sum of the areas of the figures that make up the curvilinear trapezoid aABb, in which the areas of the parts located above the #-axis are plus, those below the jc-axis, minus (Fig. 37).
The definitions of integral sum and definite integral are b. eralized to the case of an interval [a, b], where a Example 1. Form the integral sum S n for the function
>
by dividing the interval into n equal parts and chooscoincide with the left end-points of the subintervals x i+l ]. What is the lim S n equal to? n -+ CO 9 Ax. and tc/ J Solution. Here, -. Whence n n
on the interval ing [x it
gen-
naturally
points
[1,10]
that
|/
= 101 = Hence
(Fig. 38),
S n -58-L.
lim n
=
.
2
->> oo
Example 2. Find the area bounded by an and the ordinates * = 0, and x = a (a >
jc-axis,
Partition
Solution. parts
=
.
the
arc of the parabola 0).
base a into n equal
Choosing the value
of
the
tion at the beginning of each subinterval,
y
func-
we
will
have
The areas of the rectangles are obtained by multiplying each yk
Summing, we
by the base
A*=
(Fig.
39).
get the area of the step-like figure
Using the formula
sum
for the
of the squares of integers,
2>
n(n+\)(2n+\) 6
Fig. 39
= x*
t
the
140
we
[Ch.
Definite Integrals
5
find
= g'n(n-l)(2n-l) 6H*
S and, passing
'to
the limit,
S-
lim n -*
w
'
we obtain >
Sn =
lim n
-* co
(-D(2-l) = al 6n
>
Evaluate the following definite integrals, regarding them as the limits of appropriate integral sums: b
1501.
1502.
i
1503.
dx.
x*dx.
a
-2
T
10
1504.
J(0 t +g*)
J2*dx. S
and g are constant.
1505*.
8
j
x dx.
i
1506*. Find the area of a the hyperbola
curvilinear
by two ordinates: x = a and x = b 1507*. Find
trapezoid
(0
bounded
by
and the x-axis.
X
=
sintdt.
Sec. 2. Evaluating Definite Integrals by Means of Indefinite Integrals
1. A definite integral with variable upper limit. continuous on an interval [a, b], then the function
If
a is
the aritiderivative of the function F' (x)
= f (x)
2. The Newton-Leibniz formula. o
f (x)\
that
is,
a<*<6. F' (*) = /(*)
for If
th ^n
a
function
f (t)
is
Sec. 2]
_
Evaluating Definite Integrals by Indefinite Integrals
The antiderivative F
Example
1.
(x)
is
computed by finding the
_
indefinite integral
Find the integral
Find
Find the derivatives
of the following functions:
X
\={\ntdt 1510.
(
f(jc)=Td/.
1512.
/
*
1513. Find the points of the
extremum
of the function
X
y
= j!lild/
in the region
Applying the Newton-Leibniz formula,
l^~-
1515.
find the integrals:
X
1
1514.
*>0.
1516.
X
1
f
Jdt.
J
~*
^-. *
1517.
J*cos/<#. '
-.
Using definite integrals, find the limits of the sums:
1520.
lim
n-
Rl
[Ch.
Definite Integrals
142
Evaluate the integrals: 8
A
.
2*
J(jt*
P
+ 3)d*.
1522.
dx
I Jl
4
1523. cos* a da.
1536. Jl
1524.
2d*.
JK* 2
2
1537.
sin'cpdcp.
* ]'
1 1539. Jl
JF
4
1540.
1
71
4~ l
iL 8
***
1541.
f
dx
JF=3F 1831. 1543. .L
coshA:dA;.
Ins
4
1532.
\ j
2
sec a da.
f*
*'
rfjt
5 J cosh !' In
.
J
sinh*xdjc.
5
Sec. 3]
_ _ Improper Integrals
Sec. 3. Improper
143
Integrals
1. Integrals of unbounded functions. If a function f (x) is not bounded any neighbourhood of a point c of an interval [a, b] and is continuous and c
a<*
t
C-B
b
(f(x)dx =
J a
lim C
^a
b
f(x)dx
+e
C
lim
f(x)dx.
(1)
-*%+e
are finite, the improper inteIf the limits' on the right side of (1) exist and a or c thj b, gral is called convergent, otherwise it is divergent. When c definition is correspondingly simplified. If there is a continuous function F (x) on [a, b] such that F'(x) f(x)
=
=
=
when x
then
c (generalized antiderivative),
*
F(a).
If
a<*<6
|/(*)|
and
f
(2)
F(*)dx converges, then
the
in-
a
tegral (1) also converges (comparison test).
and
lim
X-+C
when x-+
then
c,
for
1)
f
(x)
\
x m
c
\
oo,
oo,
A
^
0,
i.
e.,
f(x)~
the integral (1)
1
If
converges,
the function / (x)
2) for
*
the
when
continuous
is
^ I
m^>l
then we assume
\f(x)dx= J
lim b ->
oo
(3)
\f(x)dx J
and depending on whether there the respective integral Similarly,
is
is a finite limit or not on the called convergent or divergent.
b
lim
f(x)dx=
a -^
I'
C \
m<
integral (1) diverges. 2. Integrals with infinite limits.
<
^A
I/WK^W
(f(x)dx
f
J
oo
f(x)dx=
)
a
and the
integral
of (3),
b
oo
and
right
[p(x)dx
lim a->
oo
converges,
\f(x)dx. J
then
the
infe-
a gral (3) converges as well. If
/W^rO
oo,
then
1)
gral (3) diverges.
and for
lim
m>
1
f (x)
xm
= Ajt
the integral
(3)
t
A^Q,
converges,
i.e., 2) for
/(x)~-4 when
m
the inte-
144
__ Definite Integrals
Example
&_
f 2 J x
[C/i.
5
1.
lim
M
f 4? + e-*oJ* "f"* *2
e -K> J
-i
=
lim
lim (I-lU e-*oV (l-l e
e->oVe
/
e
and the integral diverges. Example 2. 6
QO
= Example
3.
lim lim (*rT^2= 2 &->< ft-oo J
(arc tan 6
1+^
arc
tan
())
=
.
2
Test the convergence of the probability integral
(4) *r
Solution.
We
put C
00
1
00
e~ x
*
dx=
C
2
dv+
C
e~ x dx. *
1
The first of the two integrals on the right 2 the second one converges, since e~x
dx=
not an improper integral, while
is
lim
when
(
x^\
e~ b
+ e~
and l
)=e~
hence, the integral (4) converges. Example 4. Test the following integral for convergence:
w r j
^^
y
J?
'
Solution.
When x -++&>, we have
Since the integral
converges, our integral (5) likewise converges. Example 5. Test for convergence the elliptic integral
dx
l
'
t
Sec. 3]
145
Improper Integrals
Solution. The point of discontinuity of the integrand the Lagrange formula we get
where
*<*,
Hence,
for
*-+
I
is
x=l. Applying
we have
V
1
1
/
/T ^
2
\\-xj
1
'
Since the integral 1
i
v dx
ff nr^y converges, the given integral
(6)
converges as well.
Evaluate the improper integrals
(or establish their divergence):
xlnx a
P AY
1552.
-^.
f1
1560.
I a
i
*L a
oo
1553.
J.
1561.
Jcotjcdx
146
_ 00
1562.
kx
\e-
dx
GO
f -arc tan*
_
Definite Integrals
(*>0).
^
d*-
1565.
1566
(Ch.
5
-
1564.
Test the convergence of the following integrals:
-
100
1567.
f
5
j)
.-
**..,_
/*+2 */*+*
1
.
1571.
C
J
./*
..
^/l-jc
1570.
1574*. Prove that the Euler integral of the
fiist
kind
(beta-
function)
B(P, q)=* converges when
p>0
and
q>0.
1575*. Prove that the Euler integral of the second kind (gam-
ma-function)
converges for
p>Q.
Sec. 4. Change of Variable in a Definite Integral
a
If a function f (x) is continuous over and *=-q>(0 It a function continuous together with its derivative cp' (t) over a<^<6, where a=*9(a) and &=cp(P), and /[
Sec. 4]
of Variable in
Change
a Definite Integral
147
then
Example
1.
Find x*
We
Solution.
V~a*x*dx
(a
> 0).
put
asin
t\
dt.
Then
P=
?
arc sin
= arc sin l
=y
and, Therefore,
.
we
consequently,
we
shall
can
take
a
= arcsinO = 0,
have
IL a
C x2
2
VV
x2
=a
4
a2
dx=\
sin
2
2 1
a 2 sin 2
J/~a
jl
JL
2
Z
f sin
2
/
cos
2
/
d/ =
1
a cos
t
dt
IL t 2
f sin 2/
d/ =
-^-
C -^-
(1
Jtfl
1576.
Can the substitution
Transform the following indicated substitutions:
A:
= COS/
definite
be
made
integrals
in the integral
by means
of
the
8
1577.
f(x)dx,
.
x
= arc tan/.
i
1578
.
ly=,
1581. For the integral b
148
__ Definite Integrals
[Ch.
5
indicate an integral linear substitution
which the limits
as a result of
of integration
would be
and
1,
respectively.
Applying the indicated
evaluate
substitutions,
the
following
integrals:
1583. a
1584.
x \Ve -\dx,
n
o
n 2
1586.
Evaluate
the
following
integrals
by
means
of
appropriate
substitutions:
1587.
f^=
A~
i5 8 9.
fi^^d*. *
1590.
x
*
'
2 2
1538.
____
8
J
f J 2*4-
%=.
1^3*+
1
Evaluate the integrals: a
1591.
C J
dx y
xV
x2
1593.
e + 5x4-\
tyax
J
x*dx.
o 271
1594.
fg-f 53 cos x
.
J i
1595. Prove that
o
if
f(x)
is
an even function, then
Integration by Parts
Sec. 5]
But
if
f(x)
1596.
is
Show
an odd function, then
that 00
00
00
~'T=
-00
1597.
Show
149
dx
-
that !L 2
1
arc cos
x
J
1598.
Show
x
o
o
that
T C
f(s\nx)dx =
T f
Sec. 5. Integration by Parts functions If the interval [a, b], then
u
u
(x)
(x) v' (x)
and
dx
Applying the formula
v (x)
are
continuously
= u (x) v(x) for
t;
integration
differentiate on
(*) u' (x) dx.
by
parts,
the
(1)
evaluate the
following integrals: *L 00
2
1599.
^xcosxdx.
1603. oe
e
1600.
Jlnxd*.
1604.
J
1
1
1601.
J*V*dx. o
1602.
Je*sinxdx.
1605. o
e- aJC cos6A:dA:
(fl>0).
__
150
[Ch. 5
Definite Integrals
1606**. Show that for the gamma-function (see the following reduction formula holds true:
From
+ =
T (n n\, \) that for the integral
this derive that
Show
1607.
2
= = [sm /= sin" xdx n
is
1575)
a natural number.
2
n
f
n
if
Example
[
cos
71
the reduction formula n
n ~*
n
holds true.
Find /, if n is a natural Devaluate I 9 and 7 10 1608. Applying repeated
number. Using the formula obtained,
.
integral (see
Example B(p,
<7)
parts,
evaluate the
=
where p and 9 are positive 1609*.
by
integration
1574)
integers.
Express the following integral
in
terms of
B
(beta-
function):
m= if
m
sin
and n are nonnegative
w
x cos
n
integers.
Sec. 6. Mean-Value Theorem
1. Evaluation of
If
integrals.
f(x)^F(x)
^f(x)dx^^F(
66
f(x) and
where
m
is
the interval
the smallest [a,
b]>
for
a<*<6
M
(1)
and,
besides,
(p(*)^0, then
6
j
9
W ^.
(2)
a
a
and
then
X )dx.
mJq>(x)d*
a<*<&,
a
a
If
for
b
b
is
the
largest
value of the function /
(x)
on
Mean-Value Theorem
Sec. 6]
In particular,
if
151
m (ba) <.[f(x)dx*^M
(6
(3>
a).
a
The inequalities
(2)
and
may
(3)
be replaced,
respectively,
by
their equiva-
lent equalities: b
b
a
and b
-a), a
where c and
Example
are certain 1.
numbers lying between a and
b.
Evaluate the integral 5L
M 2
Solution. Since
(Xsin'x^l, we
have
Jl
<
2
2'
that
i/l K 2
'
is,
1.57
2. The mean value
of a function.
1.91.
The number b
is
called the
mean value
of the function / (x)
on the interval
1610*. Determine the signs of the integrals without evaluating
them:
n b)
152
[Ch. b
Definite Integrals
Determine (without evaluating) which of the following
1611. integrals
is
greater: i
i
V\+x*
a)
dx
or
J
dx\
1
1
2
2
x sin x dx
or
e x*dx
or
b) ^
J x sin
2
x dx\
2
c)
\
Find the mean values
of the functions
on the indicated
inter-
vals:
1612. /(x) 1613. f(x) 1614.
=x =a
0
2 ,
f(x)=sm
2
x,
1615.
1616. Prove that ( J o
dx lies
r
1/2+^
Find the exact value
0.70.
between
^2
of this integral.
Evaluate the integrals: n 4
4
1617.
+ jfdx.
1620*. JT
+
1618
.
2
1
r
djf
J gqpjj.
2JI
1619.
f
1622. Integrating by parts, prove that 200JI
10071
43
0.67 and -4= V^2
The Areas
Sec. 7]
of
153
Plane Figures
Sec. 7. The Areas of Plane Figures
1. Area in rectangular coordinates. If a continuous curve is defined in f(x) [f(x)^Q], the area of the rectangular coordinates by the equation y curvilinear trapezoid bounded by this curve, by two vertical lines at the
a Fig.
X
b
40
=
a and x 6 and points x is given by the formula
by
a
segment
of
the jc-axis
(Fig. 40),
b (1)
Example straight
1.
lines x
Compute
=
\
and
.v
the 3,
area bounded by the parabola and the x-axis (Fig. 41).
b
y-fi(*) Fig. 42
Solution. The sought-for area
Fig. 43
is
expressed by the integral
y
X = -~ ^
X
the
154
Definite Integrals
Example the
2.
Evaluate the area
(Ch.
bounded by the curve # = 2
y*
*/
5
and
/-axis (Fig. 42).
Solution. Here, the roles of the coordinate sought-for area is expressed by the integral
axes
are changed
and so the
=
where the limits of integration y l 2 and i/ 2 =l are found as the ordinates of the points of intersection of the curve with the t/-axis,
X
a Fig. 44
In the
curves
/
Fig. 45
more general case, if the area S is bounded by two continuous and y = f 2 (x) and by two vertical lines x = a and x = b, where
= M*)
/i(*XM*) wn
n
a<*<&
(Fig. 43),
we
will then have:
(2)
Example
3.
Evaluate the area 5 contained between the curves
y
=
If
set 1
=2 of
x 2 and
l /
equations
=JC 2
(3)
(3)
simultaneously, we find the of formula (2), we obtain
and # 2 =1. By virtue
=
=
the curve is defined by equations in parametric form x q>(/), y y(t\ the area of the curvilinear trapezoid bounded by this curve, by two
then
The Areas
Sec. 7]
a and #=&), and
vertical lines (x integral
Plane Figures
of
by a segment
155
of the
*-axis
is
expressed
by the
f, are determined from the equations and 6 =
where a
and
/,
= (p(/j)
equations
Due
Solution.
quadrant and
we /
2
xc
y l
parametric
= a cos = bsint. c
/,
i
it is sufficient to compute the area of a the result by four. If in the equation jc aco
symmetry,
multiply
and then x
put x
first
= 0.
to the
then
I
\
its
a,
we
get the limits of integration
/,
= = y and
Therefore, ji
2
-i-S
=
(b
sin
a
(
sin /)
dt=ab(
sin
2 /
dt
=^
S nab. 2. The area in polar coordinates.
and, hence,
If a curve is defined in polar coordinates by the equation r~f (
Fig. 47
Fig. 46
which
correspond
to
the
values
cp,
=a
and
cp,=
p,
is
expressed
by
the
integral
Example
5.
Find
(Fig. 47).
the
area
contained
inside
Bernoulli's
lemniscate
156
__
Definite Integrals
Solution.
By
virtue of the
[Ch. 5
curve we
of the
symmetry
quadrant of the sought-for area:
Whence S = a2
_ determine
first
one
.
=
z
1623. Compute the area bounded by the parabola y 4x x and the x-axis. 1624. Compute the area bounded by the curve y \nx, the e. ;t-axis and the straight line x x (x 1625*. Find the area bounded by the curve y 2) 1) (* and the x-axis. 1626. Find the area bounded by the curve y* x, the straight 8. line y=l and the vertical line x 1627. Compute the area bounded by a single half-wave of the and the Jt-axis. sinusoidal curve 1628. Compute the area contained between the curve y ianx,
=
=
=
=
y=smx
=
the x-axis and the straight
line
x=~
.
=
1629. Find the area contained between the hyperbola xy m*, and x 3a (a>0) and the x-axis. the vertical lines 1630. Find the area contained between the witch of Agnesi
=
x^a
u= x- + a
y
2
and the x-axis.
2
1631. Compute the area of the figure bounded by the curve 8 and the y-axis. the straight line y 1632. Find the area bounded by the parabolas y**=2px and
=
y~x*, x
2
= 2py.
1633. Evaluate the area
and the straight line f/ = 1634. Compute the area
bounded by the parabola y = 2x
x*
x.
of a segment cut off by the straight 2 2x from the parabola y = x 1635. Compute the area contained between the parabolas y*=*x*,
line
y=3
.
*/=Y and 1636.
the straight line y
Compute the area contained
and y = 4
y=
1637.
Agnesi
1638.
between
the
parabolas
2
|x
.
Compute the area contained z between the witch
y=
and the
= 2x.
1
and the parabola
x
f^.
Compute the area bounded by the curves y**e*> straight line
x=l.
of
_
The Areas
Sec. 7\
of
Plane Figures
1639. Find the area of the figure
~
= f2
and the straight
l
__
157
bounded by the hyperbola
linex = 2a.
1640*. Find the entire area bounded by the astroid
1641. Find the area between the catenary
y = a cosh the (/-axis and the straight line
,
= ~(e + 2
f/
!)
=
x 1642. Find the area bounded by the curve a*y* 1643. Compute the area contained within the curve
2
(a*
2 ;c ).
~
1644. Find the area between the equilateral hyperbola x* y* the x-axis and the diameter passing through the point (5,4).
= 9,
1645. Find
the
and the ordinate Find
1646*.
between
area
x=l
the curve y
(x>l). bounded
area
the
the
x-axis, 9
the by J
asymptote x = 2a (a>0). 1647*. Find the area between the strophoid
and
~i,
2
cissoid
r/ y
X = 2ax o
its
y*
=x
(
x ~~ a)2
2Jfl^~~
X
and
asymptote (a>0).
its
1648. circle
and
2
jt
the area of the two parts into which divided by the parabola if =2x. 2 2 the area contained between the circle x y
the
Compute
+
2
r/
=--8 is
1649. Compute the parabola x*=l2(y 1). 1650. Find the area contained within the astroid jc
1651. Find the cycloid
the
= acos
8
/;
y= b
sin
+
= 16
8
/.
bounded by the x-axis and one arc
area
f
\
x = a(t # = a(l
of
sin/), cos/).
1652. Find the area bounded by one branch of the trochoid
tefr***/and a tangent
to
it
at its lower points.
158
5
(Ch.
Definite Integrals
1653. Find the area bounded by the cardioid cos2/),
--
sin20.
1654*. Find the area of the loop of the folium of Descartes
*-r+T"
*-
a (1-f coscp). 1655*. Find the entire area of the cardioid r 1656*. Find the area contained between the first and second turns of Archimedes' spiral, r acp
=
(Fig. 48).
1657. Find the area of one of the leaves of the curve r acos2cp. 1658. Find the entire area bound2 2 a sin4cp. ed by the curve r 1659*. Find the area bounded by the curve r asin3cp. 1660. Find the area bounded by Pascal's limagon
=
=
=
Fig. 48
1661.
Find
the
= 2 + cos
r
area
and the two half-lines
cp.
bounded by the parabola
9=4-
and 9
1662. Fin'd the area of the ellipse r
= y. r = -r 1
+ e cos
1663. Find the area bounded by the curve
=
r
r
=a
sec
2
^
(e
cp
= 2acos3cp
a. lying outside the circle r 1664*. Find the area bounded by the curve x*
and
+ y* = x* + y*.
Sec. 8. The Arc Length of a Curve
1. The arc length
in rectangular coordinates.
y=f(x) contained between two
The arc length s of = a and x~b
points with abscissas x
2 8
2 8
a
curve
is
' ' Example I. Find the length of the astroid * +/'*-= a (Fig. 49). Solution. Differentiating the equation of the astroid, we get
The Arc Length
Sec. 8]
For this reason, we have
of a
Curve
159
for th* arc length of a quarter of the astroid:
Whence s = 6a. 2. The arc
length of a curve represented parametrically. If a curve is
=
s
where
t
l
and
t2
=
"
are values of the parameter that correspond to the extremities
of the arc.
49
Fig
Example
2.
Fig. 50
Find the length
one arc
of
=
x a(t [ \ j/=:a(l Solution.
We
have
of the
cos/)-
cosf) and -^
-^=a(l
= asin/.
/=2a The limits of integration of the arc of the cycloid. If a curve is defined the arc length s is
^=
and
*
2
cycloid (Fig. 50)
sin/),
= 2ji
by the equation
r
sln~-d/=
C
correspond
= /(cp)
Therefore,
to
in polar
the
extreme points
coordinates, then
P
where a and p are the values the arc.
of
the
polar
angle
at
the extreme points of
160
[Ch.
Definite Integrals
Example
The
3.
Find the
entire curve
is
the
length of
entire
described by a point as
cp
curve
r
=
asin-|-
ranges from
5
(Fig. 51).
to 3ji.
Fig. 51
Solution.
She curve
We
have
r'
= a sin
2
-^- cos
,
o
o
therefore the entire arc length of
is
8JI
s=
J J/a*
1665.
y*
8JI
= x*
sin
-|
+o
Compute the
cos'
sin*
-f-
-|-
arc
length
of
=a d
sin
j
2
-f-
rfq>
=^
semicubical
from the coordinate origin to the point x = 1666*. Find the length of the catenary y =
A
1667.
x=0
to
Compute
B(b,h). the arc length of
x=l.
the parabola y
curve y
of
the
of
the curve y
*=K8.
*=
1670. Find the to
1671.
between
length of the curve y
arc
to
= e*
= lnx
= 2}/"x
from
lying between
from x
= /3
= arc sin (e~*)
from
= In secy,
lying
jc=l.
Compute the t/
=
and
j/
=
arc length -5-
of the
curve x
.
o
1672. Find the arc length of the curve x
=1
from the
(6,a) to the point
1668. Find the arc length the points (0,1) and (l,e). 1669. Find the arc length to
parabola
4.
acosh-^-
vertex
.
= e.
= ^-y
2
-^\ny
from
Volumes
Sec, 9]
of Solids
161
1673. Find the length of the right branch of the tractrix
a+vV-t/
2
from y = a to
y
t/=&(0<6
1674. Find the length of the closed part of the- curve
= x(x
a
3a)
.
1675. Find the length of the curve to
t/
= ln ( coth ~-J
from
x=b (0
=
to
/
= 7\
1677. Find the length of the evolute of the ellipse
1678. Find the length of the curve
x=- a (2 cos/ y = a(2 sin/ 1679. Find the length of the
cos 2/)
2
-y
which
is
cut
off
I
/
first
1680. Find the entire length of 1681. Find the arc length of r = asec
f
sin 2/).
turn of Archimedes' spiral
the cardioid r that part of
by a vertical
line
= a(l + coscp). the
passing
parabola
through
the pole. 1682. Find the length of the hyperbolic spiral rq>= 1 from the point (2,'/ 2 ) to the point C/,,2). ae m v, 1683. Find the arc length of the logarithmic spiral r r a. lying inside the circle
=
=
1684. Find
r=l
to
the arc
length
r = 3.
of
the curve
= -g-
(r
+
from j
'
Sec. 9. Volumes of Solids
.
1. The volume of a solid of revolution. The volumes of solids formed by the revolution of a curvilinear trapezoid [bounded by the curve y&f (x) fhe AT- ax is and two a and x vertical lines x b\ about the x- and '(/-axes are J
=
6
-
1900
t
162
Definite Integrals
[Cft.
5
expressed, respectively, by the formulas: b
b
Vx =ji
1)
J a
y*dx\
2)
V Y =2n
J
xr/dx*).
a
Example 1. Compute the volumes of solids formed by the revolution of a bounded by a single lobe of the sinusoidal curve # = sinx and by the segment O<;*
Solution.
a)
V^^-ji
b)
Vy=2n
\
xsinxdx=2jt(
xcosx + sinx)Jc ~2ji t
.
The volume of a solid formed by revolution about the t/-axis of a figure bounded by the curve x=g(y), the (/-axis and by two parallel lines y = c and t/
= d,
may
be determined from the formula
from formula
obtained x and y.
(1),
given
above, by
interchanging
the
coordinates
If the curve is defined in a different form (parametrically, in polar coordinates, etc.), then in the foregoing formulas we must change the variable of integration in appropriate fashion. In the more general case, the volumes of solids formed by the revolution about the x- and (/-axes of a figure bounded by the curves /! /! (x) and y 2 f z (x) a and x b are, respectively, [where f\(x)^f z (x)] and the straight lines x equal to
=
t
= =
a
and b
Example circle x*
2.
+ (y
Find the volume of a torus formed by 2 a 2 (6^a) about the x-axis (Fig. 52).
&)
=
the rotation of the
*) The solid is formed by the revolution, about the (/-axis, of a curvilinear a, x trapezoid bounded by the curve y b, f(x) and the straight lines x and 0. For a volume element we take the volume of that part of the solid formed by revolving about the 0-axis a rectangle with sides y and dx at a distance x from the (/-axis, Then the volume element dVy=2nxydx, whence
=
=
b
V K =2JI
{
xydx.
=
=
Volumes
Sec. 9]
Solution.
We
of Solids
163'
have I/,
=6
I^a
2
x 9 and y 2
= t>+ Va*x
2
Therefore,
(the latter integral
is
taken by the substitution
x=asi
K
-a
x a Fig
52
The volume of a solid obtained by the rotation, about the polar axis, of a sector formed by an arc of the curve r F((p) and by two radius vectors ipr-=a,
=
=
3
Vp = ~ 2
C JT \
r 8 sin cpd
q>.
a
This same formula is conveniently used when seeking the volume obtained by the rotation, about the polar axis, of some closed curve defined in polar coordinates.
Example
r
= asin2(p
3.
Determine the volume formed
by the rotation
about the polar axis.
Solution.
= 2.--n\
rsinq>d(p
=
8
yJia
C
sin
8
.L 2
= ^ Jia 3
8
f
J
sin
4
cos 8
cp
dcp
=
^
lOo
jia
8 .
2(p sin q> dcp
=
of
the curve
164
__
\Ch. 5
Definite Integrals
2. Computing the volumes of solids from known cross-sections. If S S(x) the cross-sectional area cut off by a plane perpendicular to some straight line (which we take to be the x-axis) at a point with abscissa *, then the volume of the solid is is
where *, and x 2 are the abscissas of the extreme cross-sections of the solid. Example 4. Determine the volume of a wedge cut off a circular cylinder by a plane passing through the diameter of the base and inclined to the base at an angle a. The radius of the base is R (Fig. 53). Solution. For the *-axis we take th? diameter of the base along which the cutting plane intersects the base, and for the (/-axis we take the diameter of the base perpendicular to it. The equation of the circumference of the base is
*2
+ =R 2
j/
The area
2 .
of the section
ABC
at
a distance
x from the origin
is
2
S(x) = area A ABC = -^ ABBC = -^yy tana =^- tana. 1
1
r/
Therefore, the sought-
,
for
volume
of the
wedge
is
R
R
y=
2~
f y2
tanad*=tana
(*
(R
1
1 *)<& = y tana R
.
1685. Find the volume of a solid formed by rotation, about the x-axis, of an area bounded by the x-axis and the parabola 1686. Find the volume of an ellipsoid formed by the rotation of the ellipse ^r
+| =
1687. Find the
the x-axis', of
-
8
l
volume
about the x-axis.
formed by the rotation, about an area bounded by the catenary y = acosh the of a solid
xa.
,
and the straight lines 1688. Find the volume of a solid formed by the rotation, about a the x-axis, of the curve j/=sin x in the interval between x xr-axis,
=
and x = n. 1689. Find the volume of a solid formed by the rotation, about the x-axis, of an area bounded by the semicubical parabola if = x s the x-axis, and the straight line x== 1. 1690. Find the volume of a solid formed by the rotation of the same area (as in Problem 1689) about the {/-axis. 1691. Find-, the volumes of the solids formed by the rotation of an area bounded by the lines y = e*, x = 0, y = about: a) the x-axis and b) the y-axis. 1692. Find the volume of a solid formed by the rotation, about the t/-axis, of that part of the parabola j/ 2 = 4ax which is cut off by the straight line x = a. ,
Sec. 9]
_ _ Volumes
of Solids
165
1693. Find the volume of a solid formed by the rotation, about the straight line x a, of that part of the parabola y*=4ax which is cut oft by this line. 1694. Find the volume of a solid formed by the rotation, about the straight line y p, of a figure bounded by the parabola
=
=
2 t/
= 2p*
and the straight
line
*
= --
.
1695. Find the volume of a solid formed by the rotation, about the x-axis of the area contained between the parabolas y x* L
=
and
y= Y*.
1696. Find the volume of a solid formed by the rotation, about the x-axis, of a loop of the curve (* 4a)tf =ax(x 3j)
1697. Find the of the cyssoid y*
volume A8
= ^ _x
of a solid
about
its
generated
by the rotation
asymptote x = 2a.
1698. Find the volume of a paraboloid of revolution whose base has radius R and whose altitude is //. 1699. A right parabolic segment whose base is 2a and altitude h is in rotation about the base. Determine the volume of the resulting solid of revolution (Cavalieri's "lemon"). 1700. Show that the volume of a part cut by the plane jc 2a off a solid formed by the rotation of the equilateral hyperbola x* tf^c? about the *-axis is equal to the volume of a sphere of radius a. 1701. Find the volume of a solid formed by the rotation of a bounded by one arc of the cycloid x=-a (/ sin t), figure y=^ a (\ cos/) and the x-axis, about: a) the x-axis, b) the y-axis, and c) the axis of symmetry ot the figure. 1702. Find the volume of a solid formed by the rotation of acos 8 /, y bsm*t about the //-axis. the astroid * 1703. Find the volume of a solid obtained by rotating the cardioid r a(l -hcostp) about the polar axis. 1704. Find the volume of a solid formed by rotation of the acos 2
=
=
=
=
=
+
166
of
[Ch. 5
Definite Integrals
1708. A circle undergoing deformation the points of its circumference lies on
moving so that one the y-axis, the centre
is
+
describes an ellipse ^^-=1, and the plane of the circle is perpendicular to the jq/-plane. Find the volume of the solid generated by the circle. 1709. The plane of a moving triangle remains perpendicular to the stationary diameter of a circle of radius a. The base of the triangle is a chord of the circle, while its vertex slides along a straight line parallel to the stationary diameter at a distance h from the plane of the circle. Find the volume of the solid (called a conoid) formed by the motion of this triangle from one end of the diameter to the other. 1710. Find the volume of the solid bounded by the cylinders z 2 z* a*. a and y 1711. Find the volume of the segment cut off from the ellip"'
+ =
=
u
2
2
= * by the plane x = a. paraboloid |21712. Find the volume of the solid bounded by the hyperbo-
tic
loid of
+
z
^ -f
one sheet
1713. Find the
^-=1 and
rj
volume
the planes 2 X2
of the ellipsoid ^2
=
U2
Z
and z
li.
+ ^ + ^2"=
!
10. The Area of a Surface of Revolution
Sec.
The area
of
formed by the rotation, about a and x f(x) between the points x
a surface
arc of the curve y the formula
=
the
= b,
is
an expressed by
x-axis, of
b
b
Vl+y'*dx (ds
=
2
is
(1)
the differential ol the arc of the curve).
Zfta Fig. 54
If
surface
of the curve cbtained from formula
the equation
$x
is
is
(!)
represented differently, the area of the by an appropriate change of variables.
Sec.
The Area
10]
of
a Surface of Revolution
Example 1. Find the area of a surface formed by 2 x-axis, of a loop of the curve 9i/ ;t(3 x)* (Fig. 54). Solution. For the upper part of the curve, when
167 rotation, about
=
/
= -- (3
mula
x)
y~x. Whence
the differential of the arc
(X*<3,
ds=
X ~^
r_dx.
2 the area of the surface
(1)
V
the
we have Fromfor-
x
9
= 2n
(
J
Example of
the
2.
\($6
Find the area x
cycloid
=a
of a surface
s\nt)\
(t
2
y = a(l
V
formed by the rotation of one arc cost) about its axis of symmetry
(Fig. 55).
Solution. The desired surface is formed by rotation of the arc OA about the straight line AB, the equation of which is x na. Taking y as the independent variable and noting that the axis of rotation AB is displaced relative to the #-axis a distance na, we
=
will
Y
have
Passing to the variable
(na
/,
we obtain
da
d
y
n =
4na
2
(
(
nsin
/sin
y + sinf sin
~-
Fig. 56
dt-j
The dimensions
of a parabolic mirror AOB are indicated required to find the area of its surface. 1715. Find the area of the surface of a spindle obtained by rotation of a lobe of the sinusoidal curve about the
1714.
in Fig. 56.
It
is
ys'mx
#-axis.
1716. Find the area of the surface formed a part
of
about the
the
tangential curve
t/
= tan*
by the rotation from Jt = to
of
A:==^-,
jc-axis.
1717. Find the area of the surface formed by rotation, about the x-axis, of an arc of the curve y e-* 9 from x to x
=
=
168
\Ch. 5
Definite Integrals
1718. Find the area of the surface (called a catenoid) formed
about the x-axis from by the rotation of a catenary # = acosh x = to x = a. 1719. Find the area of the surface of rotation of the astroid 3 x'/'-i
-a
/'/
3 3 '
about the y-axis.
1720. Find the JC
= -^/
2
y
area
the surface of rotation of the curve
of
y=\
from
In (/about the *-axis
to
y = e.
1721*. Find the surface of a torus formed by rotation of the 2 x* b) ^=a* about the *-axis a). (y 1722. Find the area of the surface formed by rotation of the
+
circle
ellipse
^+^ =
about:
1
1)
the *-axis, 2) the y-axis
(a>6).
1723. Find the area of the surface formed by rotation of one arc of the cycloid x cos t) about: a) the a(t sin/) and r/ a(l x-axis, b) the y-axis, c) the tangent to the cycloid at its highest
=
=
point. 1724. Find the area of the surface formed by rotation, about the j^-axis, of the cardioid
x t/
= a (2 cost = a(2sin/
cos
2/),
\
sin2/).
/
1725. Determine the area of the surface formed by the rotation 2 2 a cos2
of the lemniscate r
=
=
Sec. 11. Moments. Centres of Gravity. Guldin's Theorems
1. Static moment. The static moment relative to the /-axis of a material A having mass m and at a distance d from tha /-axis is the quantitv
point
Mi~md. The
static
moment
relative to the /-axis of a system of n material points in the plane of the axis and at distances
with masses m,, m 2 ..., m n lying d lt d 2 ..., d n is the sum ,
,
n
M i=
m A2 =
/
(1)
!
where the distances of points lying on one side of the /-axis have the plus sign, those on the other side have the minus sign. In a similar manner we
define the static moment of a system of points relative to a plane. If the masses continuously fill the line or figure of the x#-plane, then the static moments about the x- and /-axes are expressed ^respectivex and not as the sums (1). For the cases of ly) as integrals and geometric figures, Itie density is considered equal to unity.
M
My
Sec.
is
Moments. Centres of Gravity. Guldin's Theorems
II]
In particular: the arc length,
for
1)
the curve *
= *(s);
we have
y=y(s) whare t
L
= V(dx)* + (dy)*
is
M Y =.
(s) ds\
x
(s)
2) for a
plane figure bounded by th3 curve y a and y b we obtain
=
bounded by the
Find the
static
straight lines:
Solution. Here, y
(2)
= y(x),
ihz Jt-axis and
two
t
b
= b II
x\y\dx.
(3)
the x- and
/-axes of a triangle
a
a 1.
ds
Fig. 58
b
Example
s
the differential of the arc);
Fig. 57
vertical lines x
parameter
L
y
(ds
the
169
moments about
~-f-^ a b
~
.
)
=
l,
x
= 0, = //
Applying formula
(Fig. 57)
(3),
we
obtain
ab*
and
2. Moment
of inertia. The moment of inertia, about an /-axis, of a imfe2 point of mass m at a distance d from the /-axis, is the number l t -=-tnd . of n material of an a The moment of inertia, about /-axis, system points with masses m lt m 2t ..., n is the sum rial
m
170
__
[Ch. 5
Definite Integrals
where d lf d 2 ..., d n are the distances of the points from the /-axis. In the case of a continuous mass, we get an appropriate integral in place of a sum. Example 2. Find the moment of inertia of a triangle with base b and altitude h about its base. Solution. For the base of the triangle we take the x-axis, for its altitude, the y-axis (Fig 58). Divide the triangle into infinitely narrow horizontal strips of width dy t which play the role of elementary masses dm. Utilizing the similarity of
we
triangles,
obtain j = bL dm
h
y .
n
,
dy
and
Whence
3. Centre of gravity. The figure (arc or area) of mass
M
coordinates of the centre of gravity of a plane are computed from the formulas
moments of the mass. In the case of geometnumerically equal to the corresponding arc or area. For the coordinates of the centre of gravity (x, y) of an arc of the plane curve y connecting the points A[a f (a)] and B [6, f (b)], f (x) (a^x^b), where
MX
ric figures,
and
My
are the static
the mass
M
is
=
t
we have B
B
t>
b (
-A\yds
2 y 1^1 +(y') dx
a
b
S
2
'}
The coordinates
dx
of the centre of gravity (x, y) of the curvilinear trapezoid
x^b, Q^y<^f(x) may
be computed from the formulas
b
~~
X
b
-.(y^dx
^xydx a
*~~
-
O
f
y
a_ ^
,3
b
where
S=f y dx
is
the area of the figure.
a
a
There are similar formulas for the coordinates of the centre of gravitv 6 y of volume. Example 3. Find the centre of gravity of an arc of the semicircle ,,2^n*- i,i^n\ (pig
59)
Sec
Moments. Centres
11]
We
Solution.
Guldin's Theorems
of Gravity
171
have
and
Whence
f,
'
yds--
xJ ..
,.
yV-.v
2
^= y a
J
2
Hence,
a
4. Guldin's theorems. Theorem 1. The area of a surface obtained by the plane curve about some axis lying in the. same plane
intersecting
it
circumference of
is
to circle
equal the
rotation of an arc of as the curve and not the product of the length of the curve by the described by the centre of gravity of the arc of
the curve.
Theorem 2. The volume of a solid obtained by rotation of a plane figure about some axis lying in the plane of the figure and not intersecting it is equal to the product of the area of this figure by the circumference of the circle described by the centre of gravity of the figure.
Fig. 59
1727. Find the static moments a segment of the straight line x o
lying between the axes.
y o
about
the
coordinate
axes of
__
172
Definite Integrals
1728. Find the static
about
its
moments
of a rectangle,
[Ch. 6
with sides a and
&,
sides.
1729. Find the static moments, about the x- and t/-axes, and the coordinates of the centre of gravity of a triangle bounded by the straight lines x + y Q. a, x Q, and y 1730. Find the static moments, about the x- and (/-axes, and the coordinates of the centre of gravity of an arc of the astroid
=
=
_t_
*
=
_i^
-If/*
>
==aT
>
lying in the first quadrant. 1731. Find the static moment of the circle r
= 2asin
about the polar axis. 1732. Find the coordinates of the centre of gravity of an arc of the catenary
y = a cosh
~
=
x=
a to x a. Find the centre of gravity of an arc of a circle of radius a subtending an angle 2a. 1734. Find the coordinates of the centre of gravity of the arc of one arch of the cycloid
from
1733.
x
= a(t
sin/); y
= a(\
cos/).
1735. Find the coordinates of the centre of gravity of an area bounded by the ellipse 2 -|-=l and the coordinate axes ( 1736. Find the coordinates bounded by the curves
of the centre of gravity of
1737. Find the coordinates of the centre of gravity bounded by the first arch of the cycloid x = a(t sin/), r/ = a(l cos/) and the jc-axis. 1738**. Find the centre of gravity of a hemisphere lying above the ;q/-plane with centre at the origin.
of
an area
an area
of radius a
1739**. Find the centre of gravity of a homogeneous right circular cone with base radius r and altitude h. 1740**. Find the centre of gravity of a homogeneous hemi-
sphere of radius a origin.
lying
above the jo/-plane with centre
at the
&?c. 12]
Applying Definite Integrals
1741. Find the its
moment
to Solution of Physical
Problems
173
inertia of a circle of radius a about
of
diarneler.
1742. Find the moments of inertia of a rectangle with sides a and b about its sides. 1743. Find the moment of inertia of a right parabolic segment with base 26 and altitude ft about its axis of symmetry. 1744. Find the moments of inertia of the area of the ellipse 2 JC
U2
^ + ^=1
about
its
principal axes.
1745**. Find the polar moment of inertia of a circular ring and R t (R
dicular to its plane. 1746**. Find the moment of inertia of a homogeneous right circular cone with base radius R and altitude H about its axis. 1747**. Find the moment of inertia of a homogeneous sphere about its diameter. of radius a and of mass 1748. Find the surlace and volume of a torus obtained by rotating a circle of radius a about an axis lying in its plane and at a distance b (b>a) from its centre. 1749. a) Determine the position of the centre of gravity of
M
2
t
2
T
an arc of the astroid x -\-i/ T = a* lying in the first quadrant. b) Find the centre of gravity of an area bounded by the curves 2 = 2px and x* = 2py. t/ 17f>0**. a) Find the centre of gravity of a semicircle using Guldin's theorem. b) Prove by Guldin's theorem that the centre of gravity of a triangle is distant from its base by one third of its altitude Sec.
12. Applying Definite
Integrals to the
Solution of Physical Problems
1. The path traversed by a point. If a point is in motion along some curve and the absolute value of the velocity o~/(/) is a known function of the time t, then the path traversed by the point in an interval of time is '. *
Example
1.
The velocity
of a point
o
Find the path ing
the
during
s
= 0. 1/
covered by the point ol motion.
commencement this interval?
8
in
is
m/sec.
7=10
the interval of time sec followis the mean velocity cf motion
What
174
[Ch. 5
Definite Integrals
Solution.
We
have: 10
p
t*
.-Jo.irt-o.i
T = 250
metres
and
=-=25 2. The work of the x-axis,
of a force. then the work
If
m/sec.
a variable force
X=f(x)
acts in the direction
over an interval [x ly x z ]
of this force
is
A= 2. What work has to be performed to stretch a spring 6 cm, if kgf stretches it by 1 cm? Solution, According to Hook's law the force X kgf stretching the spring kx, where k is a proportionality constant. by x m is equal to 100v. we get and l kgf, 100 and, hence, X 0.01 Putting x Whence the sought-for work is
Example
a force of
1
=
X= m
0.08
0.06
A=
=
=
X=
100 x dx
= 50 x
= 0. 18
2
j
kgm
3. Kinetic energy. The kinetic energy of a material velocity v is defined as
point of mass
m
and
mv*
The
mv m
2%
kinetic ...,
mn
energy of a system of n material points with masses having respective velocities t; lf v 2 ..., v n is equal to ,
,
To compute the
kinetic energy of a solid, the latter is appropriately partitioned into elementary particles (which play the part of material points); then by summing the kinetic energies of these particles we get, in the limit, an integral in place of the sum (1). Example 3. Find the kinetic energy of a homogeneous circular cylinder of density 6 with base radius R and altitude h rotating about its axis with angular velocity CD. Solution. For the elementary mass dm we take the mass of a hollow cylinder of altitude h with inner radius r and wall thickness dr (Fig. 60). We have:
Since the linear velocity of kinetic energy is
the
mass dm
is
equal
to
t;
=
/-co,
the elementary
Sec.
12]
Applying Definite Integrals
to
Solution of Physical Problems
175
Whence 2
r 9 dr^=
nco 6/?
4
fc
4. Pressure of a liquid. To compute the force of liquid pressure we use Pascal's law, which states that the force of pressure of a liquid on an area S at a depth of immersion h Is
where y
is
the specific weight of the liquid.
Example 4. Find the force of pressure experienced by r submerged vertically in water so that its diameter is
radius
water surface (Fig
a
semicircle of with the
flush
61).
Solution, We partition the area of the semicircle into elements strips parallel to the surface of the water. The area of one such element (ignoring higher-order infinitesimals) located at a distance h from the surface is ds
The pressure experienced by
where y
is
^ 2xdh = 2 this
V'*
element
h2
dli.
is
the specific weight of the water equal to unity. the entire pressure is
Whence
r
=2
The velocity of a body thrown vertically upwards with velocity V Q (air resistance neglected), is given by the
1751. initial
C J
ITS
Definite Integrals
\Cfi.
5
formula
=
.
ff'.
where t is the time that elapses and g is the acceleration of gravAt what distance from the initial position will the body ity. be in t seconds from the time it is thrown? 1752. The velocity of a body thrown vertically upwards with initial
velocity
v
(air
resistance
allowed
for)
is
given
by the
formula
where
t is the time, g is the acceleration of gravity, and c is a constant. Find the altitude reached by the body. 1753. point on the x-axis performs harmonic oscillations
A
about the coordinate origin;
its
velocity
given by the formula
is
where t is the time and t; co are constants. Find the law of oscillation of a point if when / = it had an abscissa * = 0. What is the mean value of the absolute magnitude of the velocity of the point during one cycle? 1754. The velocity of motion of a point is v = te~"' Qlt m/sec. Find the path covered by the point from the commencement of motion to full stop. ,
1755. A rocket rises vertically upwards. Considering that when the rocket thrust is constant, the acceleration due to decreasing
weight of the rocket
increases by the law
find the velocity at any instant of time /, is zero. Find the altitude reached at lima
/==^~^ if /
=
the
(a
initial
ftf
>0),
velocity
/
r 1756*. Calculate the work that has to be done to pump the water out of a vertical cylindrical barrel with base radius R and altitude H. 1757. Calculate the work that has to be done in order to pump the water out of a conical vessel with vertex downwards, the radius of the base of which is R and the altitude H. 1758. Calculate the work to be done in order to pump water out of a semispherical boiler of radius R 10 m. 1759. Calculate the work needed to pump oil out of a tank
=
through an upper opening (the tank has the shape of a cylinder with horizontal axis) if the specific weight of the oil is y, the length of the tank H and the radius of the base R. 1760**. What work has to be done to raise a body of massm from the earth's surface (radius R) to an altitude ft? What is the work
if
the body
is
removed
to infinity?
Sec.
12]
Applying Definite Integrals
What work *2
=10 cm? 1762**.
and
Problems
177
=
Two
1761**. on the
lie
to Solution of Physical
electric charges * 100 CGSE-and e l==200 CGSE x-axis at points * and *, 1 cm, respectively. will be done if the second charge is moved to point
A
length
p=10kgfcm
=
cylinder with a movable piston of diameter /
= 80cm
2 .
is
filled
with
What work must be done
steam
at
to halve the
a
D = 20 cm pressure
volume of
steam with temperature kept constant (isoihermic process)? 1763**. Determine the work performed in the adiabatic expan3 of air sion and pressure (having initial volume u =l 2 8 10 m ? p _=l kgf/cm ) to volume u, 1764**. A vertical shaft of weight P and radius a rests on a bearing AB (Fig. 62). The frictional force between a small part a of the base of the shaft and the surface of the support in contact with it is F==fipa, const is the pressure of the shaft where p on the surface of the support referred to unit area of the support, while pi is the coefficient of friction. Find the work done by the frictional force during one revolution of the the
=
m
ft
=
shaft.
1765**. Calculate the kinetic energy of a and radius R rotating with angular velocity G> about an axis that passes through its centre perpendicular to its plane. 1766. Calculate the kinetic energy of a right circular cone of mass rotating with angular velocity CD about its axis, if the radius of the base of the cone is R and the altitude is H. 1767*. What work has to be don? to stop an iron sphere of 2 me'res rotating with angular velocity w = 1,000 rpm radius R j about its diameter? (Specific weight of iron, y = 7.8 s/cm .) 1768. A vertical triangle with base 6 and altitude h is submerged vertex downwards in water so that its base is on the surface of the water. Find the pressure of the water. 1769. A vertical dam has the shipa of a trapezoid. Calculate the water pressure on the dam if we know that the upper base 20 m. a 70 m, the lower base 6=50 m, and the height h 1770. Find the pressure of a liquid, whose specific weight is y. on a vertical ellipse (with axes 2a and 26) whose centre is submerged in the liquid to a distance h, while the major axis 2a of the ellipse is parallel to the level of the liquid (h^b). 1771. Find the water pressure on a vertical circular cone with radius of base R and altitude H submerged in walei vertex downwards so that its base is on the surface of the water.
disk of mass
M
M
=
=
=
178
Definite Integrals
[C/i.
5
Miscellaneous Problems
=
1772. Find the mass of a rod of length / 100 cm if the linear density of the rod at a distance x cm from one of its ends is
6
= 2 + 0.001 x
2
g/cm.
1773. According to empirical data the specific thermal capacity of water at a temperature is
= 0.9983 What from
/C (0^/<100) 5.184xlO- + 6.912xlO5
2
7
/
quantity of heat has to be expended to heat to 100 C?
/
.
g
1
0C
of water
2
of of
1774. The wind exerts a uniform pressure p g/cm on a door width b cm and height h cm. Find the moment of the pressure the wind striving to turn the door on its hinges. 1775.
What
is
the
force
of
attraction
of
M
a material
and mass on a material point of mass a straight line with the rod at a distance a from one length
/
rod of
m
lying on of its ends?
1776**. In the case of steady-state laminar (low of a liquid through a pipe of circular cross-section of radius a, the velocity of flow v at a point distant r from the axis of the pipe is given by the formula
difference at the ends of the pipe, |i is is the pressure the coefficient of viscosity, and / is the length of the pipe. Determine the discharge of liquid Q (that is, the quantity of liquid flowjng through a cross-section of the pipe in unit time). 1777*. The conditions are the same as in Problem 1776, but the pipe has a rectangular cross-section, and the base a is great compared with the altitude 26. Here the rate of flow u at a point M(x,y) is defined by the formula
where p
Determine the discharge
of
liquid Q.
1778**. In studies of the dynamic qualities of an automobile, use is frequently made of special types of diagrams: the velocities v are laid off on the Jtr-axis, and the reciprocals of corresponding accelerations a, on the (/-axis. Show that the area S bounded v 2t and v l and v by an arc of this graph, by two ordinates v by the *-axis is numerically equal to the time needed to increase the velocity of motion of the automobile from v l to v 2 (accelera-
=
tion time).
=
Sec. 12]
Applying Definite Integrals
1779.
a
A
downward
of the
horizontal
beam
vertical load
beam, and
of
to Solution of Physical
of length
uniformly
support
Problems
179
/ is in- equilibrium due to distributed over the length
reactions
A and
fi(yl==5==-y-j
t
M
directed vertically upwards. Find the bending moment x in a cross-section x, that is, the moment about the point P with abscissa x of all forces acting on the portion of the beam AP. 1780. A horizontal beam of length / is in equilibrium due to support reactions A and B and a load distributed along the kx, where x is the distance length of the beam with intensity q from the left support and k is a constant factor. Find the bend-
=
ing
moment Note.
Mx
in cross-section x.
The intensity
of
load distribution
is
the
load
(force) referred to
unit length.
1781*. Find the sinusoidal current
during a cycle
T
quantity
of heat released
in a conductor
by an alternating
with resistance
/?.
Chapter VI
FUNCTIONS OF SEVERAL VARIABLES
Sec. 1. Faiic Notions
1. The concept of a function of several variables. Functional notation. variable quantity 2 is called a single-valued function of two variables jc, y, if to each set of their values (x, //) in a givm range there corresponds a unique value of z The variables x and y are called arguments or independent variables. The functional relation is denoted by
A
*
Similarly,
we
Fxample
= /(*,
y).
define functions of three or more arguments. Express the volume of a cone V as a function of
1.
eratrix x and of its base radius y Solution. From geometry we know that the
where h
is
the altitude of the cone. But h
This
is
y
volume
**
y
2 -
of a
cone
its
is
Hence,
the desired functional relation. of the function z^f(x.y)
The value point is
P
that
(a.b).
is,
gen-
when
x=^-a
and y
at a bt
denoted by / (a,b) or f (P) Generally speakthe geometric representation of a func-
ing,
tion like
z
f
in
(x,y)
nate system X, Y. Z
Example
2.
Find/
is
(2,
Fig. 63
Solution.
Substituting
r=2
and t/=
3,
we
find
a
rectangular
a surface (Fig.
3)
and/1,
coordi63). if
Sec.
Basic Notions
1)
Putting
*=1
~
and replacing y by
,
we
[SI
have
will
1
SL\
*' thai
is,
/(l.
2. Domain
)=f(*.
0).
of definition of a function. By the domain of definition of a function ? f(x, y) we understand a set of points (*, r/) in an jq/-plane in which the given function is defined (that is to say, in which it takes on definite real values) In the simplest cases, the domain of definition of a function U a finite or infinite part of the jo/-plane bounded by one or several curves (the boundan, of the domain).
=
Similarly, for a function of three variables u f(x, y, z) the definition of the function is a volume in At/z-space. Example 3. Find the domain of definition of the function
domain
of
1
>
<
2 x2 function has real values if 4 or x* + y 2 4. y inequality is satisfied hy the coordinates of points lying inside a circle of radius 2 with centre at the coordinate origin. The domain of definition oi the function is the interior of the circle (Fig 64).
Solution. The
The
latter
Fig. 64
Example
4.
Fiy
Find the domain z
Solution.
The
first
term
of definition of the function
= arc sin
or |
function
*
when
+
|/
xy
of the function
The second term has
when
65
real
defined for
is
values
.The domain
if
xr/^O,
of
1<~ in
i.e.,
definition
of
or
two
cases:
the
entire
|
is
shown
in Fig. 65
and includes the boundaries
of the
domain.
Functions
182
of Several Variables
[Ch. 6
3. Level lines and level surfaces of a function. The level line of a function 2 f(x, y) is a line / (*, y)-C (in an *r/-plane) at the points of which C (usually labelled in the function takes on one and the same value z
=
drawings).
The iace
level surface of a C, at the z)
=
/ (x, y,
function of three arguments u~f(x, points of which the function takes value u~C. Example 5. Construct the the function z x*y. Solution. The equation of
=
has the form x 2 y Putting
C = 0,
1784. Find /(1/2, 3), /(I, -1),
z
__x
y
i
2,
~ -j ....
level
lines
of
the level lines
.
we
get a family
1782. Express the volume V of a regular tetragonal pyramid as a function of its altitude x and lateral edge y. 1783. Express the lateral surface S of a regular hexagonal truncated pyramid as a function of the sides x and y of the bases and the altitude z.
Fig. 66
Find
1,
or y
is a sura constant
z)
t
on
lines (Fig. 66).
of level
1785
=C
y
if
1
x,
f(
f(y,x),
y),
2
1
,
if
f(*,y)
1786. Find the values assumed by the function
at points of the
parabola y
=
z ,
and construct the graph
function
1787. Find the value of the function Z 2
at points of the circle x 1788*. Determine f(x),
1789*. Find f(x, y)
if
rrr
-
+y*=R if
2 .
of the
Sec.
__ Basic Notions
1]
183
= V7+/(l/x 1). Determine the functions / y= 1. Letz = */(-V Determine the functions f and
1790*. Let * z
if
z
=x
and
when
1791**.
z
if
when x=\. 1792. Find and sketch lowing functions: a)
b)
z==Y\ x z=\-\-V
2
domains
the
of
definition of the fol-
2
y
\
(x
= 2 - x + arc cos z-l/l-^ + z = arc sin X 2 = V 7IT 2 =
c) z
d)
//;
2
e) f) '
;
r
g) h)
1793.
Find the domains
of
the following functions of three
arguments: a)
u=-\x +
b) u
y
+
z\
= ln(xyz):
c)
w=--arc sin jc+arc sin #
d)
u = V\x
2
2
y
+ arc sinz;
z\
1794. Construct the level lines of the given functions and determine the character of the surfaces depicted by these functions: a)
z^x + y;
b) 2
= ^+y
c) z
= x*-y*;
d) z
=
z
=
2 ;
e)
O^
1795. Find the level lines of the following functions: a) z b) 2
= ln(*'+f/); = ar
d) z
= /(y
a*);
1796. Find the level surfaces of the functions pendent variables: x y-\-z; a) u * ' ' b) u = c) u = J
= +
of three inde-
_ _
[Ch. 6
Functions of Several Variables
184
Sec. 2. Continuity
1. The limit of a function.
= /(*, y) there e > 2
is
P
A number A
is
1
called the limit of a function
approaches the point P (o. is a 6 > such that when < Q < 6, where Q = |/\x the distance between P and P', we have the inequality as
the
point
(x, y)
I
ft),
if
a)*
for
+ U/
any 6)
1
y)A\
/(*,
In this case we write lim X-K2 y-*b
2. Continuity and continuous at a point
points of discontinuity.
P
(a,
b)
A
function z=f(x, y)
is
called
if
lim f(x x-+a
9
0)
= /(a,
b).
A function that is continuous at all points of a given range is called continuous over this range A function /(AT, y) may cease to be continuous either at separate points (isolated point of discontinuity) or at points that form one or several lines (lines of discontinuity) or (at times) more complex geometric objects. Example 1. Find the discontinuities of the function
_*+> Solution. zero. But **
The function -r/
function has for
=
its
or y
will be meaningless if the denominator becomes 2 x is the equation of a parabola. Hence, the given
discontinuity the parabola y
x2
.
1797*. Find the following limits of functions: a)
c)li n
liin(*?^)sinl;
M;
e)lin-L;
1798. Test the following function for continuity: ft x
/)
= / V\x* I
if
when * 2 -r-// 2
1799. Find points of discontinuity of the a) z
=
ln;
c)
e-
functions:
Sec. 3]
Partial Derivatives
1800*.
Show
that the function
SL
when when
-{ is
185
** jt
+
=^=0,
= =
continuous with respect to each of the variables x and y sepabut is not continuous at the point (0, 0) with respect to
rately,
these variables together. Sec. 3. Partial Derivatives
t. Definition of a partial derivative. example, y constant, we get the derivative
If
f(x,y), then assuming, for
z
which
is called the partial derivative of the function z with respect to the variable x. In similar fashion we define and denote the partial derivative of the function z with respect to the variable y It is obvious that to find partial derivatives, one can use the ordinary formulas of differentiation. Example 1. Find the partial derivatives of the function
y
we
Solution. Regarding y as constant,
dz
1
1
get
=
1 t
dx
,
tan
x
,
cos 2
dz
we
2.
Find the
2.
x
2*
y
partial derivatives of the following function of three
a
= *y z + 2*
3y
+ 2 + 5.
?
Euler's theorem. A function if for every real factor k
degree n
2x'
y
/
1
arguments:
Solution.
.
I/sin y
have
will
y
Example
y
y
y
Similarly, holding x constant,
x
f (x,
y) fs called a
we have
f (kx, ky) --=
ttf
homogeneous function of
the equality (x,
//)
A
[Ch. 6
Functions of Several Variables
186
rational integral function will be
and the same degree. The following relationship holds of degree n (Euler's theorem): xfx (x, y)
Find the 1801. z
=
homogeneous for a
if
all
homogeneous
its
terms are of one
differentiable function
+ yfy (x, y) = nf (x, y).
partial derivatives of the following functions: 1808. 2 3axy.
=
1809. 2
=
1810. 2
=
1811. 2
=
1806.
1807. z
= arc tan
1814. Find /;(2,
and fy (2,
1)
1815. Find /;(!, 2, 0),
/i(l, 2, 0),
/(*, y,z)
Verify Euler's theorem ples 1816 to 1819: 1816. f(x,y) 1817. z
if
1)
xy
f(x,y)
ft
(l, 2, 0)
+L
.
if
= ln(xy-\-z).
on homogeneous functions
= Ax + 2Bxy-Cy 3
t
in
Exam-
1818. /
.
= J-
1820.
)
,
where
r
=
:
2 1821. Calculate
1822.
Show
that
1823.
Show
that
1824.
Show
that
1825.
Show
J826. Find
that
and y 57,
= 2,
if
x+
,
g+fj+S-0,
g+g+g-1,
-.(,.
y),
2
it
if
= ln(*'
= u = (x-y)(y-z)(z- X
if
2
).
if
-*+J=J.
Sec. 4]
_
Total Differential of a Function
1827. Find z--^z(x y)
knowing that
t
(J2
=
X 2 -4-
2
= slny
J/ ^and z(x, y)
_
187
when *=1.
Through the point M(l,2, 6) of a surface z = 2x*+y* drawn planes parallel to the coordinate surfaces XOZ and YOZ. Determine the angles formed with the coordinate axes by the tangent lines (to the resulting cross-sections) drawn at their common point M. 1829. The area of a trapezoid with bases a and b and alti1828.
are
tude h
is
equal
to
S=
l
/,(fl
+ &) A.
Find
g, g,
g
and,
using
the drawing, determine their geometrical meaning. 1830*. Show that the function
0,
has partial derivatives fx (x, y) and fy (x, y) at the point (0,0), although it is discontinuous at this point. Construct the geometric image of this function near the point (0, 0). Sec. 4. Total Differential of a Function
1. Total
z
= /(*i
y)
increment of
a
The
function.
total
increment
of
a
function
the difference
is
Az-Af (x,
+ Aj/)-f (*,
#)--=/(*+ Ax,
y).
a function. The total (or exact) differential of a function z f(x, y) is the principal part of the total increment Az, which is linear with respect to the increments in the arguments Ax and A//. The difference between the total increment and the total differential of
2. The
total differential of
+
is an infinitesimal of higher order compared with Q At/*. \^&x* function definitely has a total differential if its partial derivatives are continuous. If a function has a total differential, then it is called differenThe differentials of independent variables coincide with their incret table. ments, that is, dx=kx and dy=ky. The total differential of the function z /(x, y) is computed by the 'formula
the function
A
=
dz
.
=^ dX +
,
d2=
dz
.
,
dy
-
d-y
Similarly, the total differential of a function of three arguments u is computed from the formula .
du
Example
1.
= du -3- dx .
dx
,
-j'
du ^- d y dy ,
du
.
dz. + -rdz .
For the function f(x
find the total increment
t
and the
y)=x* + xyy total
2
differential.
=/
(x, y, z)
[Ch. 6
Functions of Several Variables
J88 Solution.
f(x+Ax,
= 2x
y
+ Ay) = (x -f Ax)
Ax + AA
2
+x
-
= [(2x + y) A* + (x
2
-f (x
+ A*) (y
Ay + y Ax + Ax2 Ay 2y) A^l
-f-
Ay)
2y Ay
+ (A* + AA> Ay
(y -f
Ay
a
Ay)
2 ;
=
Ay
(2x + y) A* -{-(* 2y) Ay is the total differential of Here, the expression d/ = 2 2 infinitesimal of higher order is an the function, while (A* -f AJC* ) comared with VAx 2 +Ay 2 compared Example 2. Find the total differential of the function
AyAy
.
Solution.
3. Applying For
sufficiently
Q= y Av 2 -f
Ay
2 ,
the total differential of a function to approximate calculations. small |AA:| and |Ay| and, hence, for sufficiently small we have for a differentiate function z f(x t y) the approxor
=
Az^dz
imate equality
dz
.
3.
Example
The
Solution,
we
=
altitude of a cone is // 30cm, the radius of the base the volume of the cone change, if we increase by
= 10cm. How will 3mm and diminish R fl
H
by
The volume
1
mm?
of the
cone
is
V = -~-nR 2 H. The change o
in
volume
replace approximately by the differential
AV =^ dV =
-j
n (2RH dR + R* dH) =
= lji(
2.10.30.0.1
3
+ 100.0.3) =
s 01 Example 4. Compute 1.02 approximately. Solution. We consider the function z^x^. considered the increased value of this function Ay 0.01. The initial value of the function z
lOjiss
31. 4 cm*.
-
=
=
In x 8 Hence, 1.02
-
01
The desired number may be when jc=l, y = 3, Ajc = 0.02, s l =l,
Ay = 3-1.0.02+
1- In 1-0.01
=0.06.
^ 1+0.06=1.06. =
1831. For the function
f(x,y) x*y find the total increment total differential at the point (1, 2); compare them if 0.1 f Ay 0.2. a) Ax=l, A//-2; b) A* 1832. Show that for the functions u and v of several (for
and the
=
=
example, two) variables the ordinary rules of differentiation holcb v
du
udv
Sec. 4}
_
Find the
total
Total Differential of a Function
_
differentials of the following functions:
1841.
z- In tan i. x
1842. Find df(l, 1836. z
1837.
K
= si
= ln(x*+y*). = lnl+i. /*,
1838. z
1844.
1839.
1845.
= arc tan
if
~*'
= u = u=
1846. w
-2-+
=
arctan^.
1847. Find d/
+ arctan-*. 1848.
tj)
1),
1843 - "
2=yx?.
1840. 2
189
(3, 4, 5)
if
=
One
=
side of a rectangle is a 10 cm, the other & 24cm. a diagonal / of the rectangle change if the side a is increased by 4 and b is shortened by 1
How
will
mm
mm?
the change and compare it with the exact value. A closed box with outer dimensions 1849. and 6 cm is made of 2-mm-thick plywood.
Approximate
10 cm, 8 cm, Approximate the
volume
of material used in making the box. 1850*. The central angle of a circular sector is 80; it is desired to reduce it by 1. By how much should the radius of the sector be increased so that the area will remain unchanged, if the original leng:h of the radius is 20 cm? 1851. Approximate: 2
a)
(1.02)'-
c)
sin32-cos59
calculating
(0.97)
sin 60
;
b)
.
.
(when converting degrees into radius and three significant figures; round off the
take
last digit).
Show
that the relative error of a product is approximasum of the relative errors of the factors. 1853. Measurements of a triangle yielded the following data: side a=-100m2m. side fc angle
1852.
tely equal to the
ABC
C
601.
To what
degree
side c? 1854. The oscillation from the formula
of
period
= 200m3m,
accuracy can
T
of
a
we compute the
pendulum
is
computed
190
_ _
[Ch. 6
Functions of Several Variables
where
is
/
the length of the
of gravity. Find the error, a result of small errors A/
1855.
pendulum and g is the acceleration when determining T, obtained as a
= and Ag = (J in measuring / and and P (x, The distance between the points P (* ,
g.
V
equal to Q, while the angle formed by the vector P P the x-axis is a. By how much will the angle a change dx, y dy)? point P(P is fixed) moves to P l (x
y)
with
is
if
the
+
+
Sec. 5. Differentiation of Composite Functions
1. The case of one independent variable* If z f(x, y) is a differentiate function of the arguments x and y, which in turn are differentiate functions of an independent variable /,
then the derivative of the composite function z puted from the formula dz
____i
dt
'"dxdt^dydt
= /[
^(01 ma X
be com-
'
In particular, if / coincides with one of the arguments, for instance then the "total" derivative of the function z with respect to x will be:
x,
dzdy
Example
1.
Find
~,
if
where * = cos/, y
y,
Solution.
e'*+
2
From formula
^3(-sinO +
Example
2.
* 8X+
(1)
^-2^
Find the partial derivative z
Solution.
t*.
we have:
= e*y
t
~
and the
total derivative
,
if
where y
jZ^syety.
From formula
(2)
we obtain
2. The case
of several independent variables. If z is a composite function of variables, for instance, z where *=q> (u v) f(x,y), y=ty(u t v) (u and v are independent variables), then the partial derivatives z with respect to u and v are expressed as
several
=
independent
t
dz dy ~
Sec. 5]
and
_
Differentiation of Composite Functions
~~-~
_
dzdy
dz__dzdx
191
*
In all the cases considered the following formula holds: .
dz
dz
-j-
.
dx
dx
dz
.
.
+ dy 5- dy y
(the invariance property of a total differential).
Example p
Find ^- and ^du dv
3.
z=:f(x,
if
,
x=uv, y=
where
y),
Solution. Applying formulas (3) and
we
(4),
.
get:
and
Example
Show
4.
that the
function
z
Solution.
ment
x*
The function
+ y*=t,
(x*
+y
2
satisfies
)
the
equation
=
*
dx
=
dy
depends on x and y via the intermediate
cp
argu-
therefore,
dz__dzdt___
2
and
dzdt
\z
.
,
.
rt
,
Substituting the partial derivatives into the left-hand side of tion,
we
get t/
that
is,
*~^
i)2
the function z satisfies the given equation.
1856. Find
~
if
2
=
,
where x = e t
y
1857. Find
u
t
y = \nt.
if
= lnsin-=, f
1858. Find
where x = 3t\ y = yt'+l.
y
if j
u
= xyz,
where
^=/ +l, y=ln^, I
z
the
equa-
192
_ _ Functions of Several Variables
^
1859. Find
if
where x =
r,
I860. Find
if
z
1861. Find
= uv
g
1862. Find
g
where u =
,
~
and 2
= arc ~
and
~ and j z = f(u, v), ~ ow
and
= arc tan
1865. Find
~
z
1866.
Show
= #*.
if
where a = A:
~
1
,
where x = usmv,
where
w
= x^/ + ^.
where x =
3*), cp
sin
ip,
R
=R
e
then
=
1868.
where
Show
=a
if
cos
and
*_a.
if
= /(*.
//
if
|
-
1867. Find
v^e
1 t/
if
= /(u),
that
*/
where y =
t
oy ,
and
and
tan
if
1863. Find
1864. Find
= cos #,
sin x, v
if
= xy
2
[Ch.
y> *).
that
where
y=
(
if
/ is a differentiable function,
dz
dz
then
cos
9 cos \|>>
sin 9,
_
Sec. 6]
1869.
Show
_
that the function
w = f(u, where u = x + at, v
= y + bt
dw
dT= a Show
satisfies the
1871.
, ,
dw
dt
that the function
=J
+
equation
Show
v),
satisfy the equation
dw 1870.
19S
Derivative in a Given Direction
that the function
equation x-^ + y -^ = j 1872. Show that the function
satisfies the
satisfies the
(A-
The side
y
)
^-
+ xy-^ =
m
increases at the rate of a rectangle x -^20 the other side f/ 30 decreases at 4 m/sec. What the rate of change of the perimeter and the area of the rect-
1873.
=
of 5 m/sec, is
2
2
equation
m
angle? 1874.
What
is
The equations
of
motion
of a material
point are
the rate of recession of this point from the
coordinate
origin? 1875.
Two boats start out from A at one time; one moves northwards, the other in a northeasterly direction. Their velocities are respectively 20 km/hr and 40 km/hr. At what rate does the distance between them increase? Sec. 6.
Derivative in a
Given
Direction
1. The derivative of a function function z
7- 1900
= /(#,
and
the Gradient
in a given direction.
y) in a given direction /
= PP,
is
of
a Function
The derivative
of a
where f(P) and If
the function z
where a
(Ch. 6
Functions of Several Variables
194
is
are values of the function at the points P and differentiate, then the following formula holds:
/ (P,) is
the angle formed by the vector
/
with the x-axis (Fig. 67).
Y e^
P(*,y)
Fig. 67
In similar fashion we define the derivative in a given function of three arguments u f(x, y z). In this case
=
du
du = du cos a + du 5- cos p + 3- cos v, dz dx dy ,
/ for a
,
r
'
dl
direction
t
T
(2)
where a, P, Y are ^ ne angles between the direction / and the corresponding coordinate axes. The directional derivative characterises the rate of change of the function in the given direction. 2 2 at the point 3
=
dz -.
dx
= 4*; .
(dz\ =4; A 3\dxjp
dz
Here,
sina
Applying formula
(1),
we
= sin
120
=
get
The minus
sign indicates that the function diminishes at the given point and in the given direction. 2. The gradient of a function. The gradient of a function z f(x, ij) is 3 vector whose projections on the coordinate axes are the corresponding par-
=
Derivative in a Given Direction
Sec. 6]
tial
195
derivatives of the given function:
dz
,
,
dz (3)
The derivative
of the given function in the direction / the gradient of the function by the following formula:
is
connected
with
~ That is, the derivative in a given direction is equal to the projection of the gradient of the function on the direction of differentiation. The gradient of a function at each point is directed along the normal to the corresponding level line of the function. The direction of the gradient of the function at a given point is the direction of the maximum rate of increase of the function at this point, thlft
on
its
is,
when /=grad
z the derivative
-^
takes
greatest value, equal to
In similar fashion u -=/(*, y, z):
we
define the gradient
du
. ,
du
of
function
a
. ,
du
of
three
variables,
.
The gradient of a function of three variables at each point is directed along the normal to the level surface passing through this point. Example 2. Find and construct the gradient of the function z~x*y at the point P(I, 1).
X2
J
X
Fig. 68
Solution.
Compute
the partial derivatives and their values at the point P.
_
i
dxjpdz ^T"
Hence, grad 7*
z
= 2t+J
(Fig. 68).
s \y'* .
9.2
_ _ Functions of Several Variables
196
1876. Find
the
=
x* derivative of the function z xy2y* the direction that produces an angle
at the point P(l, 2) in of 60 with the x-axis.
= x*
1877. Find the derivative of the function z at the point Af(l, 2) in the direction from
point
tf (4,
[Ch. 6
+
6).
_
+
1 xy* 2x*y point to the
this
=
+
1878. Find the derivative of the function z lnYx* y* the point P(l, 1) in the direction of the bisector of the
at first
quadrantal angle.
1879. Find the derivative of the function u = x* 3yz + 5 at in the direction that forms identical the point Af(l, 2, 1) angles with all the coordinate axes. 1880. Find the derivative of the function u = xy + yz -)- zx at
the point M(2, 1, 3) in the direction from this point to the point N(S, 5, 15). eP + e*) \n (e* 1881. Find the derivative of the function u at the origin in the direction which forms with the coordinate axes x, y, z the angles a, p, y, respectively. 1882. The point at which the derivative of a function in any direction is zero is called the stationary point of this function. Find the stationary points of the following functions:
=
a)
+
z-=x*
= x* + y*= u 2y*-{ z*xyyz
b) z c)
1883.
Show
that the derivative of the function z
+ =
any point of the ellipse 2x* y* C* along the ellipse is equal to zero. 1884. Find grad z at the point (2, 1) if at
1885. Find grad z at the point
(5,
3)
if
1886. Find grad u at the point 1887. Find the magnitude and
(1,
2,
3),
point
(2,
2,
1)
direction
taken
normal
to the
u=xyz.
if
of
grad
u
at the
if
1888. Find the angle between the
*=ln-j-
=
at the points
A
(1/2,
1/4)
gradients
and 5(1,
1).
of
the
function
Sec. 7]
_
Higher-Order Derivatives and Differentials
1889. Find the steepest slope of the surface
=
_
197
+
x' z 4y* at the point (2, 1, 8). 1890. Construct a vector field of the gradient of the following functions:
= je-f y:
a) z
= x* + y
c) z
z \
Sec. 7. Higher-Order Derivatives and Differentials
1. Higher-order partial derivatives. The second partial derivatives of a function z /(*, y) are the partial derivatives of its first partial derivatives. For second derivatives we use the notations
32z
d (dz\
Derivatives of order higher than second are similarly defined and denoted. If the partial derivatives to be evaluated are continuous, then the result of repeated differentiation is independent of the order in which the differentiation is performed. Example 1. Find the second partial derivatives of the function
z~ arc tan
.
y Solution. First lind the dz
first
==
partial
_J_
dz^ dy~~
Now
m
derivatives:
j__
y
1_ J ,
differentiate a second time:
^L = d .
dz z
f
y
\-
2 *y
.
dxdy
We
note that the so-called so-
different
"mixed" partial
derivative
may
be
found in a
el way, namely:
d /
dxdy~~dydx
dx(
2
y*
_
198
f(x
t
We
= d(d*z)
d*z
and, generally, z
function function:
similarly define the differentials of a function z of order higher than
two, for instance:
If
[Ch. 6
Higher-order differentials. The second differential of a of this y) is the differential of the differential (first-order)
2. z
_
Functions of Several Variables
= /(x,
differential of
dz = d(d n -
l
z).
y), where x and y are independent variables, then the second the function z is computed from the formula
(1)
Generally, the following symbolic formula holds true:
formally expanded by the binomial law. z f (x, (/), where the arguments x and y are functions of one or eral independent variables, then it
is
If
=
"- * If
x and
i/
sev-
+2
are independent variables, then d 2 jt
becomes identical with formula (!) Example 2. Find the total differentials
= 0,
d 2 y = Q, and
of the first
formula
and second orders
(2)
of
the function z
Solution. First method.
We
=
2;t
2
3xyy
2 .
have
*-<*-*
I*-*.
Therefore,
dz
= fr
Further we have
~ ^-4 djc
whence
it
2
'
J*!L-_ 3
2
^l '
djcd^""
dy*
follows that
Second method. Differentiating we find
Differentiating again and remembering that dx and dy are not y, we get
x and
= (4dx
3dj/)
dx
(3d*
dependent on
+ 2d(/) dy =4dx*6dx dy2dy
z .
Higher-Order Derivatives and Differentials
Sec. 7]
if
Find
if
1894. Find
if
1893
.
^
= arc tan A
1
Find
1895.
.
.
xy
if
1896. Find all second partial derivatives of the function
Find
1897.
^~-
dxdy dz
if
u 1898. Find
,
if
z
1899. Find
f(0,
=
0),
= sin (xy).
= (l 1900.
Show
/^(O, 0)
f xt/ (Q, 0),
4
that
if
= arc sin 1901.
1902*.
Show
Show
that
=
y
if
that for the function
if
199
_ _
[Ch. 6
Functions of Several Variables
200
= 0]
[provided that f(0, 0)
we have
rxy(' 0)=-1, 1903
-
&$"=
Find
z
g
where
z
y(x,
z
1906.
u
if
= f(x,
= f(u,
v),
r
Show
a)*
= arc tan
+ (y
= lny
,
satisfies the
b)\
(x,
t)
=A
+
sin (akt
2 d^u_ ~~~ a d^u 2
Show
dx
sin
Kx
u(x //
,
'
that the function (x-x
i
,
cp)
equation of oscillations of a string dt*
(where * conduction
Laplace equation
that the function
u
1909.
y),
that the function
Show
satisfies the
= <((x,
Laplace equation
= Y(x
1908.
z),
that the function
/
where
y,
where u
u
1907.
f(u, v),
y).
Show
satisfies the
0)=+1.
= xy.
where u = x* + y*, v 1904. Find
/;,(0,
y
'
z
>
,
z
>
'>-
)*
+ (y-y n )*
e
a are constants)
satisfies
^a'/^-u^U*^
the equation of heat
_
Higher-Order Derivatives and Differentials
Sec. 7]
Show
1910.
that the function
_
201
are arbitrary twice differentiable functions, satiscp and \|) the equation of oscillations of a string
where fies
Show
1911.
satisfies
that the function
the equation
x 1912.
Show
satisfies the
1913.
2
+ 2wy dxdy + Jy ,
5-,2 dv
r-
,
2
x-, dy*
l
= n0.
that the function
equation
Show
-7
'
d su
that the function z
= f[x + y(y)]
satisfies the
equa-
= u(x,
which
tion d22
dz
~
dx dx dy
1914.
Find u-^u(x, y)
dzd 2z 2 dy dx
'
if
dTSy^^ 1915.
the form of the function u
Determine
satisfies the
equation
1916. Find d*z
if
2
1917. Find d u
if 11
1918. Find d*z
.
if - rn \l^
1919.
=
(f} \^/F
Find dz and d*z z
=u
v
whpr^ vviitivx
/ t>
-r
y2
j\r
-1- /y* . |
y
if
where u
=~
,
v
= xy.
y),
_ _
202
Functions of Several Variables
1920. Find d'z
z
= ax,
where u
v),
v
= by^
v
= ye*.
if
= f(u,
1922. Find d*z
6
if
z~f(u, 1921. Find d*z
[C/i.
where u = xey
v),
,
if
z
= e x cos y.
1923. Find the third differential of the function
= * cos y + y sin x.
z
Determine all third partial derivatives. 1924. Find df(l 2) and d*f(l, 2) if 9
= x + xy + y* 2
f(x, y)
4\nx
\Q\ny.
2
1925. Find d /(0, 0,0)
f(x
Sec. 8.
+
9
y,
=x z)
if
z
Integration of Total Differentials
t. The condition for a total differential. For an expression P (x, y)dx-}~ Q(* y)dy> where the functions P (x, y) and Q (x, y) are continuous in a
simply connected region (in D) the total ficient that
D
together with their first partial derivatives, to be some function u (x, y), it is necessary and suf-
differential ol
aq^ap ~~ dx
Example
is
a total
t.
Make
sure that the expression
differential of
some function, and
Solution. In the given case,
=
1,
'
dy
find that function.
P = 2x + y Q t
x+2y.
Therefore,
,5 = --
and, hence,
where u It
is
is
the desired function.
given
that
-=
But on the other hand and Finally
we have
2jt
+ #;
therefore,
= x + y' (y) = x + 2y, whence
q>' (y)
= 2y,
(p(f/)
=
=
Sec.
ti\
_
2. The
203
case of three variables. Similarly, the expression
P(x,
where P
_
Integration of Total Differentials
z)dx
y,
+ Q(x,
z)dy
y,
+ R(x
t
z)dz,
y,
R(x, y, z) are, together with their first partial functions of the variables x, y and 2, is the total differential of some function u (x y, z) if and only if the following conditions (x, y, z),
Q(x,
y, z),
continuous
derivatives,
t
are fulfilled:
dQ^W dR^dQ dP^dR '
dx
Example
2.
Be sure
'
dy
dy
dz
dx
dz
that the expression
the total differential of some function, and find that function. f 2 3jc Solution. Here, p 1, 3x, R 2yz+\. We establish 30 the fact thai is
=
dQ dx
+
=
Q=z +
= dP = O
.
dy
dR = dQ -r
=
dz
dy
dP c.
rr
= OR = r
V.
/
dx
dz
and, hence,
~ where u
We
is
the sought-for function.
have
hence,
u= On
2
(3x
+ 3y
\)dx
= x* + 3xy
+
z).
the other hand,
du -
-
3
~ = dtp
dz
whence
y^
=z
2
and ~P
dz
= 2f/z+l.
two variables q>(#, 2) tion for total differential find q>:
of
whose is
The problem reduces partial derivatives are
fulfilled.
We
that
x
is,
y(y,
= ^2 + 2 + C, 2
e)
And
finally,
to finding the function
known and
the condi-
_ _ Functions of Several Variables
204
Having convinced
are
total
differentials
[C/i.
6
yourself that the expressions given below of certain functions, find these functions.
1926.
ydx + xdy.
1927.
(cosx+3x*y)dx + (x'y
1
)
1928-
1930. 1931.
-dx^dy. y x dx + *
-
*
y
r
*
z
dy.
1932. Determine the constants a and 6 in such a the expression 2 z 2 (ax + 2xy + y ) dx-(x + 2xy + by*) dy
should
be
a
total
differential of
some function
z,
manner
and
find
that
that
function.
Convince yourself that the expressions given below are some functions and find these functions.
total
differentials of
1933. 2
+
+ 3z)dx+(4xy + 2y + 8xy* + 2) dx +
z)dy
+ (3x
l)dy
(x
1934. (3x 2y* 1935. (2xyz3y*z
y
2)dz.
2
y
xdx + ydy + zdz 1938*. Given the projections of a force on the coordinate axes
v _.
v_
y
to
where A, is a constant. What must the coefficient K be for the force to have a potential? 1939. What condition must the function f(x, y) satisfy for the expression f(x,
y)(dx
+ dy)
to be a total differential?
1940. Find the function u
if
du = f(xy) (ydx + xdy).
Sec. 9]
_
_
Differentiation of Implicit Functions
Sec. 9. Differentiation of Implicit Functions
205
1. The case of one independent variable.
If the equation f(x, y) 0, where differentiate function of the variables x and y, defines y as a function of x, then the derivative of this implicitly defined function, provided
/
that
a
is
y)
(*
y) ?= 0,
f' (x, y
may
be found from the formula
dy dx
f' x
=
(**y)
f'y(x,y)'
found
are
derivatives
Higher-order
by successive differentiation
formula
of
a) Example
1.
Find
-
dx
-~
and
Solution. Denoting the
if
dx 2
left-hand
side
of this
equation by
f (x, y),
we
find
the partial derivatives
f'u
(x
t
y)-=3(x
Whence, applying formula
z
+y
(1),
z 2
(
2y
.
)
we
get
find the second derivative, differentiate with respect to x the first derivative \vhich we have found, taking into consideration the fact that y is a functiun of x'
To
x -~ dx
y J
dx
F
(x,
2
dx\
y
y J
x
y J
~
(
)
y J
\
2
y
'
2
if
2. The case of several independent variables. Similarly, 0, where F (x, y, z) is a differentiate function y, z)
x,
y and
z,
Fz
(x
y,
z)
if
of
the equation the variables
defines z as a function of the independent variables x and y and then the partial derivatives of this implicitly represented 0, function can, generally speaking, be found from the formulas t
^
'
dK
F'z
(x,
dlJ
y, z)
F'g
(x, y,
z)
another way of finding the derivatives of the function ating the equation F (x, y, z)=0, we find
Here
is
dF
Whence
it
is
_,
,
dF
J
,
dF
,
rt
possible to determine dz, and, therefore, dz TT-
dx
.
dz
and 3dy
.
z:
different^
_ _ Example
we
[Ch. 6
Functions of Several Variables
206
Find
2.
and
-T-
if
j-
Solution. First method. Denoting the left side of this equation find the partial derivatives
F'x
(x, y,
z)
= 2x,
Applying formulas dz
_
d*~
F'x(x> F'z (x
t
(2),
y. *)
F'
y (x, y, z)
we
40-z+l, F z
(x,
y,
2)
^ ~"
2)
(x,
//,
z),
= 6z-0.
get
_
2x
~~ y,
=
F
by
Jy(*.
dz^ dy~
'
6*
z)
y
^
(x>
!/,
^
1
z
4//
62
/
z)
Second method. Differentiating the given equation, we obtain 2x dx
4f/
Whence we determine dz
t
dy
+ 6zdz
that
is,
zdy + dy = 0.
y dz
the total differential of the implicit func-
tion: \
z}dy
4//
~ dy we Comparing with the formula dz = -Q- dx -\,
2x
dz
3. A system
6z
y
dy
F(x, G(x,
f
and v as functions
u
y,
62
y
system of two equations
of implicit functions. If a
\ defines u
\4yz
dz '
dx
see that
t
i>)
y, u,
o)
= 0, =
of the variables
x and y and the Jacobian
dF_ dF_
D(F, G) D(u, v)'
dudv
dGdG du dv
then the differentials of these functions (and hence their partial derivatives as well) may be found from the following set of equations . dF dF dF dx ^ dv =0, ^+ -z- dyy + -^~ du ^ + ^dx dv
'dF
,,
,
dy
,
Example
3.
3dx
-r~
dy
,
^du dG -^--3y dv du *
l
.
.
.
t
/Q
'
The equations
define u and v as functions of x and ^ w; find
and -,,dx dx dy dy
rr?
.
.
Differentiation of Implicit Functions
Sec. 9]
Solution.
First
207
method. Differentiating both equations with respect to #
we obtain du
dv
.
whence __
?ff
dx~~ Similarly
we
u ~^~y x y
dv __ u-\-x dx~~ x y
'
'
find
_
v
dy~~
x
du
+y
*
dy~~~ x
y
we
Second method. By differentiation differentials of all four variables:
x du
_
dv '
y
two equations that connect the
find
du 4- dv = dx + dy, + u dx + y dv + v dy =4).
Solving this system for the differentials du and dv,
we obtain
xy Whence
_ dx
'
x
y
(}y__-f-x
dx~~xy
f
x
t)f/
_
dv
y
v -}-x '
'
dy~~x
y
4. Parametric bles x and y
is
representation of a function. If a function represented parametrically by the equations z
= z(u,
z
of the
varia-
v)
and
then the differential equations
be found from the following system
may
of this function
of
dx
dx
Knowing dz ^~=pr and 3-
dz
dx
the differential
.
dy
~^. ^
dz^p
du
dx dv
dz
dz
xdu
dv
,
,
.
5~du-\--5- dv,
d*
+ qdy,
.
dv.
we
find the partial derivatives
_ _
208
[Ch. 6
Functions of Several Variables
Example
The function
4.
arguments x and y
of the
z
is
defined by the
equations _.
,
dz
,
dz
Find ^- and 3ox
.
dy
we
Solution. First method. By differentiation connect the differentials of all five variables:
dx
From
the
first
= du + dv
find three equations that
,
two equations we determine du and dv: 20 dx
"^
dy
_
,
2u dx
dy
'
'
2(v
2(u
u)
w)
Substituting into the third equation the values of du and have: d dy
-
rfy
just found,
we
= 6wu (u '
Whence 3-
=
3au,
3-
^jc
dt/
=TT 2
(w-fy). v ;
Second method. From the third given equation we can find
+ 3^; *=3Jf dx ^ dx Differentiate the respect to y:
two equations
first
the
first
system we
dy
.
we
da
dv a'
dy
u v
find
= _l__
dy~~2(u Substituting the expressions
__
dx~ u
~
dv_
v)'
and
~
^ (5) '
with respect to x and then with
dy
v
dx~~ v the second system
first
l
dy
find du___
From
= 3'Jf + 3t,'f!.
f
dx
From
f dy
1
dy~~2(vu)' into formula (5),
we obtain
_
Sec. 9]
_
209
Differentiation of Implicit Functions
1941. Let y be a [unction of x defined by the equation y*
Find ma *
dy dhj di' d?
1942. y
Show
that
and ana is
^
dx"'
a function defined
^=
1943. Find
and explain the
% ^
1944. Find
by the equation
if
y=\+y x
and
g
.
y
if
result obtained.
=
A;
and (g) (g) \ax J K~\ \ax jx=i
1945. Find
z
if
Taking advantage of the results obtained, show approximately the portions of the given curve in the neighbourhood of the point
*=1. 1946.
The function y In
dx
1947.
and
Find
dx
c,. j Flnd
5i
and
2
and
9
s
-1-
2//
^
variables x and y
and
^ dy
4
8
z
3;q/z
2r/
is
defined by
+ 3 = 0.
%-
~ dx
A:
Find
if
dx z
^2
1949. Find
1950.
arc tan
function z of the
x ,
|?T7 =
dx*
1948. The the equation
^2
defined by the equation
is
cosy
The function
and
for the -j-
if
-|-
y cos 2
+zcosx= 1.
defined by the equation
z
is
x*
+y
2
system
z
of
2
xy
= 0.
values *==
1,
//
= 0,
z
= l.
__
210
dz 1QR1 1951. KinA
Fmd,
1952. /(*, y,z) 1953.
equation
Functions of Several Variables
l>
,
= 0.
= (p(x, y), = 0. ty(x, 2
f/)
d* 2
&*
dz
1955. 2
is
if if
,
where y Find ~.
is
a
function of x
defined by the
if
a function of the variables x and # defined
equation 2x*
[Ch. 6
Show
2 Find dz and d z,
1954.
d*Z ,
__
+ 2y*
-4-
8xz
z*
z
+8-
by the
.
Find dz and d z for the values x = 2, f/^=0, 2^=1. 2 1. What are the 1956. Find dz and d 2, if In z=jc + i/ 4-2 first- and second-order derivatives of the function 2? 1957. Let the function 2 be defined by the equation 2
where
1958.
an arbitrary differentiate
is
Show
constants.
Show
is
satisfies the
02,
y
bz)
1960.
y=
xq> (z)
f(y y) ,
Show
+
&, c are
Q,
an arbitrary differentiate function of two arguments, equation
a-+b - = 1959.
a,
that the function 2 defined by the equation
F(x where F
and
function
that
o|>
= 0.
that
(2)
Show
the
that
function
satisfies the
1
xfx +y~ = z. 2
defined
the
by
equation
equation
d*zf(
1961. The functions y and 2 of the independent variable x are 2 2 2 2 2 2 defined by a system of equations * 2 ^0, x 32 4. 2#
+#
+
+
=
_ _
Sec. 10]
Change
2H
of Variables
1962. The functions y and z of the independent variable x are defined by the following system of equations:
Find dy,
dz, d*y, d*z. 1963. The functions u and v of the independent variables x and y are defined implicitly by the system of equations
Calculate du
du
d2u
d*u
dzu 2
~dy
for x=--Q,
y
'
dv dv dx' d~y
62 v '
dz v
d*v
~d\*'
dxdy
j
dy*
1.
1964. The functions u and v of the independent variables and y are defined implicitly by the system of equations
Find du, dv, d*u,
d*v.
The functions u and
v of the variables x and defined implicitly by the system of equations
1965.
x
.v
= (p(w,
v),
y are
//=-i|)(w, v).
(b du du du 51' cty' d~x> fy'
^ and g? x = M cos u, y w sin u, 2 = x = u + v, y = u and ~ v,z = uv. b) Find = c) Find dz, where r and are functions of the variables 1967. e = F(r, 1966. a)
Find
,
y
if
if
\l
jc
,
.
,
cp)
x and y defined by the system of equations
Find
2
ax
and
.
dy
1968. Regarding z as a function of x
x = a cos
Sec.
q)
cos
i|),
and
y, find
y = b sincpcosij), z
~
and ~,
if
=
10. Change of Variables
When
changing
variables in differential
them should be expressed
in
expressions,
the derivatives in
terms of other derivatives by the rules of
entiation of a composite function.
differ-
[Ch. 6
Functions of Several Variables
212
\. Change of variables in expressions containing ordinary Example 1. Transform the equation x2 y
dx putting
*
=
derivatives.
y-
Solution. Express the derivatives of y with respect to x in terms of the derivatives of y with respect to /. We have
dx
dy
dy
dt
dt
dx
_
dt
dt
'
t*
dt
Substituting
the
expressions
equation and replacing x by
the
of -r-
we
,
derivatives
found
just
the given
into
get
or
dt*
Example
2.
ruij -"'
Transform the equation x
g + ^y_g =0
.
taking y for the argument and x for the function. Solution. Express the derivatives of y with respect to x in derivatives of x with respect to y.
terms of the
dx^Jx'' dy
"~
'
dx*
dx{ dx \
^
}
2
dyl dx \dx~~ \ dy J
-
fdx\ (Ty)
dx
-
'
fdx\* (dy)
dy
Substituting these expressions of the derivatives into the given equation, will
have
d*x dj
1 '
dy
\dy
L=o
dx
dy
'
we
Sec.
or,
10]
_ _
213
of Variables
Change
finally,
Example
3.
Transform the equation
~xy'
dx
by passing to the polar coordinates ,v=rcoscp, Solution. Considering
dx = cos
cp
f/
= rsinq>.
as a function of
r
dr
cp,
d// = sin
r sin cp dcp,
(1)
from formula dr
cp
+ r cos
cp
(1)
we have
dcp,
whence dr .
.
.
.
d//_sm cp dr -f cos cp 3J T dr
cos
dcp __ ~~
cp
-
sin cp r
,
r
r sin cp T d(pr
cos
cp
3- -f
r cos cp ^
_rcos
d
-- r sin
dr coscp
or,
-
r
--
dr cp
Putting into the given equation the expressions for sin
,
\
cos (p Y
dcp
cpH-
rcoscp
r sin q)
x,
rsm
//,
and
-^
,
we
will
have
cp
rsincp*
cp
after simplifications,
2
Change of variables in expressions containing Example 4. Take the equation of oscillations of
and change
it
new independent
to the
partial derivatives. a string
variables a and p, where
a = .v
at,
Solution. Let us express the partial derivatives of u with respect to x and t terms of the partial derivatives of u with respect to a and p. Applying the formulas for differentiating a composite function
in
du___du
du__du da
du dp
da,
di^dadT^d^dT
9
du dp
^dadx^dfidi
we get du
du
,
^T-=^( da
d/
^'^
du__du
^
a)
du
+ ,
du -35"
dp
Q
a (du is
\dp
^'
_^_,^f
du\ 3~
I
day
1
_ _
214
Functions of Several Variables
..
[C/i.
6
v
~
same formulas:
Differentiate again using the
dt~~dadt
dt 2 ~~dt
d2u
d 2 u\,
x
2 (d u
,
d*u
dz u
dx*~dx\dx)
da\dx
_d
2
u
d'u
d*u
/)ft 8
'
rtn
'
'
/)ft
/}R2
we
Substituting into the ^iven equation, 2
2
/d a
d2 u
-=
for
5.
2 Transform the equation x 2 ^- + y
the
x
y
new independent *
d*u\
= 0.
r
Example
d2 u
Jd u
d u\
d u
have
will 2
2
2
=z
taking
,
w
and
variables,
=
2
-g-
for z
x
,v,
the
new
function.
and Solution. Let us express the partial derivatives ypartial
derivatives
^~
and
.
^
To do
On
x
dx v
t
dy
oy_
-^^,
dx
dz ^
.
2
the other hand,
dw
dw =
:;
du
du
dw ~\
dv
dv
Therefore,
dw -3-
du
,
du
dw
dx
dv = + -5dv x ,
,
dz
2
z
= 2
or
dw
,
i^ w_f^x__m dy\^,^x
&*_
.
Whence
dw
1
dw\
,
.
z
2
and, consequently,
dz^_ dx~~
2
dw ~~ 1 dw \ J z x*dv J ~~du \x
/
terms of the
differentiate the given relation-
this,
ships between the old and new variables:
u
~ in
dw
.
Sec.
Change
10]
215
of Variables
and dz __ z^
dw
2 dy~~y dv
'
Substituting these expressions into the given equation,
we
get
or
Transform the equation
1969.
2
x j--
putting x=--e*. 1970. Transform the equation
putting
A:
cos/.
T
rm 1971
69
Transform the following equations, taking y as the ar-
gument:
The tangent of the angle [A formed by the tangent line and the radius vector OM of the point of tangency (Fig. 69)
1972.
MT is
expressed as follows: tan u=^
216
_ _ Functions of Several Variables
Transform
this
by
expression
to
passing
1973. Express, in the polar coordinates x the formula of the curvature of the curve
1974.
[C/i.
polar
= r cos
6
coordinates:
q>,
y=
r sin
cp,
Transform the following equation to new independent
variables u and v: dz
dz
if
u = x
v =x +y 2
t
*-*3&
'
2 .
1975. Transform variables u and v:
the
following dz
if
=n
equation
new independent
~
dz
.
to
= x, 0=-p
11
1976. Transform the Laplace equation
,
dM_ n ~-"
dx*~*'dy
z
to the polar coordinates A:
= rcoscp,
y
= rsm
1977. Transform the equation
X
2
^2_ y ^__ ~ U> dx* 2
2
ar/
u=*xy and
putting
y
=
.
1978. Transform the equation dz
dz
,
yTx- K jy=(y-^ z
>
by introducing new independent variables
and the new function w=\nz (x + y). 1979. Transform the equation dx*
taking u
>=
= x + y,
for the
v
new
=^
for
function.
the
new independent
variables
and
Sec. 11]
_
The Tangent Plane and the Normal
to
a Surface
_
217
1980. Transform the equation
putting u
= x+y,
v
=x
w = xyz,
y,
where
w=w(u,
v).
11. The Tangent Plane and the Normal to a Surface
Sec.
1. The equations of a tangent plane and a normal for the case of explicrepresentation of a surface. The tangent plane to a surface at a point to (point of tangency) is a plane in which lie all the tangents at the point various curves drawn on the surface through this point. The normal to the surface is the perpendicular to the tangent plane at the point of tangency If the equation of a surface, in a rectangular coordinate system, is given in explicit form, z function, then f (x, y), where f (x, y) is a differentiate z ) of the surface is the equation of the tangent plane at the point (x f/ it
M
M
z-*o=/i(* z
0o)(X-*o)
.
,
+ /i(*o,
)0
,
r
-0o).
and X, K, Z are the current coordinates
Here, t/ ) f (x the tangent plane. The equations of the normal are of the form
where
,
Z
F,
.Y,
Example the surface z
1.
(i) of
the point of
are the current coordinates of the point of the normal. Write the equations of the tangent plane and the normal
=
2
at the point
y
M (2,
the
of
M
=v '
dx
to
1,1).
Solution. Let us lind the partial derivatives their values at the point
M
(fo\ '
given
function and
=2
\dxjM
*~ Whence, applying formulas or 2x-|-2f/
10
z
= ^-i- =
-, which
2. Equations it
representation implicitly,
and F
(X Q
,
which
t/
,
z
)
is
(1) is
and
(2),
we
will
the equation
have
of the
z
1
=2(*
2)
+2
tangent plane and
(r/-|- 1)
^
==
the equation of the normal.
of the tangent plane and the normal for the case of implicof a surface. When the equation of a surface is represented
= 0,
the corresponding equations will have the form
_ _ Functions of Several Variables
218
which
[Ch. 6
the equation of the tangent plane, and
is
XXQ F'X
(**,
Yy
_ *
y
ZZQ
_
Q
Z 0) F'z (*0. J/0.
(*0> 00. *0)
Fy
)
which are the equations of the normal. the equations of Example 2. Write s 8
the tangent plane and the normal to a. z a at a point for which x the surface 3;q/z 0, z/ Solution. Find the z-coordinate of the point of tangency, putting x 0, a 8 whence z a. Thus, the z* a into the equation of the surface: .j/
=
=
=
=
point of tangency
is
F
Denoting by
M (0,
and
Applying formulas
(3)
+ a=:0,
a).
the left-hand side of the their values at the point Af:
which
and
we
(4),
~r~
=
n
'
wn
^
equation,
~2 5
the
Qj/
cn are
a
e q ua ^ions of
^ ne
)
/?sina,
of
paraboloid
the normal.
and
revolution
z
= x*+y
2
the
the equaindicated
at
the
point
;
cone
b) to the c) to
the
i
points: a) to '
find
get
1981. Write the equation of the tangent plane tions of the normal to the following surfaces at
0-
we
the equation of the tangent plane,
is
x
or
=
=
,
(x, y, z)
partial derivatives
or ^-(-z
a,
=
the
= at the x*+y* + z = 2Rz
^ + -^
y-
2
sphere
point at
the
(4, 3, 4);
point
(ffcosa,
/?).
1982. At
what point
of the ellipsoid ~2
y2 _4. "" ^ f_4.^ __ f.2
a
2
b
2
c2
1
*
does the normal to it form equal angles with the coordinate axes? 1983. Planes perpendicular to the A:- and #-axes are drawn z* 169. through the point (3, 4, 12) of the sphere x* y* Write the equation of the plane passing through the tangents to the obtained sections at their common point M. 1984. Show that the equation of the tangent plane to the central surface (of order two)
M
+
ax
2
+ by
2
2
-\-cz
=k
+ =
Sec. 11]
_
The Tangent Plane and the Normal
M (x
at the point
Draw
#
,
,
z
to
the
+
Draw
1986.
which cuts
off
_
219
a Surface
has the form
)
surface x x to the 4
to
+ 2tf + 3z = 21
2
2
= 0.
^a+fi + 'T^l
to the ellipsoid
tangent
a
planes
tangent
plane
on the coordinate axes. equal segments 2 2
z* 1987. On the surface je 2*^=0 find points at which the tangent planes are parallel to the coordinate planes. s 1988. Prove that the tangent planes to the surface jq/z a of the tetrahedron of constant volume with the planes form coordinates. r f 1989. Show that the tangent planes to the surface }/ x-\-\/ y
+y
=m
+
-\
is
\/
z^Ya
cut
on the coordinate axes, segments whose sum
off,
constant.
Show
1990.
that the cone
^-i-f!
= -^
and the sphere
are tangent at the points (0, b,c). 1991. The angle between the tangent planes drawn to given surfaces at a point under consideration is called the angle between two surfaces at the point of their intersection. At what angle does the cylinder x*-\-y* R* and the sphere
=
(x-R)
2
2
!-
-#
2
intersect at the point
Affy,
^-^,
OJ?
1992. Surfaces are called orthogonal if they intersect at right angles at each point of the line of their intersection. Show that the surfaces x*+y* z* r t (sphere), y xiany 2 (** (plane), and z y*)ian*-ty (cone), which are the coordinate surfaces of the spherical coordinates r, cp, tj?, are mutually ortho-
=
gonal. 1993.
z^xf(~ * \
Show )
=
+ =
+
that all the planes tangent lo the conical surface
at the point
M
(,v
,
//
z
,
),
where x
+ 0,
pass through
/
the coordinate origin. 1994*. Find the projections of the ellipsoid /'
+z
1
on the coordinate planes. 1995. Prove that the normal revolution z
= /(/^ +
a
2
f/
)
(/'
xy
at
+ 0)
1=0 any point
of
the
intersect the axis
of
surface of rotation.
220
_
Functions of Several Variables
_
[C/i.
6
Sec. 12. Taylor's Formula for a Function of Several Variables
of all orders f (x, y) have continuous partial derivatives (rc+l)th inclusive in the neighbourhood of a point (a, b). Then Taylor's formula will hold in the neighbourhood under consideration:
Let a function
up
to the
-a) + f'y (a, b)(y-b)]
+
where
In other notation, )
+ -Jy
[/tfX
if
...+[^^ or
-j
df
(x, y)
+
2 rf
/ (x, y)
+
.
.
.
The particular case of formula (1), when a b Q, is called Maclaunn's formula. Similar formulas hold for functions of three and a larger number of variables. A3 Example.' Find the increment obtained by the function f (x, y) when passing from the values 1 to the values *,-- -{-//, 1, y
=
*=
1
Solution. The desired increment may be found by applying formula calculate the successive partial derivatives and their values at
First
given point
(1, 2):
= 3- 1+3.2=9, = /;il,2)=-6.4 + 3.1 -21 fxx (\, 2) = 6-l=6 /;
x
f
(1,2)
f
^(1,2)= -12.2= -24,
f
(2).
the
Taylor's Formula for a Function of Several Variables
Sec. 12]
are
All subsequent derivatives into formula (2), we obtain:
~
2
[/i
-64-2/z k- 3-f
1996.
powers
Expand h and
of
2 fc
f (x i fe
+ ^y
24))
(
ft,
y
zero.
identically
8
-
[/i
+ k)
G
+ 3/z
2
+
fc.
in a series
of
2
221
these
Putting
3/ife
_
-0-f *'(
positive
results
12)]
integral
if
=
2
+
+
* 1997. Expand the function f (x, y) 6x 2xy 3y* 4 by Taylor's formula in the neighbourhood of the point 2/y
(-2,
1).
1998. Find the increment received by the function ^-x*y when passing from the values x=l, {/=! to
4*
tegral
the
yz
neighbourhood
of
(1, 1, 1).
Expand
powers
of
f (x /?,
t
k,
/ (x, y, z)
2001.
f(x, y, z)
z-l-4 by Taylor's formula in
3#
the point 2000.
= Jc'-fy + 2* +2xy 1
Expand the function
1999.
f(x,y)~
=--=
ft,
x
*
/)
in a series
of
positive in-
if
/,
2
z-|
fr,
//-J
and
// 4 z
2
2A:e
-
2jr//
Expand the following function
2yz.
a
Maclaurin's series
in a
Maclaurin's series
in
up to terms of the third order inclusive: /(.Y,
2002.
//)
= ?* sin//.
Expand the following function
up to terms
of order four inclusive: / (x, //)
= cos x cos y.
2003. Expand the following function in a Taylor's series in the neighbourhood of the point (1, 1) up to terms of order two inclusive: /(*.
{/)
= {/*
2004. Expand the following function in a Taylor's series in the neighbourhood of the point (1, 1) up to terms of order three inclusive:
_ _
222
[Ch. 6
Functions of Several Variables
2005. Derive approximate formulas (accurate to second -order terms in a and P) for the expressions
if
and |p| are small compared with unity. 2006*. Using Taylor's formulas up to second-order
|a|
terms,
approximate
^O98;
1/T03;
a)
2 01 -
b) (0.95)
.
an implicit function of x and y defined by the + y = 0, which takes on the value z= 1 for x= 1 equation and y=l. Write several terms of the expansion of the function z in increasing powers of the differences and y 1. 2007. z z
is
9
2xz
x\
Sec.
13. The
Extremum
of
a Function
of Several
Variables
1. Definition of an extremum of a function. We say that a function f(x,y) has a maximum (minimum) f (a, b) at the point P (a, b), if for all points P' (x, y) different from P in a sufficiently small neighbourhood of P the inequality /(a, b) > f(x, y) [or, accordingly, /(a, b) < f (x y)] is fulfilled. The generic term for maximum and minimum of a function is extremum. In similar fashion we define the extremum of a function of three or more t
variables.
2. Necessary conditions for an extremum. The points at which a differentiate function f (x, y) may attain an extremum (so-called stationary points) are found by solving the following system of equations:
0)-0,
t' x (x.
f't/
(x
t
y)-Q
(1)
(necessary conditions for an
extremum). System (I) is equivalent to a single 0. In the general case, at the point of the extremum equation, df(x, #) P (a, b), the function f (x, y), or df (a, ft) = 0, or df (a, b) does not exist. 3. Sufficient conditions for an extremum. Let P (a, b) be a stationary point of the function f(x, y), that is, df (a, &)- 0. Then: a) if d*f (a b) < for dx z dy*>Q then /(a, b) is the maximum of the function f(x, //); b) if d z f(a, ft)>0 for d* 2 -}- di/ 2 then /(a, b) is the minimum of the function 0, 2 /(* 0); c ) if d /(a, ft) changes sign, then f (a, b) is not an extremum of /(v, //). t
+
t
>
The foregoing conditions are equivalent
= f'y (a,
ft)
-0
Then: I) if P(a, ft), namely (or
C>0);
2)
A=f xx
and
if
then the question (which is to say,
>
A
0,
(a,
then
maximum,
a
A
<
0,
then
ft),
B~fxy
the if
an extremum
(or is
to the
ft),
C
<
following: let
C = /^(ci,
has
function
A<
there
(a,
an
0),
f[ (a,
We
ft).
form
extremum at a minimum,
and
no extremum
at
P
(a
t
b)----
the
the point if
A
>
A==0. remains open
ft);
3)
if
of the function at P (a, ft) requires further investigation). 4. The case of a function of many variables. For a function of three or more variables, the necessary conditions for the existence of an extremum of it
Sec.
13]
_
The Extremum of a Function
of Several Variables
_
223
are similar to conditions (1), while the sufficient conditions are analogous to the conditions a), b), and c) 3. Example 1. Test the following function for an extremum:
Solution. Find the partial derivatives and form a system of equations
(1):
or r
**
\ xy
we
Solving the system
Let us find
(
(2,
P 4 (_2,-1).
P,(-l,-2);
1);
second derivatives
tiie
d2 z adx 2
=c
6.v,
d2z 3 T-
dxdy
For the pomt
= 6, A^=4C
2
P
t
:
= 36
<
144
0.
6r/,
for
c = 6x
dy
each stationary point.
=6. B =
A = (g} 2
\dx Jp
d*z T-22
= ry
B2
and form the discriminant A=^/4C 1)
/
get four stationary points:
Pt
P,(l,2);
+ *_5-0, 2 = 0.
=12, C=(g) = (fL\ \dy J p, \dxdyjp, 2
l
Thus, there
2) For the point P 2 4 --12, B^6, C-12; the function has a minimum. This minimum function for A -2, y~\' :
no extremum at the point P,.
is
A
= 144
is
36
equal
> 0,
>
/I
0.
the value
to
3012^28. 6; A = 36 ^-6, 12, C^ ^- 12, B = 6, C= 12; A = 144
At P 2 of
the
)
i
3) For the point P 9 no extremum. 4) For the point P 4
:
:
At the point
P4
fi---
the function has a
maximum
equal to
2 ma x
144
36
< 0.
>
0,
^
There
A
< 0.
6-f-30-{-
4- 12 ---28 5*.
of
Conditional extremum. In the simplest case, the conditional extremum
a function /(A,
attained <|)(jr,
on
//)
is
a
maximum
condition
that
its
or
//),
minimum
this function which is related by the equation the conditional extremum of a funcwe form the so-called Lagra
arguments
(coupling equation). To find given the relationship q> (A-,
w)
tion /(A-,
the
of
are
=
function F(A-,
where X
is
y)-=f(
an undetermined
multiplier, and
we seek
the ordinary
of this auxiliary function. The necessary conditions for the to a system of three equations:
with three unknowns to
x,
t/,
X,
determine these unknowns.
from which
it
is,
generally
extremum
extremum reduce
speaking,
possible
_ _
[Ch. 6
Functions of Several Variables
224
The question of the existence and character of a conditional extremum is solved on the basis of a study of the sign of the second differential of the Lagrange function: -
yi for the given
system
dx 2
y
dxdy
of values of x, y,
dy
2
h obtained from
(2)
or the condition
that dx and dy are related by the equation
<
Q, and a Namely, the function / (x y) has a conditional maximum, if d*F conditional minimum, if d 2 F 0. As a particular case, if the discriminant A of the function F (x, y) at a stationary point is positive, then at this point there is a conditional maximum of the function / (x, y), if A (or C 0), and a conditional minimum, if A 0) (or C In similar fashion we find the conditional extremum of a function of three or more variables provided there is one or several coupling equations (the number of which, however, must be less than the number of the variables) Here, we have to introduce into the Lagrange function as many undetermined multipliers factors as there are coupling equations. Example 2. Find the extremum of the function t
:
>
<
>
>
z=:6
4*
<
3y
provided the variables x and y satisfy the equation
x*-\-y*=\ Solution. Geometrically, the problem reduces to finding the greatest and 6 4.v plane z 3y for points of its intersection with the cylinder ji 2 -f// 2 =l We form the Lagrange function least values of the e-coordinate of the
F(x, y)--=6
We
have
T=~
following system
4
= ***-
+ 2>jr,
-
4x-3f/-l-M* 3
+ 2X#.
The
of equations:
: i
Solving this system
we
find i
-
5
^-"2"'
X
_
4
'-~5~'
and 2
____^_ ~" 2
1
___
^~""5"
Since
dx 2 it
follows that
-*"'
dxdy
=0,
f
2
2
-|-{/
1).
necessary conditions yield the
Sec.
The Extremutn
13]
54
~__
x
and
-=-
o
2
3 = --, o
Function of Several Variables
thend 2 /7 >0,
and,
225
the function
4^3
consequently, 5
=
=
-- and f/ -=O D and, consequently, the function at this point has a conditional
has a conditional z
F
then d
/
of a
minimum
at this point.
If
K-
-^ Z
x
,
,
Thus,
= 6 + i5 + -=ll,
z max
A
6. Greatest and smallest values of a function. rentiable in a limited closed region attains its greatest (smallest) value either at a sta-
function
that
is
diffe-
tionary point or at a point of the boundary of the region.
Example
3.
Determine the greatest and
smallest values of the
function
in the region
A'<0, [/<0, x + y^z
3
Solution. The indicated region is a angle (Fig. 70). 1) Let us find the stationary points: I
z' K
|
7 {J
~ 2xy 1=0, ^ 2y x -}-1^0; \-
whence x-= 1, //-= At A1 the value to test for
tri-
70
M
and we get the point 1 the function ZM
1;
of
=
It
(
1,
is
not absolutely necessary
an cxtrcmum
1)
Let us investigate the function on the boundaries of the region. we have 2 [/ 2 -f-f/, and the problem reduces to seeking the When A greatest and smallest values of this function of one argument on the interval 2)
=
3^//^0.
Investigating,
=
(2 sm ) v _
(3,0);
When
find
get
(Zsm) y =*
x-[-y
=
z
that
(2g r ) x=0
T" at thc 3
A-
P oint
we
find
that
(z &m ) x
+
,,^^^~r
we
(~ V 2
will
3
we
the
at
point
3);
(0,
l^)
x z -\-x. Similarly,
3 or //-=
=6
l
at the point (0,
When //~0 we point
we
at the point
'
(
(2g r ) v=0
=6
at
the
)
have /
that
find
z
= 3A
3 -^
2
-f-9A'-j-6.
3 \ ~
,
J
;
Similarly
(2 gr )
+ =
3 metres coincides with (z gr ) x =o anc (^r).jf=o- ^ n * ne straight line jc ^ we could test the Function for a conditional extremum without reducing to a function of one argument. 3) Correlating all the values obtained of the function z, we conclude at the stationary 1 that z gr 6 at the points (0, 3) and (3, 0); z sm ^
=
point
8-1900
M.
=
226
_ _
[Ch. 6
Functions of Several Variables
maximum and minimum
Test for
the following
functions of
two variables:
2013.
= (x \)*+2y*. = (x I) 2y*. 2x y. z = x* + xy + tf = z *y (6 xy)(x>0 2 = x +y 2x*+4xy 2y z =x
2014.
z=l
2008. z
2
2009. z 2010. 2011. 2012.
9
= 2 =
2015. z 20,6.
4
4
.
(
Find the extrema 2 2017. a = x
of the 2
+ +z
=
2018.
z
xy
^++
Find the extrema
functions
following
2
f/
+
+x
of
three variables:
2z.
-(^>0, y>0, z>0).
of the following implicitly represented func-
tions:
2019*. x* 2020. x
9
+ y* + z* 2x+4y 62 11=0. y*3x + 4y + z* + z 8 = 0.
Determine the conditional extrema 2021. z
= xy
for
+ 2y z = x* + if
2022. z^=x
for
2023.
for
2027.
= cos + cos W= 2y + 2z ^=^ + +2 u = xtfz*
2028.
u^xyz
2
2024. z 2025.
2026.
A:
2
2
-=
+
a
A:
8
f/
f/
for// for
for
of the following functions:
*
A: 2
A:
+ -
= ~-.
r
+
for .v-f-y
-
+ z=
12(jc>0,
provided x-(~j/+e=5,
xy+yz+zx=8.
2029. Prove the inequality
if
x^zQ Hint: Seek the
maximum
of the
function u
= xyz
provided
Finding the Greatest and Smallest Values of Functions
Sec. 14]
in
227
z=l+x + 2y
2030. Determine the greatest value of the function the regions: a) x>0, y^*0, x y^l\ b)
+
2031. Determine the greatest and smallest values of the func2 2 #* tions a) z y in the region x?+y *^l. x*y and b) z 2032. Determine the greatest and smallest values of the func-
=
=
tion
z
= sinx-h sin y-f- sin (x + y)
in
the
O^jt^-2.-
region
2033. Determine the greatest and smallest values of the func1 x' tion z 3xy in the region y*
=
+
^
0^x^2,
14. Finding the Greatest and Smallest Values of Functions
Sec.
Example t. It is required to break up a positive number a into three nonnegative numbers so that their product should be the greatest possible. Solution. Let the desired numbers be x, y, a x We seek the maxi//.
mum
the function / (x, y) xy(a to the problem, the closed triangle x^O, y^zQ, x of
According
+ y^a
x y). function
/ (x,
is
y)
considered inside a
(Fig. 71).
Fig. 71
Solving the system of equations
f'(x,
we
will
have the unique stationary point
(-TTI
triangle. Let us test the sufficiency conditions.
>
(/)
=
20,
T We
(x, y)=a2x2y, f IXIf
^
or
* ne
j
have f" yy'(x,
/)
=
of
the
_ _
228
[Ch. 6
Functions of Several Variables
Consequently,
And
so
at
f-g-,
~)
the function reaches a
the contour of the triangle, this
maximum
to say that the product will be greatest, a3 greatest value is equal to is
maximum.
Since f(x, */)=0 on
will be the greatest value, if
x
=y=a
x
//
which
= --, and
the
.
-^=-
Note
The ploblem can
also be solved
extremum, by seeking the maximum that x
+ y + z = a.
2034.
From among
all
of the
by the methods of a conditional function u xyz on the condition
rectangular
=
parallelepipeds
with
a
given volume V, find the one whose total surface is the least. 2035. For what dimensions does an open rectangular bathtub given capacity V have the smallest surface? 2036. Of all triangles of a given perimeter 2p, find the one that has the greatest area. 2037. Find a rectangular parallelepiped of a given surface S
of a
with greatest volume. 2038. Represent a positive number a in the form of a product of four positive factors which have the least possible sum. on an x^- plane, the sum of 2039. Find a point (x, y), the squares of the distances of which from three straight lines = Q, f/ = 0, x y+l=0) is the least possible. (x 2040. Find a triangle of a given perimeter 2p, which, upon being revolved about one of its sides, generates a solid of greatest volume. 2041. Given in a plane are three material points P, (x lf # ), P*( x z' #2)* ^i(*> #3) w ith masses m lf m 2 m 3 For what position of the point P(x, y) will the quadratic moment (the moment of inertia) of the given system of points relative to the point P 2 P2 P2 P 3 P 2 ) be the least? (i.e., the sum m.P.P 2 3 2042. Draw a plane through the point c) to form (a, b a tetrahedron of least volume with the planes of the coordinates. 2043. Inscribe in an ellipsoid a rectangular parallelepiped of greatest volume. 2044. Determine the outer dimensions of an open box with a given wall thickness 8 and capacity (internal) V so that the smallest quantity of material is used to make it.
M
t
,
+m
+m
.
M
t
Sec.
14]
Finding the Greatest and Smallest Values
of
229
Functions
2045. At what point of the ellipse w2
r2
-Ta2
does the tangent line to angle of smallest area? 2046*. Find the axes
2047.
Inscribe in surface.
a
z
'
form with the coordinate axes a
it
of
5x
+ TT-1 b tri-
the ellipse
2 -f-
8xy + 5tf
given
= 9.
sphere
a
cylinder
having
the
greatest total
2048. The beds of two rivers (in a certain region) approxi2 = 0. y mately represent a parabola y = x* and a straight line x It is required to connect these rivers by a straight canal of least length. Through what points will it pass? 2049. Find the shortest distance from to the straight line x __ ~~ 1
_
y
~~ 3
the
point
M(l,
2,
3)
2 '
2
2050*. The points A and B are situated in different optical media separated by a straight line (Fig. 72). The velocity of
*1
Fii<.
72
Fig. 73
light in the first medium is v l9 in the second, v 2 Applying the Fermat principle, according to which a light ray is propagated along a line AMD which requires the least time to cover, derive the law of refraction of light rays. 2051. Using the Fermat principle, derive the law of reflection of a light ray from a plane in a homogeneous medium (Fig. 73). 2052*. If a current / Hows in an electric circuit containing a .
resistance
/?,
proportional
then the quantity of 2
to /
/?.
heat
released in unit time is / into
Determine how to divide the current
230
_ _
currents / t
#i
[Ch. 6
Functions of Several Variables
,
72 ,
so
#a
R*> possible?
/,
by means
that
the
of three wires,
generation of
whose resistances are
heat would be the
least
Sec. 15. Singular Points of Plane Curves
1. Definition of a singular point.
=
is
f(x, #) at once:
2. Basic
called a singular point
if
A
At
point Af (x c y Q ) of a plane curve coordinates satisfy three equations ,
its
M (*
let
the
a node (double point) (Fig. 75); is either a cusp of the first kind (Fig. 76) or of c) if second kind (Fig. 77), or an isolated point, or a tacnode (Fig. 78).
the
types of singular points.
a singular point
,
#
),
second derivatives
be not
all
equal to zero and
then: a)
b)
if if
A>0, A<0,
A = 0,
then then then
M M M
is
A = /1C-B 2
,
an isolated point (Fig.
74);
is
Fig. 74
Fig. 75
When solving the problems of this section it is always necessary to draw the curves. z x* has a node if a ax* 0; an Example 1. Show that the curve y isolated point if a 0. 0; a cusp of the first kind if a 2 z Solution. Here, f (x, y)z=zax Let us find the partial derivatives and equate them to zero:
=
<
+ x'y
fx
(x,
t/)
=
.
+
=
>
Singular Points of Plane Curves
Sec. 15]
system has two solutions:
This
coordinates of the point
Hence, there
is
a
N
do not
0(0, 0) and the
satisfy
unique singular point
(0,
N
231
(
equation
--~a
Oj,'btit
the
0).
Fig. 78
Fig. 77
Fig. 76
t
of the given curve.
Let us find the second derivatives and their values at the point 0: ,
4=20, 0=0,
a =0 fa>0
Pig.
Fig. 80
79
Fig. 81
Hence, if
a>0,
if
a
if
2 {/
=x
< 0,
a--^0, 8
or
#=
then then then
A<0 A
>
Aj^O. y which
Y^\
about the x-axis, first kind (Fig. 81).
is a node (Fig. 79); and the point is an isolated point (Fig. 80); and The equation of the curve in this case
exists only is a tangent.
when Jc^O;
the curve
Hence, the point
M
is
is
will
be
symmetric
a cusp of the
232
Functions of Several Variables
Determine the character
the
of
[Ch. 6
singular points of the follo-
wing curves: 2053. y* 2054. (y
=
x* -\-x\
2055.
=* ay=aV-x'.
2056.
jtyjt
2057. x*
x
+y
2
8
5
.
)
2
9
f/^O.
3axy = Q (folium
2061.
of Descartes).
= (cissoid). = a*(x y (lemniscate). + = (a x)x (strophoid). + = ftV (a>0, 6>0) (x* + y*)(x a)* 3
2058. y*(a x) 2 2059. (x* y*) 2060. (a x)y*
jc
2
z
)
2
(conchoid).
Consider three cases: 1)
a>6,
a
2)
= 6,
a<6.
3)
2062. Determine the change in character of the singular point 2 curve y (x b) (x a)(x c) depending on the values of
=
of the a,
6,
c(a
are real).
Sec. 16. Envelope
t. Definition of an envelope.
The envelope of a family of plane curves curve (or a set of several curves) which is tangent to all lines of the given family, and at each point is tangent to some line of the given family. 2. Equations of an envelope. If a family of curves is
a
f(x
dependent on
a)=0
y,
t
a single variable parameter of the latter are found
metric equations
a has an envelope, then the parafrom the system of equations
= 0, a) = o.
t
f(x. y, a)
\
&(*.
y.
Eliminating the parameter a from the system the form
D(x,
0)
(1)
(1),
we
= 0.
get an equation of (2)
It should be pointed out that the formally obtained curve (2) (the *>ocalled "discriminant curve") may contain, in addition to an envelope (if there is one), a locus of singular points of the given f amily, which locus ts not part of the envelope of this family. When solving the problems of this section it is advisable to make
drawings.
Example. Find the envelope
*cosa+f/sina
of
the family of curves
p
= 0(p = const,
p>0).
Sec. 16]
233
Envelope
Solution. The given family of curves depends on the parameter a. the system of equations (1):
J \
Solving the envelope
system
for
x
*cosa + y sin a x
sin
x and y
a+y t
cos
p = 0,
a
= 0.
we obtain parametric equations
= pcosa,
r/
of the
=
Squaring both equations and adding, we eliminate the parameter
frig.
Form
a:
82
Thus, the envelope of this family of straight lines is a circle of radius p with centre at the origin. This particular family of straight lines is a family of tangent lines to this circle (Fig. 82).
2063. Find the envelope of the family of circles
2064. Find the envelope of the family of straight lines
(k is a variable parameter).
2065. Find the envelope of a family of circles of the same radius R whose centres lie on the jc-axis. 2066. Find a curve which forms an envelope of a section of length / when its end-points slide along the coordinate axes. 2067. Find the envelope of a family of straight lines that form with the coordinate axes a triangle of constant area S. 2068. Find the envelope of ellipses of constant area S whose axes of symmetry coincide.
Functions of Several Variables
234
2069. Investigate the character
the "discriminant curves" a constant parameter):
of
of families of the following lines (C
=
=
;
=
+ x) (y
d) strophoids (a
is
8
cubic parabolas y (x C) 2 b) semicubical parabolas t/ (x 2 C) (x c) Neile parabolas y* a)
[Ch. 6
C)*;
;
=*
8
C)
2
x).
(a
Fig. 83
2070.
The equation
trajectory of a shell fired from a v at an angle a to the horizon
of the
with initial velocity point resistance (air disregarded) is
Taking the angle a as the parameter, find the envelope of all trajectories bf the shell located in one and the same vertical plane ("safety parabola' ) (Fig. 83). 1
17. Arc Length of a Space Curve
Sec.
The dinates
where
differential of equal to
an arc
of a
space curve in rectangular Cartesian coor-
is
current coordinates of a point of the curve.
z are the
x, y,
If
are parametric equations of the space curve, then the arc length of a section of
it
from
t
=
t
l
to
t
=
t2
/a
=
f j
is
M
1/7
,
,
(-df)
The Vector Function
Sec. 18]
of a Scalar
235
Argument
In Problems 2071-2076 find the arc length of the curve:
=
2071. x
t,
y=/
2
2-~-'
,
from
to
t
= 2.
= 2 cos y = 2 sin z = -|- from == to = = *' cos y = e* sin z = e from = to arbitrary t from JC==0 to x==6 y = 4~' 2 = 4~ * = 3(/, 2jcy = 92 from the point (0, 0, 0) to M (3, 3, 2). = aarcsin~, z = -|-ln ^j from the point 0(0,0,0)
2072. x
/,
t
/
1
t,
it.
t
2073.
A:
2074.
/,
f
/,
.
-
f
2075. 2076.
f/
M(*
to the point
,
j/
z
,
).
2077. The position of a point for any time by the equations
Find the mean velocity Sec.
=
/
of
(f>0)
f
motion between times
f
=
l
is
defined
and ^=10.
18. The Vector Function of a Scalar Argument
1. The derivative of the vector function of a scalar argument. The vector a (0 may be defined by specifying three scalar functions ax (t) a (t) and a z (t) which are its projections on the coordinate axes: function a
9
t
y
The derivative of the vector function a-=a(t) with respect to the scalar argument t is a new vector function defined by the equality
da The modulus
a(t
+ M)-a(t)_da x (t)
.
day (0
.
da f
(t)
of the derivative of the vector function is
da dt
The end-point
of the variable of the radius vector
r=r(/) describes
in
space
the curve
r=x(t)l+y(t)J+*(t)*. which
is
called the hodograph of the vector r.
The derivative
-~
is
a vector, tangent to the
hodograph at the corre-
sponding point; here,
dr [
\
where
s is
dt
[_ \
ds
dt
'
the arc length of the hodograph reckoned from
For example,
Up
some
initial point.
236
__ If
the
parameter
Functions of Several Variables
the time, then -jT
is
t
=
tf
is
_
[Ch. 6
the velocity vector of the
=w
=
is * ne acceleration vector of the and JTS "^T of vector the r. extremity 2. Basic rules for differentiating the vector function of a scalar argument.
extremity
of the vector r,
= m-~-,
2)
-77-
(ma)
3)
-77-
(cpa)==-~-a
4,
7)
where
+ (p-^-
,
m
is
a constant scalar;
where q>(0
a scalar function of
is
<*,_..+..:
a-~=0,
if |
a
|-= const.
Example 1. The radius vector time defined by the equation
r=i Determine the trajectory Solution.
From
Eliminating the
(1)
time
of
a
of
moving point
of
motion, the velocity and acceleration.
t,
we
find that the trajectory of 1
~~
u
z '
differentiating,
we
find the velocity
and the acceleration
The magnitude
any instant
(1)
4"" 3 (1),
at
we have:
x
From equation
is
4t*j+3t*k.
line:
We
/;
of the velocity
note that the acceleration
\*L
\dt*
is
is
constant and
= ]/(_ 8) +6 2
2
is
=iO.
motion
is
a straight
Sec.
_
18]
Show
2078.
where r
The Vector Function
l
the
that
of
a Scalar Argument
vector
and r 2 are radius vectors
_
= (/* 2 equation two given points,
rr,
of
237
r,) is
/,
the
of a straight line. 2079. Determine which lines are hodographs of the following vector functions:
equation
a)
= at -f c\ = a/ + r r
c)
2
b)
ft/;
d)
= a cos -f b sin r = a cosh -f 6 sinh
r
t
t\
/
/,
a, ft, and c are constant vectors; the vectors a and b are perpendicular to each other. 2080. Find the derivative vector-function of the function a(t)a (/), where a(t) is a scalar function, while a(/) a(t) is a unit vector, for cases when the vector a(t) varies: 1) in length only, 2) in direction only, 3) in length and in direction (general case). Interpret geometrically the results obtained. 2081. Using the rules of differentiating a vector functisn with respect to a scalar argument, derive a formula for differentiating a mixed product of three vector functions a, 6, and c. 2082. Find the derivative, with respect to the parameter t, of the volume of a parallelepiped constructed on three vectors:
where
=
a = i + tj+t z k,
2083.
The equation
of
r
motion
is
= 3/cos/~j-4/sinf,
is the time. Determine the trajectory of motion, the the acceleration. Construct the trajectory of motion and velocity and the vectors of velocity and acceleration for times, / = 0,
where
/
.
2084. The equation of motion
is
Determine the trajectory of motion, the velocity and the accelWhat are the magnitudes of velocity and acceleration and / = and what directions have they for time = -y? 2085. The equation of motion is
eration.
r= i cos a cos at -\-jsmacostot + k sincof, where a and o are constants and / is the time. Determine the trajectory of motion and the magnitudes and directions o! the velocity and the acceleration.
-
_
Functions of Several Variables
238
2086. The equation of motion
of
shell
a
\Ch. 6
air re-
(neglecting
is
sistance)
where V Q {V OX v oy v oz } is the initial velocity. Find the velocity and the acceleration at any instant of time. 2087. Prove that if a point is in motion along the parabola ,
,
2
z/=
,
z
manner
in such a
on the x-axis remains constant
that
the
of
projection
[-= const],
velocity
then the accelera-
tion remains constant as well.
2088. A point lying on the thread into a beam describes the spiral
of
a
screw being screwed
where 9 is the turning angle of the screw, a is the radius of the screw, and h is the height of rise in a rotation of one radian. Determine the velocity of the point. 2089. Find the velocity of a point on the circumference of a wheel of radius a rotating with constant angular velocity co so that its centre moves in a straight line with constant velocity V Q .
Sec. 19. The Natural Trihedron of a Space Curve
At any nonsingular point
M
(;c,
//,
sible to construct a natural trihedron
z) of a
space curve r
consisting
of
three
r(t)
is
it
pos-
mutually perpen-
dicular planes (Fig. 84):
M A4jA4
containing the vectors
1)
osculating plane
2)
normal plane
3)
rectifying plane MMjAIj, which
MM*M
S,
2
,
which
is
perpendicular to is
At the intersection we obtain three straight lines; 1) the tangent 2) the principal normal MAf 2 of which are defined by the appropriate vectors: 1)
2) 3)
T=-rr
(the vector of the tangent l'me)\
^-gfx^p
(
NBXT (the
the vect
r
f tne
binormal);
vector of the principal normal)]
The corresponding unit vectors
T
o
and
-r-^
the vector
JV
;
3)
;
~
and
at
perpendicular to the first
MM^
all
TT
two planes.
the binomial
MM
9t
The Natural Trihedron
Sec.
19]
may
be computed from the formulas
dr
t= dF
v
;
of
a Space Curve
239
ds
=
dt ds
If X K, Z are the current coordinates of the point of the tangent, then the equations of the tangent have the form f
X-x = Y
Z
y==
'
3C
*
z
(I)
'2
if
Normal Rectifying
plane
plane
Osculating plane
Fig. 84
where of
Tx
--
;
Tv=
~
the line and the plane
T 2 = -~
,
we
;
from the condition
get an equation
of the
Z
2)
of
perpendicularity
normal plane:
= 0.
T x Ty 7\ by B x
(2)
Bv
B z and A^.
(1) and (2), we replace get the equations of the binomial and the principal normal and, respectively, the osculating plane and the rectifying plane. Example t. Find the basic unit vectors T, v and p of the curve If
in
Wy,
equations
^V^,
,
at the point
i
at this point.
Solution.
and
,
,
= 1.
Write the equations mal
,
we
We
have
of
the tangent, the principal normal and the binor-
[Ch. 6
Functions of Several Variables
240
Whence, when
1,
we
get dt
=
dt
X
2 3 ^= 026 1
dt i
J
>'
1
2
3
662 Consequently, T
Since for
= <+2/+3A
P=
3/-ay+*
V
?=1 we have *=1, y=l, 2=1,
~
'
-1U-8/+9*
follows that
it
"~
""
2
1
3
are the equations of the tangent,
x\_y ~~
l_z ~~ 3
3
1
1
are the equations of the binomial and
*-l
z-1
y-1
9
8
11
are the equations of the principal normal. If a space curve is represented as an intersection of F(JC, y, z)
then in place
of
the vectors
-^-
= 0, and
G(x, TT-Z
y, 2)
two surfaces
= 0,
we can take
the vectors dr{dx, dy, dz}
and d 2 r {d*x, d*y, d z z}; and one of the variables x, y, z may be considered independent and we can put its second differential equal to zero. Example 2. Write the equation of the osculating plane of the circle *2 at its point
M(l,
1,
+ + z = 6, 2
2
J/
x
+ y + z^Q
(3)
2).
Solution. Differentiating the system (3) and considering x an independent we will have
variable,
x dx
+ y dy + z dz
--=
0,
and dx*
+ dy + y d*y + dz + z d*z .= 0, 1
2
d 2 (/ Putting
x=l, y=\>
z~2,
we
get
The Natural Trihedron of a Space Curve
Sec. 19]
Hence, the osculating plane
dx
{dx,
t
241
defined by the vectors
is
-
and
0}
|o,
2
yd*
,
jdx*\
or 1,
{1,
and
0}
1,
{0,
1}.
Whence the normal vector of the osculating plane J
B=
=
1
1
-1 and, therefore,
its
equation
is
lj
k
1
is
-l(x-l that
as
is,
it
should be, since our curve
is
located in this plane.
2090. Find the basic unit vectors
x^l at the point
f
= -g-
T, v,
y=sin/,
cosf,
p of the curve z
=
t
*
2091. Find the unit vectors of the tangent and the of the conic spiral
principal
normal
at an arbitrary point. Determine the angles that these lines with the z-axis. 2092. Find the basic unit vectors r, v, p of the curve
y
=
x*,
z
make
= 2x
2. at the point x 2093. For the screw line
y
= asmt,
z
= bt
write the equations of the straight lines that form a natural trihedron at an arbitrary point of the line. Determine the direction cosines of the tangent line and the principal normal. 2094. Write the equations of the planes that form the natural trihedron of the curve x*
one
2
-1- 1/
+ * = 6, 2
x
2
if
-1-
z
2
-4
points M(l, 1, 2). the equations ot the tangent line, the normal z t* the and t*, /, osculating plane of the curve * plane y at the point (2, 4, 8).
at
of
2095.
its
Form
M
=
=
=
_ _
[Ch. 6
Functions of Several Variables
242
2096.
Form
and binormal
the equations of the tangent, principal normal, an arbitrary point of the curve
at
Find the points at which the tangent to to the plane x 10 = 0. 3y-\- 2z
this curve is parallel
+
2097. Form equations of the tangent, the osculating the principal normal and the binormal of the curve
at the point / at this point.
= 2.
Compute the
direction cosines of the binormal
2098. Write the equations of the tangent and plane to the following curves: a)
x = R cos 2
b)
z=x*+y*, x = y *
C)
2
/,
y = R sin /cos/,
+ y + z = 25, 2
2
z
= Rsmt
for
/
x+z=5
=
=
2/3,
at the point (2,
normal Find the equation y = x at the coordinate origin. 2100. Find the equation of the osculating plane
*
normal
the
?-;
at the point (1,1, 2);
2099
z=x
plane,
of
3).
plane to the curve
the
z
if,
*,
(/
= <>-',
2101. Find 2
* 2 b) * 2
a)
c)
JC
to the
curve
=
= ty2
0. at the point / the equations of the osculating plane to the curves:
2
+y + 2 = 9, x y = 3 at the point (2, 1, 2); = 4y, x' = 24z at the point (6, 9, 9); + z = a y fz = 6 at any point of the curve (x 2
2
2
2
2
2
2
2
2
,
QJ
y ot z
).
Form
the equations of the osculating plane, the principal normal and the binormal to the curve
2102.
z
y 2103.
Form
= x,
*
2
=z
at the point (1,
1,
1).
the equations of the osculating plane, the princi-
normal and the binormal to the conical screw-line A; = /COS/, j/=/sin/, z~bt at the origin. Find the unit vectors of the tangent, the principal normal, and the binormal at the origin. pal
Sec. 20. Curvature and Torsion of a Space Curve
1. Curvature.
By
the curvature of a curve at a point
number *
/(
R
=
lim AS-+O
JL, As
f
M
we mean the
Curvature and Torsion
Sec. 20]
of
a Space Curve
243
the angle of turn of the tangent line (angle of continence) on a curve MN, As is the arc length of this segment of the curve. segment R is called the radius of curvature. If a curve is defined by the equation r=r(s), where s is the arc length, then
where
(p
is
of the
For the case
we have
parametric representation of the curve
of a general
(1)
2. Torsion. By mean the number
torsion (second curvature) of a curve
at
a
point
M
we
-1 r-l-ihn As-*o As Q
the angle of turn of the binormal (angle of contingence of the where second kind) on the segment of the curve MN. The quantity Q is called the radius of torsion or the radius of second curvature. If r=r(s), then is
drd 2 rd sr ds ds 2 ds 3
dp_
ds
where the minus sign
is
direction, and the plus If /=/(/), where t
taken when the vectors
when
sign,
and v have
the
an arbitrary parameter, then
is
1
=
dr d 2r dV d/d?" d7 -
(2)
Q
~d,
dt
Example
1.
Find the curvature and the torsion of the screw-line r
Solution.
We
same
not the same.
= i a cos
t
-\-j
a sin
/
+ k bt
t
+ja cos t + kb
(a
> 0).
have
~= _/ a
_
sin
t
2
d r
==
/
a cos
/
= _/ a sin/
/a
sin
/,
cos/.
Ja
Whence k
J a sin a cos
t
/
a cost b -a sin
/
=
i
ab sin
/
jab cos
t
+ a*k
Hence, on the basis
formulas
of
1
JDL
R ~(
we
(2),
+ 6i)
a
"fl'
/.
get
a
VaT+b*^ i
fl
and
(1)
+a
1
and 1
~
Q
a*b
a 2 (a 2
b
=
+ b*)~~ a + b 2
2
'
Thus, for a screw-line, the curvature and torsion are constants. 3 Frenet formulas:
dr
_
v
^v '
7s~"~R
2104. Prove
___ ~~
5s
T
iP
^P_
v
'
ds~~~~o"'
~~~~R~T~~Q
the curvature at all points of a line is a straight line. 2105. Prove that if the torsion at all points of a curve is zero, then the curve is a plane curve. 2106. Prove that the curve that
zero, then the line
if
is
x=\+3t + 2t is
a
2 ,
y
= 22t + 5t\ z=lt*
plane curve; find the plane in which it lies. Compute the curvature of the following curves:
2107. a)
x = cost, y = s'mt, 2 2 2 z = l, y -| //
b) x*
2108. curves: a)
b)
z
= cosh
= e'cos/,
=
at the point / 0; at the point (1, 1,
Compute the curvature and
y = e sint, z x^acosht, y asiuht.
je
/
-2x + z = Q
1).
torsion at any point of the
i
e*\
z
= at
(hyperbolic screw-line).
2109. Find the radii of curvature and torsion at an arbitrary point (x, y, z) of the curves: 2
x 9 b) x
a)
= =
2110. Prove that the tangential and normal are expressed by the formulas acceleration v2 dv VOif T
w
WT-^T,
Wv-
components
of
R
where v is the velocity, R is the radius of curvature of the T and v are unit vectors of the tangent and principal trajectory, the to curve. normal
Sec. 20]
2111.
=
_
_
la cost
tion w. 2112.
Curvature and Torsion of a Space Curve
A
+ja
sin
in
is
point t
+ btk
The equation
Determine,
at
trajectory and acceleration.
times 2)
the
/
of
24S
uniform motion along a screw-line r =* with velocity v. Compute its accelera-
motion
is
and /=!: 1) the curvature of the tangential and normal components of the
Chapter VII
MULTIPLE AND LINE INTEGRALS
Sec. 1. The Double Integral
in
Rectangular Coordinates
1. Direct computation of double integrals. The double integral of a continuous function f (x, y) over a bounded closed region S is the limit of the corresponding two-dimensional integral sum
f (x,
y)dx dy =
lim max A*i max Ar// -
(1)
c
&y k = yk+l yk and the sum is extended over those xg, and k for which the points (*/, y k ) belong to S. 2. Setting up the limits of integration in a double integral. We distinguish two basic types of region of integration.
where A*
-
t
values of
x,
= Xf +l
i
x
o Fig. 85
1)
The region
of integration
Fig. 86
5
(Fig. 85)
is
bounded on the
left
and
right
while the vaFig. 85). In the region S, the variable x varies from x l to x riable y (for x constant) varies from 9, (x) to j/ 2 q> 2 (x). The integral (1) ma>
^=
=
The Double Integral
Sec. 1]
247
in Rectangular Coordinates
be computed by reducing to an iterated integral by the formula *a
f(x, y)dy,
J (S)
where x
is
held constant
when
\
calculating
/(x, y) dy.
2) The region of integration S is bounded from below and from above by the straight lines y y and y yt(y z >yi), and from the left and the (y)], right by the continuous curves x = ^> (y) (AB) and x = 2 (y) (CD) [t|? 2 (y) each of which intersects the parallel y = Y (y* Y y t) at only one point l
(Fig. 86). As before,
^^
i|?
l
^ <
we have $2
Vl
JJ/(*.
y)dxdy=\dy
(S)
here, in the integral
j/i
\
f(x, y) dx
we
(U)
f (x.
J i|?,
y)dx
t
((/)
consider y constant.
If the region of integration does not belong to any of the above-discussed types, then an attempt is made to break it up into parts, each of which does belong to one of these two types. Example 1. Evaluate the integral
/
D
Solution.
Example
2.
Determine the limits
of integration of the integral
(x, (S)
y)dxdy
248
_ _ Multiple and Line Integrals
2
[Ch. 7 2
=
* 1 the region of integration 5 (Fig. 87) is bounded by the hyperbola y 2 and x 2 (we have in view the region constraight lines * taining the coordinate origin). Solution. The region of integration ABCD (Fig. 87) is bounded by the 2 and x 2 and by two branches of the hyperbola straight lines if
=
=
and by two
=
x-
y=yT+* that
is,
it
belongs to the
first
and
type.
We
=
(/--1 have:
dx
f(x, y)dy.
J
J
Evaluate the following iterated integrals:
21
2113.
2114
35
\dy\(jf + 2y)dx.
2117.
^pTv?31
2118.
Jdy
J
rrfr.
dcp
j a
o
sin
Jt_
2115
2
X 2dU
8
COS
(p
'
00 Write the equations of curves bounding regions over which the following dduble integrals are extended, and draw these regions: 2
2121.
2-t/
Jrf/
f(x, y)dx.
X+9
3
2122.
J
JdxJ IX 4
2125.
f(x, y)dy.
10-y
dy o
JdxJ/(*. y)dy. K25-JC-
3
3
2123.
2X
3
2124.
00 ^dx
f(x, ^)dx.
2126.
f(x, y)dy.
X+2
2
y
$
1
d*
-ix*
f(x, y)dy.
Set up the limits of integration in one order and then in the other in the double integral
JJ/(*. y)dxdy (S)
for the indicated regions S.
Sec.
The Double Integral
1]
2127.
C(0,
S
is
a rectangle
S S
is
a triangle
is
a
in Rectangular Coordinates
with vertices 0(0,
0),
249*
4(2,0), 5(2,
1),
1).
2128. 2129.
C(0,
with vertices 0(0,
4(1,
0),
trapezoid with vertices 0(0,
0),
A
0),
(2, 0),
5(1, 5(1,
1).
1),
1).
2130.
C(2,
7),
2131. 0),
(0,
S
parallelogram with
a
is
D(l,
5).
S
a
is
whose
circular sector arc end-points are
vertices
045 A
4(1,
2),
5(2,
4),
with centre at the point and 5 f - 1, 1) (Fig. 88).
(1, 1)
Fig
89
right parabolic segment 405 bounded by the a segment of the straight line 54 connecting and parabola the points 5(~-l, 2) and 4(1, 2) (Fig. 89). 2133. S is a circular ring bounded by circles with radii r=l and /?-=2 and with common centre 0(0, 0). 2134. S is bounded by the hyperbola if x? \ and the circle
2132.
S
is
a
504
x
2
2
|
^9
(the region containing the origin is meant). 2135. Set up the limits of integration in the double //
(x,
integral
y)dxdy
(S)
if
the region S
b) * c)
x
2
2
h//
defined by the inequalities
is
2
e)
;
|-
//*
*;
Change the order
//
2
=
2a;
:a.
of integration in the following
double integrals:
\2X
2136.
f(x, y)dy.
2137.
JdJ/(x.
y)dy.
Multiple and Line Integrals
250
2138.
\dx
$
\
f(x, y)dy.
2141. \
d*
dy
\
f(x
t
y)dx.
-
VTZx^tf
a
2139.
-
[Ch. 7
*
f(x, y)dy.
J
2142.
\dy
V*-*
$/(*, yi ~2
2
20
2140.
Jdx
f(x
J
9
y)dy.
RVT A
2143.
J
-V
d*J/(x,
o
sin
2144.
\dx
t/)dy+
f(x. y)dy.
o
x
/)d/.
J/(jc,
Evaluate the following double integrals: ( where S is a triangle with vertices 0(0, [ xdxdy
2145.
t
0),
(S)
A(\,
1),
and B(0,
1).
A(2,0)X Fig. 90
2146.
^xdxdy,
where the region
Fig. 91
of integration
Sis bounded
(S)
by the straight line passing through the points A (2, 0), fi(0, 2) of a circle with centre at the point C(0, 1), and
and by the arc radius
1
(Fig. 90).
The Double Integral
Sec. 1]
*
2147.
Vr a ,
\
\
JJ
2
in Rectangular Coordinates
===, where S x*y*
is
251
a r part of a circle of radius
(S)
a with centre at 0(0, 2148.
lying in the
first
where S
a
y dx dy
V**
$ $
0)
2
t
is
quadrant. triangle with
vertices
(S)
A
0(0,0), 2149.
-1), and
(I,
fl(l,
1).
y*dxdy, where S
\j !/"*#
triangle with vertices
a
is
(S)
0(0,
4(10,
0),
rr 2150.
ed
J J (S)
1),
and
fl(l,
1).
-
dxdy, where S
ey
by the
parabola
y*
is
=x
a curvilinear triangle
and
OAB
bound-
x = Q, (/=!
the straight lines
(Fig. 91).
2151.
ff^Ti, -
the parabola 2152. extend:
y=7f and
Compute the
is
a parabolic segment
the straight line y
integrals
\dx
[ j
j
= x.
_rt
2
tfsmxdy; c)
$
2
bounded by
and draw the regions over which they
i-f-cosjc
at
a)
where S
$/
x* sin'
,
1
"
b) COS*
When solving Problems 2153 to 2157 the drawings first. 2153. Evaluate the double integral
it
is
abvisable to
make
(5)
if
S
line
is
a region
x = p.
bounded by the parabola
y*
= 2px and
the straight
2154*. Evaluate the double integral
^xydxdy, extended over the region S, which 2 2 and an upper semicircle (x 2) #
is
+ =
bounded 1.
by
the #-axis
_ _ Multiple and Line Integrals
252
(Ch. 7
2155. Evaluate the double integral
dxdy
/25=5
f
(S)
where S
the area of a circle of radius a, to the coordinate axes and lies in the gent 2156*. Evaluate the double integral is
which first
circle
is
tan-
quadrant.
^ydxdy, (S)
where the region S
is
bounded by the axis
of abscissas
and an
of the cycloid
= R(t y = R(\
x
sin/),
cos/).
2157. Evaluate the double integral
(S)
in which the region of integration nate axes and an arc of the astroid
2158. Find the
region
mean value
S
is
bounded by the coordi-
of the function f(x, y)=--xy
2
in the
SJO^Jt^l, 0
Hint.
The mean value
of a function
f(x, y)
the region
in
5
is
the
number
2159. Find the mean value of the square of the distance of 2 2 from the coordia point (x, y) of the circle (x nate origin.
M
Sec.
2.
Change
^R
af+y
of
Variables in a Double Integral
1. Double integral in polar coordinates. In a double integral, when passing from rectangular coordinates (x, y) to polar coordinates (r, cp), which are connected with rectangular coordinates by the relations #
we have
= /'Cos(p,
y
r sin
ip,
the formula
^{ (S)
f (*>
y)dxdy=(( (S)
(r
cos
q>,
r sin
cp)
r dr
e/cp,
(1)
Sec. 2]
_
Change
of Variables in a Double Integral
_
253
a integration (S) is bounded by the half-lines r the curves r and r r 2 ((p), where r l (q>) r,(cp) r z f
the
region
r=^p(a
of
=
and
and and
a^rp^p,
P
F
C f
r) r
(q),
dr
=
dcp
/-J
a
(S)
<
F
f
\ dcp
r) r rfr,
(cp,
r, (cp)
r 2 (cp)
where F
(cp,
In evaluating the
r sin (p).
r)~/(rcos(p,
\
integral
F
((p,
r)rrfr
'i (
we hold
the quantity (p constant. the region of integration does not belong to one of the kinds that has been examined, it is broken up into parts, each of which is a region of a If
given type.
2. Double if
integral in curvilinear coordinates.
the more general case,
In
double integral
in the
(x,
y)dxdy
from the variables x, y to the variables u, v, which it is required to pass are connected with x, y by the continuous and differentiate relationships A-^cp(w, i),
between the points region
S
f
of
the
of
t|?(K,
I/
v)
both directions, continuous) correspondence
that establish a one-to-one (and, in
and the points
region S of the .v//-plane if the Jacobian
the
of
some
UV- plane, and
w
*^
D(u retains a constant
f
dx dy du da
y)
,
V'
dx dy dv dv
t
sign in the region 5, then the formula
(^
y) dx
dy
-
I [cp (u,
$
v),
(w,
v)
]
|
/
|
du du
(6)
holds true
The limits of the new integral are determined from general rules on the basis of the type of region S' Example 1. In passing to polar coordinates, evaluate
where the region 5
is
origin (Fig 92). Solution. Putting x J/'l
x
2
radius
a circle of
rcoscp,
if^
Y"\
//
(r
/?
rsincp, cos
2
cp)
=
!
we
with centre
at
the coordinate
obtain:
(r sin cp)
2
= ^1
r8
Multiple and Line Integrals
254
Since the coordinate r in the region to 2jt, it follows that varies from
S
varies
2Jt
ff
Pass to
to
1
for
any
q>,
and
q>
1
y\x* coordinates r and cp and set up the limits of to the new variables in the following
polar
with
integration
from
(Ch. 7
respect
integrals:
2160.
2162.
K,
2161.
y)dy.
Jd*$/(]/J
JJ/(x, y)dxdy, \-'/
where S
a triangle
is
bounded by the straight
lines
(/
= #, y~
x,
2163.
*
-i
2164.
^f(x, y)dxdy, where S
is
bounded by the lemniscate
(S)
Fig. 92
2165. Passing to polar coordinates,
calculate the double inte
gral (S)
where S
C(f,0)
is
a semicircle
(Fig. 93).
of
diameter a with centre at the poin
_
Sec. 2]
Change of Variables in a Double Integral
_
255
2166. Passing to polar coordinates, evaluate the double integral
(S)
extended over a region bounded by the circle jc* y* = 2ax. 2167. Passing to polar coordinates, evaluate the double
+
in-
tegral
(S)
integration S is a semicircle of radius a with centre at the coordinate origin and lying above the #-axis. r 2168. Evaluate the double integral of a function f(r,
where the region
of
=
=
=
V a*-x*
a
Jdx
Vx' +
J
tfdy.
2170. Passing to polar coordinates, evaluate
(S)
where the region S
is
a
loop of the lemniscate
2171*. Evaluate the double integral
ll
T/^R;
(S)
extended over the region S bounded by the ellipse
x x* -j
+ ^^ u
passing to generalized polar coordinates: X
=
^wo
r
Y
,
y
2172**. Transform c
p*
\dx\f (x y)dy t
o
(0
ax
and c:>0) by introducing new variables u
256
_
__
Multiple and Line Integrals
2173*.
= x + y,
Change the variables u
v
= xy
[Ch. 7
in the integral
i
i
\dx\f (x,y)dy. 2174**. Evaluate the double integral
(S)
where S
is
bounded by the curve
a region
-_ k
b2
Hint.
Make
~~fi 2
'
2
the substitution br sin
cp.
The area
of
y
>,
Sec. 3. Computing Areas
1. Area
in rectangular coordinates.
a plane region S
is
(S) If
the region
S
defined
is
by the
a^x^b,
inequalities
q>
(x)
^y ^
\|)
(x)
,
then b
S
=
op
\dx
X)
J
a
is
(
cp
dy.
(x)
2. Area in polar coordinates. If a region S in polar coordinates r and defined by the inequalities a^cp^p, / (cp)^/' (q>), then
F
P
S
= ffrdcpdr=
C
a
(S)
6/9
q>
P)
C
/-dr.
/(q
2175. Construct regions whose areas are expressed by the in-
_
tegrals x+2
2
a))
dx J
j
d; dy;
b)
j
dy
d*. J
Evaluate these areas and change the order of integration. 2176. Construct regions whose areas are expressed dy the tegrals ji
arc tan
a)
] n_
T Compute
2
*
d(f
sec
J
2
rdr\
o
b)
J
n
"T" these areas.
a(i+coscp)
^9
\ a
in-
Sec
_ _ Computing Areas
3]
2177.
Compute
the area bounded by the straight lines
x
257
y,
2178. Compute the area lying above the x-axis and bounded 3a. 4ax, and the straight line x-[-y by this axis, the parabola y*
=
=
2179*.
Compute
the area bounded by the ellipse
2180. Find the area
bounded by the parabolas
f z y =10x4- 25 and y
=
6x +
9.
2181. Passing to polar coordinates, find the area bounded by the lines 2 2 z Q. x x -[-y* 4x, 2x, x, y y
-y =
=
=
=
2182. Find the area bounded by the straight line r cosq)=l circle r~2. (The area is not to contain a pole.) 2183. Find the area bounded by the curves
and the
r
= a(14coscp)
and
r
acosq>(a>Q).
bounded by the
2184. Find the area (
*
,
~*~
V 4
line
(/!V__ " ~ **_//? "9 J
'
9
4"
2185*. Find the area bounded by the ellipse
(x2y
2
!-3)
4 (3* -\-4y-
2
1)
-
100.
2186. Find the area of a curvilinear quadrangle bounded by 2 z the arcs of the parabolas x --=ay x* a,x, y by y* $x(Q<.
=
y
=
=
y
new variables u and x2
= uy,
if
u,
and put
vx.
2187. Find the area of a curvilinear quadrangle bounded by the arcs of the curves if=ax, if a, xy bx, xy $(Q<.a<.b,
=
=
0
new variables u and u,
9
-1900
v,
y*~vx.
and put
=
(Ch. 7
Multiple and Line Integrals
258
Sec. 4. Ccmputing Volumes
The volume V
a cyltndroid bounded above by a continuous surface of low by the ph-ne 2 0, and on the sides by a right cylindrical surface, which cuts out of the ju/-plane a region S (Fig. 94), is equal to *
= /(*,
y),
be
2188. Use a double integral to express the volume of a pyravertices 0(0, 0)", A(\, 0, 0), fl(l, 1,0) and C(0, 0, 1) ol the limits Set integration. 95). up (Fig.
mid wiih
C(0,0,1)
Fig. 94
Fig. 95
In Problems 2189 to 2192 sketch the solid expressed by the given double integral: 2189.
f J
dx
x
f (1 J
Z-X
2190.
y)dy.
_
2193. Sketch the solid V a*
a
tegral
f
dx
- x*
(
YC?
2191.
value
is
x)dy.
2192.
whose volume
tfy* dy\
whose volume
is
expressed by the in-
reason geometrically to find the
of this integral. 2184. Find the volume of a solid bounded by the elliptical 2 x + y=\, and the coordi2x* -f f/ 1, the plane paraboloid z nate planes. 215. A solid is bounded by a hyperbolic paraboloid z x* tf and the planes (/ 0, e 0, x=l. Compute its volume.
=
+
=
=
_
Sec. 5]
2196.
planes
*/
A
solid
= 0,
is
= 0,
2
th
Computing
_
Area* of Surfaces
2 bounded by the cylinder x
y = x.
Compute
its
=a
2
+-z
2
259
and the
volume.
Find the volumes bounded by the following surfaces:
= y\ x* -\-y* --='*, 2 = 0. = = 2J/T, * + 2 = 6, 2 = 0. y"x, y = x* +y\ y = x z //=!, 2 = 0. x -}-*H-2 = a, 3*4-*/ = a, ~x4-tj=^a, + -1, y = *' = 0, 2 = 0. = 2ax, 2 = a*, 2 = p* (a > p). x if
2197. az 2198.
2199.
2200. 2201.
2202.
f/
2
,
*/
= 0,
2
= 0.
,
2
-h
In Problems 2203 to 2211 use polar and generalized polar coordinates. 2203. Find the entire volume enclosed between the cylinder 2 2 2 2 a x 2 -\-y a and the hyperboloid x -f- if 2* 2204. Find the entire volume contained between the cone 2 2 2 z* = a 2 and the hyperboloid x 2 -f- if 2(* +if) 2 x + /* f 2205. Find the volume bounded by the surfaces 2a2
=
.
=
.
=
**
2
4-//
2
2
=a
2
,
2
= 0.
volume
2206. Determine the
of the ellipsoid
2207. Find the volume of a solid bounded by the paraboloid 2 2 2 -= x 2 H- // 2 and the 3a -f- 2 (The volume lying sphere * 4inside the paraboloid is meant.) 2208. Compute the volume of a solid bounded by the jq/-plane t 2 2 2 2 the cyl inder x -*- y 2a^, and the cone x -f if = 2 2209. Compute the volume of a solid bounded by the jq/-plane, 2 2 2 the surface 2 =-ae~ <*'J4 " >, and the cylinder x + y = /? 2210. Compute the volume of a solid bounded by the *f/-plane,
^
202
=
.
=
.
J
.
the paraboloid
z
=^+
2
5
,
= and the cylinder ^-h| 2
2211. In what ratio does the hyperboloid 2 2 2 divide the volume of the sphere x 4- t^ 2212*. Find the volume of a solid bounded 4-
2~.
2
jc
^3a
2
2
-h//
2
2
= a*
?
by the surfaces
Sec. 5. Computing the Areas of Surfaces
The area o of a smooth single-valued surface z on the jci/-plane is the region S, is equal to
-JJ
= f(x
t
y),
whose projection
Multiple and Line Integrals
260
[Ch. 7
%+j+=
2213. Find the area of that part of the plane
=
1
lies between the coordinate planes. 2214. Find the area of that part of the surface of the cylin2 2 der x 4 y R 2 (z^O) which lies between the planes z mxand
which
z
=
= = nx(m>n>Q). 2215*. 2
Compute 2 2
=z
the
area of that
part
which is situated y bounded by the plane y -\-z-a.
cone x
2216.
Compute 2 2
y cylinder 2 2 2 2 x y +z =-a 2217. Compute
+
the area
=ax
+
x
of that
which
part
of
out
of
cut
is
of the
in the
first
the it
surface of the octant and is
surface of the by the sphere
.
sphere x
2
2
y
-\-
+
z
2
the area
=a
2
of that
part
of the y2
cut out by the surface -^
surface
of
the
..2
+ ^=1.
2218. Compute the area of that part of the surface of the 2 2 ax paraboloid y + z = 2ax which lies between the cylinder tf and the plane x= a. 2219. Compute the area of that part of the surface of the 2 2 2ax which lies between the xy-plane and the cylinder x + y
=
cone x
2
2
-\-y
=z
2 .
2220*. Compute the area of that part of the surface ol the 2 2 = 2ax. cone x 2 y 2 which lies inside the cylinder x 2 z f/ 2221*. Prove that the areas of the parts of the surfaces of the 2 2 z 2 2az arid x 2az cut out by the cylinparaboloids x + y y 2 2 der x -\-tf are size. of R equivalent 2222*. A sphere of radius a is cut by two circular cylinders whose base diameters are equal to the radius of the sphere and which are tangent to each other along one of the diameters of the sphere. Find the volume and the area of the surface of the remaining part of the sphere. 2223* An opening with square base whose side is equal to a is cut out of a sphere of radius a. The axis of the opening coincides with the diameter of the sphere. Find the area of the surface of the sphere cut out by the opening. of 2224*. Compute the area that part of the helicoid
=
+
=
=
=
e
= carctan
ders
x
2
X 2
-i-y
which
=a
2
lies
in
the
first
octant
between the cylin-
and
Sec. 6. Applications of the Double
Integral
in
Mechanics
1. The mass and static moments ot a lamina. If S is a region in an jq/-plane occupied by a lamina, and Q (x, y) is the surface density of the lamina at the point (x, y), then the mass of the lamina and its static
M
Sec. 6]
_
_
Applications of the Double Integral in Mechanics
M
moments x and double integrals A4
MY
relative to the
--=
H
Q
(x, y)
dx
and t/-axes are
x-
Mx= J
dy,
jj
(S)
yQ
J
expressed
261
by the
dx dy,
(x, y)
(S)
M Y =^
J*e(x,
y)dxdy.
(1)
(5) If
the lamina
homogeneous, then Q
is
2. The coordinates
centre of gravity of a lamina, then
-
My _i_
y
where
M
is
tive to the
formulas
(1)
const.
(x, y)
of the centre of gravity of a lamina.
M
-M ~_ //
M
M x My
coordinate axes (see 1). we can put Q=l.
the
3. The moments lamina relative
of
to the x-
/X=
S S
and
,
t/-axes are,
y'Q (x, y) dx dy,
of
its
is
moments
The
moments
static
rela-
homogeneous, then in of inertia 01 a
respectively, equal to
/r=
*2 Q
J J
(*. y)
*x dy.
(2)
(S)
(S)
The moment
are
lamina
a lamina.
of
inertia
the
'
the mass of the lamina and If
is
(x, y)
x d
.
J
'
C
If
inertia of a lamina relative to the origin
is
(3)
Putting Q(X, //)-=! in formulas inertia of a plane iigure.
2225. Find the
mass
(2)
and
(3),
of a circular
we
get the geometric
lamina
of radius
proportional to the distance of a point density and is equal to 6 at the edge of the lamina. is
2226.
OB = a
A lamina has the OA = b, and its
shape of a right
moments
R
if
of
the
from the centre
triangle
with
legs
and
density at any point is equal to the distance of the point from the leg 0/4. Find the static moments of the lamina relative to the legs 0/4 and OB. 2227. Compute the coordinates of the centre of gravity of the sin* area OmAnO (Fig. 96), which is bounded by the curve // and the straight line OA that passes through the coordinate origin
and the vertex
A
(-^
,
Ij
of a sine curve.
2228. Find the coordinates of the centre of gravity of an area
bounded by the cardioid r = a(\ + cosij)). 2229. Find the coordinates of the centre
of gravity of a circular sector of radius a with angle at the vertex 2a (Fig. 97). 2230. Compute the coordinates of the centre of gravity of an area bounded by the parabolas // 4.x f 4 and if 2x4-4. 2231. Compute the moment of inertia of a triangle bounded 2 relative to the #-axis. 2, y 2, # by the straight lines x y
=
+
=
=
=
[Ch. 7
Multiple and Line Integrals
262
2232. Find the ters
d and
moment
D(d
a)
of inertia of an annulus with diamerelative to its centre, and b) relative to
diameter. 2233. Compute the moment of inertia of a relative to the axis passing through its vertex the plane of the square. 2234*. C)mpute the moment of inertia of the parabola if = ax by the straight line x its
square with side a perpendicularly to
segment cut
a
=a
straight line
//
=
relative
to
oil
the
a.
Fig. 96
2235*. Compute the moment of inertia of an area bounded by 4 and the straight line x-\-y the hyperbola xy 5 relative to the straight line x y. 2236*. In a square lamina with side a, the density is proportional to the distance from one of its vertices. C:>mpute the moment of inertia of the lamina relative to the side that passes through this vertex. 2237. Find the moment of in?rtia of the cardioid r a(l cos
= =
=
=
+
=
Sec. 7. Triple Integrals
1. Triple integrals in rectangular coordinates. The triple integral of the /(*, y, <) extended over the region V is the limit of the corresponding threefold iterated sum: function
62
= lim
max max max
\x\ A'/j
- o ~> o
Az/c -> o
2 2 2^ f
i
fa
(*t> &/>
d
z
'
Sec. 7]
2o3
Triple Integrals
Evaluation of a triple integral reduces to th? successive computation ordinary (onefold iterated) integrals or to the computation double and one single integral. three
Example
Z d*d/dz, JJ*y V
/=.$ where the region V
We
is
defined by the inequalities
have X
1
2 '
Example
one
Compute
1.
Solution.
of the
of
7T
dy
Evaluate
2.
2
2
AT
extended over the volume of the ellipsoid
t
Z
// --
-
-|-
2
=
-j-
1
.
Solution. a
^
J
x2
dxdy
dz
-
a
x2
dA-
J J
J
-a
(V)
S yz
is
V
We
^ + -^^1 --
the area of the ellipse
S vvzz -=Jib I/
x*S ^
yz
dx
t
-a
(5^.)
M^
\\here
dydz=
^^
j^2
2
JC
= cons ^
an(^
=*n \-V r a
l-~-c
2
a-
therefore finally get
-a
(V)
2. Change
of variables in a triple integral.
.
0.
If
in tha triple integral
2)dxdydz
from the variables AT, //, z to the variables it is required to pass u, ay, which are connected with *, //, z bv the relations x ^(u v, w), y = ty(u,v,w), z = X(". u ^)i where the functions cp, i|), x are: 1) continuous together with their paitial first derivatives; ,
t
2)
in one-to-one (ind,
tween the points
some region
V
in
both directions, continuju?) correspondence beV oi xf/z-space and the points of
of the region of integration
of l/l/l^-space;
[Ch. 7
Multiple and Line Integrals
264 3) the functional
of these functions
determinant (Jacobian)
dx dx
dx
~dv
dw
dii
D (u,
v,
dy dy dy da dv dw
w)
dz
dz
dz
dw
da dv
we can make
region V, then
retains a constant sign in the
use
the for-
of
mula
= $ J
\f(x,y,2)dxdydz
(V) \
IT ( w
\ f
\
L?
w )>
>
ty( u
>
y
*
w )<
In particular, 1) for cylindrical coordinates
X get
/
cp,
rcosrp,
spherical coordinates
the radius vector) (Fig. 99),
x
we have
r,
//
h (Fig. 98), where z^-//,
rsinrp,
r\
2) for r
du dv dw.
Fig. 99
Fig. 98
we
\
/
Example
r
2
3.
= r cos
cos
i|)
cp,
ap,
r (
is
the
longitude,
q),
z
\\>
the
latitude,
where
cos 9,
f/
= /-cosi|3 sin
/
sin
v|\
i[).
Passing to spherical coordinates, compute
JSJ (V)
where V
is a sphere of radius R. Solution. For a sphere, the ranges of the spherical tude), \|) (latitude), and r (radius vector) will be
coordinates
fp
(longi-
Sec
We
265
Triple Integrals
7\
therefore have
f f f
Vx
2
z
-\-y
3. Applications sional A'//z-space
-\-z*dxdydz=\
of
dcp
dty f r A'COS \|?dr
f
The volume
triple integrals.
= Ji# 4
of a region of
.
three-dimen-
is
The mass
of a solid
occupying the region V
M
-----
C f f
Y
z)
v
(
is
d
(V)
where \(x,y,z) is the density of the body at the point (*,//,*). The slal/c moments of the body relative to the coordinate planes are
M yy = "(V)'
Myz =
(*
(*
I '
)} 0')"
MZX
~-
'
\
\
\
Y
(A
f/
2)
f/
dx dy dz.
(V)
The coordinates
gravity are
of the centre of
M
f
_ J ~~
_
M
'
'
Al
then we can put Y (*, y> z) = If the solid is homogeneous, mulas for the coordinates of the centre of gravity. The moments of inertia relative to the coordinate axes are
T-
J J $
(y*
+ **) Y (*,
U, 2)
dx dy
I
in
the
for-
dz;
(V)
(V) l
=J
J J
(*
(V)
Putting of inertia
of
Y(*0
^i^21
in
* nese
formulas,
we
get the
the body.
A. Evaluating triple integrals Set up the limits of integration in the triple integral J J (V)
for
the indicated regions V.
^f(x,y,
z)dxdydz
geometric
moments
266
Multifile
V
2240.
is
a tetrahedron Jr
X
V
2241.
is
V
2242*.
2243.
V
is
+ ^fl
2
2
J/
= 0,
2
= 0.
1
2
,
= 0,
= /f.
2
volume bounded by the surfaces
a
000
2245.
= 0,
cone bounded by the surfaces
a
Compute the following 2244.
X
bounded by the surfaces
a cylinder
is
bounded by the planes
yJrZ =\,
JC'
[Ch. 7
and Lire Integrals
integrals:
+ +-2+1
*
V
djt
dy
J
J
a
2246
.
fd*
00
I
]
('
J
1-X
1
2247.
rfy
.1
.)
dx
1-JC-t/
dy
J
xyzdz.
J o
o
p
2248. Evaluate
d* dy dz '
1)3
where V is the region ol integration planes and the plane x-\~y-[z\. 2249. Evaluate
bounded
by the coordinate
r r r (V}
where V (the region 2 paiaboloid
2cu^x
of 2
-\-y
integration) is the and the sphere X?
common
of the
part
2
+ y* + 2 ^3a
2 .
2250. Evaluate
(V)
where spheres
V (region x
2
+ y*
\-z*
of
integration)
^R'
and
x
2
is
the
+ \f + z
2
common
^ 2Rz
part
of
the
gee.
__ Triple Integrals
7]
267
2251. Evaluate
^zdxdydz, (V)
where V
is
a
volume bounded by the plane j. +
half of the ellipsoid
-+
z
=
and the upper
-J.==l.
2252. Evaluate
(V)
V
where
Xz is
2 IJ
+ "^r
the interior of the ellipsoid ~^r
2253. Evaluate
V
where 2
2
=
(jc
region
(the 2
2
hi/
)
of
is
integration)
and the plane
bounded
the
by
cone.
= h.
z
2254. Passing to cylindrical coordinates, evaluate
where V 2
jc
the surfaces x* is a region bounded by a -=z and containing the point (0,0, R). -| // 2255. Evaluate 2
2
J
21
-
jc
a
j
transforming it to cylindrical coordinates. Evaluate
first
225(5.
-
-
di/
dz,
J
transforming it to cylindrical coordinates. 2257. Evaluate
first
VR*-X*
R
Vfla-jir*-0a J
\dx -/? first
transforming
\
-/f^TJa it
dy
J
(A:
+
o
to spherical coordinates.
+y*
-\-
z*
= 2Rz
9
[Ch. 7
Multiple and Line Integrals
268
2258. Passing to spherical
evaluate the integral
coordinates,
(V)
where V B.
2
is
the interior of the sphere x -\-y
Computing volumes by means
2259. Use a triple integral bounded by the surfaces
to
2
+z
2
triple integrals
of
compute the volume
of a solid
2260**. Compute the volume of that part of the cylinder 2 2 x 2 -f tf = 2ax which is contained between the paraboloid* + y = 2az and the xy-plane. 2261*. Compute the volume of a solid bounded by the sphere 2 2 2 2 2 2 x +y +z =a and the cone z ---x + /y" (external to the cone). 2262*. Compute the volume of a solid bounded by the sphere 2 2 z x 2 +y +z = 4 and the paraboloid x +if=-3z (internal to the paraboloid). 2263. Compute the volume of a solid bounded by the xy-plane, z 2 2 2 z 2 a (internal ax and the sphere x the cylinder x -f- z to the cylinder). 2264. Compute the volume of a solid bounded by the paraboloid
+y =
=
+y
+ -~ = 2 -i
-
and the plane
C. Applications to
of
triple integrals
mechanics and physics
M
2265. Find the mass
0
<*/
Fig.
and the plane
(l(x,
y,
z)
point
the den-
if
y,
(x,
= x-\-y-\-z.
z)
is
2266. Out of an octant of the 2 2 2 2 c z x .> 0, x y sphere cut a solid OABC */^0, bounded by the coordinate planes
+ + z^O
100
-+- =
sity
the
of a rec-
Q^x^a,
tangular parallelepiped at
x--=a.
1
(a
c,
&
(Fig.
100).
<
,
Find the mass
of this body if the density at each point (x, y, z) is equal to the 0-coordinate of the point. 2267*. In a solid which has the shape of a hemisphere 22*0, the density varies in proportion to the
Improper Integrals Dependent on a Parameter
Sec. 8]
260
distance of the point from the centre. Find the centre of gravity of the solid. 2268. Find the centre of gravity of a solid bounded by the 2 2 4x and the plane x=2. paraboloid // +2z 2269*. Find the moment of inertia of a circular cylinder, whose altitude is h and the radius of the base is a, relative to the axis which serves as the diameter of the base of the cylinder. 2270*. Find the moment of inertia of a circular con^ base, a, and (altitude, /i, radius of density Q) relative to the diameter of the base. 2271**. Find the force of attraction exerted by a homogeneous cone of altitude h and vertex angle a (in axial cross-section) on a material point containing unit mas^ and located at its vertex. 2272**. Show that the force of attraction exerted by a homogeneous sphere on an external material point does not change if the entire mass of the sphere is concentrated at its centre.
=
on a Parameter.
Sec. 8. Improper Integrals Dependent
Improper Multiple Integrals 1. Differentiation with respect to a parameter. In the case of certain imposed on the functions / (.v, a), f'a (x, a) and on the corresponding improper integrals \vc have the Leibniz rule
restrictions
(.v,
a) dx
=
a
Example
fa
(A-,
a) dx.
differentiating with respect to a parameter, evaluate
By
1.
\
'i
~~
>
dx
\
(a
> 0,
p
> 0).
Solution. Let
Then
~ da
Whence F equation.
(a,
We
p)
=
have
Whence C(p)
2a
1
-
Ina + C(p). To
0=
= -^-lnp.
~
In P
Hence,
find
+ C(P).
"2a*
C(p), we put a =
in the latter
Multiple and Line Integrals
270
2. Improper double and a) An infinite region. If region 5, then we put
triple integrals. a function f (x, y) is continuous in
y)
\{f(x, U
[Ch. 7
dx dy
=
lim
"S\
( f (x, y)
an unbounded
dx dy,
(1)
W
where a is a finite region lying entirely within S, where a -+ S signifies that we expand the region o by an arbitrary law so that any point of 5 should enter it and remain in it. If there is a limit on the right and if it does not depend on the choice of the region o then the corresponding improper inte,
gral
is
called convergent
otherwise it is divergent. is nonnegative [f (x, //)
,
If the inU'grand / (,v, vergence of an inirioper integral
is
it
y)^Q], then
on
the right of (1) lo exist at least for the region 5. b) A discontinuous function. If a function / (x, y) uous in a bounded closed region S, except the point (*,
for
the con-
and sufficient for the limit one system of regions o that exhaust
necej-sary
y)dxdy=\\m
is
everywhere contin-
P
(a,
b),
then we put
f(x,y)dxdy.
(2)
where S 6 is a region obtained from S by eliminating a small region of dia meter e that contains the roint P. If (2) has a limit that does not depend on the tyre of small regions eliminated from 5, the improper integral under is called convergent, otherwise it is divergent. /(A, //)^>0, then the limit on the ripht of (2) is not dependent on the type of regions eliminated from S; for instance, such regions may be circles
consideration If
with centre at P.
of radius
The concept of
of
improper double integrals
is
readily extended to the case
triple integrals.
Example
2.
Test for convergence
dxdy
,
3)
is}
where S
is the entire j/-plane. Solution. Let a be a circle of radius origin. Passing to polar coordinates for
If
p<
then
then
1,
lim
o
-
a
lim 7 (a) O -* S
7(o)=
r
P
*
= lim Q
/(a)
=00 and
with
Q
p^
1,
centre
at
the
coordinate
we have
the integral diverges. But
if
p>
1,
~f QC
and
the
integral
converges.
For
p=l
we
have
Sec. 8]
_ Q
2JT
/
Improper Integrals Dependent on a Parameter
(a)= Cdcpf -f^l-==jiln(l+Q r * J
J
o
o
2
lim/(a)
);
~r
Q -*
oo
for
p>
= oo,
that
_
is,
271
the
integral
diverges.
Thus, 'the integral
2273. Find
(3)
/' (*),
converges
1.
if
2274. Prove that the function +
u
=
00
r
*f'
2 \ J ^
+
w
j
,2
,
(l/
dz
2)
QC
satisfies
the Laplace equation
*u dx*
is
'
~c)y*
F
2275. The Laplace transformation defined by the formula
Find d)
d ~U + ^ *-Q
F(p),
if:
= cos p/. /(/)
a)
/(O-l;
b)
(p)
/(/)=e;
2276. Taking advantage of the (ormula
compute the
integral \
xn
2277*. Using the formula
evaluate the integral
~
l
for the
\nxdx.
c)
function /(/)
/(/)
= sinp/;
272
_
Multiple and Line Integrals
Applying differentiation with respect
__
\Ch. 7
to a parameter,
evaluate
the following integrals: GO
2278.
"
00
e
2279
.
rno^ 228
-
"-""*
f
~**~ e ~'X
(
ax
arc tan
dx
>
0,
>
0).
smrnxdx (a>0, p>0). .
dx
e-<"d*
2282.
(a
-
(o^O).
Evaluate the following improper integrals: GO
QC
2283. x_
2284.
2285.
\dy\ev
dx.
y
\\
c) c/
*,
^ 4 ~r
2
,
where S
is
a region defined
by the inequali-
/
(5)
ties
#^5
2287.
1,
The
Euler-Poisson
integral
defined
by
the
formula
00
may
also be written in the
uate / by multiplying coordinates. 2288. Evaluate CO
form
I=\c-^dy.
Eval-
these formulas and then passing to polar
00
GO
dz
273
Line Integrals
See. 9]
Test for convergence the improper double integrals: 2
2289**.
where S
55 \nVx*-\-y dxdy,
a
is
circle
*
2
+
2 f/
(S)
U
2290.
where S
TTiriva
a
is
defined by
region
the ine-
(6')
quality x
2
2291 *.
2292.
tf^\
-|
I
I
$,
("exterior" of the circle). .
J_
2
d xd
CCr JJJ
.
2
t
(x T~
y
where S
,
*
^\ \
,
z
a square
is
where V
,
a
is
<
A; |
|
1
,
\y\^l.
defined
region
by
the
)
(V)
inequality x
2
-\-
if
z
-\-
2
^
("exterior" of a sphere).
1
Sec. 9. Line Integrals
1. Line integrals of the (Irst type. Let / (x, y) be a continuous function = (p (A-) [a<: \ <;&] be the equation of some smooth curve C. */ Let us construct a system of points M, (A-,-, //,) (/~0, 1, 2, .... n) that break up the curve C into elementary arcs Al _,M = As and let us form the and
I
integral
max
2^
sum S n
As/ -^ 0,
is
AV
^ s r ^ ne
#/)
'
-
-
t
t
im it of this sum,
when n
-* oo
and
called a line integral of the first type n
lim
"->>ac (cfs
is
2=
/
/(xf,
r/,-)
As/=
\
/ (x,
i/)
ds
c
i
the arc differential) and
is
evaluated from the formula
b '
\
J
f (A
!/) rfs
C In
the
case -
'
"
of rt
],
\
J a
f(x
parametric
we have
t
cp (,v))
y~\
+ (q>'
representation
of
2
(A-))
the
c/.vr.
curve
C:
.Y
= q>(/),
Also considered are line integrals of the first type of functions of three variables f (x, y, z) taken along a space curve. These integrals are evaluated in like fashion A line integral of the lirsl type does not depend on the direction of the path of integration; if the integrand / is interpreted as a linear density of Hie curve of integration C, then this integral represents the mass of the curve C.
274
Multiple and Line Integrals
Example
1.
Evaluate the
[Ch. 7
line integral
where C is the contour of the triangle ABO with vertices A (1, and 0(0, 0) (Fig 101). Solution. Here, the equation AB is (/=l x, the equation and the equation OA is */ = 0. We therefore have
AB
2. Line uous
BO
0),
OB
B is
(0,
1),
x
= 0,
OA
integrals of the second type. If P (x, y) and Q (x, y) are continand f/
=
functions
B
A Fig.
x varies, then the corresponding
X
101
line integral
of the
second type
is
expressed
as follows:
P
(x,
y)dx + Q
(x,
y)dy =
C
[P
(x,
cp
(*))
+
(x)
Q
(x, q> (x))\ dx.
a
In the more general case when the curve C is represented parametrically: y ty(t), where / vanes from a to 0, we have
=
=
y)
Similar formulas hold space curve.
dy
+
for
a
'
[P
(
Une
(0,
integral of
(9(0,
the
second
type
(OJ
taken
over a
A line integral of the second type changes sign when the direction of the path of integration is reversed. This integral may he interpreted mechanically P (x, y), Q(x, y) } along the as the work of an appropriate variable force { curve of integration C Example 2. Evaluate the line integral Cyi
____
____
C is the upper half clockwise. Solution. We have
where
2 y dx
275
Line Inlegtah
Sec. 9]
+ x* dy =
2
[b
sin
2
the
of
t-(a sin
*--=a cost,
ellipse
/)
+a
2
/
-b cos /] dt
ab* f sin 8
1
dt
=
+ a*b
jt
3. The case
of a total differential.
traversed
sin*
oo
cos 2
n
=
=b
y
/
dt
= -t a&.
?t
the
If
C cos 8
a
of
integrand
line integral
second type is a total differential of some single-valued function U^--U(x, */), that is, P (x, y)dx-\-Q(x, y)dy-=dU(x, y], then this line integral is not dependent on the path of integration and we have the Newton-Leibniz formula (x 2 yj
of
the
,
P(x, y)dx
where
y
)
is
In particular,
if
(x lt
}
y)dy = U(x 2
+ Q(x,
the initial and (* 2 #,) is the the contour of integration C ,
,
yJ
U
terminal is
(x l9
point then
closed,
(1)
the
of
path
(2)
of integration C is contained entirely within some S and 2) the functions P(r, (/) and Q (x, y) together with tlvir partial dcnviitives of the first order are continuous in 5, then a is the necessary and sufficient coiditioi foi th? existence of the function identical" fulJilment (in S) of the equality If
1)
the
contour
simnlv-connectH
reaio.i
U
2_^ dX
(3) (d)
~dy
If conditions one and two are not ful(see integration of total differentials) does not guarantee the existence of a jillcd, the presence of condition (3)
single-valued lunction U, and formulas (1) and (2) may prove wrong (see Problem 23:2) We give a method of finding a function U (x, u) from its total diflerential based on the use of line integrals (which is yet another method of integrating a total differential). For the contour of integration C let us take a broken line P P (Fig 102), where P (.v y ) is a fixed [oint and 0, y and dy (x, y) is a variable point. Then along P P, we have y We get: we have dx and along P t
M
U
(x
t
y)-U
II
M
1
(J
,
~
M
(x
,
y
)=
P
(x,
y)
dx
+ Q (x,
y)
dy = y (x.
Similarly, integrating with respect to
P C P 2 M, we have x
u
U(x.
y)-U(*v
*/
)-$Q(* y)dy+^P(x, ,
Vo
x*
y)dx.
y)dy.
[Ch. 7
Multiple and Line Integrals
276
+ 2y) dx+ (2x 6y) dy = dU. = O, y = Q. Then
Find U.
Example 3. (4x Solution. Let A-O U(x,
y)
=
or
+ 2y) dx + C= U
where C
(0,
0)
is
2
-i-c,
3//
an arbitrary constant.
Y
Uo
--4/I
X
XQ
102
Fig.
4. Green's formula for a plane. If C is the boundary of a region S and the functions P (x, y) and Q (x, y) are continuous together with their firstorder partial derivatives in the closed region S-j-C, then Green's formula holds:
(S)
here
t'^e
remain
C
circulation about the contour
chosen so that the region S should
is
to the left.
5. Applications
of line integrals.
=
1
)
(p
An y
area
dx=
bounded by the closed contour C
(p
is
xdy
(the direction of circulation of the contour is chosen counterclockwise). The fo lowing formula for area is more convenient for application:
2)
Z
The work (x,
y
t
X=
X(x, y, of a force, having projections accordingly, the work of a force field),
z) (or,
2),
Y=Y
(x, y,
along a path
C
z), is
_
Sec. 9]
__
277
Line Integrals
expressed by the integral
A= ^Xdx + Ydy + Zdz. c
If the force has a potential, i.e., if there exists a function potential function or a force function) such that
(a
dV dx
=A
dU
dU j
TT
,
L
-7
,
A
*j)
//2,
=
where
(v l5
(*J,
'/J.
(*i,
l/ lt
Xdx + Ydy + Zdz^
J (*|. {/,,
,
dU
the initial and (x 2
Zj) is
,
= U(x
is
tli2
2293.
xyds, where
J
C
2t
y tt
z 2 )-t/ (x lt y,
z } ),
terminal point of the path.
A. Line Integrals of the First
Evaluate the following
equal to
2-J
z2)
is
Z.)
J
Zi)
f/ 1?
(x, y, z)
dz
dy
then the work, irrespective of the shape of the path C, (*.
U=U
Type
line integrals:
the contour of the square
is
|
x\
+
\
y\
=a
(fl>0). s
2294.
\
r
c
K
--^=
,. 2 A:
if
|-
,
connecting the points 0(0,
^xyds, where
2295.
C
where
a segment of the straight line
is
\
\-
C
and A
0)
(I,
a quarter
is
2).
of the
c
lying
the
in
2296.
first
+
=
l
fi>
quadrant.
where C
ifds,
ellipse ^i
is
the
first
arc of the cycloid x
=a
(t
sin /)>
c
a
cos
(1
/).
}x
2297.
2
+y
2
ds,
where
c
circle
x
----
a (cos
/
2
2298.
^ (x
(-/sin/), y 2
2 -\-
y
d$,
)
C
is
an arc
= a(smt
where C
is
of the involute of
the
tcost) I0^/=^2ji].
an arc
of the logarithmic spi-
c ral
r^ae m v(m>Q) 2299.
niscate r
2300.
+ y) JU c = a cos2
2
where C
is
(0, a)
to the point
0(
oo, 0).
the right-hand loop of the lem-
2
J
(A:
\-y)ds,
c q/2
A
from the point
>
where
C
is
an
arc
of
the
curve
JK
=^
Multiple and Line Integrals
278
2301.
\
%2
(*
.
~T~
y
i
i
# = acos/ y = as\nt, f
2
2302.
J t/"2#
is
the
where
C
is
turn of the screw-line
first
= bt.
z
2
-I-
C
where
*2
.
[Ch. 7
z ds y
the
x
circle
2
+ y +z = a 2
2
2 ,
c
x=y. 2303*. Find the area of the
cylinder y
=
surface
lateral
o
x
2
bounded by the planes
z
the parabolic
of
= 0, x~0, 2 = x, = 6. //
ae cos?, 2304. Find the arc length of the conic screw-line C x sin /, z ae from the point 0(0, 0, 0) to the point A (a, 0, a). 2305. Determine the mass of the contour of the ellipse 1
y = ae* 2 2-
+
=
yi -T2-
= l,
if
l
the linear density ot
it
at each
point
M (x,
y) is
equal to \y\. 2306. Find the mass of the first turn of the screw-line A; = a cos/, y = asmt, zbt, if the density at each point is equal to the radius vector of this point. 2307. Determine the coordinates of the centre of gravity of a half-arc of the cycloid
x = a(t
sin
/),
y = a(\
cost)
[0
2308. Find the moment of inertia, about the z-axis, of the turn of the screw-line x = a cos/, !/ bt. asin/, z 2309. With what force will a mass distributed with uni2 2 form density over the circle x -f y 2 a z 0, act on a mass m located at the point A (0, 0, &)?
=
=
first
M
=
,
=
B. Line Integrals of the Second
Type
Evaluate the following line integrals: 2310.
2
J (x
2xy)dx-\- (2xy+tf)dy, where
AB 2 parabola y = x from the point 2311.
^
(2a
y)dx
\
-xdy,
^4(1,
where C
to
1) is
an
AB
is
the
point
arc
an arc
of
of the
B(2, the
4). first
c
arch of the cycloid x a(t a(l sin/), # cos/) which arc runs in the direction of increasing parameter
=
=
2
/.
taken along different paths emanating J 2xydxx dy OA from the coordinate origin 0(0, 0) and terminating at the point 2312.
A
(2, ia)
1)
(Fig. .103):
the straight line
Om A\
Lim
Sec. 9]
parabola OnA,
b) the
279
Integrals
the axis oi
symmetry
of
which
is
the
of
which
is
the
(/-axis;
c)
the parabola
the axis of
OpA,
symmetry
x-axis;
broken line
d) the e)
OBA\ OCA.
the broken line 2
2313.
2xydx ^x dy
J
as in
Problem 2312.
OA
(x+u)dx
(x
x*
+ y*
i/)dy
counterclockwise.
Fig.
^tfdx + x*dy, where C
2315.
103
the upper half of the ellipse
is
c
x^acost, y = 2316.
bs\r\t traced clockwise.
cosydxsmxdy
\
taken along the segment
bisector of the second quadrantal point A is 2 and the ordinate of d
x!f(l'
2317.
dll}
^*
(f
=
,
where C
angle,
B
is
is
if
the
AB
of the
abscissa oi the
2.
the right-hand
loop
ol
the
a
lemmscate co$2(p traced counterclockwise. 2318. Evaluate the line integrals with respect to expressionswhich are total differentials: r*
(
-
(a. 4)
8)
(2.
a)
rt
xdy -\-ydx,
^
(2,
f (1.
J
c)
i)
$ (0, 0)
1)
1)
ydx d)
b)
(O r
2)
1
(i.
xdx + ydy,
2)
xdy ,
(along a path
that
does
not
intersect the
_
Multiple and Line Integrals
280
(x, y)
dx + dy x ~ry
f J
e)
f)
p a th
ong a
^at
[C/t.
not
(joes
intersect
7
the
= 0),
straight line A;-f-y (* 2
(a i
___
J/ 2 )
,
q>
J (*;,
2319. Find the antiderivative functions evaluate the integrals:
integrands and
0)
(3,
a)
of the
(x<
J
+ 4xy') dx 4
(6x
V
4
5#
)
dy
,
(-2, -1) o)
(i,
* J ~~~ y b)
! x
\
j
-1)
(0,
straight line y (,
C)
1)
x
^
y)
\
(the integration path does not intersect the
= x),
(x+Wx + ydy
(the integration
j a.
i)
the straight line y A\
f
I
,
J
(0,
\V
X !
x*
0)
2320.
=
+ y*
path does not intersect
x),
yA J
fix
I
y
/
\
I
\}^xZ + y
z
fat
J
Compute /=.
f
xdx + yfy
taken clockwise along the quarter of the
X*
ellipse -5
the first quadrant. 2321. Show that if f(u) is a continuous closed piecewise-smooth contour, then
W
2
+ ^=!
that
lies in
function and
c
2322. Find the antiderivative function a)
b) c)
du = (2x du = (3* 2 du =
d)' dtt
U
if:
+ 3y)dx + (3x4y)dy\
=
2xy + y
2 )
dx
2
(x
2xy
+ 3y*) dy\
C
is
a
Sec
281
Line Integrals
9]
Evaluate the curves: 2323.
line
taken along the following space
integrals
z)dx+(zx)dy + (x
^ (y
C
where
tj)dz,
is
a
turn
c of the screw- line i
/
= asin/,
corresponding to the variation 2324.
of
the parameter
ydx + zdy + xdz, where
(p
c I
\
;t
/y
^
C
is
/
from
to
the circle
= /?cosacos/, = /?cosa sin = ^sina (a = const), /,
traced in the direction of increasing parameter.
2325.
xydx + yzdy
(
+ zxdz,
where
OA
is
an
arc
of
the
OA circle
situated on the side of the A'Z-plane where //>0. 2326. Evaluate the line integrals of the total (,
a)
4.
zdz,
xdx-\-ydy
] (1,
0,
-3)
(i.
i.
i)
(3,
4.
5)
b)
x
differentials:
8)
//
z
dx f 2 x dy -
xd\
r
\-
xy dz
,
\-\idij-\-zdz
j
rA'.|.y"
J
-
+2
'
(0, 0, 0)
-M ^v d)
yte + mv + 'y**
f J (i.
xyz ,
(the
integration
path
is
situated
D
in the first octant).
C. Green's
Formula
2327. Using Green's formula, transform the line integral /
=
2
\f*
-\
if
dx + y [xy +
In
(jc
4
c
where the contour
C bounds
the region S.
K?T?)]
dy,
_ _ Multiple and Line Integrals
282
[C/t.
7
2328. Applying Green's formula, evaluate /
=
2 (x?
if)
-t-
dx +
(x
+ y)
2
dy,
where C
is the contour of a triangle (traced in the positive direcverlices at the points A (I, 1), fl(2, 2) and C(l, 3). with tion) Verify the result obiained by computing the integral directly. 2329. Applying Green's formula, evaluate the integral
x*y dx
+ xif dy,
c
where C 2330.
chord
is
Find
y
the circle x*
is
+ if = R*
traced counterclockwise.
A parabola AmB, whose axis is the #-axis and whose AnB, is drawn through the points A (1,0) and 8(2,3). (x
+ y)dx(x
y)dy directly and by applying Green's
AmBnA formula. 2331. Find
$
e* y [y*
dx
(1 -f xtj)dy\,
\-
if
the points
A and B
AmB on the #-axis, while the area, bounded by the integration path AmB and the segment AB, is equal to S. 2332*. Evaluate Consider two cases: lie
^ifc^f.
when when
the origin is outside the contour C, the contour encircles the origin n times. b) 2333**. Show that if C is a closed curve, then
a)
where
s is the arc length and n is the outer normal. 2334. Applying Green's formula, find the value of the integral
I
= (j)[xcos(X, n)+ysm(X,
n)]ds,
c
where ds
is the differential of the arc the contour C. 2335*. Evaluate the integral
taken along the contour of a
A
(1,
is
traced counterclockwise.
0).
fl(0
f
1),
C(-l,
0)
and n
is
the outer normal to
square with vertices at the points 1), provided the contour
and>(0,
2S&
Line Integrals
Sec. 9]
D. Applications of the Line Integral
Evaluate the areas of figures bounded by the following curves: a cos/, y bs\nt. 2336. The ellipse x acos 3 /, #-=asin 8 /. 2337. The astroid jt x a (2 cos/ cos 2/), 2338. The cardioid y a (2 sin/
=
= =
=
=
sin 20-
A
2339*.
of
loop
the
folium
(a>0).
Descartes x*
+if
3zxy
=Q
= axy.
2340. The curve (x + y)* 2341*. A circle of radius fixed circle of radius
integer, find the area by some point of the
R
of r
of
(cardioid). A circle of
2342*.
r
R and
is
rolling without
outside
sliding along a n is an Assuming that
it.
bounded by the curve (epicycloid) described moving circle. Analyze the particular case radius
a fixed circle of radius
r
R and
is
rolling
inside
without sliding along D
Assuming that
it.
is
an
integer, find the area bounded by the curve (hypocycloid) described by some point of the moving circle. Analyze the particular r>
when
case
r
=j
(astroid).
2343. A field is generated by a force of constant magnitude F Find the work that the field does in the positive jt-direelion when a material point traces clockwise a quarter of the circle 2 2 x -^-y ^=R lying in the first quadrant. 2344. Find the work done by the force of gravity when a material point of mass m is moved ironi position A (JC P // l? zjz 2 ) (the z-axis is directed vertically upto position B (x 2 // 2 ,
,
wards). 2345. Find the work done by an elastic force directed towards the coordinate origin if the magnitude of the force is proportional to the distance of the point fiom the origin and if the point of application of the force traces counterclockwise a quarter of the ellipse 2346.
^s4-^i=l Find
the
lying in the potential
first
and determine the work done by the = 0, K:=0. Z-=rng (force a) X rial
is
point
B(* b)
Uv
moved
from
quadrant.
of a force R {X, Y, Z\ force over a given path if: of gravity) and the mate-
function
position
A
y l9
zj
where
jx
(x lt
to
position
*i)'-
x=
?,
K=-^. Z=-*
f
= const
and
Yx* 4 if f (Newton attractive force) and the material point moves from position A (a, b, c) to infinity;
r
-\-
_ _ Multiple and Line Integrals
284
X=
c)
force),
x* 4-
Sec.
2
#
=
=
and
k*x, Y k*y, Z the initial point of the
+ z = /? 2
2 ,
[Ch. 7
where k
k*z,
= const
(elastic
path is located on the sphere while the terminal point is located on the sphere
10. Surface Integrals
1. Surface integral of the first type. Let function and z=-cp(*, y) a smooth surface S.
The
surface integral of the first type
'(x,
//,
z)dS= n
is
f (x,
//,
2)
be a continuous
the limit of the integral
sum
lim -> 00
fas
I
.3
is the area of the /th element of the surface S, the point (*/, y lt belongs to this element, and the maximum diameter of elements of partition tends to zero. The value of this integral is not dependent on the choice of side of the surface S over which the integration is performed. If a projection a of the surface S on the jo/-plane is single-valued, that to the z-axis intersects the surface S at only is, every straight line parallel one point, then the appropriate surface integral of the first type may be calculated from the formula
where AS/ z/)
(*' y)
S
dx
dlJ-
(a)
Example
where S
is
1.
Compute
the surface integral
the surf ace of the cube
0
Let us compute the sum of the surface integrals over the the cube (z=l) and over the lower edge of the cube (z 0):
00
2. Surface
integral
is
of
00
00
The desired surface
upper edge
obviously
three
times
greater
=
and
equal to
=
If P P(x, //, z), Q Q (*, y, z), continuous functions and S + is a side of the smooth surface S characterized by the direction of the normal n {cos a, cos p, cos Y}. t'hen ihe corresponding surface integral of the second type is expressed as follows:
R = R(x,
integral of the second type.
y, z) are
P dy dz + Q
dz
dx+ R
dx
dy=
f f
(P cos a
-f
Q
cos p
+ R cos Y) dS.
Sec. 10]
285
Surface Integrals
When we pass to the other side, S~, of the surface, this integral reverses sign. then the direcIf the surface 5 is represented implicitly, F (x, y, z) 0, tion cosines of the normal of this surface are determined from the formulas
=
1 OF COSBQ^ COSa==-pr^-, D dx
dF
1
-FT^ D dy
,
COS VT
=
OF
1
-rr- -T
D
,
dz
where
and the choice
should
of sign before the radical of the surface S.
with the side
be
brought into agreement
=
=
P (.v, //, z), Q Q (x, //, z), If the functions P continuously differentiable and C is a closed contour bounding a two-sided surface S, we then have the Stokes' formula 3.
formula.
Stokes'
R = R(x,
y, z) are
(j)
C
dQ\ = rr\fdR 3-- 5- 1 dz \
\
JJ l\dy 5
c s
a
J
Id?
dR\ Tdx
+ -3 ^\dt
fi cos a l
+ ^ ,
j
dP\
fdQ
T-
3-2-
\dx
1
dy J
1
cos v y
_
dS, |
where cos a, cos p, cosy are the direction cosines of the normal to the surface S, and the direction of the normal is defined so that on the side of the normal the contour S is traced counterclockwise (in a rigiit-handed coordinate system).
Evaluate the following surface integrals of the 2347.
8
$$ (*
4
tf)dS, where S
z
first
the sphere x +//
is
type: 2
2
-{-z
= a*.
6
2348.
5$
Vx
s.
2
-\-tfdS
where 5
is
the
surface
lateral
the
of
s
+ g_?6 i= =0 [O^z^bl
cone
Evaluate the following surface 2349.
\ \
integrals
of
the second type:
yz dydz -\-xzdz dx-\- xydxdy, where 5
is
the
external
s
side of the surface of a tetrahedron
y
=Q
t
2
2350.
= 0, x+y + z = a.
Nzdxdy, where S
t
is
bounded by the planes
the external side of the ellipsoid
x dydz-\-y*dzdx + z*dxdy,
2351. o
where S
is
+? = 2
the 2
a (z side of the surface of the hemisphere +// 2352. Find the mass ot the surface of the cube Os^z 1, if the surface density at the point Is equal to xyz.
O^y^l,
<
0,
A:
external
^0).
O^x^l,
M (x,
y, z)
_ _ Multiple and Line Integrals
286
[C/i.
7
2353. Determine the coordinates of the centre of gravity of a
2 2 homogeneous parabolic envelope az^je + #
Find
2354.
the
moment
(0
inertia of a part
of
=V
the lateral
of
< <
x 2 -f y 2 [0 z about the z-axis. surface of the cone z ft] 2355. Applying Stokes* formula, transform the integrals: 2
(x
a)
- yz) dx + (y
2
zx)
dy +
2
(z
xy) dz\
c b)
(j)
ydx-\-zdy + xdz.
c
Applying Stokes' formula, find the given integrals and verify the results by direct calculations: 2356.
(f (y
+ z)dx + (z + x)dy + (x + y)dz,
where
C
the circle
is
c
2357.
x)dy +
z)dx^(z
(y
2
2
JC
-|-//
(x
y)dz, where
C
is
the ellipse
X + 2=l.
=1,
+ (x-{-y)dy + (x + y + z)dz, where C is the curve = acos/, z = a (sin + cos/) [0
2358. ()xdx c
/
//
2
2
z
A
/IflC with vertices
2360. In what case /
(a, 0, 0), is
B
C (0,
(0, a, 0),
0, a).
the line integral
= $ Pdx + Qdy + Rdz c
over any closed contour
equal to zero?
11. The Ostrogradsky-Gauss Formula
Sec.
If
Q
C
=Q
thMr
5
is
a
closed
R=
(A% y, z), first partial
smooth surface bounding the volume V, and P
P
(x, y, z),
are functions that are continuous together with derivatives in the closed region V then we have the Ostro/? (v,
(/,
e)
t
gradsky- Gauss formula
where crsa, cos p, surface S
cosy are the direction cosines
of the outer normal to the
Applying the Ostrogradsky-Gauss formula, transform the following surface integrals over the closed surfaces S bounding the
>ec.
11]
_
The Ostrogradsky-Gauss Formula
/olume V(t*osa, cosp, lormal to the surface 2361. JJ
_
287
cosy are direction cosines of the outer S).
xydxdy+yzdydz -\-zxdzdx.
s
2362.
J J 6
x 2 dy dz +
y* dz
dx
+z
2
d* dy.
2363.
Using the Ostrogradsky-Gauss formula, compute the following surface integrals:
2o65.
x*dydz + y*dzdx + z*dxdy, where S
J
the external
is
s
side of the surface of the
2366. of a
\ \
V
cube
O^x^a, O^r/^n, O^z^a.
xdydz + ydzdx + zdxdy, where 5
the external side
is
pyramid bounded by the surfaces x + y-{-z
= a,
x
= Q,y = Q,
z=-0. 2367.
x*
dydz-\-if dzdx
= z* dxdy,
where 5
the
is
external
^
side of the sphere x
2368
2
~a cos p + 2 2
2
f
2
^(jc cosa-t y
// 2
2
-|-z
2
.
cos
y)
d5,
where S
is
the exter-
o
nal total surface of the cone
2369. Prove that direction, then
if
S
is
a
closed
surface
and
/
is
any
fixed
where n
is the outer normal to the surface S. 2370. Prove that the volume of the solid surface S is equal to
= -3
V bounded by
the
M
where cose, cosp, cosy are the direction cosines normal to the surface S.
of
the outer
Multiple and Line Integrals
288
[Ch. 7
Sec. 12. Fundamentals of Field Theory
1. Scalar and vector of the point
surfaces f
t/,
(x,
A
fields.
= /(/>) = /(*, = C, where y, z)
z),
C=
defined by the scalar function is a point of space. The const, are called level surfaces of the scalar scalar field
P
where
is
(x, y, z)
field.
=
A vector field is defined by the vector function of the point a a(P)~ a(r), where P is a point of space and r=xi-\-yj+zk is the radius vector of the point P. In coordinate form, a a v j-\-a z k, where a x ~a x (x, y, z), ax i a and a z a z (x, //, z) are projections of the vector a on the y ay(x, y, z), coordinate axes. The vector lines (force lines, flow lines) of a vector field are found from the following system of differential equations
+
=
~~~'
dx__dy _dz
A
scalar or vector
'ry; -
that
field
where
"
V = ^3-+y^- + ^y
is
not depend on the time is called nonstationary.
does
t time, depends on the t. The Gradient. Th vector
it
if
stationary; ~
it
Hamiltonian operator
the
t
or
(del,
called
is
nabla),
is
=
called the gradient of the field U f (P) at the given point P (ci. Ch. VI, Sec. 6). The gradient is in the direction of the normal n to the level surface at the point P and in the direction of increasing function U, and has length equal to
dn~~ If
the direction
is
\dx
given
by the unit vector
= (the derivative of the function 3. Divergence and rotation.
+ ay j+a z k
the scalar
is
The rotation
U
in
-
cos
/
- cos p +
a+
the direction
The divergence
diva-^ +
(curl) of a vector field
{cos a, cos p, cos Y},
/).
of a vector field
\
\
S
is
i
the integral
an dS =
a (P)
~ a^i
\~
^
+ ^^Va. a (P) = ax + ay j + az k
4. Flux of a vector. The flux of a vector field a(P) in a direction defined by the unit vector of the normal S
cosy
is
the
vector
da y
da z
to the surface
-
then
\
\
s
an dS
\ \
(ax cos
a -|- a
y
cos p
through a surfaces ujcosa, cos p, COSY}
+ az cos Y) dS.
S
If S is a closed surface bounding a volume V, and n is a unit vector of the outer normal to the surface S, then the Ostrogradsky -Gauss formula holds,
which
289
Fundamentals of Field Theory
Sec. 12] in vector
form
is ff
(\
r\
r*
div
5. Circulation of a vector, the work a along the curve C is defined by
vector
f a
dr =
C
a s ds
\
V
C
a dx dy
dz.
The
of a Held.
integral of the
line
the formula
a x dx
-f-
a dy -f a z dz y
(0
C
and represents the work done by the field a along projection of the vector a on the tangent to C).
curve
the
C
(a s is
the
If C is closed, then the line integral (1) is called the circulation of the vector field a around the contour C. If the closed curve C bounds a two-sided surfaces, then Stokes' formula holds, which in vector form has the form
adr=
f f
rotadS,
/i
where n is the vector of the normal to the surface S; the direction of the vector should be chosen so that for an observer looking in the direction of n the circulation of the contour C should be counterclockwise in a right-handed coordinate system.
6. tial
Potential
and solenoidal
fields.
The
vector
iield
a(r)
is
called poten-
if
U,
where Uf(r) is a scalar function (the potential of the field). For the potentiality of a field a, given in a simply-connected domain, it be non rotational, that is, rota = 0. In it is necessary and sufficient that that case there exists a potential U defined by the equation
dU ~a x dx-}- a v dy -f- a 2 dz. If
the potential
U
is a
single-valued function, then
a dr
\
U
(B)
U (A);
AB in
particular, the circulation of the vector
A
a = 0;
If
a
is
equal to zero:
m adr=Q.
vector field a (r) is called solenoidal if at each point of the field div in this case the flux of the vector through any closed surface is zero. the field is at the same time potential and solenoidal, then div (grad U)=.Q
and the potential function
U
"Er++5-<>.
is
or
harmonic;
AU=0
'
that
where
is,
it
satisfies the
Laplace
A= *'=;+>+> isthe
Laplacian operator
2371. Determine the level surfaces of the scalar field
where
r
U = F(Q), 10-1900
\fx*+y*-\-z*.
where
What
will the level surfaces be of a field
290
_ _ Multiple and Line Integrals
[Ch. 7
2372. Determine the level surfaces ot the scalar field
U = arc sin
-
2373. Show that straight lines parallel to a vector c are the vector lines of a vector field a(P) c where c is a constant vector. 2374. Find the vector lines of the field a CD*// 4 CDJC/, where CD is a constant. 2375. Derive the formulas:
=
t
=
a)
+ C K) = C gTad(UV) = Ugrad grad (t/ = 26/ grad grad(C
l
i/
2
1
gradf/
+C
2
gradV, where C,
and
C2
are constants; b)
2
c) j\
)
d)
V grad(/ U AfU\ =-* grad
e)
grad
(^ J
=
cp' (t7)
grad U.
2376. Find the magnitude and the direction of the gradient 3 x* z of the field U if 3xyz at the point A (2, 1, 1). Deterthe mine at what points gradient of the field is perpendicular to the z-axis and at what points it is equal to zero. 2377. Evaluate grad f/, if U is equal, respectively, to: a) r,
=
b)
r\
c)
j
,
+ +
d) /(r)(r
= /?+^qr?).
=
2378. Find the gradient of the scalar field U cr, where c is What will the level surfaces be of this field, and what will their position be relative to the vector c?
a constant vector.
2379. Find the derivative of the function
given point P(x, y, this point. 'In
z)
x
U = ^ + y^ + ~ala
in the direction of the radius vector r of will this derivative be equal to the
what case
magnitude of the gradient? 2380. Find the derivative
of
the
function
U=
in the di-
rection of /{cosa, cosp, cosy}. In what case will this derivative be equal to zero? 2381. Derive the formulas: a)
div^aj + C^^Cjdivaj + Cjjdivajj, where C and C l
constants; b) div (i/c) c)
= grad /, where c is div((/a) = grad U-a+ (/diva.
2382. Evaluate
di
2383. Find div
a
where
r
=
for the
central
2
are
a constant vector;
vector
field
a(P)
= /(r)~
,
Sec.
Fundamentals of Field Theory
12]
291
2384. Derive the formulas:
+ C a = C rota + C rota where C and rot(/c) = grad U-c, where c is a constant vector; rot (Ua) = grad U a + U rot a. rot(C a l
a)
1
2
2)
l
1
f
2
,
C
t
2
are
constants; b) c)
2385.
Evaluate the divergence and the rotation of the vector respectively, equal to: a) r\ b) re and c) f(r)c, where c
a
if
is
a constant vector.
a
is,
2386. Find the divergence and rotation of the field of linear velocities of the points of a solid rotating counterclockwise with constant angular velocity o> about the z-axis. 2387.' Evaluate the rotation of a field of linear velocities co r of the points of a body rotating with constant angular
=
velocity
=
=
+
,
=
+
a) the lateral surface of the cone 0
j
force
F=
^
of a
mass w, located
point of
at the coordinate
through an arbitrary closed surface surrounding this point. 2394. Evaluate the line integral of a vector r around one turn of the screw-line * = /?cos/; y = Rsmt\ z = lif from / =
origin,
to
=
2n. 2395. Using Stokes' theorem, evaluate the circulation of the vector a zk along the circumference x* + if x*tfi R 2 z=-0, 2 2 jt taking the hemisphere z J/"/? if for the surface. 2396. Show that if a force F is central, that is, it is directed /
=
+j+
=
\
=
towards a fixed point
and depends only on the distance r from where f(r) is a single-valued continuous point: function, then the field is a potential field. Find the potential U
F = f(r)r,
this
of the field.
2397. Find the potential by a material point of mass nates:
a=
equation 10*
~r. Show
U
m
of a gravitational field generated located at the origin of coordi-
that the potential
U
satisfies the
Laplace
_ _ Multiple and Line Integrals
292
(Ch. 7
2398. Find out whether the given vector field has a potential find if the potential exists:
U
and
a)
b) c)
a a a=
2399. Prove that the central space field lenoidal only is
U,
when
= ~, f(r)
2400. Will the vector field a constant vector)?
where k
is
a = r(cxr)
= /(r)rwill
be so-
constant.
be solenoidal (where c
VIII
Chapter
SERIES
Sec.
1.
Number Series
1. Fundamental concepts. A number series 00
a,+a t +...+a +...= tt
is
called convergent
if
its
has a finite limit as n
>
(1)
tt
sum
The quantity
oo.
l
S=
lim S n n -+ oo
is
then called the sum
number
of the series, while the
is
partial
2a n-\
called the remainder of the series.
If
Sn
the limit lim n -*
does not exist (or
is
QO
infinite), the series is then called divergent. If a series converges, then lim a n Q (necessary condition for convergence).
n-*oo
The converse
not true. For convergence of the series (1) it is necessary and sufficient that such that for n any positive number e it be possible to choose an and for any positive p the following inequality is fulfilled: is
N
(Cauchifs test). The convergence or divergence subtract a finite
number
of
2.
Tests of convergence
a)
Comparison
test
*!
I.
of
series
a
is
not
violated
if
for
>N
we add
or
terms.
its
and divergence
If
+ *,+ ..
of positive series. after a certain n n
=
+*,!+. ..^
,
and the
series
(2)
converges, then the series (1) also converges. If the series ( J) diverges, then (2) diverges as well. It is convenient, for purposes of comparing series, to take a geometric progression: 00
2 aq n=o
n (a
*
0),
294
[Ch. 8
Series
which converges
which
is
a
|^|
for
\q\^\, and
diverges for
the harmonic series
divergent series.
Example
The
1.
series
++
+
+
converges, since here a
=
1
J_
"~n-2 n
2*'
while the geometric progression 1
n=i
whose ratio
Example
is
=
<7
,
The
2.
converges.
series
diverges, since of
b)
its
In
ln_3
ln_2
general term
is
/i
greater than
the
term
corresponding
the harmonic series (which diverges).
Comparison
test
II.
there
If
exists a finite
and
nonzero
limit
lim ? y. b n
n if
a n -^b n ), then the
(in
particular,
the
same time. Example 3. The
series (1)
and
series
diverges, since
1"
1
whereas a series with general
Example
4.
The
term
n
n J
2
diverges.
series
_J_
1
_J
converges, since
while a series with general
term
^
converges.
(2)
converge or diverge
at
Sec. 1]
c)
__ Number
D'Alembert's
Let
test.
>
an
295
Series
certain
a
(after
n)
and
there be
let
a limit
-l =
lim n
Then the it
not
is
<
series (1) converges
if q the series
known whether
Example
5.
an
-> GO
q.
and
1,
diverges
if
q
>
If
1.
=
<7
1,
then
convergent or not.
is
Test the convergence of the series
1+1+1+ 2 2 '
22
s
Solution. Here,
and
?-H =
lim
2
Hence, the given series converges. d) Cauchy's test. Let of^^O (after a certain n) and lim n
Then
(1)
converges
if
q<\,
let
there
-> OD
n
V
and diverges
if
q>\. When q=l
t
remains open. e) Cauchy's integral test. If a n f(n), where the function f (x) rnonotomcally decreasing and continuous for jc^a^l, the series
of the convergence of the series
be a limif
n/~ =
=
the question is
(1)
positive, and the
integral 00
"
/ (x)
dx
converge or diverge at the same time.
By means
converges of
series
if
of the integral test
p>
may
1,
be
and diverges tested
by
it
may
be proved that the
The convergence of a large number if p<^\. comparing with the corresponding Dirichlet
series (3)
Example
6.
Test the following series for convergence
--^"
-U--L + -L+ Solution.
We
* 1
i
-
r "' i
have 1 ~
Dirichlet series
__1_1
J^ 4/i
2
'
__
296
[Ch. 8
Series
=
Since the Dirichlet series converges for p 2, it follows that on the basis of test II we can say that the given series likewise converges. 3. Tests for convergence of alternating series. If a series
comparison
+ |fll+...,
l+.-. composed
values of the
of the absolute
(4)
terms of the series
converges, But if (1) con-
(1),
then (1) also converges and is called absolutely convergent. verges and (4) diverges, then the series (1) is called conditionally (not absolutely) convergent. For investigating the absolute convergence of the series (1), we can make use [for the series (4)] of the familiar convergence tests of positive series. For instance, (1) converges absolutely if
<
lim
n
-> oo
\
n
In the general case, the divergence of
gence of
But
(4).
n (4)
?2_J
lim
if
-
GO
I
/KI<
lim
or
0-n
n
I
>
1
(1)
or
f
not
does
1.
from
follow
lim /\a n n -> oo K
\>
1,
the
diver-
then not only does
diverge but the series (1) does also. Leibniz test If for the alternating series
*!-* conditions
the following
are
(*^0)
+*3- **+ fulfilled:
1)
(5)
^ b ^b
bl
2
s
^.
.
.
;
(5)
bn
lim
2)
n
then
-
=
oc
converges.
In this case, for the
remainder
of
Rn
the series
the evaluation
holds.
Example
7.
Test for convergence the series
Solution. Let us
form a series of
the
absolute
values
o!
the
terms of
this series:
Since
.
lim
lim I
1
n
-
oo
,
the series converges absolutely.
Example
8.
The
series
converges, since the conditions of the Leibniz converges conditionally, since the series 14-
diverges (harmonic series).
-*
4-
4-4
test are fulfilled.
This series
__
Sec
Number
/\
Series
297
Note. For the convergence of an alternating series it is not sufficient that general term should tend to zero. The Leibniz test only states that an alternating series converges if the absolute value of its general term tends to zero monotonically. Thus, for example, the series its
diverges despite the fact that its general term tends to zero (here, of course, the monotonic variation of the absolute value of the general term has been
S 2k = S'k + S"k
violated). Indeed, here,
and
S k = cc(S k
lim k -
Sk
lim
limit fe
exists
is
a partial
and
,
sum
finite (S k
is
where
of the a
is
harmonic
partial
sum
series),
whereas the
o f the convergent geo-
-*
metric progression), hence,
S 2fe =oo.
lim k
-> 00
On
the other hand, the Leibniz test is not necessary for the convergence series: an alternating series may converge if the absolute value of its general term tends to zero in nonmonotonic fashion Thus, the series
of
an alternating
.22
~3 a
_
42
~~'""(2n
1)'
converges (and it converges absolutely), although the Leibniz test is not fulfilled: though the absolute value of the general term of the series tends to zero, it does not do so monotonically. 4. Series with complex terms A series with the general term c n a n
=
+
00 z
-]-ib n (i
1)
converges
and only
if,
if,
the
series
with
real
terms
2
an
n=i 00
and
2&,, conver g e n
a*
same time;
the
SC
B
n=i
=S= n
The series (6) definitely converges series
whose terms are the moduli 5. Operations on series. a)
that
A
is,
convergent series
and
of the
may
8 i
be
+'2Xn =
(6)
i
is
called absolutely convergent,
terms of the series multiplied
+a
1
(6),
+...+fl n + ...=S
t
if
the
converges.
termwise by
if
a,
then
in this case
\
any
number
fc;
__
298
Series
b)
sum
the
By
two convergent
(difference) of fli
_
[Ch.
m
series
+ flt+-..+fl n +...=5 + *,+ .. .+*+...=$,
(7)
If
(8)
*!
we mean
a series
+ (a
bl)
(a,
c)
The product
of
l
b z)
If
verges d)
terms verges
cn
+
the series
= a,bn + a A-I +
.
.
+ (a n
.
(7)
and
bn )
.
.
.
=S,
S2
.
the series
is
(8)
+
+ c>+...+cn +... +0M = 2
cl
where
8
*>
(9)
t
)
the series (7) and (8) converge absolutely, then the series (9) also conabsolutely and has a sum equal to S S 2 If a series converges absolutely, its sum remains unchanged when the of the series are rearranged. This property is absent if the series conconditionally. .
{
Write the simplest formula of the the indicated terms:
term
/ith
of the series
using
2401.
1+1+J + I+...
2404.
i+4 + j+^+...
2402.
1+1+1+1+"-
2405.
2403.
1+1 + 4 + 4+.. + + + ~4
2406.
4+*.+ ^+... f+{ + + +
2407.
.
-
2409.
1
2410.
1+
1
H-
1
- 14 ~ 1
1
.
.
.
+...
-I-
In Problems 2411-2415 it is required to write the first 4 or 5 terms of the series on the basis of the known general term a n .
2411. a n
2412.
2413
=
|q^.
2414. a n
=
2415 \* din*. a
=
t^. flrt
=
Test the following series for convergence by applying the comparison tests (or the necessary condition):
2416.
1
1
+1
!
+ ...+(
1)-'
+
...
Number
2418.
+ +
0419
J___1_ 444^ ^ 4 ^ 6
.
2
2422.
++
+
2421.
'
'
4^
-I-
'
2i
'
'
'
'
+
+
...
-= + -=-f -= + ^3-4
.
... 4'
1/2-3
02
2423. 2
299
Series
]fn
(n
+ \)
O"
OJ
+ + y+... i-
+=;-+..
.
-'
4 J/3
3 V'2
Using d'Alembert's
(n
test
lest,
the
gence: 2427.
98 -*'
-l= 2 1 !
1
1-5
+
''-} ^ 1-5-9 -
"
following
2 "~'
+ 4 4-^- +...-'
K
1-1)
'
i
'
j
'
-
1-5- 9. ..(4/i-3)
Test for convergence, using Cauchy's test:
Test for convergence the positive series:
2431.
1+1 + 1+...+1+... 1 + 1 + 1+ ..+ _ 1
2432.
.
((H
2433
-
r4+4T7 + 7no+
[
)2
)
r/lo/
^TtjO
+...
+(3n-2M3n+l
24.i+f -^+... +sSFI+ 2435.
1
...
^ + 1- + ^+...+^+... ^
5
.'^
7
i
i
I
I
J_ i
i
_X.
9 ^
*
j_
i
*
__________ l
bones
for
conver-
300
Series
2439.
-+ +
+...
2440.
1+
+
+...
[Ch.
++...
2441> 2442.
1++.+ 1, 4
1-3.5
-3,
1-jT
-
...
4^M2~*"
1.3.5.. .(2/1-1) '
""
4-8-12...4n
2445. ""
1000. 1002. 1004... (998 1-4-7. ..(3n-2)
2.5-8...(6 n !. 5.9...
2447
-7)(6g-4)
+2n)
"
(8_H)(8n -7)
2
...
I,
'- 4
1 '
4'9
^--^ 1.3.5.7.9 ....^ '""
2449
2450.
2451.
1-4-9.
..
""*
I
|
'
1-3-5-7- 9... (4n
arcsin.
2455.
sin.
2456.
3)
!,
i=J
2452 152.
Ulnfl+l). ^ \
n J
2457.
,. ~
.
.
^-
ri'lnn-ln In n
00
2453.
yin^i.
2458. /J=2
2454
-
E -'
2459
-
STTOTJ
.
8
Number
Sec. 1]
2464.
Series
.
cos-2-) ^
2469. Prove that the series 1)
converges for arbitrary q
t
p>l, p
if
and
for
?>1,
if
p=l; p=l.
and for
(_n-i
i
2470.
l-
2471.
1
2472.
l_ + _... +
2473.
l_
2474
+ -....ui--f...
-- - + /2 K3
l
+
..
+ 3_... + (^S +
"
'
...
+
.
^- -. +(-!)-
2475.
__
2476.
-- -- -- ---^--h...+
+
.
+ _. ..+(_!) ^
7=4
2|^2
1
1
.^+...
I
9
a
7=4
3^3
4^41
1
,
n+\
n
/
{) (n
947Q "'
l_k!-J-lL_ 7
7.9~t"7.9.11
-i
/
"TV
ii-' U
+
\)
1-4. 7... (3/1-2) 7.9-11. ..(2/1 5)'
+
302
[Ch. 8
Series
na
sin
2481.
2482.
(_l)"l^,
(
ly-'tan
n^i
n
^
Vn
2483. Convince yourself that the d'Alembert test for converdecide ide the quesl question of the convergence of the
gence does not series
2a
where
>
whereas by means of the Cauchy
test it is possible to establish that this series converges. 2484*. Convince yourself that the Leibniz test cannot be applied to the alternating series a) to d). Find out which of these series diverge, which converge conditionally and which con-
verge absolutely: a)'
___
_J
l
_1_
1
1
3
v
'
^\ d)
i_
+y *~
+ 21
j
.j; r
.
3F
_
.
3
"
|
3
"L
32
h 5
.
3?+
__u "
'
'
'
3*
_ _ T _l + T T + TT T +... !
J
!
]
i
.
i
!
.
Test the following series with complex terms for convergence: CO
00
^2+#.
2485.
2488.
00
2486,
00
(2 '
X" 7
1=1
V.
3
1) ''.
2489. =i
__
Sec.
Number
1]
Series
V
-
{/
= X-?
r(2-Q + 1" l
[
'
[n(3-20-3ij
fa 2493. Between the curves
and y=-^ X
and
303
the right
to
intersection are constructed segments parallel an equal distance from each other. Will the sum the lengths of these segments be finite? 2494. Will the sum of the lengths of the segments mentioned
of their point of to the t/-axis at
of
Problem 2493 be
in
curve y
= X
curve \)^-\ x
finite if the
is
by the
replaced
? 00
Form
2495.
Does
sum
the
^-^ and
the series
of
sum converge?
this
00
2496.
Form
the
anc*
* es *
and
difference
2497.
]T
2
r
Does
the
formed
series
by
the
subtracting
series
00
TT
=
divergent
series
i
QC
n
the
for convergence.
'*
rt
n
of
sef i es
-f
21 n
i
converge?
i
sum converges while
2498. Choose two series such that their their difference diverges.
CC
2499.
Does
Form
the product
of the
QC
V ~nVn
series
and V.OTTM2
tt
n
this product converge?
2500.
Form
the series (l
+1 + 1+
..
.
+ J~-f-
.
.
.
V.
Does
this series converge?
2501. Given the series
1+1 1+
...+
( -
:
^+...
Estimate
the error committed when replacing the sum of this series with the sum of the first four terms, the sum of the first five terms. What can you say about the signs of these errors? 2502*. Estimate the error due to replacing the sum of the series
y+ by the sum
2!
(2")
of its first
+ ~3\('2J + " + 5H2") n terms.
+"
_ __
304
[Ch. 8
Series
2503. Estimate
the
due to replacing the sum
error
the
of
series
1
i
by the
sum
JL -L-L-L-l.
4-
^2M3M
of its first
4-
"+/i!^"'
n terms. In particular, estimate the accu-
racy of such an approximation for n=10. 2504**. Estimate the error due to replacing
the
sum
of the
series 1
sum
+ 22 + 32+
+^i+
'
n terms. In particular, estimate the accuan approximation for n= 1,000. 2505**. Estimate the error due to replacing the sum of the
by the
of its first
racy of such series 1
by the
sum
+ 2(i)'43()V. of
its
first
.
n terms. 00
Zl
_ I)""
-
-
1
does one have
n=i
take to compute decimals?
to
2507.
sum
its
How many
two decimal
to
terms of
the series (2/1
have to take to compute
its
sum
to
to
places?
-MIS*
three
does one
two decimal places?
to three?
to four?
2508*. Find
the
2509. Find the
sum
sum
the
of
series
-L +
gL + jL
+
..
.
+
of the series
Sec. 2. Functional Series
1. Region of convergence. The
set of values of the
argument x
for
which
the functional series /.<*)
converges
is
+ /.(*)+.. .+M*)+...
called the region of convergence of this series.
S(*)
= n
where S n
(x)
vergence,
is
of the series.
fl
(x)
+f
called the
z
lim
Sn
(1)
The function
(x),
-* QO
(x)+ ...+f n (x), and x belongs to sum of the series; R n (x) = S(x)S n
the (x) is
region of conthe remainder
Sec. 2]
In
_ _
305
Functional Series
the
cases,
simplest
it
(1), to
convergence of a series holding x constant.
is sufficient, when determining the region of apply to this series certain convergence tests,
n
Diverges
,%f?
.
-3-101
Determine the region
1.
X
104
Fit*.
Example
Diverges
of
of the series
convergence
x+\
" (x+\y ~ (X+\Y " ~ (*+iy TT" 2-2* "~T^r t""'~ n- 2" + 1
t
t
'
we
Solution. Denoting by u n the general term of the series,
lim
M -*|
I
\U n
Using d'Alembert's
we
get the
we have
'
that
l,
is,
harmonic the
*'* * lim I'+H Jx+l\ 2 n^*2 n + (n+\) |jc|
series
that
series
1
1
+ -~
(c n
and a are
gence)
|
x
a |
+ c, (A-a) + c
real
<
3
1< x <
or
To +...,
converges (conditionally). Thus, the scries converges when 2. Power series. For any power c
3
if
is,
+ TT+-Q-+.. o 2
2,
test)
have
'
assert that the series converges (and converges
oo<^<
if
will
1
l
we can
test,
if
absolutely),
=
\
11
2
(*
-t
oo
the
series diverges, 104).
(Fig.
When x=l
which diverges, and when
which
(in
if
x=
3
with the Leibniz
accord
3^*<1. series
fl)'+
.
.
.
+c n (x
B fl)
+
.
.
.
(3)
numbers) there exists an interval (the interval of convercentre at the point x a, with in which the series (3)
R with
>
a R the series diverges. In special cases, the converges absolutely; for \x and oo. At the end-points of radius of convergence R may also be equal to a the interval of convergence x R, the power series may either converge or diverge. The interval of convergence is ordinarily determined with the help of the d'Alembert or Cauchy tests, by applying them to a series, the terms of which are the absolute values of the terms of the given series (3). Applying to the series of absolute values \
=
the convergence tests of d'Alembert and Cauchy, we get, respectively, for the radius of convergence of the power series (3), the formulas
= lim n -
rt*
and
# = Hm -
"/|c,,|
too
'
However, one must be very careful in using them because the limits on the right frequently do not exist. For example, if an infinitude of coefficients cn
306
__ Series
\Ch. 8
vanishes [as a particular instance, this occurs if the series contains terms with only even or only odd powers of (x a)], one cannot use these formulas. It is then advisable, when determining the interval of convergence, to apply the d'Alembert or Cauchy tests directly, as was done when we investigated the series (2), without resorting to general formulas for the radius of convergence. z
If
= x + ty
is
a
complex variable, then
z-Zo) (cn = a n + ib n
=
for the
series
power n
Z
+...+C n (Z-Z Q +...
(4)
)
there exists a certain circle (circle of convergence) z z inside which the series converges z |>fl the series diverges. At points lying on the cirabsolutely; for Iz cumference of tne circle of convergence, the series (4) may both converge and diverge. It is customary to determine the circle of convergence by means of the d'Alembert or Cauchy tests applied to the scries -z
|z
|
,
z
Jt
-f/f/
)
=
with centre at the point
,
whose terms are absolute values of the terms of the given series. Thus, for example, by means of the d'Alembert test it is easy to see that the circle of convergence of the series ""
""
2-2 2
1-2
~~"'~~
3.2 s
~""'
n .2*
is determined by the inequality |z-f 1 |<2 [it is sufficient to repeat the calculations carried out on page 305 which served to determine the interval of convergence of the series (2), only here x is replaced by z]. The centre of the circle of convergence lies at the point z 1, while the radius R of this circle (the radius of convergence) is equal to 2. 3. Uniform convergence. The functional series (1) converges uniformly on some interval if, no matter what e 0, it is possible to find an N such that does not depend on x and that when for all x of the given interval we have the inequality R n (x) where R n (x) is the remainder of the e,
=
>
\
|
given series. If \fn( x )\*f* c n
2
cn
converges,
(rt=l,
then
the
2,
<
...)
n>N
a^x^b
when
functional
series
(I)
and
number
the
converges
series
on the interval
n-\
V
absolutely and uniformly (Weierstrass' test). series (3) converges absolutely and uniformly on any interval lying within its interval of convergence. The power series (3) may be termwise differentiated and integrated within its interval of convergence (for fl|
The power
*-) +
+c n (xa)+
2
then
for
.
.
l
-a)+. ..+ncn (xa)*- +. cQ
dx+
.
.
.
any x of the interval of convergence of the series
d (xa) dx+
c2
2
(xa) dx+
.
.
.
+
cn
.
.
=/(*), (3),
(5)
we have
=f (x),
(xa) n dx+...=
(6)
_
Sec. 2]
_
Functional Series
307
number # also belongs to the interval of convergence of. the series (3)J. Here, the series (6) and (7) have the same interval of convergence as the
[the
series
(3).
Find the region
of
convergence
of the series:
2510.
(-D-'.
2511. n-
n-
+I
2512.
-jbn
l)"
(
M=
25.9.
1
252
-
2-(jzrk-
n-i
I
00
O(T
|
O
1
3D
M x l^A-
v~^ sin (9n
^'oMI^fl
v^
2/1-4^'*'
>
O^vQI
n=i
ri
1 *
I
=o nr
2514.
X
25.5**.
"
2 sin
-
=
X^i. /l
2523.
=
2516.
2517.
2522
-J-
(-l)"
fI
i
V^
/I=l
*- nMn *.
.
X
2524*.
2525.
2-p-
Find the interval of convergence of the power series and test the convergence at the end-points of the interval of convergence: 2531.
2532. 30
^
2528.
yin
I
X ^rr. n-1
2533.
2
2529.
.
n=i
2530.
2534.
'
-_^".
2535.
308
Series
[Ch.
CO
OCQA 25
V>, ^
00
n
f
V""
s ;-= \2rt-f-ly
V
1
r* .
0**1 > 2551. ^-
(
x
l
7^ n
.
2n4
n=i
CO
+ 5)* n ~
^-rr 4
CO
2537. CO
2553.
(-!) n=i
2 "' V X (3
,n 2539.
t
^-' 2554.
2541.
^1
*
a=i
2556.
2542**.
2^
nl
xnl
.
n=i
2557.
+l
21 /I
=
l)"
(
>
l
(AT-2)" /j"
2544*.
2V=1
2558.
2545.
l(
2(-l)"" -^|r. n
T
2559*.
n=o X(A;
2549.
2 ^S^
2562.
Y.
(3ft
n=
2550.
V n"
(*
+
3)".
2563.
2)".
Y(
- 2)
3
>" .^-. ! (n+l) 2"+ l
1)"
8
309
Functional Series
Sec. 2]
Determine the circle 2564.
of
convergence:
V W.
V <=gC
2566.
Aarf
^arf
fl.
n=i
n=o
.
J
00
2565. 2568.
2567
(1-M0 2 ".
'
'
+ 2i) + (1 + 20 (3 -h 20 2 +
(1
.
.
+
.
+ (l+ 2569.
+ r=-, + (l-lHI-20
l
+ (1-00-20. ..(l-/ii) 9*70 257 2571. Proceeding from the definition of uniform prove that the series
convergence,
..+*"+...
.
does not converge uniformly in the interval ( 1, 1), but converges uniformly on any subinterval within this interval.
sum
Solution. Using the formula for the get, for
|jc|<
of a
geometric
progression,
we
1,
Within the interval ( 1, 1) let us take a submterval [ 1+ct, 1 a], where a is an arbitrarily small positive number. In this subinterval |jc|s^l a, |1
xl^a
and, consequently, I
*,.(*)
<
I
_ at"*
/I (
1 .
;>
To prove the uniform convergence of the given series over the subintervai it is possible to choose a, 1 a], it must be shown that for any e > an N dependent only on e such that for any n > N we will have the ine-
[1+
quality
I
Rn
(x)
I
Taking any e>0,
is,
j
Thus>
puttin g
a)
n>N
n
t
\R n (x)\
is
indeed
<
-
+ > 1
[since ln(l
ea
ln
ln(ea)_
when
let
a)
(n+l)ln(l In (1
-
x of the subinterval under consideration. (1 a)"* 1 w+l
for all
Ns=s
f
ln(l less than
\ a) e
1,
we
are
a)<0] and
convinced that
x of the subinterval the given series on any sub-
for
all
1+a, 1 aj and the uniform convergence of interval within the interval ( 1, 1) is thus proved. As for the entire interval ( 1, 1), it contains points that are arbitrarily
(-
close to Jt=i,
and since
Hm R n (x) = \im X-+1
X-+1
-- =00,
J/I+ .
1
1
X
no mattei
how
large n
is,
310
_ __ Series
[C/i.
8
points x will be found for which R n (x) is greater than any arbitrarily large N we number Hence, it is impossible to choose an N such that for n e at all points of the interval ( would have the inequality R n (x) 1, 1), and this means that the convergence of the series in the interval ( 1, 1) is not uniform.
>
<
\
|
2572. Using the definition of uniform convergence, prove that: a) the series
converges uniformly in any
finite
interval;
b) the series x*
converges
(-1.
throughout
uniformly
the
of
interval
convergence
i);
c) the series
1+F + F+- -+;?+... in
converges uniformly positive number;
the
interval
(1-1-8,
where 8
co)
is
any
d) the series 2
+ (x*- *) +
4
jc
(x
)
6
8 jc ) -f-
(jc
x 2n+2 )
+ (x " 2
.
.
.
-h
.
.
.
1, converges not only within the interval ( 1), but at the extremities of this interval, however the convergence of the series in
(1,
1) is nonuniform. Prove the uniform convergence
of the functional series in the
indicated intervals: 00
2573
on the inlerval
-
I"
1
'
!]
on
2574.
over *he en tire number scale.
2^ ~~2~ 00 I
2575.
Applying
sums
I
(
)"~
T=
termwise
on the interval
differentiation
[0,
and
integration,
of the series: Y2
V3
y
y
y.3
y5
\n
+ ^ + ^+. ..+?-+... x x -*- + --...+(-l)'-*L + ., , + + +... + _ + ...
2576. x
2577.
2578
.
v
yLfl
l
1].
t
find
the
__ Taylor's Series
Sec. 3]
v8
2579.
s
+ -' .- +(-!)"+2x + 3x* + ... 4
1
l-3*
2
-[5x
2582. 1.242-3A;
4
+
+---
... 2
...+(/+
3-4A: 4-
Find the sums of the 2583
r 2 "" 1
*-
2580. 2581.
311
!)*""'
4
series:
+++ ++
-
^ + l+!+...+^L i+...
2586.
/r
Sec. 3. Taylor's Series
1. Expanding a function in a power series. If a function f (x) can be of the point a, in a series of expanded, in some neighbourhood a, then this series (called Taylor's series) is of the form powers of A:
\xa\
(I)
When holds
a
the Taylor scries is also called a Maclaurin's series. Equation (I) the remainder term (or simply remainder) of the
\xa\
when
if
Taylor series
as n
*
oo.
To evaluate fin (x)
the remainder, one can
make
=
<*-
f(n
*
l) l
fl
+
(Lagrant*e's form).
Example
f if
We
X ) =,
smh
n
even, and
(
is
x,
=
(H)
/
0<0<1
where
(2)
=
find
f" (x)
)l
the function f (x) cosh x in a series of powers of x. the derivatives of the given function f (x) cosh x, sinh x, ...; generally, f (n) (x) cosh x, f" (.v) cosh x, sinn A, if n is odd. Putting a 0, we get /(0) (A-) 1,
Expand
1.
Solution.
fl
use of the formula
= =
=
/'(0)=0, T(0)-l, /'"(0) = 0, ...; generally, / >(0) ")=-0 if n is odd. Whence, from (1), we have: (II
=
=
1,
=
if
n
is
even, and
(3)
To determine d'Alembert
the
test.
interval
We
of
convergence
of
the
series
have
lini
n->
oo
(2/1
(3)
we apply the
312
[Ch. 8
Series
for any*. Hence, the series converges remainder term, in accord with formula
.cosh 9*,
R n (x)=
>9>
1
,
it
sinh9*,
~
=^e'*',
X I"-*'** and therefore \R n (x)\^
n
if
is
odd, and
if
n
is
even.
*
i
e1
=
.
.
A
series
with the general term
in
with
accord
the
necessary
Y\ n
^ \
1
made immediately evident with
any x
(this is therefore,
|sinh9*|
1
I
test);
QO
follows that
-
converges for d'Alembert's convergence,
interval
has the form:
n 1 A +
R n (x)= Since
the
in (2),
the help of
condition
for
n+l
lim
and consequently lim #(*) = ()
for
any
x.
This signifies that the
sum
of the
/2->00
series (3) for
any x
is
indeed equal to cosh*.
2. Techniques employed for expanding in power Making use of the principal expansions I.
*
= !++*!+.
series.
(_oo<*
..+fj+...
y
II.
III.
IV.
and also the formula for the sum of a geometric progression, it is possible, in many cases, simply to obtain the expansion of a given function in a power series, without having to investigate the remainder term. It is sometimes advisable to make use of termwise differentiation or integration when expanding a function in a series. When expanding rational functions in power series it is advisable to decompose these functions into partial fractions.
On
x=
the boundaries of the interval of convergence (i.e., when 1 it converges absoas follows: for m 1 it and lutely on both boundaries; for diverges when for 1 it l; diverges on both bounconditionally converges when x *)
and x=l) the expansion IV behaves
=
daries.
>
>
m^
m^O
x--\
313
Taylor's Series
Sec. 3]
Example
Expand
2.
in
powers
of x *) the function
3
Solution. Decomposing the function
into
fractions,
partial
we
will have
Since
(4)
and 2
V^
^^ ... _
i
^
v
^^
\\j 1
n
A>
t
(5)
n =o it
follows that
we
finally get
(6) /i
n=Q
=o
The geometric progressions and
of
|*|
hence,
3. Taylor's series two variables /(x,
point
(a,
b)
(4)
n n-Qv
and
formula
(5)
converge,
(6)
holds
when
respectively, for
|x|<-j
|x|
i.e.,
= = 0, is
1
when
for a function of two variables. Expanding a function into a Taylor's series in the neighbourhood of a //)
has the form
...
fc If a notation
<
the
Taylor
series
is
then called
a
Maclaunn's
series.
(7)
Here the
as follows: ,
y)
(X
x=a !l=t> ~
d*f(x,y)
f
a)
dx*
+2
*)
dy*
+
~W
2
and so
Here and henceforward we mean "in positive integral powers".
forth.
314
(Ch. 8
Series
The expansion
occurs
(7)
if
the remainder term of the series
fe=l
as n
* oo.
The remainder term may be represented
in the
form
n+i
where
0<0<
1.
Expand the indicated functions in positive integral powers of *, find the intervals of convergence of the resulting series and investigate the behaviour of their remainders: 2587. a
x
2589. cos(* 2 2590. sin *.
(a>0).
9*a sin cin^r-i-M * -r--r Jooo.
,-,>*
.
+ a).
/^
1
of the principal expansions I-V and a geometric the expansion, in powers of *, of the following write progression, functions, and indicate the intervals of convergence of the scries:
Making use
2592.
3
2593.
X2
*7 ~4X
2594. xe~*
2598. cos
-
7^rJT2 5
J
*.
2599. sin 3*
.
,.
o
.
2
x
+ * cos 3*.
2600.
.
x
2595. e \
2601.
2596. sinh*.
2602. In
* .
1
2597. cos2*.
2603. In
(1
,v
-f*
2*
2
).
the
Applying expand following functions in powers of *, and indicate the intervals in which these expansions differentiation,
occur:
2604.
(l+*)ln(l+*).
2606. arc sin*.
+
_ +*
2
2607. In (x V\ ). various the functions in techniques, expand given Applying powers of * and indicate the intervals in which these expansions occur:
2605. arc tan*
2
2
2608. sin *cos *. 2609. 2610. 2611.
+O~
8
(1
2613
'
2614
-
-
cosh8
*'
Sec
315
Taylor's Series
3}
2615. ln(x
2
+ 3x + 2).
in/i_L.^/*v + *)** rMl
2618. J.
2616.
f'Jl
dx.
\
o
X
2619. 2617.
^L_
[
"-* z ^"
*
V
^
1
*4
Write the first three nonzero terms of the expansion of the following functions in powers of x: 2620. tan*. 2623. sec*. 2621. tanh*. 2624. In cos*. 2622. e^ x 2625.
where
e
is
2627.
A
the
eccentricity
and 2a
is
the
major axis
of
the
ellipse.
tenary J line
heavy string hangs, under its own weight, in a ca= acosh and H is the horizontal where a = // J ,
a
q
tension of the string, while q that for small *, to the order string hangs in a
2628.
powers
-f
*
2
^
8
.
2*
2
5*
2
in
a
series
of
of *-| 4.
2629.
powers
function
the
,
V
y^a
parabola
Expand
the weight of unit length. Show 4 of * it may be taken that the
is
of
s
4*
f(*)-5*
2
3* +2.
in a series of
Expand f(x+h)
h in a
powers of * series of powers of *
in a
series of
2630.
Expand In*
2631.
Expand
2632.
Expand
Xz
in a series of
powers
of
1.
1.
*+l.
j
2633.
2634. 2635.
Expand
X2
~Y~
oX
in a series of ~\~ JL
Expand
^
Expand
e* in a
,
,
4jc
2636. Expand ]/* in 2637.
Expand cos*
2638.
2 Expand cos *
2639*.
Expand In*
of
powers
*-f4.
powers of x series of powers of * + 2. a series of powers of * 4. 7
in a series of
in a series of
in a series of
powers
of
*
powers
of
* j
in a series of
powers
_^
of
.
^^.
T-J
|-2.
.
316
[Ch. 8
Series
2640.
-*
Expand
in a series of
of
powers
.
2641.
What
is
the
the
of
magnitude
error
we put appro-
if
ximately
^2 + 1 + 1 + 1? To what degree of accuracy if we make use of the series
2642. ber
j
,
we
will
-
by taking the sum
of
its
five
first
2643*. Calculate the number ing the function arc sin A: in a ple 2606). 2644. How many terms do
^
.
-* O
u
x=l?
to three decimals
we have
num-
..,
terms when
series of
cosAr=l
calculate the
powers
of
x
by expand(see
Exam-
to take of the series
|j+...,
in order to calculate cos 18
2645.
How many
to three decimal places? terms do we have to take of the series
*
+...,
to calculate sin 15 to four decimal places? 2646. How many terms of the series
have to be taken to 2647.
How many
find the
number
e
to four
decimal places?
terms of the series In
do we have
to take to calculate
cimals?
In 2
to
two decimals? to 3 de-
__
2648. Calculate \/7 to two decimals by expanding the funcx in a series of powers of x. tion l/S 2649. Find out the origin of the approximate formula
+
r
\/ (f-{-x&a
a = 5,
+ ~-
(a>0), evaluate
it
by means
and estimate the_ error.
2650. Calculate J/I9 to three decimals.
of
Y^3
t
putting
317
Taylor's Series
Sec. 3]
2651. For what values of x does the approximate formula 2 JC
,
yield an error not exceeding 0.01? 0.001? 0.0001? 2652. For what values of x does the approximate formula sin
X&X
yield an error that does not exceed 0.01? 0.001? '/*
2653. Evaluate
.
to four decimals.
y^-dx 1
2
^e~* dx
2654. Evaluate
to four decimals.
1
2655. Evaluate
^
l/^ccosxdx to three decimals.
1
2656. Evaluate
dx to three decimals.
J
2657. Evaluate
_
j^l+^'dx
decimals.
to four
1/9
2658. Evaluate
^yxe*dx
to three decimals.
2659. Expand the function cos(x y) in a series of powers x and y, find the region of convergence of the resulting series and investigate the remainder. Wiite the expansions, in powers of x and y, of the following functions and indicate the regions of convergence of the series:
of
2663*.
2660. sin x> sin y. 2 2661. sin(* +y*).
,
664 *
\n(lxy+xy). arctan
j/_
l
2662*.
\=*!!. \+xy
2665. f(x, y)
wers of
ft
2666. function
*:=
1
4-
ft,
2667. 2668.
~*y
and /(*,
= ax*-\-2bxy + cy*.
Expand f(x+h,y-\k)in po-
k.
y)
= x*
2y*
when passing from
+ 3xy.
Find
the values
the
#=1,
increment f/
=2
to the
of
this
values
y=2 + k. Expand Expand
2. the function e x+y in powers of x -2 and y the function sin(x+y) in powers of x and
+
__ Series
318
[C/i.
8
Write the first three or four terms of a power-series expansion x and y of the functions: x 2669. e cosy.
in
1
2670.
*'-
(H-*)
Sec. 4. Fourier Series
let
1. Dirichlet's theorem. conditions in an interval 1)
is
We
uniformly bounded;
is
say that a function
satisfies the the function
f (x)
b)
if,
in this interval,
that
is
\f(x)\^M when
(a,
a
b,
Dirich-
where
M
constant; 2) of
all
has no more than
them
the function
are
a the
of
number
finite
kind
first
has a finite limit on
f (x)
of
points
of
[i.e.,
at
each
the
0)=
left f (g
discontinuity and g discontinuity Urn f (I e) and a ~"
finite limit
on the right /(-{-0)= lim /(
+ e)
(e>0)J;
e ->o
has no more than a finite number of points of strict extrenium. Dirichlet's theorem asserts that a function /(*), which in the interval of this interval at ji, Ji) satisfies the Dirichlet conditions at any point x ( which /(x) is continuous, may be expanded in a trigonometric Fourier series: 3)
f(x)=?+ a, cos x + b
v
sin
+a
x
2
cos 2x-\-b 2 sin
2*+
.
.
.
+a n cos nx +
+ b n sinnx+..., where the Fourier
a n and b n are calculated from the formulas
coefficients
ji
= JT
J\
(1)
ji
f(x)cosnxdx(n = Q,
1,
...);&=
2,
f
J
JI
-n
-jt
If x is a point' of discontinuity, belonging to the interval ( of a jt, n), function f (,v), then the sum of the Fourier series S (x) is equal to the arithmetical mean of the left and right limits of the function:
SM = ~ At the end-points
of the interval
2. Incomplete Fourier s=/(jc)], then in
formula
series.
*
=
If
n and X = K,
a function
/ (*)
is
even
(1)
6 rt
-0
(w
=
l
f
2,
...)
and ji
a
2
r
/"=^
J
/
= 0,
1,2,
...).
[i.
e.,
/(-
x)
=
a function
If
319
Fourier Series
Sec. 4]
odd
is
/ (x)
/(
[i.e.,
and
x)
=
an
then
/ (*)],
=Q
t/i
= 0,
2 ...)
i,
IT
bn
=~
nx dx
/ (x) sin
J\
JT
(n
=
1
2,
,
. .
.).
A
function specified in an interval (0, n) may, at our discretion, be contior an odd function; hence, in the interval ( Jt, 0) either as an even it may be expanded in the interval (0, Ji) in an incomplete Fourier series of sines or of cosines of multiple arcs. 3. Fourier series of a period 21. If a function f (x) satisfies the Dirichlet the discontinuities conditions in some interval ( /, /) of length 2/, then at of the function belonging to this interval the following expansion holds:
nued
r,
=a + y
x
ju
,
/(*)
fli
cos -J-
.
,
2nx
juc
+ &! stay +a ,
,
2
cos j-
2nx
+ b sm+.., .
.
.
.
2
nnx
.
,
.
nnx
,
where
-M
e
/
-- dx
flllX
v
f (x)
cos
.
.
= 0, /-v
(/i
t
r
1,
2,
v
...), (2)
6n
= -L
\f(x)sln^-dx(n=\,
'
2,
-i
At the points
of discontinuity of the function f (x) and at the end-points of the interval, the sum of the Fourier series is defined in a manner similar to that which we have in the expansion in the interval ( Jt, n). In the case of an expansion of the function / (x) in a Fourier series in
x-l
an arbitrary interval formulas
(2)
(a,
a-f-2/) of
length
the
2/,
limits
of
integration
in
should be replaced respectively by a and a-|-2/
functions in a Fourier determine the sum of the series
the following
in the the points of discontinuity and at the end-points of the interval (x ic, X function itself and of the sum JT), construct the graph of the of the corresponding series [outside the interval ( JT, ji) as well]: Q x when ~~ n
Expand
interval
ji,
(
ji),
series
at
=
=
< ^
Consider the special case when 9fi79 2672.
=
xr
Consider c)
fl
= 0,
6
the
=
1;
2673. f(x) 2674. f(x) 2675. f(x) 2679.
=
=x = eax = slnax.
(0,
=~
bx when special cases: 0. d) a=l, 6 2
.
.
Expand
the interval
ct
>
a)
= b = l\
2676. / (x) 2677. f(x) 2678. f (x)
the function f(x)
2ji).
a
=
1,
b)
a
=
1,
6=1;
= cos ax. = smhax. = cosh ax.
-jr-
in
a Fourier
series
in
320
_
Series
2680.
in the interval
number
g
the function f(x)
Expand
=~
in sines of
(Ch. 8
multiple arcs
Use the expansion obtained
n).
(0,
__ to
sum
the
series:
Take the functions indicated below and expand them, in the interval (0, jt), into incomplete Fourier series: a) of sines of multiple arcs, b) of cosines of multiple arcs. Sketch graphs of the functions and graphs of the sums of the corresponding seiies in their domains of definition. x. Find the sum of the following series by means 2681. f(x) of the
expansion obtained:
I+P+P+... Find the sums of the following number series 2682. f(x) = x by means of the expansion obtained: 2
.
1)
2683. f(x)
*
+7)2+ 32+
= eax
.
1
2684.
1
2^"
32
42
"!"
,
when
/(*)=
when
-^
x when x when
ji
Expand the following sines of multiple arcs:
2687. f(x) 2688.
2)
i
(x =
0<;c^~ *
functions, in the
when
0<#^^-,
when
~n < x < n.
x(n
interval
(0,
K),
in
x).
= sin-|. /(A;)
Expand the following functions, in the interval sines of multiple arcs: u when
(0,
Ji),
in co-
Sec. 4\
__
_
321
Fourier Series
2690.
when 2691. / (x)
= x sin x. cos x
when
cos x
when
< x^ ~
,
Y
2693. Using the expansions oi the functions x and x* in the in cosines of multiple arcs (see Problems 2681 and it) 2682), prove the equality interval (0,
2694**. Prove that
+
?L Ji,
(
and in
=
x}
the function /(x)
if
x\
f(~
then
,
its
n) represents an expansion
if
the function f(x)
the interval
Ji,
(
odd
is
is
it
Ji)
even and we have
is
Fourier series in in cosines of
arid
/fy-f *) =/
expanded
in
the
interval
odd multiple
(y~*)
sines
of
arcs,
then
odd mul-
tiple arcs.
Expand the following functions in cated intervals: 2695. 2696. 2697. 2698.
f(x)
= \x\
f(x)=*2x f(x)
Fourier series in the indi-
!<*
(
(0
1).
= e* (/<*
f(jc)=10
Jt
(5
Expand the following functions, in the indicated intervals, in incomplete Fourier series: a) in sines of multiple arcs, and b) in cosines of multiple arcs:
(0<*<
2699. /(*)=! * 2700. /(*) 2701.
=
1).
(Q
2702.
/M-{ 2 _-; ^n
2703.
Expand the following function
T (3
'
J
\
in
cosines
\ :
)
when -|
,
11-1900
^
of
multiple
Chapter IX
DIFFERENTIAL EQUATIONS
Sec. 1. Verifying Solutions. Forming Curves. Initial Conditions
1. Basic concepts. An equation F(x, y,
where y
Differential
Equations
of
Families of
of the type (n)
y'
t/)
= 0,
(1)
y(x) is the sought-for function, is called a differential equation of order n. The function y y(x), which converts equation (1) into an identity, is called the solution of the equation, while the graph of this function is called an integral curve. If the solution is represented implicitly, O(A, f/)~0, then it is usually called an integral Example 1. Check that the function t/ sinjt is a solution of the equation
=
=
Solution.
We
have:
and, consequently, /"
+y
sin x -f sin x
The integral */,
Cp
....
Cw
)
^
0.
=
(2)
of the differential equation (1), which contains n independent arbitrary constants C t ..., C n and is equivalent (in the given region) to equation (1), is called the general integral of this equation (in the respective region). By assigning definite values to the constants C,, ..., C n in (2), we get particular ,
integrals. if we have a family of curves C n from the system of equations
Conversely, eters Cj
= 0,
dx
-0.
....
(2)
and eliminate
the param-
*-0,
dx n
we, generally speaking, get a differential equation of type (1) whose general integral in the corresponding region is the relation (2). Example 2. Find the differential equation of the family of parabolas
y^C^x-CJ. Solution. Differentiating equation /'
= 2C
l
(*
Eliminating the parameters C, and the desired differential equation
C 2)
(3)
(3)
twice,
and
C 2 from
(/"
we
get:
= 2C,.
equations
(4) (3)
and
(4),
we obtain
Sec
It
1
is
Verifying Solutions
1
function
verify that tha
to
easy
323
converts
(3)
this
equation
identity.
2. Initial conditions. differential equation y
desired particular solution y
for the
If
= f(x,
(n)
y
y, y'
into
= y(x)
(n ~ l)
an
of a
(5)
)
the initial conditions
and we know the general solution
are given
y =
arbitrary constants
from the system
of
Cn
(5)
are determined
(if
this
which y(0)
i
Find the curve
=
=
y'(0) have:
l.
We
Solution.
of the
family
=
in
formulas
(6)
2.
i/
Putting #
possible)
C lt .... C B ), c ..... c ). .
*,
for
is
-cp(*o,
3.
C, .....
...,
equation
C n ),
equations
Example
C lt
of
'=
and
(6)
1=0,
we obtain
(7),
2-C
+ ^,
(?)
2C 2
t
,
whence (:,=(),
and, hence,
y
C,=
l
= e~.
Determine whether the indicated functions are solutions of the given differential equations: 2704. *//'== 2#, 2705. y 2706.
2707. 2708.
2709. 2710.
t2
(jc
= 5x*.
t/
= x* h
//)
.
djc -h
JK
dy
= 0,
= ^=^
.
+ = 0, = 3sinjc 4 cos*. ~? + = 0, x = C, cos
(/
^/
2
a) jc
-(X. y^* Show
2
f
X/ 2
2711. (x
8
t
/
-f
e.
that for the given relations are integrals:
11*
//
//
differential
equations
the
indicated
_ _
324
Differential Equations
2712. (x 2713. (xy-x)y"
Form
differential
C lt
=
t
+ xy' + yy' equations
=
Q, 2y' y \n(xy). of the given families
(C, C,, C, are arbitrary constants): Cx. 2714. y
= =
272l
9718 ii&. 971Q 2719.
n *.=
9799 *722.
,
fo' ? 3re P arameters >-
yr
2724.
^
2720.
& M-- = 2 + Ce"^.
2726.
Form
thej
+ay
0,- y .)'-2p*
-* y
*-.
curves
of
a parameter).
(
#//-plane.
i
4
'
+ = <:'.
2717. *'
i
y
2715. y Cx*. 2716. *>=-.2C*.
[Ch. 9
2725.
differential
= =
equation of
all straight lines in
the
;
2727. Form the differential vertical axis in the ^y-plane. 2728. Form the differential
equation of equation
of
all
with
parabolas circles
all
in
the
xy-plane.
For the given families of curves find the given initial conditions: 2729. x*y* 5. C, 0(0) 2730. 2731. y C sin 2732. y^C.e-x
= y=(C^ =
the
lines
that
satisfy
=
l
= 0,
Sec. 2. First-Order Differential Equations
P. Types of first-order differential equations. A differential equation the first order in an unknown function yt solved for the derivative y' is the form t
y'
where f(x, y) consider the
= f(*.
of of
(i)
the given function. In certain cases it is convenient to variable x as the sought-for function, and to write (1) in the is
form
x'=e(*,y)>
$--.
(i')
j
where
gfr
Taking into account that (1)
and
(!')
may
0'
=
^
and
*'
=
j^
the
differential
equations
be written in the symmetric form
P(x, y)dx+Q(x, t/)
where
P
and Q
(2)
y) are knowri functions. By solutions to (2) we mean functions of the form t/ cp(jc) ihat satisfy this equation. The general integral of equations (1) (x, y)
(x,
=
or
and
x=ty(y) (I'),
or
equation
where
325
First-Order Differential Equations
Sec. 21
C
is
(2),
is
form
of the
an arbitrary constant.
2. Direction
The
field.
set of directions
tana = /(x,
y)
called a direction field of the differential equation (1) and is ordinarily depicted by means of short lines or arrows inclined at an angle a. Curves f(x, y) k, at the points of which the inclination of the field has a constant value, equal to k, are called isoclines. By constructing the isoclines and direction field, it is possible, in the simplest cases, to give a
is
Fig
rough sketch of the
field of
intagral
105
curves, regarding
the latter
which at each point have the given direction of the field. Example 1. Using the method of isoclines, construct the
as
field of
curves integral
curves of the equation y'=*x. Solution. rection field,
The family
By constructing the isoclines x~k (straight lines) and the diwe obtain approximately the field of integral curves (Fig. 105).
of parabolas
the general solution. Using the method of isoclines, make approximate constructions of fields of integral curves for the indicated differential equations: is
2733. y'
2734.
= =
x.
-f-
2735. y'=l-ftf
2736.
y'=
2737. y'
=
8 .
326
(Ch. 9
Differential Equations
3. Cauchy's theorem. b < y < B}
If
U\a
a function / (#, y) is continuous in some region in this region has a bounded derivative
and
(** y)> tnen through each point (* y Q ) that belongs to U there passes one and only one integral curve y y(x) of the equation (1) [cp (* ) #ol4. Euler's broken-line method. For an approximate construction of the
f'y
,
M
(* integral curve of equation (1) passing through a given point replace the curve by a broken line with vertices M,-(x/, #/), where
f
=
2.
(i'0,
Using Euler's method
find (/(I),
if
y(0)=l
(/i
1.
for the
2
We
t/
),
we
(one step of the process),
/i
/)
Example
,
2,
...).
equation
'
-0.1).
construct the table:
Thus,
/(!)=
1.248.
For
the
sake
of
comparison,
the
exact
value
is
i
T ss 1.284
e
Using Euler's method, find the particular solutions to the given differential equations for the indicated values of x: 2738. y' = y, y(0)=l; find y(\) (A-0.1). 2739.
y'-x + y, /(!)-
2740. ^' 2741.
= -X_,
t/(0)
1;
find y(2),
= 2;
-, y(Q)=l;
find
= 0.1). = 0.1). (A = 0.2).
(A
(/t
Differential Equations with Variables Separable
Sec. 3]
Sec.
First-Order
3.
Differential
Equations
with
327
Variables
Separable.
Orthogonal Trajectories
1. First-order equations with variables separable. is a first-order equation of the type
An equation with
variables
separable
X (x) Y
(y)
Dividing both sides of equation -
= f(x)dx
y'
= f(x)g(y}
dx
+ X, (x) Y, (y) dy = Q
(1)
(i)
(!')
by g(y) and multiplying
Whence, by integrating, we
get
the general
by
dx,
we
get
integral of equa-
tion (1) in the form
Similarly, dividing both sides of equation (!') by X, (x) we get the general integral of (!') in the form
Y
(y)
and integrating,
=
=
then the function y is ti Q (r/ )=0, yQ we have directly evident) a solution of equation (1) Similarly, the straight a and will be the integral curves of equation (!'), if a and b are, respectively, the roots of the equations X, (*)() and Y (*/) 0, by the ieft sides of which we had to divide the initial equation. Example 1. Solve the equation If
for
also (as lines x
some value y
is
y-b
=
'
3>
In particular, find the solution that satisfies the initial conditions
Solution. Equation
(3)
be written in the torm
may
dx~~
x
Whence, separating variables, we have
and, consequently, In |
where the arbitrary constant
y
|
=
In
|
x\
taken antilogarithms we get the general solution In C,
is
+ ln C in
|t
logarithmic form. After taking
f where
C=
When
C,.
dividi dividing by y ila contained in the formula
we could' lose = 0.(4) for C '
the solution
=0. but
the
latter
is
328
(Ch. 9
Differential Equations
Utilizing the given initial sired particular solution is
conditions,
we
get
C = 2;
hence, the de-
and,
J2
y ~~
'
x
Certain differential equations that reduce to 2 separable. Differential equations of the form
with
equations
variables
=
reduce to equations of the form (1) by means of the substitution u is the new sough t-for function 3 Orthogonal trajectories are curves that intersect the lines of the given a right angle. If F (x, y, #') (x, y> ort=0 ia is a parameter) at family is the difierential equation of the family, then
where u
=
O
is
the differential equation of the orthogonal trajectories. Example 2. Find the orthogonal trajectories of the family of ellipses
Solution Differentiating the equation (5), tion of the family
Whence, replacing
if
by
^7,
we
find
the
duerential
106
get
the
differential
equation
orthogonal trajectories ~~
~~
integrating,
we have
i
*x*
(family
'
x
it'
of
equa-
0.
(/'
Fig.
we
parabolas) (Fig. 106).
of
the
329
Differential Equations with Variables Separable
Sec. 3\
4. Forming differential equations. When forming differential equations in geometrical problems, we can frequently make use of the geometrical meaning of the derivative as the tangent of an angle formed by the tangent line to the curve in the pos'tive x-direction. In many cases this makes it possible straightway to establish a relationship between the ordinate y of the desired curve, its abscissa x, and the tangent of the angle of the tangent line (/', that is to say, to obtain the difleiential equation. In other instances (see Problems 2783, 2890, 2895), use is made of the geometrical significance of the definite integral as the area of a curvilinear trapezoid or the length of an arc. In this case, by hypothesis we have a simple integral equation (since the desired function is under the sign of the integral); however, we can readily pass to a differential equation by differentiating both sides. Example 3. Find a curve passing through the point (3,2) for which the segment of any tangent line contained between the coordinate axes is divided in half at the point of tangency. Solution. Let (x,y) be the mid-point of the tangent line AB. which by hypothesis is the point of tangency (the points A and B are points of intersection of the tangent line with the y- and *-axes). It is given that OA 2y and OB Zx. The slope of the tangent to the curve at (x, y) is
M
=
M
OA
dy_ dx~ This
is
y x
OB~
'
the differential equation of the sought-for curve. Transforming, d\
we
get
dy _ ~
~x
~T~~y~
and, consequently, In
\nx-\-\ny Utilizing the initial condition, curve is the hyperbola xy 6.
Cor xy
C.
we determine C = 3-2
Solve the differential equations: 2 2 2742. tan A: sin y d* + cos x cot ydy
6.
Hence,
the
desired
= Q.
=
2743. xy'~ // {/'. 2744. xyy' =-. = a(l 2745. // jq/' x fan ydx 2746. 3c 2747. y' tan * //.
\x\ =
+*V). x
+ (l
e
)
sec*
ydy = Q.
Find the particular solutions of equations that indicated initial conditions: x 0. 2748. (1 +e ) y y' = e*\ //= 1 when jt ! when x 2749. (xy* + x) dx-\-(x* yy)dy=* 0; // 2750. r/'sin x
= y\ny\ y~l
when
= = *= -|.
satisfy
= Q.
Solve the differential equations by changing the variables: 2751. y' = (x+y)**
=
2752. i/ (8*42//+l)'. 2753. (2x + 3{/ l)dx-{ (4x 2754. (2x y)dx (4x 2y
+
+ fo/ -
5) dij
= 0.
the
_ _
330
Differential Equations
In
Examples 2755 and 2756, pass
[C/i.
9
to polar coordinates:
2755.
2756. 2757*. Find a curve whose segment of the tangent is equal to the distance of the point of tangency from the origin. 2758. Find the curve whose segment of the normal at any point of a curve lying between the coordinate axes is divided in
two
at this point.
2759. Find a curve whose subtangent is of constant length a. 2760. Find a curve which has a subtangent twice the abscissa of the point of tangency. 2761*. Find a curve whose abscissa of the centre of gravity of an area bounded by the coordinate axes, by this curve and the ordinate of any of its points is equal to 3/4 the abscissa of this point.
2762. Find the equation of a curve that passes through the point (3,1), for which the segment of the tangent between the point of tangency and the *-axis is divided in half at the point of intersection with the y-axis. 2763. Find the equation of a curve which passes through the point (2,0), if the segment of the tangent to the curve between the point of tangency and the t/-axis is of constant length 2. Find the orthogonal trajectories of the given families of curves (a is a parameter), construct the families and their orthogonal trajectories. 2
2764. x 2 2765. t/
+ y =a = ffx. 2
2
2766. xy 2767. (x
.
Sec. 4. First-Order Homogeneous Differential
1. Homogeneous equations.
A
= a. 2
2
a)
t-f/
=a*.
Equations
differential equation
P(x y)dx+Q(x,y)dy = t
(1)
is called homogeneous, if P (AT, y) and Q (x, y) are homogeneous functions of the same degree. Equation (1) may be reduced to the form
and by means it
is
of the substitution y xu, where u is a new unknown function, transformed to an equation with variables separable. We can also apply
the substitution
Example
1.
x-yu. Find the general solution
to the equation
Sec. 4]
_
_
First-Order Homogeneous Differential Equations
Solution. Put y
we
Integrating,
=
get w
ux',
then
-f xu'
Q
=
In In
,
that reduce to
or
whence x In In
y
2. Equations
= eu + u
331
.
x
homogeneous equations.
If
'
and 6 I
!
#2^2
Ue
0,
then, putting into equation (2) x
w
+ a,
j/
= t;-fp,
where
I
the constants
a and P
are found from the following system of equations,
+ c, = 0,
a 2a
+ b$ + c = 0, t
we
get a homogeneous differential equation in the variables u and v. If u, we get an equation with variables 0, then, putting in (2) a,x 4- b^y separable.
6
Integrate the differential equations:
2768. 0' 2769.
=1
277
1.
y^-^.
-
(x-y)ydx-x*dy = Q.
=
2
find the family 2771. For the equation (x +y*) dx 2xydy of integral curves, and also indicate the curves that pass through the points (4,0) andj_l,l), respectively. 2772.
= +
+
2773. xdy -\-ifdx. ydx Vx* 2 2 0. 2774. (4x* + 3xy jf) dy f/ ) dx + (4y 3jvy 1 2775. Find the particular solution of the equation (x = x 2. 2xydy Q, provided that r/=l when Solve the equations: 2776. (2x
+
=
+
=
9777 f/./ 2/77.
-
3y*)dx+
1
2779. Find the equation of a curve that passes through the point (1,0) and has the property that the segment cut off b\ the tangent line on the r/-axis is equal to the radius vector of the point of tangency. 2780**. What shape should the reflector of a search light have so that the rays from a point source of light are reflected as a parallel
beam?
532
_ _
[Ch. 9
Differential Equations
2781. Find the equation of a curve whose subtangent is equal mean of the coordinates of the point of tangency. 2782. Find the equation of a curve for which the segment cut off on the y-axls by the normal at any point of the curve is equal to the distance of this point from the origin. 2783*. Find the equation of a curve for which the area contained between the #-axis, the curve and two ordinates, one of which is a constant and the other a variable, is equal to the ratio of the cube of the variable ordinate to the appropriate to the arithmetic
abscissa. off
2784. Find a curve for which the segment on the y-axis cut by any tangent line is equal to the abscissa of the point of
tangency. Sec. 5. First-Order Linear Differential
Equations.
Bernoulli's Equation
1. Linear equations. A
differential equation of the
)-y^Q
form (1)
(x)
f
of
degree one in y and y is called linear. If a function Q(jt)=~0, then equation ).y
takes the form
(1)
=Q
(2)
and is called a homogeneous linear differential equation. In this case, the variables may be separated, and we get the general solution of (2) in the form
y = C-e
-
P P(X) dx
J
(3)
.
To solve the inhomogeneous linear equation (1), we apply a method that called variation of parameters, which consists in first finding the general solution of the respective homogeneous linear equation, that is, relationship (3). Then, assuming here that C is a function of x, we seek the solution of the inhomogeneous equation (1) in the form of (3). To do this, we put into (1) y and y' which are found from (3), and then from the differential equation thus obtained we determine the function C(x). We thus get the general solution of the inhomogeneous equation (1) in the form is
^/x -f J = C(x).e Example
I.
Solve the equation y'
Solution.
Solving
it
we
tan **/-}- cos
x.
The corresponding homogeneous equation get:
C ^- r *^I 1
(4) is
333
Bernoulli's Equation
Sec. 5]
C
Considering
and differentiating, we
as a function of x,
dC Putting y and y' into
dC
1
cos*
.
djc
we
(4),
~
A dx
cos x
I
x
sin
fi'nd;
~ '
/,x>o2 2 x cos
get:
sin*
C
cos 2 *
cos*
,
dC -r-=
or
' ,
dx
whence
=
Ccos 2 *d*==i-* +
Hence, the general solution
of
equation
j (4)
has the form
COS* In solving the linear equation
we can
(1)
make
also
use of the substitu-
tion
uv
y
where u and v are functions of
x.
f
we require that '
(1) will
have the form
+ v'u^Q(x).
\-P(x)u]v
[u If
(5)
t
Then equation
(6)
+ P(jc)M = 0,
(7)
then from (7) we find M, and from (6) we find u; hence, from (5) 2\ Bernoulli's equation. A first order equation of the form y'
+ P (<) y ^ Q
(x)
we
find y.
y\
where a 7=0 and a 7= 1, is called Bernoulli's equation ""*. near equation by means of the substitution z the or substitution the y = uv apply directly
It
1
r/
It
t
reduced to
is
a
li-
also possible to method of variais
tion of parameters.
Example
2.
Solve the equation y'
Solution. This
is
=
y+*
VH-
Bernoulli's equation. Putting
y=^u>v,
we
ijet
u'v
+ v'u
To determine
uv
+x
the function u
y"uv
we
or
u'
~w-0 x
we have u
Putting this expression into (8),
f
:
u j
-f
v'u
require that the relation u'
be fulfilled, whence
v
we
get
x*.
=x
V^taT.
(8)
334
_ _ Differential Equations
whence we
[Cfi.
9
find v:
and, consequently, the general solution
is
obtained
in
the form
Find the general integrals of the equations: 2785.
ax
-=*. x + = x*.
2786.
2787*. (\ 2788. y*dx(2xy
Find the particular solutions that satisfy
the
indicated
con-
ditions:
+y
2789. X y'
e*
= Q\
y = b when x
1-- *
2790. y'
j-2-7
2791. y'
yianx =
= 0;
cos x
;
*/
y=
=
= a.
when x-0.
when
jt
= 0.
Find the general solutions of the equations: 2792. *l dx
+ JLx = '
Xyy*
m
2793. 2xy x 2794.
0dx +
(
+x
2795. 3xdy--=y(l sin A: 3y* smx)dx. 2796. Given three particular solutions y,
equation. Prove that the expression
What
^^
y lt y 2 of a linear remains unchanged for
the geometrical significance of this result? 2797. Find the curves for which the area of a triangle formed by the *-axis, a tangent line and the radius vector of the point of tangency is constant. 2798. Find the equation of a curve, a segment of which, cul off on the x-axis by a tangent line, is equal to the square of the ordinate of the point of tangency. 2799. Find the equation of a curve, a segment of which, cut off on the y-axis by a tangent line, is equal to the subnormal. 2800. Find the equation of a curve, a segment of which, cut off on the y-axis by a tangent line, is proportional to the square of the ordinate of the point of tangency.
any
x.
is
Exact Differential Equations. Integrating Factor
Sec. 6]
2801. Find the equation of the curve for which of the tangent is equal to the distance of the point tion of this tangent with the x-axis from the point
335
the
segment
of
intersec-
M (0,a).
Sec. 6. Exact Differential Equations. Integrating Factor
1. Exact
differential equations. If for the differential equation
P(x.y) the equality
dU
is
:p =-^-
fulfilled,
dx+Q(x
y)dy = Q
t
then equation
(1)
may
(1)
be written
in
the
=
and is then called an exact differential equation. The gen= C. The function U (x, y) is detereral integral of equation (1) is U (x, y) mined by the technique given in Ch. VI, Sec. 8, or from the formula form
(see
(x,
t/)
Ch, VII, Sec.
Example
1.
9).
Find the general integral 2
(3x
Solution.
This
+ 6Af/
2 )
an exact
is
5= i2 X y and,
dx
of the differential
equation
+ (6x*y + 4y') dy = 0.
differential
since
equation,
hence, the equation
is
of
the form
-
-J K/
= 0.
Here,
=
and
whence
U=
2
(3^
+ 6xy*) dx +
q>
(y)
= x> + 3*V +
q> (y).
= x y + 4y* (by y -f (y) = and hypothesis); from this we get y* + C*. We finally get (y) f/(r, t/)-= -f-3xV + +C consequently, x -f3^V + / = C is the sought-for Differentiating
U
with respect
to y,
q>
jt
we
6jc
l
cp'
-y
= 4(/* '(//)
4
s
find
f
q>
s
.V .
C(
,
general integral of the equation. 2. Integrating factor. If the left side of equation (l)is not a total (exact) differential and the conditions of the Cauchy theorem are fulfilled, then there exists a function U, U.(A', y) (integrating factor) such that
=
\i(Pdx+ Qdy) = dU. Whence
it
is
found that the function
The integrating
factor
u,
is
u,
satisfies the
(2)
equation
readily found in two cases:
336
Differential Equations
Example Solution.
"
.
and
Solve
2.
equation
Here P = 2*y + rV-f-^
f
,
_ d#
or
dx
+ x*y + ~
2xy
!
'
hence>
r dx
cty
J
.
,
^^^ + r
9
Q=x*+y*
"
Since it
the
[C/t.
'
ft=
Q x jB, dx
follows that 1 fdP ^ JS ----dQ\ and dx=dx Q.\dy dxj
-r
.
rffi
Multiplying the equation by
which
is
.
^.
\\
\i
= e*
an exact differential equation.
t
.
In
11 *
= ^,
-
11 = ^. r
we obtain
Integrating
it,
we
get
the
general
integral
Find the general integrals of the equations: 2802 (x + 2z/)dy = 0. t/)d^+(^ 1 2803. (V + 1/ 2x) dx + 2xydy = Q. 3 8 2804. (jc S^ 8 + 2) dx (3x'y y ) d(/ = 0.
+
2805.
+
^-^=='
2806.
.
y
y
2807. Find the particular integral of the equation
=
which
satisfies the initial condition #(0) 2. Solve the equations that admit of an integrating factor of the form fi = M*) or M' H'(y)
=
:
+ y*)dx2xydy = 0. 4-xy)dxxdy = Q.
2808. (x 2809. y(l 2810.
281
1.
(jc
cos
t/
y sin
r/) rfi/
+ (x sin
*/ -(
//
cos y) dx
= 0.
First-Order Differential Equation? not Solved for Derivative
Sec. 7]
Sec. for
7.
337
Equations not Solved
First-Order Differential
the Derivative
1. First-order differential equations of higher powers. F(x,
which
for example two equations:
is
y,
two
of degree
!/')
in y',
y'=ti(x,y)>
If
an equation
= 0,
(I)
the.i
by solving
(1)
for y'
<*.
we
get (2)
M
Thus, generally speaking, through each point (xQt (/ ) of some region of a plane there pass two integral curves. The general integral of equation (1) then, generally speaking, has the form
0(AM/, where
and
C^OM*,
C) 0> 2
{/,
C)
(x. y,
= 0,
(3)
the general integrals of equations (2). 2 are Besides, there may be a singular integral for equation (1). Geometrically, a singular integral is the envelope of a family of curves (3) and may be obtained by eliminating C from the system of equations tl> 1
OJx, or
by eliminating p =
t/'
0,
C)
= 0,
from the system
F(x
t
y,
(*, y,
0>c
F'
p)-0,
p
of
C)
=
(4)
equations
(x,y,p)=0.
(5)
We
note that the curves defined by the equations (4) or (5) are not always solutions of equation (1); therefore, in each case, a check is necessary. Example 1. Find the general and singular integrals of the equation 2 A'//'
Solution. Solving for y'
2A'j/'
-f
y
Q.
we have two homogeneous
equations:
do fined in the region x
the general integrals of which are
or
Multiplying,
we
get the general integral of the given equation (2*
+y
C)
2
4
2 (A-
+ xy) -
or (a
we
(It
family of parabolas). Differentiating the general find the singular integral
may
integral with respect
be verified that y-(-jc=0
to
C and
eliminating C,
+ Jt-O. is
the solution of this equation.)
Differential
_
Equations
(Ch. 9
by differentiating possible to find the singular integral with respect to p and eliminating p. 2. Solving a differential equation by introducing a parameter. If a firstorder differential equation is of the form It
is
+ 2xp
also
#=
then the variables y and x 1
from the
be determined
may
system of equations
dq>
where p = t/' plays the part of a parameter. Similarly, if y = ty( x #') tnen x and y are determined >
from the system
of equations
Example
Find the general and singular integrals
2.
of the
equation
2
y=y' -xy'+^. Solution.
we
Making the substitution t/'=p,
rewrite the equation in the
form
Diffeientiating with respect to x and considering ft
dp
dp
p a function of
x,
we have
p-*p-p-*fx +* or
-p(2p
x)**(2p
x),
or-j-
into the original equation,
=
l.
we have
Integrating
,
we
get p
= x + C.
Substituting
the general solution or
=y
Differentiating the general solution with respect to C and eliminating C, we x^ x* obtain the singular solution: */ -r-. (It may be verified that */ -r- is the
=
=
solution of the given equation.) If we equate to zero the factor 2p x which was cancelled out, we get x x2 into the which is the and, putting p given equation, we get y=-j t
pay
,
same singular
solution.
Find the
general
(In Problems curves.)
2812.
2812
and singular integrals and 2813 construct the
(/''-^'-M^O. 2
2813. 4y'
9JC-0.
of
the
field
equations: of
integral
_
Sec
8]
_
The Lagrange and Clairaut Equations
=
Q. 2814. yy'*(xy+l)y' + x 2 2815. yij' 2xy'+y Q. 2816. Find the integral curves
339
=
M
that pass through the point
(
y
0,
the
of
equation
y'*
+y* =
1
.
)
Introducing the parameter y' = p, solve the equations: 2820. 4y = x*+y'\ 2817. x=smy'+lny'. 2818. y 2819.
= y'*e>". = y'*
/2
2
,*
9 91
'
Sec. 8. The Lagrange and Clairaut Equations
1. Lagrange's equation. An equation
of
the form
= *
= tf
(1)
called Lagrange's equation Equation (1) is reduced to a linear equation in x by differentiation and taking into consideration that dy^pdx: is
f
(p)] dp. If
p^q>(p),
then from (1) and
(2)
we
(2)
get the general solution in parametric
form:
parameter and f(p)^ g(p) are certain known functions. Besides, a singular solution that is found in the usual way. 2. C'lairaut's equation. If in equation (l)p^
where p there
is
may
a
be
raut' s equation
+
^(C) (a family of .straight lines). general solution is of the form y-Cx is also a particular solution (envelope) that results by eliminating the parameter p from the system of equations Its
There
I
x~
\
lJ:=
Example. Solve the equation
Solution. Putting ing
dy by pdx, we
y'^p we
+ i.
have //^2pv
+
(3)
;
get
p dx = 2p dx +
2.v
or
Solving this lineai equation, we will have
*=l
dp
~
different! a Ting
and replac-
__
340
__
Differential Equations
^_
[C/t.
9
Hence, the general integral will be
=l(l
To
find the singular integral,
we form
f/-2px + in the usual
the system
l,
= 2*-
way. Whence
*=
1
2
and, consequently,
Putting y into (3) we are convinced that the function obtained Is not a solution and, therefore, equation (3) does not have a singular integral.
Solve the Lagrange equations: 2822.
y-' =
2824. y
= (1 +y'
+ r=FT
2823. , ,' Find the general and singular integrals of the tions and construct the field of integral curves: 2826. y xy' +y'*. 2827. y xy'+y'. r
2828.
2829.
by
_
= = = Xy' + V \-\-(y') y = xy' + j,.
ij
Clairaut equa-
2
.
2830. Firid the curve for which the area of a triangle formed a tangent at any point and by the coordinate axes is con-
stant.
2831. Find the curve it the distance of a given point to any tangent to this curve is constant. 2832. Find the curve for which the segment of any of its tangents lying between the coordinate axes has constant length /, Sec. 9. Miscellaneous Exercises on First-Order Differential Equations
2833. Determine the types of differential equations cate methods for their solution:
= = 2ju/ + *'; y'
b) (*-*/)//' c)
d) y'
(f/-
8)
U
=
and indi-
Miscellaneous Exercises on First-Order Differential Equations
Sec. 9]
= (* 4+ y sin (x* xy) y' = #';
i)
j)
x cos f/'
k)
+ 2xy')dx + + (y* + 3*y dy = 0; m) (x' 3xy) dx + (* + 3)dy = 0;
1)
1 ;
{/'
341
(x'
)
2
n)
Solve the equations: 2834. a) (x
ycos^
b) jcln
2835.
dy
xrfx=(^
y'^dy.
2836. (2xy*y) dx-\-xdy=-Q. 2837. xy' -\- y -^ xtf \nx. 2838. y xy' 2839. f/ or/' jc#' 2840. ^( 2841. (1
= =
2842.
+ +
y'-y^^l.
2843. y^-((/'-l 2xe>)y'. 2844. //'-|-//cosAr=sin.tcosx. 2848. (x'y 2849.
x*
-\-y-l) dx
\
2845
'
2846
'
(l-^
^'-T-^ =
-
2847. y' (xcosy-'r a sin
2
= 1.
(xy + 2x3yQ)dy = Q.
= y'
2850. ./"
d.v
= (x*y f 2)
+ y dy J-| dJf = 0. 286 =* X 2862. *. = cosA:. f/y'-hf/
2852. 2dx
2853..
.
.
i ,'
J
,
2854.
2863.
2855. xdy-\ ydx^ifdx. !. 2856. //' (j: |- sin //)
=
2857.
H? = -P + P
/ ff
= 2^'H-l/l
/ //'==
j (1 +lny
2864. (2e
d*- (x
1
2859. x
*/" -h
Zxyy'
2860.
^^H-\t
ln^).
)
dy x
y')
^-
+ 2/'
x
+y
0.
= 0.
2865. 2868.
dx =
^ dydx =
/'
=2 f
^
(;(/'
=
.
t
.
4 f
x
S
J
2858. ^'
1
(-y'
v 4- 2
\
*
y-1 J
-1- 1 )
2867. a(xt}' 2868. xdy- y dx
= y* dx.
342
[Ch. 9
Differential Equations
2869. (x*
1
)/
At,
dy +
(x*
+ 3xy V^-\) dx = 0.
2870.
^
5
+
2871. J/oM1 dy+ (x y 2872. xyy'* (x* +y*)y'
= xy' + (3* + 2;q/ = 3p
2873. y 2874. 2875.
Ya*
+x
2
)
dx = Q.
.
2
y'Jdx
+ f*
1
2xy
3y*)dy=--Q.
2
2
-t-4*/
.
2
for the indicated
initial
con-
ditions:
2876.
#= y'=^\ y
2881.
jc=l.
= \\ y=\ for *=1. = 0. y = 2; y = 2 for = = = for 0. ^ + !) !; = cos = for * = 0. #' + for x = 0. x*\ y = 2y = y' = = for * = 0. y'+y 2x\ y
2877. e*- y' 2878. cot ;q/' 2879. e^(^ 2880.
for
*/
jt
-f
A;;
{/
2882. for x = 0. 2883. xy'=y\ a) //==! for jc=l; b) y for x-0. 2884. 2xy' y\ a) y=l for jc=l; b) y z 2 for x-0; b)y=l forjc=-0; 2885. 2xyy'-\-x 0; a) # = Q for je= 1. c) y 2886. Find the curve passing through the point (0, 1), for which the subtangent is equal to the sum of the cooidinates of the point' of tangency. 2887. Find a curve if we know that the sum of the segments* cut off on the coordinate axes by a tangent to it is constant and \
=
=
= =
y-0
equal to 2a. 2888. The
sum of the lengths of the normal and subnormal equal to unity. Find the equation of the curve if it is known that the curve passes through the coordinate origin. 2889*. Find a curve whose angle formed by a tangent and the radius vector of the point of tangency is constant. 2890. Find a curve knowing that the area contained between the coordinate axes, this curve and the ordinate of any point on it is equal to the cube of the ordinate. 2891. Find a curve knowing that the area of a sector bounded by the polar axis, by this curve and by the radius vector of any point of it is proportional to the cube of this radius is
vector.
2892. Find a curve, the segment of which, cut off by the tangent on the x-axis, is equal to the length of the tangent.
Miscellaneous Exercises on First-Order Differential Equations
Sec. 9]
2893. Find
the curve,
between the
contained
which the segment
of
axes
coordinate
of the
divided
is
343
tangent
into half by
the parabola if =--2x. 2894. Find the curve whose normal at any point of it is equal to the distance of this point from the origin. 2895*. The area bounded by a curve, the coordinate axes, and the ordinate of some point of the curve is equal to the length of the corresponding arc of the curve. Find the equation of this curve if it is known that the latter passes through the
point (0, 1). 2896. Find the curve for which the area of a triangle formed vector of the point of by the x-axis, a tangent, and the radius 2 is to a constant and tangency equal 2897. Find the curve if we know that the mid-point of the segment cut off on the x-axis by a tangent and a normal to the curve is a constant point (a, 0). .
When forming
first-order differential equations, particularly in phvsical frequently advisable to apply the so-called method of differenwhich consists in the fact that approximate relationships between tials, infinitesimal h.crements of the desired quantities (these relationships are accurate to infinitesimals of higher order) are replaced by the corresponding relationships between their differentials. This does not affect the result. Problem. A tank contains 100 litres of an aqueous solution containing 10 kg of salt. Water is entering the tank at the rate of 3 litres per minute, and the mixture is flowing out at 2 litres per minute. The concentration is maintained uniform by stirring. How much salt will the tank contain at the end of one hour? Solution. The concentration c of a substance is the quantity of it in unit volume. If the concentration is uniform, then the quantity of substance in volume V is cV. Let the quantity of salt in the tank at the end of t minutes be x kg. The quantity of solution in the tank at that instant will be 100 / litres,
problems,
is
it
+
and, consequently, the concentration
During time
c=
kg per
QQ
litre.
solution flows out of the tank (the solution contains 2cdt kg of salt). Therefore, a change of dx in the quantity of salt in the tank is given by the relationship
This ing,
is
2dt litres
dt,
the
of
the sought -for differential equation. Separating variables
we obtain
ln* =
21n(100+0 + lnC C
or
*
(100-M)
C= x
The constant C 100,000.
100,000 -
At
is
the .
..
found from the expiration of
ft f =^ 3.9 kilograms of
and integrat-
u
salt.
1
fact
one
'
that
=
xvh^n f 0, the tank
hour,
\
10,
will
that
is,
contain
_ _
344
[Ch. 9
Differential Equations
2898*. Prove that for a heavy liquid rotating about a vertical axis the free surface has the form of a paraboloid of revolution. 2899*. Find the relationship between the air pressure and the 2 altitude if it is known that the pressure is 1 kgf on 1 cm at 2 at an altitude of 500 metres. sea level and 0.92 kgf on 1 cm 2900*. According to Hooke's law an elastic band of length / increases in length klF(k const) due to a tensile force F. By how much will the band increase in length due to its weight if the band is suspended at one end? (The initial length of the band is /.) 2901. Solve the same problem for a weight P suspended from the end of the band. When solving Problems 2902 and 2903, make use of Newton's law, by which the rate of cooling of a body is proportional to the difference of temperatures of the body and the ambient medium. 2902. Find the relationship between the temperature T and the time f if a body, heated to T degrees, is brought into a room
=
W
at constant
temperature
(a degrees).
2903. During what time will a body heated to 100 cool off and during the first to 30 if the temperature of the room is 20 20 minutes the body cooled to 60? 2904. The retarding action of friction on a disk rotating in a liquid is proportional to the angular velocity of rotation. Find the relationship between the angular velocity and time if it is known that the disk began rotating at 100 rpm and after one
minute was rotating at 60 rpm. 2905*. The rate of disintegration of radium is proportional to the quantity of radium present. Radium disintegrates by one half in 1600 years. Find the percentage of radium that has disintegrated after 100 years. 2906*. The rate of outflow of water from an aperture at a vertical distance h from the free surface is defined by the
formula
where c0.6 and g is the acceleration of gravity. During what period of time will the water filling spherical boiler of diameter 2 metres flow out of cular opening of radius 0.1 in the bottom.
m
it
a hemithrough a cir-
2907*. The quantity of light absorbed in passing through a thin layer of water is proportional to the quantity of incident light and to the thickness of the layer. If one half of the original quantity of light is absorbed in passing through a three-metrethick layer of water, what part of this quantity will reach a depth of 30 metres?
Sec.
345
Higher-Order Differential Equations
10]
2908*. The air
resistance to a body falling with a parachute to the square of the rate of fall. Find the limit-
is
proportional ing velocity of 2909*. The is covered with
descent.
bottom of a tank with a capacity of 300 litres a mixture of salt and some insoluble substance. Assuming that the rate at which the salt dissolves is proportional to Ihe difference between the concentration at the given time and the concentration of a saturated solution (1 kg of salt per 3 litres of water) and that the given quantity of pure water dissolves 1/3 kg of salt in 1 minute, find the quantity of salt in solution at the expiration of one hour. 2910*. The electromotive force e in a circuit with current i, resistance /? and self-induction L is made up of the voltage drop
Rl and the electromotive force of self-induction the current
=
and
i
Sec.
10.
/
time
at
/
= 0.
when
if
e^Esmat
(E
and
L^.
Determine
are constants)
o>
Higher-Order Differentia) Equations
1. The case
of direct integration.
If
then
n
i
Miles
2. Cases of reduction of order. contain y explicitly, for instance,
then, assuming y'
we
p,
I.
If
differential
a
equation does not
get an equation ot an order one unit lower;
F(x, p
Example
I)
p')-0.
t
Find the particular solution
of
the equation
that satisfies the conditions
^ Solution. Putting
Solving
we
get
the
= 0,
f/'
=
#'=p, we have
latter
equation as
when
x
/ = p', a
linear
= 0. whence
equation
in
the
function p,
346
Differential Equations
From
the fact that
y'=p =
[Ch
when x = 0, we have
=^
0,
i.e.,
9
Cj==0.
Hence,
or
___ ____ 2
dx~~
'
whence, integrating once again, w? obtain
Putting
y~
solution
is
when x
0,
we
find
C 2 = 0. Hence, *2
y^ 2)
a
If
y'=p,
we
p-?->
y"
get
desired
particular
-
differential equation does not contain
then, putting
the
x explicitly,
an equation
of
an
for
instance,
order
one unit
lower:
Example
Find the particular solution
2.
provided that
/=!,
Solution. Put y'
We
#' = =p t
when then
#
From
it,
we
the fact that f/'=p
equation
= 0.
tf^p-- and
have obtained an equation
the argument). Solving
of the
our equation becomes
of the Bernoulli
type in p (y
is
considered
find
=
when r/=l, we have C 1 =
1.
Hence,
P=y Vy^ or
Integrating,
we have arc cos
Putting
y=l
x
and *=0, we obtain C 2 = 0,
=C
2,
whence
= cosx
or t/==secx.
Sec.
_
__
347
Higher-Order Differential Equations
10]
Solve the following equations: 291
1
.
2912. 2913.
/=1
2920. yy"
.
2921.
y"=\-y'\
2914. xy"
i/'
\
yif-y'
2922.
= 0.
= jfy' + y".
/"=
2923. (x
-f-
+ y') = 0.
(I
-p
.
!)
(x-[- 2) y'
+ 2915.
// = {/".
*f/"
2925
y'+T^" =^"-
'
'
+ t/")=a/".
2918.
t/'(l
2919.
xy + x0' =
t
!
+
= 0.
= r/'ln^. *
2924.
= 0.
2916.
+ x+ 2
+
1+x
2926. xy" /"= 2927. y""-f y"*=l. solutions for the indicated initial con-
l.
Find the particular ditions:
2928.
(1
2*/'
-M')z/" l
2929.
= 0;
=
l+t/' 2j/t/"; r/=l, 2930. ytf + i/"=*y"\ y=*l, 2931. xy" 0, t/ t/' y';
=
=
=
Find the general integrals 2932. 2933. 2934.
2935.
for
for
\ JC
x=
0.
= 0.
following equations:
f/
=
+ y" =
l;
i/V=l;
l,
xif=V\
(/'
=
!
= i. = 0. for
forjt
0=1, y'=\ for x=l; y=l y= Jt
-\-y'
;
for
A:
=e
s .
+ -^'y' = 2 + \nx;y = -^, y' = tor = -i, y'-l for jc= 1. = 2 for x = 0. y'y'' + y'(y\)=*Q; = 2, = = for ^ = 0. 3/Y {/-(-i/" -F 1; y= 2, y' = = foi x = 0. 0; (/=!, -t-y"-2//'' = and = for * = ^j/' + /'* yy" = 0; j/=l for
2939. y"(\-\-\nx)
2941.
= 0.
satisfy the indicated conditions:
2
2940.
*
forx=l.
1
for
of the
=3
y"-yy"=ify'. J = 0. f/i/"-hf/' -y'lnj/
2937. yy" 2938.
j/'
yy^Vy^ify'-y'if. yy' = y" +
Find solutions that 2936.
= 0, = y' = y'
j/
\
/-n-ln;
(/
<
2942. 2943.
2944.
l
1
/'
J/
-|-
jc
/
1,
348
(Ch. 9
Differential Equations
2949.
= Q\ y = 0, y' = 2 for = 2. = 0; y=l, #' = 2 for * = 0. 2jr-3/ = 40 t/= 1, -0 for = 0. = 0; #=1, y' = for * = 0. 2yy"-\-y* I/"-*/' -*/; y=-l, y' = l for x=l.
2950.
/ + e>y-2^' = 0;
z
+
2945. 2y' (y' 2946. y'y*+yif
2
*/'
i
2947.
A:
Gx)-y" I
2948.
jc
*/'
;
l
*/'*
2
i
-
i
//-I,
y'
=e
for
^--
= 0; = 0, 0' = for*=l. = (H-00')/ (l+0'")0'; 0=1, 0'=1 for x = (*+l)0" + x0 = 0'; 0=-2, 0'=4 forx=l.
2951.
H-//y"-f
2952.
1
y'
ff
2953.
Solve the equations: 2954. y' 2955. 0'
= =
/2
2956.
y"
j
= 4y
//
.
+
2957. yy' y" --= y'* y"* Choose the integral curve passing through x Q. the point (0, 0) and tangent, at it, to the straight line y 2958. Find the curves of constant radius of curvature. 2959. Find a curve whose radius of curvature is proportional to the cube of the normal. 2960. Find a curve whose radius of curvature is equal to the .
+ =
normal. 2961. Find a curve
whose radius
of curvature
is
double the
normal. 2962. Find the curves whose projection of the radius of curvature on the //-axis is a constant. 2963. Find the equation of the cable of a suspension bridge on the assumption that the load is distributed uniformly along the projection of the cable on a horizontal straight line. The weight of the cable is neglected. 2964*. Find the position of equilibrium of a flexible nontensile thread, the ends of which are attached at two points and which has a constant load q (including the weight of the thread) per unit length.
2965*. A heavy body with no initial velocity is sliding along an inclined plane. Find the law of motion if the angle of inclination is a, and the coefficient of friction is p,. (Hint. plane.)
The
2966*. is
frictional force is
We may
consider
to the
that
proportional square if the initial velocity
motion
where
ji/V,
^V is the force of reaction of the
the
of the is
zero..
air resistance
velocity.
in free fall
Find the law
of
Sec. 11]
_
_
Linear Differential Equations
349
2967*. A motor-boat weighing 300 kgf is in rectilinear motion with initial velocity 66 m/sec. The resistance of the water is proportional to the velocity and is 10 kgf at 1 metre/sec. How long will it be before the velocity becomes 8 m/sec? Sec. 11. Linear Differential Equations
1. Homogeneous
equations.
=
The
functions 0i q>i(x), f/ 2 if there are constants C,
==(P/iW are called linearly dependent not alt equal to zero, such that !/
= f
q> 2 (A:)
C lf
t
...,
...
Cn
otherwise, these functions are called linearly independent. The general solution of a homogeneous linear differential equation
+ P, (x) e/<"- + = coefficients P,-(x) '>
//<>
w
;
th continuous
!,
(/
where
are linearly ...,/ (/,, y tt (fundamental system of solutions).
2. Inhomogcneous
.
.
.
+ P n (x) y = ....
2,
independent
The general
equations.
is
n)
(1)
of the
solutions
form
of
equation
(1)
solution of an inhomogeneous
linear differential equation (2)
with continuous coefficients
where and Y
P,-
and the right side
(x)
has the form
f (x)
the general solution of the corresponding homogeneous equation (1) particular solution of the given inhomogeneous equation (2). If the fundamental system of solutions (/,, y % ..... y n of the homogeneous equation (1) is known, then the general solution of the corresponding inhomogeneous equation (2) may be found from the formula f/
is
is
a
y
=C
l
(x) y,
where the functions Cj(x) ing system of equations: '[ (*)
+ C'
t
(*'
+C
=
(*)y*
:;w
!,
t
2,
(x)
y2
+
.
.
.
+C n (x) ya
.... n) are
+
+ C'n (x)
,
determined from the follow-
/,
= 0,
+...+c;(o;=o. 0)
(the method of variation of parameters). Example. Solve the equation (4)
_ _
350
Differential Equations
[C/i.
9
Solution. Solving the homogeneous equation
*ir+if'=o
we
f
get (5)
Hence,
it
be taken that
may
yl
and the solution
of
equation
Forming the system the equation
is
(4)
(4)
and
(3)
= \nx may
taking
t/"+~ =
and # 2 =
1
be sought in the form
into account
that the reduced form of
we obtain
jt,
Whence
+
and
>l
and, consequently,
y
=
where A and B are arbitrary constants.
2968. Test the following systems
of
functions for linear rela-
tionships:
x
a) x, b)
ing
x
+
2
2
x\
1,
d) x,
x+1,
x +2;
2969.
Form
a linear
a) b) c)
d)
g)
h)
x'\
2
9
e *, e**\
sin *, cos A:, 1; 2 2 sin x, cos *, 1.
homogeneous
differential
equation, know-
fundamental system of equations:
= sin x,
y 2 = cos x\ ==xe*\ y 2 y^e*. 2 y^x* # 2 = *
y
e*
f)
;
c) 0,
its
x\
e) *,
1;
2x
,
l
<
f/,
= ^x
y*
= ^x
sin ^,
f/ 8
= ^ cos
A:.
2970. Knowing the fundamental system homogeneous differential equation
find its particular solution
y that
of solutions of a linear
satisfies the
initial
conditions
_
Linear Differential Equations with Constant Coefficients
Sec. 12]
351
2971*. Solve the equation
its
knowing
particular solution y
=
-^.
l
2972. Solve the equation
x*(\nx-l)y"-xy' 4-y = 0, particular solution y =x.
knowing its By the method of variation of parameters, solve the following inhomogeneous linear equations. }
=
a
2973. **(/" 3;t xy' 2 2974*. x*y" + xy' y=x 2975. y'" -f (/'-sec A:. Sec.
.
12. Linear Differential
.
Equations of Second Order
with Constant Coefficients
t. Homogeneous equations. A second-order linear equation with constant and q without the right side is of the form
coeflicients p
U) k
If
l
and
fc
2
are roots of the characteristic equation .Q
then the general solution three ways: 1)
2) 3)
of
equation
(/-CVV + C.e*-* 6, and = *,; y-eV(C, + CV) *, -^(^cnspx-HC^mpA) if
kz
(1)
is
(2)
t
\vritten in one
are real and
/?,
^
of the following
k.\
if
(/
2. Inhomogeneous
equations.
if
*,=a + p
and * a -=a
The general solution
of a
pi (p 76 0).
linear inhomoge-
neous differential equation y"
+ py' + w=fW
(3)
be written in the form of a sum:
may
where y
the general solution
of the corresponding equation (I) without determined from formulas (1) to (3), and Y is a particular solution of the given equation (3). The function Y may be found by the method of undetermined coefficients in the following simple cases: a e *P n (x), where P n (x) is a polynomial of degree n. 1. f (x) If a is not a root of the characteristic equation (2), that is, (p (a) 96 0, then we put Y e**Q n (x) where Q n (x) is a polynomial of degree n with undetermined coeflicients. If a is a root of the characteristic equation (2), that is,
is
and
side
=
Y=x
2.
=
e
Qn
/ (*)
\\here r is the multiplicity of the root * [P n (*) cos bx+ Q m (x) sin bx\.
(x) t
=
a(r=l
or r
= 2).
352
_
__
[Ch. 9
Differential Equations
If
i
cp(a
^ 0,
bi)
then we put
Y = e ax
[S N (x) cos bx
+ T N (x) sin bx],
where S^(x) and Tu(x) But if cp(a &/) = 0, then
are polynomials of degree
K = x re ax where r=l). is
fc
2
r
the
is
[Stf (x) cos 6*
+ T N (x) sin to]
the roots a
of
multiplicity
N-max
bi
{n,
m},
,
second-order
(for
equations,
In the general case, the method of variation of parameters (see Sec. 11) used to solve equation (3). 2 Example 1. Find the general solution of the equation 2y" y' y 4xe *. 2 & l=u has roots fc,~l and Solution. The characteristic equation 2&
=
The general solution
.
(first
type)
Y
homogeneous equation
e* + C 2e The right side of the given equation is/ (x) = Y e zx (Ax + B), since n=l and /=0. Difleren(x). Hence, twice and putting the derivatives into the given equation, we is
==C
f/
=4xe zx =t ax P n tiating obtain:
the corresponding
of
%,** (4 A X
2
1
=
.
+ 45 + 4^) __ e i* (2Ax + 25 + A)
2 e *
(Ax H- B)
4xe zx
.
zx
and equating the coefficients of identical powers of x arid Cancelling out e the absolute terms on the left and right of the equality, we have bA=4 and 4 28 744-5fl
= 0,
whence 4
=
and 5
-=-
o
2A Thus, K^
f
oH
-g-*
anc^
*
)
=
-.
Jo
ne gei] eral solution of the given equation
f
Find the general
+ =
is
solution of the equation y* xe* y 2y 2 characteristic equation k has a double root 2/f-f- 1 x ft=l The ri^ht side of the equation is o! the form f(x)xe here, 0=1 and n=-l. The particular solution is Y =x*e* (Ax B), since a coincides \nHth 2. the double root k=-\ and, consequently, r Diilerentiating Y twice, substituting into the equation, and equating the
Example
2.
.
Solution. The
\
+
=
coefficients,
we obtain
/l
=
,
fl
equation will be written in the
= 0,
Hence, the general solution of the given
form *
Find the general solution of the equation */*-f y= 2 The characteristic equation and -j-l=r() has roots /?, The general solution ot the corresponding homogeneous equation 3, where a~0 and P = l| be*
Example fc
a
=
3,
Solution. i.
will |see
l
The right side
fe
is
of
the form
/'
Sec.
Linear Differential Equations with Constant Coefficients
12]
=
0, 6=1, P n (jc)=0, particular solution Y,
where a
#=1,
a
Q OT
= *.
(*)
To
_
353
this side there corresponds the
= 0,
fc=l, r=l). twice and substituting into the equation, we equate the coefficients of both sides in cos*, xcosx, sin*, and xsmx. We then get four 44 = 1, from which we deter0, 0, 4C = 0, -25 + 20 equations 2A + 2D 1 1 X2 X mine A cos * -f- -j- sin *. 0, C 0, D Therefore, (here,
Differentiating
=
,
= =
=
=
=
3. The
is
= C, cos + C
y
jc
x2 2
sin
x
is
principle of superposition of the sum of several functions
and K/(/
=
l,
then the
sum
3,
.
y'+py'+w^-fiW y is
the solution of equation
A:
x
+
sin *.
-7-
solutions.
If
the right side of equa-
corresponding solutions of the equations
are the
.., n)
cos
-j-
tion (3)
2,
44
K=
.
4
4
The general solution
(<
=
i.
2 ..... n).
= Y + Y +...+Y n n
l
(3).
Find the general solutions of the equations: 2982. y" + 2y' 2976. tf 5y'6y Q. 0. 2983. 2977. if 4y' 9y 2984. y" + ky 2978. 0. 2985. 2979. iT y 0. 2980. ^_2i/ +2j/ 2981. + 40' +130 0.
=
=
/
yy'^Q. = +
y=
= =
f
/
Find the particular solutions that satisfy the indicated conditions:
2987. y"5tj'-\-4y 2988. y"+ 3tf' +20
2989. 0" 2990. 0^ 2991.
= Q\ = 0; = 0,
y=5
= 8 for * = for jc0. y=5 0' = = 2 for x = 0. t
y'
1
1,
+ 40 = 0; + 20' = 0; 0=1, 0'=0 for ^ = = a, 0' = for x = 0. f
/=;
2992. 0" 2993. 0"
+ 30'=0; = = 0; = + f
ji
for x for
jc
= =
and and
= =
x
for
x=l.
type of particulai solutions inhomogeneous equations: 2994.
a)
Indicate
0"-40 = A:V
b) 0"
12-1900
+
90
the
x ;
= cos 2x\
= 3.
for
for
the given
_ _
354
Differential Equations
[Ch. 9
= sin 2x + e**\ x d) y" + 2y' + 2y = e sir\x\ - 5
c) y"
4y' -f 4y
f)
Find the general solutions of the equations:
= / ^ +# = + ^ + + = ^. + /
+
t
2995. y" jc 4i/' 4(/ 2996. *' 6. 2997. 2^' ^ 2998. 8^ 7^=14. 2999. y"y^e*. 3000. 3001. 3002. 3003. f 3004. t/' 3005. y" 2y'-{ 5y 3006. Find the solution of the equation y" 0. satisfies the conditions y=l, */'= 1 for x
/
/
.
+
.
=
=
+ 4y =
sin x that
Solve the equations: W2y
3007. 2)
^-2
2
+ x=/l o)
Consider
sinp/.
the
cases:
1)
p-co. 3008. 3009. 3010. 3011. 3012. 3013. 3014. 3015. 3016.
3017.
/ t/"
y" y"
2y'
y"
2(/'
y"-\-y'
=
8y = e = 5x + 2e x
x
^_y'^2;c-~l
8cos2*. .
3e
x .
+ 2r/ + y = ^ + ^ x x 2y + lQy=s\r\3xi-e /_4f/' + 4f/ = 2^ + if 3y' = /
.
{/"
f
f/"
.
.
3018. 3019. Find the solution to the equation that satisfies the conditions /=-, f/ /==l for ^
/
Solve the equations: 3020. (/" 3021. j/" 3022. 3023. tf 3024. 3025.
y = 2xsmx. 2x 4(/==g
sin2A:.
/ H- 9y = 2x sin
2y'=e
= 0,
tx
p
+
o>;
Sec.
_
Linear Differential Equations with Constant Coefficients
12]
= + = =
355
x
3026. tf2y' 3y x(l+e' ). 3x 2xe*. 3027. y" 2y' x 3028. */" 4y'+4y xe* 2xe-'* 3029. y" 3y 2y' (x^ l)e*. 3030*. y" y 2x cos x cos 2x. 3031. t/' 2y 2xe*(cosx sinx).
=
+ + = =
.
+
Applying the method
variation
of
following equations: 3032. y" + y ianx. 3033.
3034. 3035.
= y" + y = cot x. = -. y" 2y' y" + 2y' + y =
3036.
of
parameters, solve
<
the
.
3037. 3038. a)
-(-*/
b) y"
.
V
y = tanh
A:.
x 2y = 4x*e \
3039. Two identical loads are suspended from the end of a spring. Find the equation of motion that will be performed by one of these loads if the other falls. Solution. Let the increase in the length of the spring under the action in a state of rest be a and the mass of the load, m. Denote by x the coordinate of the load reckoned vertically from the position of equilibrium in the case of a single load. Then of
one load
where, obviously, k tion
is
x=C
and
d* -77 Tt
= :
l
cos
when
=
I/ /
= 0;
and, consequently, - 1
+C
2
I/
sin
whence C
t
=a
t.
7*7*
The
^ ~" ^ * initial
The
general
conditions yield
solu-
Jt
=u
and C a = 0; and so
3040*. The force stretching a spring is proportional to the increase in its length and is equal to 1 kgf when the length increases by 1 cm. A load weighing 2 kgf is suspended from the spring. Find the period of oscillatory motion of the load if it is pulled downwards slightly and then released. 3041*. A load weighing P 4 kgf is suspended from a spring and increases the length of the spring by 1 cm. Find the law of motion of the load if the upper end of the spring performs a vertical harmonic oscillation (/ 2sin30/ cm and if at the initial instant the load was at rest (resistance of the medium is
=
=
neglected). 12*
.
L
356
_ _
(Ch. 9
Differential Equations
m
is attracted by each of two 3042. A material point of mass centres with a force proportional to the distance (the constant of proportionality is k). Find the law of motion of the point knowing that the distance between the centres is 26, at the initial instant the point was located on the line connecting the centres (at a distance c from its midpoint) and had a velocity
of zero.
3043. A chain of length 6 metres is sliding from a support without friction. If the motion begins when 1 m of the chain is hanging from the support, how long will it take for the entire chain to slide down? 3044*. A long narrow tube is revolving with constant angular velocity o> about a vertical axis perpendicular to it. A ball inside the tube is sliding along it without friction. Find the law of motion of the ball relative to the tube, considering that a) at the initial instant the ball was at a distance a from
the axis of rotation; the initial velocity of the ball was zero; b) at the initial instant the ball was located on the axis of rotation and had an initial velocity v 9 .
Sec.
13.
Linear
Differential
Equations
of
Order
Higher
than
Two
with
Constant Coefficients
1. Homogeneous equations. The fundamental system of solutions y lt Un f a homogeneous linear equation with constant coefficients n y< + a iy n + + an _,y' +a n y = (1)
#t
>
<
.
is
.
.
constructed on the basis of the character of the roots of the
characteristic
equation Q.
(2)
1) if k is a real root of the equation (2) of multiplicity m, then to this root there correspond linearly independent solutions of equation (1):
Namely,
2)
then
if
to
m
a
i
the
p/ is a pair of complex roots of equation (2) of multiplicity latter there correspond 2m linearly independent solutions
equation (1): *x y l e cos PX, #,
=
= e*x sin px,
yt
= xe* x cos PX,
2. Inhotnogeneous equations. equation
A
y4
= xe* x sin PX,
m, of
...
particular solution of the inhomogeneous (3)
is
sought on the basis of rules 2
and 3
of Sec.
12.
_
Sec.
Euler's Equations
14]
__
357
Find the general solutions of the equations: 3058. y / 13
3045. y'"
3048. y' v 2J/" 3t" 3049. y'" 3050. y'v + 4y
= 0.
,
n
'
w'v' + ^O. '
*054.
w' 1
sss-
^T
'
-
306 >- y' v 3062 y'"iJ
ai/ = b.
^
-
3060.
3053.
n(n-l) ,.
-
= x'
MS,---..
SB:
3067. Find the particular solution of the equation
y'"+2y"-{-2y'+y = x that satisfies the initial conditions y (0) 14.
Sec.
A
Euler's
= y' (0) = y" (0)=-=0.
Equations
linear equation of the form n
y^f(x)
t
(I)
a, b, A ..... <4,,_,, A n are constants, is called Enter's equation. Let us introduce a new independent variable /, putting
where
l
ax
+ b-^e
1 .
Then
and
Euler's
is
equation
transformed
coefficients. z
Fxample 1. Solve the equation x y" Solution. Putting x^e*. we get
into
a
linear equation
+ xy' +{/ =
1.
z
==e
Tr
~Sx
dx~z==e
(~di*~~~di)'
Consequently, the given equatioi takes on the form
whence
= C, cos + C, sin + y = C, cos (In *) + C a sin (In x) + y
or
t
t
1
1.
with constant
358
_ _
[Ch. 9
Differential Equations
For the homogeneous Euler equation
Q
the solution
(2)
be sought in the form
may
= **.
(3)
found from (3), we get a characteristic equaexponent k. If k is a real root of the characteristic equation of multiplicity m then to it correspond m linearly independent solutions m- 1 k 2 Putting into (2) tion from which
y'
(/,
,
...,
we can
y
(n}
find the
t
0i If
a
= **. a
is
p*
correspond
2m
0,
= **-ln*.
*/,
= ** (In*)
,
..., y m
y 4 =*Mnx.sin(plnx) .....
2.
We
into the given characteristic equation
Solving
find
y^C Solve the equations:
3069.
3070.
*'g + 3*| + = 0. *V xy' 3(/ = 0. xY-|-x(/'4-4(/ = 0. <,
3071. 3072.
3073. tT
=
3074. 3075.
3076.
there
1
cos (p
I
+4l/=0.
equation, after
Hence, the general solution will be
3068.
it
= *a In x cos (P In *),
^m-i^*" (In*)*"
3Xy'
2
we
to
put
Substituting
it
.
Solve the equation
X *y'' Solution.
(\nx)
m, then
pair of complex roots of multiplicity linearly independent solutions (/ 8
Example
=x
^y_
.
46-1-4
= 0.
cancelling
out x k
,
we
get
the
Sec.
15]
*
_
Systems
of Differential
_
359
Equations
3077. Find the particular solution of the equation
+ y = 2x conditions y = 0,
*V that satisfies the initial
Sec.
xy'
15. Systems of Differential
Method
=
*/'
Equations
To find the solution, for first-order differential equations, that
of elimination.
system of two form
them with respect
the desired functions, have, for example,
we
first
equation of the system
(1)
to x.
Determining z from the
instance, of is,
We
solved for the derivatives
of
when *=1.
!
of
a
normal
a
system
differentiate
of
the
one of
and substituting the
value found, A.. \
/
(3)
into equation (2), we get a tion u. Solving it, we find
equation with one unknown func-
second-order
(4)
where C, and C 2 are arbitrary constants. Substituting function (4) into formula (3), we determine the function z without new integrations. The set of formulas (3) and (4), where y is replaced by \|>, yields the general solution of the system (1).
Example. Solve the system
z
3
,
f
+'-'-T* Solution.
We
differentiate the first equation
with respect to
x:
^ + 2^dx^+ 4^-4. dx*^
From
the
first
from the second
equation
we
will
dx
we determine
have -^ ax
=
-5-
&
**
^
=
/
\
-T-
+*+
(
-;
4
l+4x -75-
&
y
~
\
dy
TT"T ax
2y
and then j
Put ting
z
and j- into the equation obtained after differentiation, we arrive at a secondorder equation in one unknown y:
360
[Ch. 9
Differential Equations
Solving
it
we
find:
and then
We
do likewise
can
in
the case
of
a
system with a larger number of
equations.
Solve the systems: dy
3078.
I
=z
JJ
3085. dx r/
= 0, z =
when
je
= 0.
when
/
= 0.
3079. 3086.
3080.
dz
:
= 0, y=\
dT dx
3081.
4/_// 3087.
dy dz
3088*. dx
_
xy~ c)
3082.
2
da
__
dy __dz_ x-\-y~~~ z dy __
yz zx
t
'
dz
xif
isolate the integral curve passing through the point (1, 1, 2). dy
3089.
3083.
3084.
dz
3090.
.Sec.
Integration of Differential Equations by Power Series
16}
361
3091**. A shell leaves a gun with initial velocity u at an angle a to the horizon. Find the equation of motion if we take the air resistance as proportional to the velocity. 3092*. A material point is attracted by a centre with a force proportional to the distance. The motion begins from point A at a distance a from the centre with initial velocity perpendicular to OA. Find the trajectory. by Means
Sec. 16. Integration of Differential Equations
of
Power Series
not possible to integrate a differential equation with the help of elementary functions, then in some cases its solution may be sought in the If
is
it
form of a power
series: 00
'(* y=2 n=
n *o)
0)
-
o
The
undetermined
series (1) into the of the binomial x
=
are found by putting the coefficients c n (n \, 2, ...) equation and equating the coefficients of identical powers x on the left-hand and right-hand sides of the resulting
equation.
We
in the
can also seek the solution of the equation
form
of the
Taylor's series
y(*)
^ ^^ (*-*)"
=
y
(3)
and the subsequent derivatives y (n) (x ) where y(x Q) = y Q t/ ) y' (x ) = f (* are found successively by differentiating equation (2) and by (n- 2, 3, ...) ,
,
putting X Q in place of x
Example
Solution.
1.
Find the solution
We
put
y
of the equation
=
whence, differentiating, we get y"
= 2.\Ct + 3.2c x+...+n(n-\)cn x n -* + (n+l s
+ In + 2)(rt-t-l) <:+,*"+... Substituting y and y" into the given equation,
Collecting
powers
together, on
of x
the
left
of this
and equating to zero the
we
arrive at the identity
equation, the terms with identical of these powers, we will
coefiicients
362
Differential Equations
[Ch.
9
have
cs
=
2
=
^~
and so
forth.
Generally,
3.4-6. 7-... (k
=1,2,
3,
-
...).
Consequently,
*
(X*
c =f/ and c = y'Q Applying d'Alembert's
where for
< x < + oo
Example
2.
3.4-6.?..
.
1
\
.-3* (3fc+
1)
+ '" )*
(4)
.
l
oo
X^
X1
+ 3T4 + 3.4.6.7 + '" + test,
it
is
readily
seen
that series (4) converges
.
Find the solution
of the
equation
y'
Solution.
We
We
put
= + = Differentiating equation y' = x + y, we succes/=! + ^=1 + = 2, y'"=y\ /o" = 2, etc. Consequently,
have y =\, i^
sively find
l.
l
1
(/',
For the example
at
hand, this solution
*-l The procedure
may
x) or
be written in final form as
= 2e*
1
x.
similar for differential equations of higher orders. Testing the resulting series for convergence is, generally speaking, complicated and is not obligatory when solving the problems of this section. is
With the help of power series, find the solutions of the equations for the indicated initial conditions. In Examples 3097, 3098, 3101, test the solutions 3099, obtained for convergence. 2 3093. y' y x y 3094. y' x 1; y 2y
= + = 2 for * = 0. = + = y for x=l. = y for x==a 0' = / + *; = for * = 0. y' = x* y*\ # for (1 x)y' = l+x y\ y = \
3095. 3096.
3097.
f/
Problems on Fourier's Method
Sec. 17]
363
= for x = 0. = for * = 0. 0\ 0=1, #' + = 0; y=l, */'=0 for = + = 0; 0=1, 0'=0 for x = = 0; * = a; ~ = for = 0.
= 0;
3098*. 3099. 3100*.
/'
3101*.
y
= 0,
1
y'
Jt
*/
*/
3102.
*
Sec. 17. Problems on Fourier's Method
To
find the solutions of a linear
partial differential equation
homogeneous
by Fourier's method, first seek the particular solutions of this special-type equation, each of which represents the product of functions that are dependent on one argument only. In the simplest case, there is an infinite set of such solutions M (tt=l, 2,...), which are linearly independent among themselves in any finite number and which satisfy the given boundary conditions. The desired solution u is represented in the form of a series arranged according rt
to these particular solutions:
u=
Cn u n
(1)
.
C n which remain undetermined
The coefficients conditions.
=
Problem. A transversal displacement u u(x with abscissa x satisfies, at time *, the equation
from the
are found t)
t
initial
of the points of a string
_ ~~ a where a string).
=
2
-?
(T Q
the
is
Find the form
tensile
of the
and Q
force
string
(2)
dx*
dt*
at
time
t
the
is if
its
linear
ends
x=
density
of the
and * = / are
U
2 Fig.
fixed
u
and
=~*
at the initial
(/
Solution.
x) (Fig. It
that satisfies the
is
f
instant,
107) and
required
its
= 0,
107 the string had
the form of a parabola
points had zero velocity.
to find the
solution
u
= u(x,
t)
of
equation
(2)
boundary conditions a(0,
0-0,
!!(/,
0=0
(3)
_ _
364
\Ch. 9
Differential Equations
and the
initial conditions
=
(*,0)
We
^ *(/-*),
;<*, 0)
= 0.
(4)
seek the nonzero solutions of equation (2) of the special form
=X
u Putting
this expression into
(x)T(t).
equation
and separating the variables, we get
(2)
ZL(0--*1W a 2 7tO~~
*
(5)' l
'
(x)
Since the variables x and t are independent, equation (5) is possible only the general quantity of relation (5) is constant Denoting this constant X 2 we find two ordinary differential equations:
when by
,
=Q
)*-T(t)
and X"
(x)
+ K*X
(x)
= 0.
Solving these equations, we get
= A cos a\t + B sin aKt, (*) = C cos Kx + D sin X*.
T
(t)
X where A, B, C,
From
=
D
are
X(0) =
D
cannot be equal to
this reason,
XA
= -p
we do To
Let
constants.
arbitrary
we have
(3)
(since
sinX/
For
condition
where k
is
and
zero
at
an integer.
determine the constants. and X(/) = 0; hence, C = the same time as C is zero). us
It
will readily be seen that
not lose generality by taking for k only positive values every value h k there corresponds a particular solution
kan
kan
Ak (, cos-j-
I
whose sum obviously
.
.
equation
A k and B k
choose the constants satisfy the initial conditions
knx
.
kant\
knx
-7- + ^sm-y-J sm-j-
cos
satisfies
We
and the so that the
(2)
.
boundary conditions
sum
of the series
Since
(4).
CD
du
y
& =2* it
kan (
~\~ A
follows that, by putting
/
.
*
= 0,
sm
kant
we obtain
V
kant\
-p + ^cos-y-J
and 0)
2, 3,...).
(3).
kant
^
\
9
tj sin-y-
Ihat satisfies the boundary conditions We construct the series
f
.
(k=\
kan
,
sin
knx
-^
.
(3).
should
Sec.
_
17]
Problems on Fourier's Method
Hence, to determine the a Fourier series, .
du
..
function
A k and B^
coefficients
is
it
365
necessary to expand in
= ^-
sines only, the function u(x, 0)
in
_ x
(lx)
and
the
^ ^n
(x, 0)'
0.
at
From
familiar formulas (Ch. VIII, Sec. 4,3)
we have
/ ft
if
&
is
Ak
odd, and
Q
if
&
is
knx
.
32h
.
even; /
kast _
The sought-for solution
= *
will
2
jix
fA Osin ,
.
d*
T- J
A = 0,
be cos /
.
sin
3103*. its
ends,
A
u
sin
At the initial instant x= and * = /, had
y
and the points
,
form of the string 3104*. At the string
0
at
A = 0.
fl*
time
initial
the
/
a
form
string, attached at of the sine curve
had zero velocity. Find the
it
t.
time
= 0,
/
of
V
(2/i-f \)jtx
= 0,
receive a velocity -~
the
=
1.
points of
a straight
Find the form
of the
=
=
l are and x string at time t if the ends of the string # fixed (see Problem 3103). 3105*. A string of length /=100 cm and attached at its ends, and *=-/, is pulled out to a distance A 2 cm at point x
=
=
*=;50 cm
then released without any impulse. Determine the shape of the string at any time /. 3106*. In longitudinal vibrations of a thin homogeneous and rectilinear rod, whose axis coincides with the jr-axis, the of a cross-section of the rod with t) displacement u = u(x abscissa x satisfies, at time /, the equation at
the
initial
time,
and
is
9
c^u_ ~~ a 2
d/ 2 f where a
=
*
r
(E
is
Young's modulus and Q
is
the density of the
rod). Determine the longitudinal vibrations of an elastic horiand pulled zontal rod of length /= 100 cm fixed at the end * back at the end *=100 by A/ l cm, and then released without
=
=
impulse.
366
_ _ '
Differential Equations
[C/i.
9
3107*. For a rectilinear homogeneous rod whose axis coincides with the Jt-axis, the temperature u = u(x, t) in a cross-section with abscissa x at time /, in the absence of sources of heat, satisfies the equation of heat conduction di
where a is a constant. for any time t in a rod
Determine the temperature distribution of length
100
cm
if
temperature distribution u(x, 0)
= 0.01
A;
(100
x).
we know
the initial
X
Chapter
APPROXIMATE CALCULATIONS
Sec.
1.
Operations on Approximate Numbers
1. Absolute
The absolute error of an approximate number a which number A is the absolute value of the difference between them. The number A, which satisfies the inequality error.
replaces the exact
is called the limiting the limits a
absolute
A^/l^a + A
error.
or,
more
The exact number A A briefly, A=a
is
located within
2. Relative error. By the relative error of an approximate number a 0) we understand the ratio of the absolute replacing an exact number A (A error of the number a to the exact number A. The number 6, which satisfies
>
the inequality
\*-l ^* -,
(2)
A
is
called the
of the
limiting relative error
actual practice
A^a,
we
number
often take the
number
approximate 6=
for
a.
Since in
the
limiting
relative error.
3. Number of correct decimals. We say that a positive approximate number a written in the form of a decimal expansion has n correct decimal places in a narrow sense if the absolute error of this number does not exceed half unit of the nth decimal place. In this case, take, for the limiting relative error, the number
one
/
1
where k
is
the
first
significant /
is
known
that
6^
I ,
.
i ,
.
places
definitely
has
narrow
77:
we can
number
a.
And
conversely,
it
if
then the number a has n correct decimal
,
)
\
1)
n correct
the absolute error of the last decimal
If
unit
the
.
\10/ meaning
2(k-\in
>
1
of the
digit \-i
when n
of
decimals
the in
word.
the
In
particular,
narrow meaning
if
number a
the
^
1
/
1
\
w
"if I To) approximate number a does not exceed a for example, are numbers resulting place (such,
of
an
from measurements made to a definite accuracy), then it is said that all decimal places of this approximate number are correct in a broad sense. If there is a larger number of significant digits in the approximate number, the latter (if it is the final result of calculations) is ordinarily rounded off so that all the remaining digits are correct in the narrow or broad sense.
[Ch. 10
Approximate Calculations
368 Henceforth, we shall not otherwise (if
assume that
all digits in the initial data are the narrow sense. The results of intermediate calculations may contain one or two reserve digits. We note that the examples of this section are, as a rule, the results of final calculations, and for this reason the answers to them are given as approximate numbers with only correct decimals. 4. Addition and subtraction of approximate numbers. The limiting absolute error ot an algebraic sum of several numbers is equal to the sum of the limiting absolute errors of these numbers. Therefore, in order to have, in the sum of a small number of approximate numbers (all decimal places of which are correct), only correct digits (at least in the broad sense), all summands should be put into the form of that summand which has the smallest number of decimal places, and in each summand a reserve digit should be retained. Then add the resulting numbers as exact numbers, and round off the sum by one decimal place If we have to add approximate numbers that have not been rounded off, they should be rounded off and one or two reserve digits should be retained. Then be guided by the foregoing rule of addition while retaining the appropriate extra digits in the sum up to the end of the calculations. Example 1. 215.21 -f- 14.182 -f 21 .4-215.2(1) 14.1(8)4-21 4-= 250.8. The relative error of a sum of positive terms lies between the least and greatest relative errors of these terms. The relative error of a difference is not amenable to simple counting. Particularly unfavourable in this sense is the difference of two close numbers. Example 2. In subtracting the approximate numbers 6 135 and 6.131 to four correct decimal places, we get the difference 004. The limiting relative
correct
in
stated)
+
0.001 error
6=
is
.J.
+1
0/XM
=-4 = 0.25. j
..
,t
.
0.004
Hence,
not one of the
decimals
it difference is correct. Therefore, is always advisable to avoid subtracting close approximate numbers and to transform the given expression, if need be, so that this undesirable operation is omitted. 5. Multiplication and division of approximate numbers. The limiting relative error of a product and a quotient of approximate numbers is equal lo the sum of the limiting relative errors of these numbers Proceeding from Ihis and applying the rule for the number of correct decimals (3), we retain in the answer only a definite number of decimals Example 3. The product of the approximate numbers 25.3-4.12=104.236. Assuming that all decimals of the factors are correct, we find that the limiting relative error of the product is
of
the
6
Whence result,
if
number
the it
is
correctly, 25
final,
mth power
of
^0.01 +^0.01=^0.003. decimals of the product is three and the be written as follows: 25.3-4 12=104, or more
correct
should
3-4.12= 104 2
6. Powers and of the
of
=
0.3.
roots of approximate numbers.
an approximate number a
relative error of this number The limiting relative error of Is
the
th
the
mth root
is
of
The limiting relative error equal to the m-fold limiting
an approximate number a
part of the limiting relative error of the
number
a.
7. Calculating the error of the result of various operations on approximate numbers. If Aa lf ... Aa,, are the limiting absolute errors of the appro,
Sec.
Operations on Approximate Numbers
1]
xitnate
numbers a t
,
.
.
.
an
,
then the limiting absolute error
,
S = /(a t,
may
369
AS
of the
resuU
..., a n )
be evaluated approximately from the formula
The limiting relative
error
S
is
then equal to
,**-
*Ji/r
".++ =
Example 4. Evaluate S ln (10.3+ V^4. 4 ); the approximate numbers 10.3 and 4.4 are correct to one decimal place. Solution. Let us first compute the limiting absolute error AS in the 1 r A& \ ln (a+ V ~b), AS Aa We have general form: S
=
=
Aa=A6=i=2Q; 1^4.4 the approximate is
then equal to
=
2.0976...;
number y
is
'
^SOTT^JQ
^TT;
we
leave
equal to
we can
^
be sure
2.1,
=-
f
since the relative
-r=
^
;
error of
the absolute error
tne first decimal place. Hence,
U UUi> 0005 -
20
'
"2"
Thus, two decimal places will be correct. Now let us do the calculations with one reserve decimal: 2.517. log (10. 3+ |/4 4) =5= log 12 4-1.093, In (10 3+ V 4.4)=^ 1.093-2.303 And we pet the answer: 2 52 8. Establishing admissible errors of approximate numbers for a given error in the result of operations on them. Applying the formulas of 7 for the quantities AS or 6S given us and considering all particular differentials
=
U-M
quantities \-~Ac** or the M equal, we calculate the \dak \da k \f\ absolute errors Aa lt ... Aa^, ... o the approximate numbers a t
admissible
\
,
,
.
..
,a n
,
...
that enter into the operations (the principle of equal effects). It should be pointed out that sometimes when calculating the admissible errors of the arguments of a function it is not advantageous to use the principle of equal effects, since the latter may make demands that are practically unfulfilable In these cases it is advisable to make a reasonable redistribution of errors (if this is possible) so that the overall total error does not exceed a specified quantity. Thus, strictly speaking, the problem thus
posed
is
indeterminate.
Example
5.
The volume
off a circular cylinder
(equal
V
to
=~2 R*
2R) tan a.
at
an
by
a
angle
To what
of a
segment", that is, *a solid cut passing through the diameter of the base to the base, is computed from the formula "cylindrical
plane
a
degree
of
accuracy
should
we measure
the radius
_ _
[Ch. 10
Approximate Calculations
370
#=s:60 cm and the angle of inclination a so that the volume of the cylindrical segment is found to an accuracy up to 1%? Solution. If AV, A/? and Aa are the limiting absolute errors of the quantities V, R and a, then the limiting relative error of the volume V that
we
are calculating
is
*
3A/?
2Act
R "Nn^a
...
We
assume
3A
-^ < m 1
.
and
_<2Aa
1
.
1
100"
Whence
cm -=1 60
sin
f
mm;
2a
Thus, we ensure the desired accuracy in the answer to and the angle of inclination a to 9'. Ihe radius to 1
mm
1%
if
we measure
3108. Measurements yielded the following approximate numbers that are correct in the broad meaning to the number of decimal places indicated:
1207'14"; b) 38.5 cm; c) 62.215 kg. their absolute and relative errors. 3109. Compute the absolute and relative errors of the following approximate numbers which are correct in the narrow sense to the decimal places indicated: a)
Compute
a) 241.7; b) 0.035; c) 3.14. 3110. Determine the number of correct (in the narrow sense) decimals and write the approximate numbers: a) 48.361 for an accuracy of 1%; b) 14.9360 for. an accuracy of l%\ c) 592.8 for an accuracy of 2%. 3111. Add the approximate numbers, which are correct to the indicated decimals: 3.10 0.5; a) 25.386 + 0.49 2 41.72 0.09; b) 1.2-10 3.124. c) 38.1+2.0 3112. Subtract the approximate numbers, which are correct to the indicated decimals:
+
+
+
+
+
a)
148.1-63.871;
b)
29.7211.25;
c)
34.22-34.21.
3113*. Find the difference of the areas of two squares whose measured sides are 15.28 cm and 15.22 cm (accurate to 0.05 mm). 3114. Find the product of the approximate numbers, which are correct to the indicated decimals: a) 3.49-8.6; b) 25.1-1.743; c) 0.02-16.5. Indicate the possible limits of the results.
_
Operations on Approximate Numbers
Sec. /]
_
371
m
3115. The sides of a rectangle are 4.02 and 4.96 (accurate cm). Compute the area of the rectangle. 3116. Find the quotient of the approximate numbers, which are correct to the indicated decimals:
to
1
a)
5.684
:
5.032; b) 0.144
1.2; c)
:
216:4.
3117. The legs of a right triangle are 12.10 cm and 25.21 cm (accurate to 0.01 cm). Compute the tangent of the angle opposite the first leg. 3118. Compute the indicated powers of the approximate numbers (the bases are correct to the indicated decimals): a) 0.4158';
b) 65.2'; c)
3119. The
side
a
of
1.5
2 .
square
is
45.3
cm
(accurate to
1
mm).
Find the area. 3120.
Compute the values
of
the
roots
(the
radicands are
correct to the indicated decimals): a)
1/27715; b) j/65^2
;
c)
KsTT.
3121. The radii of the bases and the generatrix of a truncated 0.01 cm; r= 17.31 cm 23.64 cm 0.01 cm; / cone are # 3.14. Use these data to compute the 0.01 cm; ji 10.21 cm total surface of the truncated cone. Evaluate the absolute and
=
=
=
=
relative errors of the result.
3122. The hypotenuse of a right triangle is 15.4 cm 0.1 cm; 0.1 cm. To what degree of accuracy one of the legs is 6.8 cm can we determine the second leg and the adjacent acute angle? Find their values. 3123. Calculate the specific weight of aluminium if an aluminium cylinder of diameter 2 cm and altitude 11 cm weighs 93.4 gm. The relative error in measuring the lengths is 0.01, while the relative error in weighing is 0.001.
3124. to 221
current if the electromotive force is equal and the resistance is 809 ohms 1 ohm. of oscillation of a pendulum of length / is
Compute the
volts
1
volt
3125. The period equal to
where g is the acceleration of gravity. To what degree of accuracy do we have to measure the length of the pendulum, whose period is close to 2 sec, in order to obtain its oscillation period with a relative error of 0.5%? How accurate must the numbers n and g
be taken? 3126. It is required to measure, to within 1%, the lateral surface of a truncated cone whose base radii are 2 m and 1 m, and the generatrix is 5 m (approximately). To what degree of
372
_ _ Approximate Calculations
[Ch. 1Q
accuracy do we have to measure the radii and the generatrix and how many decimal places do we have to take the number n? 3127. To determine Young's modulus for the bending of a rod of rectangular cross-section we use the formula
to
*P ~~ 1 4
'
flL '
d*bs
where / is the rod length, b and d are the basis and altitude ol the cross-section of the rod, s is the sag, and P the load. To of accuracy do we have to measure the length / and the sag s so that the error E should not exceed 5.5%, provided that the load P is known to 0.1%, and the quantities d and 6 cm? are known to an accuracy of 1%, cm,
what degree
/50
s2.5
Sec. 2. Interpolation of Functions
1. Newton's interpolation formula. Let * x lt ... xn be the tabular values of an argument, the difference of which h 0,l Ax; (Ax/ *,-+, x/; i ., y n are the correspond..., n 1) is constant (table interval) and # y lt ing values of the function y Then the value of the function y for an intermediate value of the argument x is approximately given by Newton's interpolation formula ,
.
=
=
t
,
where
o=
and A#
"7
= #,
#
,
A2
t/
= A#,
=
A#
,
...
are successive finite
the polynomial diilerences of the function y. \\hen x 0, 1, ..., n), x/ U 0, 1, ., n). As (1) takes on, accordingly, the tabular values y { (/ particular cases of Newton's formula we obtain: for n=l, linear inter potation; the of To use Newton's for /i 2, quadratic interpolation. simplify formula, it is advisab'ie first to set up a table of finite differences. If y f (x) is a polynomial of degree n, then
=
.
=
& n y.
= const
and A" +1 f//=^0
and, hence, formula (1) is exact In the general case, if / (x) has a continuous derivative f ln + l} (x\ on the x lt ..., x n and x, then the error interval [a, b], which includes the points * ,
of
formula
(1)
is
| is some intermediate value between *;(/ practical use, the following approximate formula
where
= is
I, ..., n) and more convenient:
(),
x.
For
Interpolation of Functions
Sec, 2]
373
If the number n may be any number, then it is best to choose it so that within the limits of the given accuracy; in other the difference A" + {/ w words, the differences A # should be constant to within the given places of 1
^0
decimals
Example sin
1.
Find
sin
2615'
the
using
tabular
data
sin
26
= 0,43837,
27 -0.45399, sin 28 -0.46947. Solution. We set up the table
Here,
/i
= 60',
26
15'
q--
Applying formula have
sin
that
2615' = 0.43837
26
_!_ '
60'
4
(1)
and using the
+
+
0.01562
Let us evaluate the error R 2 if */ sun, then |*/ (/l '|*^l,
first
horizontal line of the table,
(0.00014) = 0.44229.
2
Using formula
we
will
we
(2)
and taking into account
have:
Thus, all the decimals of sin 2615' are correct. Using Newton's formula, it is alsc possible, from a given intermediate value of the function //, to find the corresponding value of the argument x (inverse interpolation). To do this, first determine the corresponding value q by the method of successive approximation, putting
and ?C'H
n\
2!
(i-O,
1,
2,
...).
Here, for q we take the common value (to the given accuracy!) m + >. Whence x jt cessive approximations q (m} <;< -{-
=
l
approximate the root
of
the equation
smhx
of
two sue*
374
Approximate Calculations Solution. Taking y
=
a
10
we have
4.457,
,_5-4.457 ~ 1.009
We
(Ch
0.543 5
^i.oog
0.565-0.435
0.220
2
1.009
=0.538;
= 0.538 + 0.027 = 0.565; :0, 538 + 0. 027 = 0. 565.
can thus take *
= 2. 2 + 0. 565- 0.2 = 2. 2 + 0. 113 = 2. 313.
2. Lagrange's interpolation formula. In the general case, a polynomial of 0, 1, .... n), is given degree n, which for *=*/ takes on given values yf (/ by the Lagrange interpolation formula
=
*!> (x
(x
__
x2)
.
.
.
(x
*o) (*
xn )
*,).
XQ) (x
(x
.
* 2)
.
.
.
(x
xn )
Xk ^) (X
.(X
'
'
'
'
"
^
'
3128. Given a table of the values of x and y:
Set up a table of the finite differences of the function y. 3129. Set lip a table of differences of the function y x* f 1 for the values *=1, 3, 5, 7, 9, 11. Make sure that JC 5jc all the finite differences of order 3 are equal. 3130*. Utilizing the constancy of fourth-order differences, set 10*' +2** 3jt up a table of differences of the function y x* for integral values of x lying in the range l^jt^lO. 3131. Given the table
=
+
=
log
+
1=0.000,
log 2 -0.301, 0.477, log 3
= = log 4 0.602, log 5 = 0.699. Use
linear interpolation to and log 4. 6.
log 3.1,
compute the numbers: log
1.7,
Iog2.5,
Interpolation of Functions
Sec. 2]
375
3132. Given the table sin 10
sin 11 sin 12
= 0.1736, = 0.1908, = 0.2079,
sin 13
sin 14 sin 15
= 0.2250, = 0.2419, = 0.2588. =
in the table by computing (with Newton's formula, for 2) the values of the sine every half degree. 3133. Form Newton's interpolation polynomial for a function represented by the table Fill
3134*. Form Newton's interpolation polynomial for a function represented by the table
Find y
for
3135.
A
x
= 5. 5.
function
Form Lagrange's y
for
JC
= 0.
For what x will y = is given by the table
interpolation polynomial and find the
value of
3136. Experiment has yielded the contraction of a spring (x as a function of the load (P kg) carried by the spring:
mm)
Find the load that yields a contraction of the spring by
mm.
14
376
(Ch. 10
Approximate Calculations
3137. Given a table of the quantities x and y
=
0.5 and for x 2: a) the values of y for x of linear interpolation; b) by Lagrange's formula.
Compute
by means
Sec. 3. Computing the Real Roots of Equations
1. Establishing initial approximations of roots. roots of a given equation
The approximation
/(*)=0
of the
(i)
consists of two stages: 1) separating the roots, that is, establishing the intervals (as small as possible) within which lies one and only one root of equation (1); 2) computing the roots to a given degree of accuracy on an interval [a, b] and If a function /(A) is defined and continuous of equation (1). /(a)./(6)<0, then on [a, b] there is at least one root or /' (x) This root will definitely be the only one if /' (x) when
<
>
a
it is advisable to use millimetre In approximating the root paper and construct a graph of the function y = f(x). The abscissas of the points of intersection of the graph with the x-axis are the rools of the equation /(x) 0. It is sometimes convenient to replace the given equation with an equivalent as the absequation (p (#) if (A). Then the roots of the equation are found cissas of points of intersection of the graohs y q)(x) and y ty (x). 2. The rule of proportionate parts (chord method). If on an interval [a, b] of the equation f (A) there is a unique root O, where the function f (x) is continuous on b], then by replacing the curve y f(x) by a chord [a, passing through the points [a, f (a)] and [b, f (b)], we obtain the first approximation of the root
=
=
=
4
t
/~\ (2)
To obtain
we apply formula (2) to that one of ends of which the function f (x) has values of opposite sign. The succeeding approximations are constructed in the same manner. The sequence of numbers cn (n=l, 2, converges to the .) a
the intervals
second approximation c 2 [a,
c,]
or
[c,,
b]
,
at the
.
root
,
that
.
is,
Generally speaking, we should continue to calculate the approximations c,, until the decimals retained in the answer cease to change (in accord ., with the specified degree of accuracy!); for intermediate calculations, take one or two reserve decimals This is a general remark. If the function / (x) has a nonzero continuous derivative /' (x) on the interval [a, b], then to evaluate the absolute error of the approximate root
c2 ,
.
.
_
_
crt ,
we can make
where
377
Computing the Real Roots of Equations
Sec. 3]
u=
use of the formula
min
a
(/' (x)
|.
b
and f" (x) ^ for 3. Newton's method (method of tangents). If /' (x) a^x^b, where f(a)f(b)<0, f(a)f'(a)>Q, then the successive approximaof an equation f(x)=Q are computed tions * (rt = 0, 1, 2, ...) to the root rt
from the formulas (3)
Under the given assumptions,
=
lim * n -> oo
To evaluate
where u,= min a
we can
the errors
<
1,
2,
...)
is
mono-
.
use the formula
\f'(x)\.
xQ
a=
=
x
For practical purposes
where
sequence x n (n
the
tonic and
.,
,
= a,
is
it
more convenient
= *_,
*
which yield
the
a/^.,)
i/i
to use the simpler
=
same accuracy
l,
2,
formulas
...),
(3')
as formulas (3).
=
&. f(b)f"(b)>Q, then in formulas (3) and (3') we should put x 4. Iterative method. Let the yiven equation be reduced to the form If
*
=
(4)
where \y' (x)\*^r< 1 (r is constant) for a ^x^b. Proceeding from the iniwinch belongs 4o the interval [a, b], we build a sequence of tial value * numbers x lt x 2 ... according to the following law: ,
,
If
a
rt
<6
(n
=
l,
2,
...),
then the limit
is the on/r/ roo/ of equation (4) on the interval [a, 6]; that is, xn are successive approximations to the root |. The evaluation of the absolute error of the nth approximation to x n is given by the formula
it
IS
_ xn y
\
I
^^s
*
xn + l j
Therefore,
if
xn and x n + l coincide
error for x n will be
-r
to
_r
xn
within
E,
I
then
the
absolute
limiting
.
In order to transform equation f(x)
Q to
(4),
we
replace the latter with
an equivalent equation
where the number
A,
^
is
chosen so that the function
-7-
[x
Kf
(x)]
=
f 1
Kf
(jc)
.378
_
__
Approximate Calculations
should be small in absolute value in the neighbourhood of the point example, we can put 1 X/'(x )=0]. Example 1. Reduce the equation 2x Inx 4=0 to the form (4) initial approximation to the root * = 2.5. Solution. Here, /(x)
x
lent
equation values of A,;
|l_ A,( 2\
|
The
=x
this
^
X ] initial
= 2x
In *
\nx
k(2x
number
U = 2.5 =0, equation
/'(jc)
and
4)
close
is
that
I
4;
is,
10
[Cft.
= 2-~. We
*
(for
for
the
write the equiva-
take 0.5 as one of the suitable of the to root the equation
r^O.6. .6
close to
1
reduced to the form
is
x
=x
Inx
0.5 (2x
4)
or
Example equation that of (5)
2.
Compute, to two decimal between 2 and 3.
places, the root \ of the preceeding
lies
Computing the root by th Example 1, putting x ? = 2.5.
iterative method. We make use of the result We carry out the calculations using formulas
with one reserve decimal.
*f
=2+
*,
= 2 + 4-
And so 1^:2 45 (we can Become fixed) Let us npw evaluate the
Considering that
all
^2. 448,
In 2.450
2
*4 = 2
(p(*)
y In2.458ss2.450,
+ ~ln2. 448=^2. 448. since
stop
here
error.
Here,
= 2 + ~lnx
approximations
to
and xn
the
third
decimal
place
has
lie
in the
interval
[2.4, 2.5],
we
get
Hence, the limiting absolute error remark made above,
in the
approximation to x 9
is,
by virtue
of the
A==
^-^-=0.0012 ^=0.001.
Thus, the exact root g of the equation 2 447
we can
lies
within the limits
< g < 2.449;
take g^2.45, and all the decimals be correct in the narrow sense.
of
this
approximate number will
_
Sec. 3]
Computing the Real Roots
of
Calculating the root by Newton's method.
Wee
We
Here,
>
On the interval 2 we have: /(3)f (3)>0. Hence, the conditions
Equations
/' (x) of 3 for
> 0; /(2)f(3)<0; = 3 fare(x)fulfilled.
and *
take
carry out the calculations using formulas
jtj^S
=2 4)
In 3
0.6(2-3 0.6(2-2 4592 0.6(2-2.4481 0.6(2-2 4477
= 2. 4592 = 2.4481 *, * 4 = 2. 4477 *,
(3')
with two reserve decimals:
4592;
In 2
4592
In 2. 4481
In2. 4477
= 2 4481; = 2. 4477; 4) = 2 4475. 4)
4)
At this stage we stop the calculations, since the third decimal place 2. 45. We omit the does not change any more. The answer is: the root, evaluation o! the error; 5. The case of a system of two equations. Let it be required to calculate the real roots of a system of two equations in two unknowns (to a given degree of accuracy):
=
f(*.
0=0,
,
6
.
there be an initial approximation to one of the solutions (|, r\) of system Jt x ot y = y Q This initial approximation may be obtained, for example, graphically, by plotting (in the same Cartesian coordinate system) the curves f(x, #) and by determining the coordinates of the points of interand tp (x, #)
and
let
=
this
.
=
section of these curves. a)
Newton's method. Let us suppose that the functional determinant dtf.jp)
d(x,y)
xx
=
Then by Newdoes not vanish near the initial approximation y Q1 y ton's method the first approximate solution to the system (6) has the form where a P O are the solution of the system of two. a x f/ x,
= +
,
j/,
= +P
.
,
linear equations
The second approximation
is
obtained in the very same way:
= *!+<*!, *, where a p $ v are the solution
Similarly
we obtain
of the
^1
= ^1+ Pi,
system of linear equations
the third and succeeding approximations.
380
_ _ Approximate Calculations
We
b) Iterative method.
the system of equations
(6),
[Ch
10
can also apply the iterative method to solving by transforming this system to an equivalent one * F(x.y).
=
= \ y (D(*, (
and assuming thai
|F>.y)| + |0;
(f)
y)
\F'y (x, y)\
t
+
\ v (x.
#|
1
(8)
in some two-dimensional neighbourhood U of the initial approximation (* y Q ) which neighbourhood also contains the exact solution (, rj) of the system. The sequence of approximations (x n y n ) ( = 1, 2, ...), which converges to the solution ol the system (7) or, what is the same thing, to the solution ,
t
,
x>f
(6),
is
constructed according to the following law:
= F(x If
all
(*, y n ) belong
to
U
yz )
t,
then
%
*/,-c
t
=
*
lim n
lim
,
n
-+ oo
-
y n =i\.
oo
The following technique is advised for transforming the system of equa* tions (6) to (7) with condition (8) observed. We consider the system of equations a/
(
which
is
equivalent to
(x.
provided
(6)
a.
that
Rewrite
0.
it
in the
form
Y.
y)^F(x
,
Choose the parameters functions F(v, y)
-jnd
a,
p,
O (x,
y, & such y} will be
that
the
t
y),
derivatives of the zero in the initial approximate solutions
partial
equal or close to
in other words, we find approximation; system of equations
a,
Y.
fi,
o as
of the
,v
,
t/
)
= 0,
Condition (8) will be observed in such a choice of parameters a, P, Y on the assumption that the partial derivatives of the functions / (x, y) and
given the initial approximation to the root *
= 0.8,
= 0.55.
Sec. 3]
_
Solution.
fv (x*
the Real Roots of Equations
Computing
y
)
Here,
= 1.1;
y)-=x*
f (x t
(*<>.
= 1-92, 0o)
+ y*\
t
t
= 1.6,
y ) equivalent to the initial one) ,
is
a(* + // -l) + p(*'-i/)-0, 2 2 \ Yl* + -0-i-a(*'-
J in the
i/ c )
1.
Write down the system (that 2
381
y)=x'-y- fx (x
q> (x,
=
__
/la, pi
\
VI Y. *l
/
form
For suitable numerical values of system of equations 1
Y anc* ^ choose the solution
a, p,
of
the
+ 1.60+1.920 = 0, l.la
p-0,
+ 1.925 = 0, + 1. lY 6 = 0; 1.6 Y
1
i.
y^
we put a =^0.3, p ^r 0.3, Then the system of equations
e.,
i
\
which
=x y=y x
0.3(^
2 2
+
0.5,
b^QA.
1)
0.3(A-
2 r/
0.5 (* -{-#
2
1)
y),
+ 04 (x*y)
t
equivalent to the initial system, has the form (7); and in a sufficiently small neighbourhood of the point (x Qt y Q ) condition (8) will be fulfilled. is
Isolate the real roots of the equations by trial and error, of the rule of proportional parts compute them to
by means
and two
decimal places. s x 1-0. 3138. X 4 * 05* 1.55 = 0. 3139. + 8 4* -1--0. 3140. x Proceeding from the graphically found initial approximations, use Newton's method to compute the real roots of the equations to two decimal places:
-
3141.
JC
3142. 2x
-\
_2jc In*
5-0. 4
= 0.
3143. 2 3144.
x
= 4x.
logjc=y.
Utilizing the graphically found initial approximations, use the iterative method to compute the real roots of the equations to
two decimal places: 8 3147. jc x 3145. x'-5.*M 0.1=0. 2 = 0. = cos*. 3146. 4* Find graphically the initial approximations and compute the real roots of the equations and systems to two decimals: 3151. x- In* 14 = 0. 3* -(-1=0. 3148. A:' 8 2 8 = x 3152. 2* + 3* 5 0. 3149. * +3* 0.5 = 0. 2 4 3153. 4* 7sin* = 0. 3150. * +* 2* 2 = 0.
382
[Ch. 10
Approximate Calculations
3154. x* 3155. e* '
+ 2x + e- *
= 0. = 0.
6
8
"
~~
" Lff
x*
(
157
4
+y
'
\
4
= 0,
\Qgx-l =0.
y
~~
3156. 3158.
Compute
the equation tan
3159.
jc
to three decimals the smallest positive root of
=
jc.
Compute the
roots of the equation x-tanh
x=l
to four
the
integral
decimal places. Sec. 4. Numerical Integration of Functions
1. Trapezoidal formula. For the approximate evaluation
of
b
If (x) [a,
is
a function
continuous on
into n equal parts
b]
= x + ih Xi
a, let
xn
with an absolute error
of
we divide the
[a, b]]
and choose the interval
= b,
of
-
interval of integration calculations
h==
.
be the abscissas of the partition points, and // /(*/) be the corresponding values of the integrand f(x). Then the trapezoidal formula yields y
Let
Q
(x
/
0,
2,
1,
..., n)
a
where M 2 = max f (x) when To attain* the specified accuracy |
\
is
That
h must be of the order value so that
is,
e
when evaluating
the integral,
the
in-
found from the inequality
terval h
of
V^e
.
The value
of
h obtained
is
rounded
off to the smaller
b
a== ~~ n h
be an integer; this is what gives us the number of partitions n. Having established h and n from (1), we compute the integral by taking the values of the integrand with one or two reserve decimal places. 2. Simpson's formula (parabolic formula). If n is an even number, then
should
in the notation of
1
Simpson's formula
b
h Q-
K#o + #n)
+ 4 (y\ + #3+.. +yn -\) + +0n-2)l
(3)
Numerical Integration
Sec. 4]
of Functions
383
holds with an absolute error of (4)
M 4 = max
where
when
! |
(x)
/
\
To ensure
the specified accuracy e when evaluating the interval of calculations h is determined from the inequality
integral,
the
(5)
That
the interval h
is,
to the smaller value so
is
of the order
that n
J/JF, is
The number h
is
rounded
off
an even integer.
Remark. Since, generally speaking, it is difficult to determine the interand the number n associated with it from the inequalities (2) and (5), practical work h is determined in the form of a rough estimate. Then,
val h in
after the result is obtained, the number n is doubled; that is, h is halved. one to the number of decimal If the new result coincides with the earlier places that we retain, then the calculations are stopped, otherwise the pro-
cedure
is
repeated, etc.
For an approximate calculation of the absolute error R of Simpson's quadrature formula (3), use can also be made of the Range principle, according to
which
where 2 and S are the results h and // = 2/i, respectively.
of calculations
from formula
(3)
with interval
3160. Under the action of a variable force F directed along the x-axis, a material point is made to move along the x-axis 4. Approximate the work A of a force F if a to x from x table is given of the values of its modulus F:
=
=
by the trapezoidal
Carry out
the calculations the Simpson formula.
formula and by
i
3161. Approximate
2
J (3*
4x)dx by the trapezoidal
formula
putting rt=10. Evaluate this integral exactly and find the absolute and relative errors of the result. Establish the upper limit A of absolute error in calculating for n=10, utilizing the error formula given in the text.
384
\Ch. 10
Approximate Calculations i
3162. Using the
* *
calculate
Simpson formula,
V
to four
.
J *
*
r
o
decimal places, taking n== 10. Establish the upper limit A of absolute error, using the error Formula given in the text. Calculate the following definite integrals to two decimals: 3,63.
3,68.
Ife.
(-!<,.
1
Jl
-~.
3164. C
3169.
317
-
-
>
2
3166.
[x\ogxdx.
^
8187.
Q | 7i
dr.
fcos* dX. FT \+X Jcos* ,
3172>
3173. Evaluate to two decimal a PPlyi n S the substitution
the
places
^
-7--
improper
Verify the calculations
1
ft
by applying Simpson's formula to the integral
\
+
is
chosen
,so
QO
that
7^722
<
integral
Jy
rr ~r x
J
2
where b
,
*
10
A
plane figure bounded by a half-wave of the sine curve *-axis is in rotation about the x-axis. Using the Simpson formula, calculate the volume ot the solid of rotation to two decimal places. 3175*. Using Simpson's formula, calculate to two decimal 3174,
f/=sin^ and the
places the length ot an arc of the ellipse in the first
y+
2
=
1
situated
quadrant.
Sec. 5. Numerical Integration of Ordinary Differential
1.
^
A method
of successive approximation be given a first-order differential equation
subject to the initial condition
(Picard's
y'^f(x.y) = # when x = *
/
Equations method).
Let
there
<1) .
Numerical Integration
Sec. 5]
Ordinary Differential Equations
of
_
385
The solution y (x) of (1), which satisfies the given initial condition, can, generally speaking, be represented in the form
y(x)=
lira yi (x) -* 00
(2)
t
where the successive approximations y
*//(*) are
from
determined
the
formulas
,
X
(*)=#<>+
If
the right side f(x, y)
R{ and
in this
satisfies,
is
|x
yt-i (*)) dx
defined and continuous
x
in the
neighbourhood
|
Vi)-f(x. i/JKJ-ltfi-tfil
(L is constant), then the process converges in the interval
of
I* ft
(x.
neighbourhood, the Lipschitz condition \f(x,
where
/
= min(a, ^ M/ \
]
and
successive
definitely
(2)
*
Af = max|/U,
/?
approximation
y)\.
And
the error here
is
/?
)
- y n W \
1
The method of successive approximation (PicarcTs method) is also applicable, with slight modifications, to normal systems of differential equations. Differential equations of higher orders may be written in the form of systems of
differential equations.
2. The Runge-Kutta method. Let ,
To do
it be required, on a given interval to find the solution y (x) of (1) to a specified degree of accuracy e.
we choose
this,
the interval of calculations /i=
the interval [* X] into n equal parts so that h* Xf are determined from the formula ,
X|
=x +M
(i=0,
1,
2,
< e.
..., n).
By the Runge-Kutta method, the corresponding values function are successively computed from the formulas
1
3
1900
by dividing
The partition points
/;
=
(/
(x/) of
the desired
[Ch. 10
Approximate Calculations
386
where f
=0,
1,
2,
...,
n and
(3)
*i"
=/(*/ + A,,
correct choice of the interval h
To check the
it
is
advisable
to
verify
the quantity
e= The fraction 6 should amount
a
to
few
hundredths, otherwise h has to be
reduced.
The Runge-Kutta method is accurate to the order of h 1 A rough estimate Runge-Kutta method on the given interval [x X] may be obtained by proceeding from the Runge principle: .
of the error of the
,
Rn
y*m
I
Urn
I
'
15
where /i = 2m, y 2m and y m are the results of calculations using the scheme (3) with interval h and interval 2/i. The Runge-Kutta method is also applicable for solving systems of differential equations y'
with given
= f(x,
y> z).
*'
=
y,
z)
(4)
= z = 2 when x x by the Milne method, =(1) X we in some way find
conditions / 3. Milne's method. To solve initial conditions y=^y when X values initial
yi=y(*i),
t/
.
,
Q,
y=0(*i).
/
subject to the the successive
=*/(*)
of the desired function y (x) [for instance, one can expand the solution y (x) in a series (Ch. IX, Sec. 17) or find these values by the method of successive approximation^or by using the Runge-Kutta method, and so forth]. The ap-
proximations y t and y] for the following values of successively found from the formulas
*=
\vhere
fi
= f(x
it
=
-
r/ t
(i=4,
5,
..., n)
are
+~(7"
y^ and7i = /(*i,
Hi)-
To check we
calculate the quantity (6)
_
Numerical Integration of Ordinary Differential Equations
Sec. 5]
387
does not exceed the unit of the last decimal \(y m retained in the y (x), then for f/ we take I// and calculate the next value y/ + 1 10~ w then one has to start from the berepeating the process. But if e/ ginning and reduce the interval of calculations. The magnitude of the initial interval is determined approximately from the inequality h 4 10~ m For the case of a solution of the system (4), the Milne formulas are written separately for the functions y (x) and z (x). The order of calculations remains the same. Example 1. Given a differential equation */'=*/ x with the initial condition y(0)=1.5. Calculate to two decimal places the value of the solution of this equation when the argument is x 1.5. Carry out the calculations If
e/
answer
for
-
t
,
>
,
<
combined Runge-Kutta and Milne method.
a
by
.
We
Solution.
choose the initial interval h from the condition /r*<0.01. let us take h 0.25. Then the entire interval of x to jc=1.5 is divided into six equal parts of length
To avoid involved writing,
integration from 0.25 by means of points x f (i 0, 1, 2, 3, 4, 5, 6); we denote by y f and y^ the corresponding values of the solution y and the derivative y' We calculate the first three values of y (not counting the initial one) by the Runge-Kutta method [from formulas (3)]; the remaining three values we calculate by the Milne method [from formulas (5)] 1/4. */s y& The value of // will obviously be the answer to the problem. We carry out the calculations with two reserve decimals according to a definite scheme consisting of two sequential Tables 1 and 2. At the end of Table 2 we obtain the answer. # #, x =^0, // =1.5 Calculating the value y r Here, / (x, */)
=
.
fl
=
/i
= 0.25.
A^
=i
+
(*<
= 4-D (0.3750 + 2-0. 3906 + 2.0.3926 + 0.4106) ==0. 3920; (
0)
,
*(>
= / (*o. =
/ /
(
0o) h
+~
)=s
(
0+
1
.5000)0.25-0.3730;
(0)
\ + -y- j = (- 0.125+1.5000 + 0.1875) 0.25= 0.3906; fc
/
Xo
/i
,
& (0)
h
/
*i
=
=n *o+y.
0o
\ + -|-) ^ = (0 125+1.5000 + 0.1953)0.25 = 0.3926; ^= 0.25+1.5000 + 0.3926)0.25 = 0.4106; (
= 1.5000 + 0.3920 =1.8920
(the
first
three
decimals
in
tins
approximate number are guaranteed). Let us check:
6
_
=
20 |0.3906Q.3926|_ .1 = 0. 13.
10.37500.39061""
By in
13*
we chose was rather rough. values y t and y 9 The results are tabulated
this criterion, the interval h that
Similarly
Table
1.
we
calculate
the
156
.
388
Approximate Calculations Table
1.
Calculating^,, f(x, y)
Calculating the value of j/
=1. 5000,
^=1.5000, Applying formulas
jr 2 ,
y9
We
have: f(x, y)
=
jf4 .
= 1.8920, = 1.6420, ^'
(5),
we
by the Runge-Kutta Method.
=
y t = 2.3243,
r/,
[Ch. 10
^=1.8243,
/z
y,
= 0.25,
,== 2.8084;
= 2.0584.
find
-^
= 1.5000 + 4
A OK
(2-
1.6420-1.8243 + 2. 2.0584)
= 3.3588;
1+3.3588 = 2.3588; =
02
+ ^(01
h y' t)
. hence, there
is
'
= 2.3243 +
^
3.3588-3.3590
no need
(2.3588
+ 4- 2.0584 +
1
.8243)
1
to reconsider the interval of calculations.
= 3.3590;
Numerical Integration of Ordinary Differential Equations
Sec. 5]
389
= =
We obtain i/ 4 y4 3.3590 (in this approximate number the first three decimals are guaranteed). Similarly we calculate the values of y^ and y^. The results are given in Table 2. Thus, we finally have y
(1.5)
= 4.74.
4. Adams' method. To solve (1) by the Adams method on the basis of the initial data */(x ) we in some way find the following three values /o g of the desired function y (x):
=
[these three values may be obtained, for instance, by expanding y (x) in a power series (Ch IX, Sec. 16), or they may be found by the method of successive approximation (1), or by applying the Runge-Kutta method (2) and so forth]. With the help of the numbers X Q x it * 2 x t and */ y lt y zt ys we calcu,
late
,
q lt q 2
,
2
We
.
,
?, where
= h y' = hf (x 2
2,
// 2 ),
qs
= hy' = hf (x,, 3
then form a diagonal table of the finite differences of
V'=f(x,
I/O
f/2
i/5
y)
q:
_ _ Approximate Calculations
390
The Adams method
Thus,
Adams formula
numbers
the
utilizing
we
= + --
A#,
it,
<7,
e
+ A*/
A? 2
A<7 2 ,
,
calculate, by
i
in
table of differen-
consists in continuing the diagonal
ces with the aid of the
Hhe difference table,
[Ch. 10\
A 2 ?,, A 8
means
of
situated
formula
o
+ IZ A <7,+ 2
-r^
A*<7
O
After
.
And when we know * 4 and #4
(7)
finding
diagonally in
and putting n-- 3
A//,,
we
calculate
we calculate q^~hf(x^ f/ 4 ), difference table and then fill into it the 2 A 3<7,, which are situated (together with 4 ) along finite differences A
#4
*/3
introduce
8-
*/ 4
A# 8 and
,
,
into the
4
,
=
,
.
,
W
m
].
In this
sense
the
Adams formula
Milne (5) and Runge-Kutta Evaluation of the error
is
(7)
equivalent
to the
formulas of
(3).
for the Adams method is complicated and for practical purposes is useless, since in the general case it yields results with considerable excess. In actual practice, we follow the course of the third finite differences, choosing the interval h so small that the adjacent differences A 8 <7f and A*
Example
2.
Using the combined Runge-Kutta and Adams method, calcu(when #1.5) the value of the solution of the x with the initial condition f/(0) 1.5 (see
late to two decimal places differential equation
y'y
Example
1).
We
use the values y lt y z //, that we obtained in the solution Their calculation is given in Table 1. We calculate the subsequent values //4 f/5 #6 by the Adams method (see Tables 3 and 4).
Solution.
of
Example
,
1.
,
,
The answer to the problem is #4 = 4.74. For solving system (4), the Adams formula (7) and the calculation scheme shown in Table 3 are applied separately for both functions y(x) and z(x).
Find three successive approximations to the solutions and systems indicated below.
differential equations
= =
3176. y'=jf+y*\ y(Q) Q. 3177. y' x y 2, z' y-z\ f/(0)=l, z(0) 3178. */" -#; y(0) 0, y'(0)=l.
= + + =
=
= -2.
of
the
Table
3.
Basic Table for Calculating /(*,
y4 y
= -* +
,
*/;
g
fc
by the
=y^0.25
,
Adams Method.
(Italicised figures are input data)
Answer: 4.74
Table
-1
153
Auxiliary Table for Calculating by the
Adams Method
_
Approximating Fourier
[Sec. 6]
Coefficients
_
393
/i = 0.2, use the Runge-Kutta method to approximately the solutions of the given differential equations and systems for the indicated intervals: 3179. 0' = x; 0(0) = 1.5 (
Putting the interval
calculate
3180. 3181.
=|
0'
0(1)
0';
y'=*z+
=
z'=0
1.
1
x,
(Kx<2). 0(0) = 1, z(0) =
(0
l
combined Runge-Kutta and Milne method or Applying Runge-Kutta and Adams method, calculate to two decimal places a
the solutions to the differential equations and systems indicated below for the indicated values of the argument; 0.5. 3182. 0'=x+0; 0=1 when x=--0. Compute y when x 0. Compute y when x=l. x* 3183. 0' 0; 0=1 when x 0. Compute when x 0.5. 3184. 0' 3; 0=1 when x 20 x 3185. J0' 20+z,
=
= =
I
= =
+
= + z' = x + 20 + 3z;
Compute y and 3186.
f0'
=
3187. 0*
30
=2
3188.
y=
z
when
0V +1=0; and
when x = 0.
2
when x = 0.
1
when x = 0.
1
0'
=
when jc=l.
=1.5.
2
A:(JI)
=
A:
0=1, A:
^ + |-cos2/ =
3189.
= = 0.5. = 0' z
2,
= 2,
0;
Compute y when
Find
z,
z;
when x=l.
Compute
z
when x = 0.5.
z
)2r'=y and Compute
= 2,
=
0;
x = Q,
=
Jc'
when
l
f=-0.
x' (n).
Sec. 6. Approximating Fourier Coefficients
Twelve-ordinate scheme. Let y n of
the function y
and y Q
-f/ 12
We
~f(x n
)
(n
points f(x) at equidistant equidist set
= 0,
1,
xn = -
..., of
up the tables: y* y\
yu
Sums
yz
y* y*
y\9 y> y*
(2j)
Differences (A) Q u l uz Iu U* U 6 M 4
Sums
s
s l s 2 s,
Diilerences
t
tv
t
Sums Differences
12)
be the values
the interval lU,2ji],
Approximate Calculations
394
The Fourier
may
\Ch
=
coefficients a n b n (n 1, 2, 3) of the function 0, y be determined approximately from the formulas: ,
6t
a2
=s
&, s,
+ 0.5(s,
&8
s 2 ),
= 0.50! + 0.866a + a = 0.866 (1, + *,), = a, a 2
10
= f(x)
8,
8,
(1)
where 0.866
.
=
We
have
+
f(x) zz
Other
'
30
10
schemes
are
also
used.
(an cos
nx +
Calculations
b n sin nx).
are
simplified
patterns.
Example. Find the Fourier polynomial for the function y represented by the table
From formulas
(1)
by the use
of
= f(x)
we have
= 9.7; a, = 24.9; = 13.9; 6 = 8.4; 6, 2
2
=i0.3; a a = 3.8; 6,
= 0.8.
Consequently, / (x)
^ 4.8 + (24.9 cos x + 13.9
sin x)
+ (10.3 cos2x
8.4 sin 2x)
+ Using the for
(3. 8
+ cos 3*
+ 0.8 sin 3x).
12-ordinate scheme, find the Fourier polynomials the following functions defined in the interval (0,2:rc) by the
Approximating Fourier Coefficients
Sec. 6]
tables of their values that correspond to the
395
values
equidistant
of the .
argument. 3190.
= 4300 = 5200 y = = 9.72 y, = 8.97 = 8.18 y, = 1.273 y = 0.788 y = 0.495 #, 4
t
= = 6.68 y, = 9.68 - 2.714 = 3.042 y, = 2.134
3191. y
t/,
3192.
J/
/2
f/ 4
t/ s
4 s
= 7400 - 2250 = y, /,
= 3850 = 7.42 y, = 6.81 y, = 6.22
yg
= 0.370 = 0.540 y, = 0.191 y,
y,
= 7600 = 4500 ^, = 250 0,
=5.60 / 10 =4.88 / n =3.67
y.
y,
=-0.357
= 0.437 n = 0.767
#, t/
3193. Using the 12-ordinate scheme, evaluate the Fourier coefficients for the following functions: a)
f(x)
=
(x>
first
several
ANSWERS
Chapter 1.
then a b) + b + \b\. Whence (a = & a (^ 6 \<\ab\ & ^ a Hence, |a |a \ab = |a/ + |6 3. a) -2
a=(a
Solution. Since
>|a|
and
\b\
Besides,
t
\
I
|
|
|.
|
(
|
<
b) x
I
|
|
|
|
1
1;
|.
|.
|
1;
7.
b)~oo
<*< + oo.
/2<*< +
be 2
+*
* but
16. 19.
oo
2<0, this
22. q>(*)
= 2^
1
i.
5* 2
10. i|>(x)
=
a)
T = JI,
T=
Periodic,
e)
y=
*-|-KO,
or
e.,
should either
x
15.
18.
1
a)
Even, b)odd,
= i[f (x) + / (x)
/
(-x)}
A
7 =ji,
c) periodic,
0
if
even, d)odd,
+ ![/ (x)-f (-x)}.
2n
r = -r-,
c)
2,...).
1,
2
3^ + 6x23.
if
i.
Arji
21.
x,
<*<
2^0,
x
K*<2. \
2
2 n, b) periodic, -^ u
nonperiodic. 27.
is,
oo
13. a)
l<;t<2. Solution. It 2)<0. Whence (A:+ 1) (*
Thus,
impossible.
0<*<1. 17. 20 K*<100.
e) odd.24. Hint. Utilize the identity
26.
that
-
-
14.
l
e.,
+00).
2), (~-2, 2), (2,
2<0;
^
is 1,
-4-< <
00,
(
*=0, |jc|^ K2. 2
or ^
<^< Jt
12.
b)
x'^0,
A'+l^O, Jt^*2,
oo;
.
d) periodic
C<
m q^ when = m +q qil +q /^A;^/,-!-^; 28.
= ^i^i + ^2(^ when + /2<^
/1
37.
/
-^-;0;
/
2
7-. 38.
a)
/.
y
=2 x=
29. (p[x|)(jc)]
=Q
when
2l 2
l
2Jf
;^[(p(x)]
y
1,
>
s
= 2^
(x
l
l
l
2)
30. x. 31. (x
when #
>
1,
when
+ 2) y
.
= when x = when jc< and * = 2, / > when 1
i/
i/
1
oo);
b)
and
l
x=V y +
\
y<0 whenO<^
oo,
and
x
=
39. a)
*
= -- (y
3)
Answers
=
*
c)
=
\
if
~
y lanyf
= arc tan 43. a) c)
*/
#=
y=u = arc l
a)
^=--;
d)
VlogT;
when
2
x
d)
= 2-10>
oo<#< +
f
(
oo);
x
e)
=
* = # when oo<0<0; x=/l/ when = 2x 5; b) # = 2a = cos.r, c) = u u= x 42. a) # = sin x; b) sinw, a = 3
,
2
2
.
,
y = 2(x*
!/""<
,
,
t/
c)
= -cosjc
oo);
40.
41.
.
oo<#< +
(
397
JC |
|
oo
1)
=
and
|x|
if
b) y = log (10 = y x when 0
< 1/2S;
10*), 46.
|*|>1.
if
-oo
Hint. See Appen-
\j
dix VI, Fig. 1. 51. Hint. Completing the square in the quadratic trinomial 2 we will have y y and yQ (4ac b 2 )j4a. a(x * ) where * b',2a Whence the desired graph is a parabola y ax* displaced along the *-axisby See .Appendix VI, 53. Hint. XQ and the t/-axis along yQ by
= +
=
=
=
.
The graph
is
See
Hint.
58.
2.
Fig.
a hyperbola
Appendix
y=
,
shifted
VI,
along
Fig.
the
3.
*-axis by
Hint.
61.
X Q and along 2
the t/-axis by (
x+
yQ
Taking the integral
Hint.
62.
.
\ (Cf. 61*). 65. Hint. See
Appendix VI, Fig.
part,
4.
we have y=-~ o
--13 9
/
/ '
67.Hint. See Appendix VI,
Appendix VI, Fig. 6. 72. Hint. See Appendix VI, Appendix VI, Fig. 8. 75. Hint. See Appendix VI, Fig. 19 78. Hint. See Appendix VI, Fig. 23. 80. Hint. See Appendix VI, Fig. 9. 81. Hint. See Appendix VI, Fig. 9. 82. Hint. See Appendix VI, Fig. 10 83. Hint. See Appendix VI, Fig. 10. 84. Hint. See Appendix VI, Fig 11. 85. Hint. See Appendix VI, Fig. 11. 87. Hint. The period of the function 5 sin 2x with amis T 2njn. 89. Hint. The desired graph is the sine curve y plitude 5 and period n displaced rightwards along the x-axis by the quantity 5.
Fig.
1
Hint.
71.
Hint.
Fig. 7. 73.
See
See
90. Hint. Putting
.
a=A cos
cp
and
b= A
sin
--- V
In our case,
cp,
we
will have
y=A
sin (x
cp)
,
A
where
=-
2
Hint. cos x
V& + b
2
and
(p
=
arctan(
4 = 10,
cp=0.927. 92.
= -jr-(l+cos2jc). 93. Hint. The desired graph the sum of the graphs = sinjc. 94. Hint. The desired graph the product of the graphs = sinx. 99. Hint. The function even For x>0 we determine and 3) y =\. When x at which >+<, # = 0; 2) y = is
i
yl yl
=x =x
the
and and
points
t/
is
2
is
i/ 2
y+\.
1)
\\
101. Hint. See Appendix VI, Fig. 14. 102. Hint. 103. Hint. See Appendix VI, Fig. 17. 104. Hint. 17. 105. Hint. See Appendix VI, Fig. 18. 107. Hint. Hint. 18. 118. Hint. See Appendix VI, Fig. 12. 119.
Fig. Fig. Fig.
15.
Fig.
12.
120.
See See See See
Appendix VI, Appendix VI, Appendix VI, Appendix VI, Hint. See Appendix VI, Fig. 13. 121. Hint. See Appendix 132. Hint. See Appendix VI, Fig. 30. 133. Hint See Appendix VI, Hint. See Appendix VI, Fig. 31. 138. Hint. See Appendix VI, Hint. See Appendix VI, Fig. 28. 140. Hint. See Appendix VI,
VI, Fig.
13.
Fig. 32. Fig. 33.
134. 139.
Fig. 25.
141. Hint.
Answers
398
Form
a table of values:
Constructing the points (x, y) obtained, we get the desired curve (see Appendix VI, Fig. 7). (Here, the parameter t cannot be laid off geometrically!) 142. See Appendix VI, Fig. 19. 143. See Appendix VI, Fig. 27. 144. See Appendix VI, Fig. 29. 145. See Appendix VI, Fig. 22 150. See Appendix VI, x 2 It is 1^25 Fig. 28. 151. Hint. Solving the equation for y, we get now easy to construct the desired curve from the points. 153. See Appendix VI, Fig. 21. 156. See Appendix VI, Fig. 27. It is sufficient to construct
y=
the points
(x, y)
to the abscissas x
corresponding
.
= 0,
a.
,
157. Hint.
-^
*]
the equation for x, we have * 10 \ogy Whence we get the y points (x, y) of the sought-for curve, assigning to the ordinate y arbitrary values (*/>0) and calculating the abscissa x from the formula * Bear in oo as y -+ 0. 159. Hint. Passing to polar coordinates mind that log y -* (
Solving
.
(
haver=
cos 8
>
q>
+ sin
r=" x = 0.4
:2.9,
0002;
x
x
f)
= -2,
*2
2.5;
=i=
n
166.
= 0.86. =2
= 2, ^ = 5; * = 5, a) *! ^ = 2,^ = 3;^ = 3, y = 2; c) 2
A^=9; 0.0002.
n
a)
3.6,
4; b)
I.
172. 1.
169.
A^
c)
Xl
/i
>
=
N
logAr<
a)
2
Use the formula 182.
0.
192.
oo.
183.
193.
oo.
~~.
173.
!
+2
184. 2.
194.
174.
2
0.
3.1;
y,25s
A-
=.,
yi
10; c)
n
^=2.7, y z
2
=
^
2
= 2;
x,=--2,
^2
9;
Sit
**= 167.
32.
/i
T
^
>
e
W = 99;
b)
^
f/
4
168. 6
when
x>X(N); c)\f(x)\>N when \x\>X(N). 171.
8C
l,
i
163.
will
//
165. ;
^^=
d)
> -~ V
a) c)
=
32)
= rsincp, we 161. F = 32 + ^=
32)/
Fig. ft
e)
'
Appendix VI, Fig and
.),,-'
/2
f
when
y,
_1^ 2 1=JV.
=15
*);
^=-2;
*,=_;
x
VI,
Appendix KK
(
r = e? (see = rcosq,
have
will
coordinates
polar
x=1.50;
e)
we
,
(p
1M.
.1
b)
3
= 0.6* (10
162.
b)
tancp=~
Hint. Passing to 3sin cp cos cp
160.
d)
and
YX* -f- y*
r
}
1.
175. 3.
+i = -g 2
-f
.
.
.
185. 72. oo.
195.
186. 2.
~
196.
(
170.
1.
+ l) (2/i+ 3a a
-
188. 197.
a)
a)
002;
>N
when
(e
0<*<6(AO;
176.
187. 2.
=
0;
177.
b) 2*
b) 1; c) 2; d)
J
.
178.
j.
~
.
oU
Hint.
180.0. 181.
1).
179. 0.
oo.
189. 0. 190. 1. 191. 0.
3* 2
.
198.
-1.
199.
^-
1.
Answers
1
200. 3. 201.
i-
202.
.
.
-1
203.
204.
.
399
12.
205.
-- i
206.
.
207.
.
1.
-| 208.
=:. 209.
.
x 215. 0. 216.
a)
228.
A.
1.
4o
236.
217. 3.
b) 0.
224.
A
218.
225. cos
ji.
A:.
244.
245.
251.6.252.
e*.
ji.
i-
239.
a)
e" 1
Solution,
1.
-m
2
233.
).
a)
y
241.
1.
.
247.
e
*
=
lim (cos*)
2
I
221.
0;
lim [1
1.
1.
242.
-i 4
.
249.
"*.
cos x)]
(1
X-0
X-+0
.
1.
b)
234.
.
e" 1
248.
.
.
Z
227.
240.
.
Jt.
-
214.
4
246.
0.
2
I(rt
4
-|. 250.
238.
--JL
226.
232.
I.
237.
.
2 19. 4". 220. o
.
2
230. 0. 231.
Jt
243.
22'
---.
213.
-
2
~:j4=. 235.
212.
-
j/jt
sin a.
229.
211.
3
-^s\n2\ z
222. cos a. 223.
--
210.
.
2
3
2sin a
"m
=
Si nee lim V
=
2 lim
2-1- lim
=e
*
that lim (cos*)
1.
b)
*->o
(see
follows
it
*
the
in
preceding
lim
lim (cos x)*
a),
As
Solution.
"/==-. e
7=0,
V i
case
^n
v
X-+Q
=e
.
Since lim
X-+0
= _J_, 2
2 lim
:
= Put
lim (cos*)*
1.
256.
1.
--
257.
6*^1=0,
1.
where a-^0.
Hint. Put
263. a)
267. a) 0; b)
1.
1;
b)
268.
M
= a,
-i a)
.
259.
264. a) 1;
b)
a
-1;
b)
1.
258
'
K
Utilize the identity a
Ina. Hint. -*
where
2
=
Hint
-
z
e
260. In a
262.
that
*-).o
lOloge. 255.
253. In 2. 254.
.
y
follows
it
269.
(see
1.
a)
Example
265. a) 1;
b)
-1; 1.
b)
259) 1.
270. a)
= e na a
261.
266. a)
.
6.
1; b) 0.
oo; b)
+00.
Answer*
400 271.
Solution.
but
if
x
=
when x=l;
whenx = 0;i/ =
l
t
e
x=l
x>
when
when
y
276.
281.
.
kn (fe=0,
then cos 2
=
/
&
x
If
/Swi,
x>0. 277.
^g.
1
6=0;
H
1 1
compound c)
>
x\
|
1000. 289. |x
1|<0.005;
b) |x
= 0.01;
b) 6
c) 3.
293
299.
3.
6)
a)
300.
x |
;
1
1
~
kn, where k
316.
a)
/(0)
x+ A*
= n;
b)
|
|
x
\
\
when
^* 2^
-
-
b)
A:)
is
<,\
we
/(0)1;
310. a)
an integer I
c) /(0)
313.
= 2;
x
7=
301.
y
292.
a)
1.
= 0.1;
1;
|
2;
b)
298.
1.
Hint.
0.98(09804);
1)
d)
;
Take
~
6
3.167(3.1623)
c)
have
Hint.
< 0.05;
|
a)
15. 297.
296.
b) 4; c)
2;
309. |
.
,
3.875(3.8730);
4)
1
jx
1 ~
0.985(0.9849);
Ax
<| Ax
No
295.
a)
(
U|>100;
b)
-^-=6;
10.954(10.954).
d)
y^zT
curve
a)
2|<
|x
Second, b) third.
a)
the
|*|>10;
a)
,
\&x\
cos
285
=x
279.
si.
"
0
d) 2; e) 3.
when
(x-f- AJC)
j/
proportionality factor (law of
6
290.
0.0005
l^"x)<| Ajc|/V^x.
integer; b) x inequality
1
;
-5-
i
U|>
0.0095(0.00952);
inequality |cos
1
<
+1;
1
3)
Ax+
the
is
when
291.
c)
=
when x<0; # =
= 4" of
asymptote
0.043(0.04139). 303,
then
Ax|/(V A:+
288.
.
<~
1
x
y
7)
the
is
1.03(1 0296);
=3
> 0,r
|
-i
a)
072(0.7480);
1
= 0.001.
b)
1;
c)
1.03(1.0309);
If
=
6
c)
)/"lO= l/"9Tl 2)
Q t = Q
interest);
#
1;
-!
2
where k
)
0
278.
lim <4C n n-Mo
284.
nj
v
y
* .
y=x
line
=
j/
x 2 -*oo.
fL
straight
Q/^^Qo
287.
274.
y=x
when
0=1 when 0
275.
+
o
e
the
= |x|.
*,-*---;
282.
4-
1
J/
cos*x
then
...),
272.
/=!.
273.
1
2,
1,
and
1.12(1.125);
5)
j
.
Hint.
307.
=
J/x-fAx-- /"il advantage
+kn, where
the
of
k
is
an
311. Hint. Take advantage of the l. 315. No >l=4. 314. f(0)
= 2; kind. 318. x = d) f (0)
e)
= = 0; / (0)
f)
/(0)
=
1.
317. x 2 is a discontinuity of the second 1 is a removable dis2 is a discontinuity of the second kind; x=2 is a removable continuity. 319. x= 320. x is is a discontinuity of the first kind. 321. a) x discontinuity a discontinuity of the second kind; b) x=0 is'a removable discontinuity. 322. x=0 is a remo vable discontinuity, x kn (k=\, 2, ..) are infinite discontinuities
=
=
=
323.
x=2jifcy
(fc
= 0,
.
1,
2,...)
are
infinite
discontinuities.
=
x^kn (=0, is a 1, 2, ...) are infinite discontinuities. 325. x 1 is a removable discontinuity; discontinuity of the first kind. 326. x 1 is a point of discontiauity of the first kind. 327. x 1 is a discon-
324.
x=
=
=
Answers
401
x=0
is a removable discontinuity. 329. x=l tinuity of the second kind. 328. of the first kind. 330. x 3 is a discontinuity of the first is a discontinuity is a 1 kind. 332. discontinuity of the first kind. 333. The function is continuous. 334. a) is a discontinuity of the first kind; b) the function kn (k is integral) are discontinuities of the first kind. is continuous; c) x 335. a) k (k is integral) are discontinuities of the first kind; b) x is integral) are points of discontinuity of the first kind. 337. No, since (k the function y E(x) is discontinuous at x=l. 338. 1.53. 339. Hint. Show that when * is sufficiently large, we have P ( x ) P (x ) <0.
=
x=
x=0
=
=
x~k
=
II
Chapter 341. 344.
3;
a) a)
0.21;
b)
1560;
624;
2ft
c)
+A
0.01;
b)
2
342.
.
0.1;
a)
100; c)
3;
b)
c)
0.000011. 345. a) abx; b) 3x*&x
1;
-.;
2*(2-_l)
e)
g^'-i); A*
;
jt+Ax-f-J^x
2;
354.
a)
~-
lim
A!?
A/-H)
At
=
lim
*
"^ d)
Q
^ =E =
1
AJ^-^O
sin 2 x
a)
12 "~
c)
]im f(x A*-*O
the
-
L^
0.16;
/ (x)
time
at
time
t.
at
time
t.
A =^
b)
t.
at
substance
6
21 -
Ax
temperature of
347>
+ Ax)
turn
of
angle
quantity
0238;
21 2
357.
Ax cos x cos
2
"
is
"" h;
^
(x)
f
the
is
l;
_ _ 2
"
35l
0.25.
= J^- = sec x. cos x (x+ Ax) 1 Solution. 359.
-.
T
the
is
356.
Ax
lim Ax->o cos x cos
cp
where
,
Ax
0.249;
lim
_
lim AJC-M)
201
'=
where
Ar
-
>
Ax
where
,
b)
l;
/(*+Ax)~ /(x) 2,
lim Af-M>
b)
\
Ax
^
c)
X
b)
;
rX = dt
355.
dt
_
353. a)
JP=
b)
Ar
~~
a)
IT)
348. 15 cm/sec. 349. 7.5. 350. 352. a)
346>
*
AX
+
-
c)
(x-f-
Ax)
a)
3x 2
358.
n8)=
;
Solution.
sec x.
AX b)
lim
Ax
*o
-4 x* ^ (8
+
c>
;
7=
2 V^
A ^~^
v
(8)
Ax
A*-M>
+ Ax 8 = hm Ax AA:^O Ax 8 8 [ J/ (8 + Ax) + ^/ (8 + Ax) + j/ ] = Hm = 1. 360. r(0) = -8, r(l) = 0, = 12 Ax-^o 3/(8+ Ax) + 2^/8 + Ax+4 = = = For the given function the equation x Hint. 3. 0. 361. 0, x, /'(2) = x 362. 30m/sec. 2. 364. 363. /'(*)=:/ (x) has the form 3x = 365. nx ^. 366 L 2 tanq>=3. Hint. Use the results of Example 3 8
..
.
_
2
_-
1
2
,
2
2
8
-
)
xo
and Problem
1.
1,
.
365.
3
367. Solution, a)
f'(Q)^ lim
Ll
!_=
=
lim
Ax AJC+OJ^/
AX
4- oo;
Answers
402
1^15^= Ax
/'(!)= lim
b)
r
=
lim
=
2fe+l
lim
sinAxl
l
1
=1.
Ax
A*->+o 5x
*+A
2
,
2
371.
'
,
,
3x
-l
/
.
+n-1
6ax
= x x* = x
9y2 ^ A
sin/.
y'
= 0.
390.
sin 2
xe x
391.
xV(x-f-7). x 2 e x 396. e x
sin x). 395.
^-x
cot x
387.
.
/^arc sin
.
\
3x 2 lnx.
398.
402
2
2
_ tanh2x
403
toco.h^dnh* cosh x 2
2x 2 405.
y
x 407.
x 411.
x
,
Vx
12fl6+18^
1
-|-2x arc
e
2
(1x 412.
//.
x
K
fex2 .
>X
16x(3
+ 2x
2 8 )
"~
*
420. 2
419. 2
2 sin x
V
15 cos
xln 2
sin
x sinx.
_2cos^
*
sin>
3cosx)
arc sin
....
+
414.
^
2(l+Jt )V^arctanjc
_3jari!*)l /"l
x*
/
.
8
x2
/I 1-tan
2
x
+ tan
4
x
COS 2 X
.Hint. x 3 cos x
^
cos*x
Ssin^^
c
\
I)
418.
.
I
I
c
.
y.
2 3a fax-}-b\ fax b \
424>
3
cos 4 x
L !
1
-f-
tanhx
421.^^^---
423> (1
427.
xsmh 2 x
sin 8 2/
422.
425.
2
cot x
x
+ x coshx.
sinh x
^
)
2
-3(xln* + sinh*cosM
413.
.
In
401.
.
*~ 1) .
x /
1
.
x
* (21n
397.
.
2 2 )
-.1/j/l;-!. X f
416.
8 2
^
2
v5
*>y*
393.
1
(2x
41B
O
y
^f-.
.
l
"
X2
j/~i
.
2
arcsinx-f-
;
*\ arc coshx
2
2
cos x)
(sinx
arc sinh x
406.
.
2
I)
384.
2x
388.
. '
4()4 1
X
*
1 ^1 x In 10 x
1+1H/-A.400. x x x
399.
on
.
2
.
4***
x 2 (2x
392.
x+
wy 1*
2
-
383. sin
386.
394. e* (cos x
3 sinx.
2)
T
380. OOU.
.
+ 5)
5x
377.
.
!
2
5 cos x
8
fW-4-9ft VJA -7- z-o
(x
ji
374.
JL
z
Hint. y
.
.
5
373.
.
i
9
2x 8 370.
3
2
2. 382.
7=
V
x arc tan x.
389.
rn
/2*-fl
,
l + 2x
369.
l
2 (1 2
+ 2.
o*7A OIS7. 379.
.
r,
|
1 x
376. -*
+ dx)
(c
12x 2
8 .
I
+ b(m-\-n)t
U*U.
.
2
Y 385.
3x- 4
l/x r^
1.
2
5x
/'-
c)
sin Ax ~ L^iJLJ=-l; ,.
lim
5x 4
368.
-JL 2x
375.
==00;
\\
*JL '=
mat m
372.
.
^_
lim
= sin- + 2
1
+ 2 sinx
2 V^lSsinx
lOcosx
1 '
2 1^1 > '
x 2 >^1-|- arc sinx
^ '
2
(1
+ x ) (arc tan x)
2
Answers
403
429.
X
430.
.
cos(x
5
-.
5x+l)
2
2*
x
1)
-}-
asin(ax +
433.
(2x-5) X
4 32.
,
3
434.
p).
sin(2/+q>).
X 2 COS 2 X
""*
?21i
2
435.
436.
.
x
sin
n2
s
x cos 2x 2 sin 3x 2
437.
.
^
.
438.
.
Solution.
a l
(2x)'
,-
2
V^l
"~ 2
2
=
-
lA
(2x)
439.
.
4x
2
x/V
2
442.
10xe~*
443.
-
-p^jf
T
440.
446.
sin2
+2
^
450.
.
f
t /
cos 2 In
L
453. (1
-j-
+
In x) x
3
455. Solution, y' 3 H- sin
5x2 cos 4x
456.
+
(1 -j-x
(sin
)
5x)' cos
2 -
\
sin --
f
tanx x 6
10).
cot x log
4 arc sin x)
-}-5sinx
e.
VI
+
454.
.
x\
x f -^
449.
.
]
arc
+ x In
(1
2
-
x (e
J
+x
(
452.
.
xlnx
x
1 .
1
2xi0 2 *
448.
^^
451.
x2
1
x
441.
2
2
447.
2.
.
445.
In 5.
pX t
"~
1
2 J^jc
1
2*5~* f
444.
.
.
2x \ In x-jx \ 2
1
'
+ sin
3
/
5x
I
cos 2
= 15 sin
= 3 sin
-=-
6 J)
\
x2~sin
5xcos5xcos 2
2
x
5x cos 5x5 cos 2
-
xx 3
8
5xcos
-r-
sin --
.
)
x 2 -f4x
3 457.
6
x
x' .
458.
1
459.
.
463.
.
x
4x 3 (a
465.
2x 3 ) (a
5x 3 ).
474. sin 3 x cos 2 x.
sin
479. .
4
476.
x cos 4 x
10tan5xsec 2
-
3cos*cos2*. 480. tan'*. 481.
-- 1
.
484.
sin
arc sin x (2 arc cos x -
r
2
-f
]/"l
x2
4
. '
478.
.
3/
1
sin
2/.
=
482 .
.
x
arc sin x)
xcosx 2
477.
5x.
2 . OB 485.
K a
x
sin 2
x
+p
2
AQC
-
V^x
cos 2 x
.
2
486.
'
1+x 2
1
V
487.
491.
,
1^2x
xa
492.
arc sin
,
...
V
1
25x 2 arc
sin
5x
_
Answers
404
,
494.
- -
495.
.
.
1+cos 2 * a w *lna. 502. f
e-*cos3x. 505.
504.
(1
Psinhpjf)'. 539.
542.
/?*
499..
2
506.
^
548.
-
l(/tanx
-^-
544.
.
545.
.
sinx
/
y+
I1^3 -. 5
V (0)= =^;
/'+
(0)
tion.
or
567
do not
We
sin ' x .
+ l/Tosxln a).
1 ,' =
549.
.
when x
1
have
c)
6n. 554.
553.
/I (0)=
exist. 555.
y'
xy'=y(\x)
= e" x (1 566. *
/_
/+
d) /'_ (0)
=/'+
(0)
+ (0)-0;
1,
1
a)
/'
x.
556. 2
+ ^-.
=^
557.
(0)=1; (0)
< 0;
y' (0)
does
I
f'W-
= -1,
/
xarctanhx
546.
.
=
y'
550.
^L
x2
1
{ 552.
(1
2coth2x. 541.
540.
y'=l when x>0;
a)
'-,*,.
b)
exist;
tanh 2 2x)
(1
cos2x
1
547. xarcsinhx.
1
sin 2xe
500.
x p sin pO- 503. e" sin p*.
e*(acos p/
2xMna).
+ 4 sin*
_)/.
6tanh 2 2x 543.
x y In 2 x
not
5
"""
1+*-
+
^-'a-^fn
496.
.
2xcosa + xt
1
= 0, 1.
b)
/.
(0)
e) /'. (0)
558.
=
2 ,
and
561. Solu-
= -(lx) (1 +2x)(l +3x) + 2(l+x) (1 +3x) + 3(x + l) (1 +2x). ~ x).
Since e' x
t
it
follows that y' 2
Answers
40S
-24 573 '
>-
574.
.
'n*.
575. *
.
,
2!li
+ cos x Inx
/
*
SIn
578.
.
576.
'
- sin * tan x)
(cos * In cos *
(cos x)
)
J
58
-
'
581.
r
"
.-- 4'<(2 ""
586.
.
-
590.
594. tan^. 596.
601.
tity.
.
,;
603.
2
-. 2
Yes, since the equality 604.
-
2j/)
arc tan 2
=^63
(-2, -12).
45.
623.
26'.
626. (1,
-3).
620
.
624.
627.
arc tan
.
b) d)
y
y-5 = 0;
^2y = x
x
+ 2 = 0.
l=0;
34 = 0. 13
= 0,
GA;
/
(3,0):
t/
2x
y=l
1;
x\
616.
xy
a) 0; b)
~^r 36
1=0; y = 0.
x
the
for
625.
21'.
633.
(0,20);
= 0;
+
6x Jt 2// x 2f/-fl=0
f-i o
point
(
1,
1).
6;
;
x
+
t/
2
= 0.
at the point (2, 0): (/=
639.
14;c
638.
634.
At
x-f-2; y
(1,
15);
-1^ lo
\
2^
for the point (1, 1);
7x
= 0. 637.
45-.
y=2
a)
635.
5y-f 21=0.
622.
0.
c)
;
+y
2
= **- *+ 1.628. k = ^.629.
+ y2 = 0; c) 3 = 0; e) 2x + y
1=0
= -Ilf
= 2x
632.
3 cosy
10
11
631.
r 10
x
615.
y,
1
6 08
*
8
an iden
is
-
605.
.
_=_
\nx-xx
-*-.
593.
2
y
-.
J
l-f3ju/ -f4(/
yVx*+y*
-
589.
f.
a
={
599. No. 600.
oo.
-.
602.
5
ex
tan
~
-tan^.592.
597.
1.
fir-*- ra.
588.
W.
l
a);
+x*)arctanx
585.
(arc tan x)X
-
=x
0;
1
636.
the 2\
point
(1,0):
at the point
41
=0.
Answers
-406
640.
The equation
Hint.
of the
tangent
Hence, the tangent
^-{-^-=1. 2y Q
is
ZXQ
crosses the x-axis at the point A (2x 0) and the y-axis at B (0, 2f/ the midpoint of AB, we get the point (* y Q ). 643. 40 36'. 644. at the at bolas are and intersect point (0, 0) tangent
The para-
,
arctan-i=^88' 648
652
'
'
the
at
= 2jia
/1+4JT 2 t
--
~.
arc tan
=a fl
-^-in/sec.
x
2
The
.
2u g/sina
+g
To determine system. The range on
velocity
/
the
;
Sf =
'
S,
= a;
cm /sec;
0;
656.
2jt.
q>
.
is
the
of
slope
3
657.
-
The equation
660.
range
2 8
5,,= -^;
of the trajectory *
sin
\sy=x tan a
2a
The
.
vector
velocity
cm /sec
9
is
velocity,
cosa
v
axes
the
~ '
~
are
and
The magnitude
j-
the velocit y vector
of
the velocity
CO8
1
cm
40
of
he 2
360
at
B
the
g,
density
x2
1
v
681.
y
5* g/crn,
is
+~~210,Y
4 .
+
=
^
u = 5;
x*) .
A
2
671
+ 2).
i*
cosh
682.
a vl //
=
6 79. y"'
= 6.
680.
a
~64sin2x
684. 0;
1; 2;
=
2.
of the rod
the density 2 cos 2x 669.
2 arc tan x
672.
.
f
is 0,
X
674 (1
The mass
666.
the density at x* 668. e (4^
2
3(1
_
y cm/sec
m'/'sec. 665.
56x 6
667.
60g/cm. 2(1 ~"* 2) .
673.
JT
M
at
is
670.
Q ,
- '6. 8 cm/sec, the area, at a rate diagonal increases at a rale ol 2 664. The surface area increases at a rate of 2ji m /sec, /sec
the volume, at a rate of 0.05 is
is
directed along the tangent to the
is
~ 663.
.
the trajectory, eliminate the parameter t from the given is the abscissa of the point A (Fig. 17). The projections
Hint.
of
=
655
'
I
tan|i=
2_|_p2.
2
uj
tan^i
;
u
g
angle
S,
-y +2(p
2
}/~l-|-4ji
2
658. 15 cm/sec. 659.
K
654.
--
rt
;
~ n -~
>
647.
(1,1).
an
T
'
653.
*
point
Finding
).
,
685.
+
/'"(3)
The velocity
The acceleration, a 0; 0.6. 686. The law 0.006; the is x acoscof; the velocity at time / is point M, 2 acD cos(of. Initial velocity, 0; awsincof; the acceleration at time / is aw 2 velocity when * initial acceleration: is -aco; acceleration when The maximum absolute value of velocity is aa>; the maximum * is 0. 2 n (n) (n+l value of acceleration absolute is aw 687. y n\a 688. a) n\ (1 \ x)~ is
of
4997; motion of
4.7.
=
;
=
.
b)
(
_ ir +
~
i .
689
.
= =
.
Answers
407
d)
2 n jc
691
2
^(OJ^-tn
692.
1)1
a) 9/
8 ;
b) 2/
2
+ 2;
XU+*-);
b)
sin
693. a)
---. $
<
-
-
=.
696.
.
708.
-o
711.
x=\ 718.
721.
725.
^b) 3
a)
and
=
AJC
No.
y
719.
dy= 45
^~^-
2
=l-
697.
f- .712. Ar/
714
d//
=
-
cf(/
= 0.009001;
=. A:
M+
2
(AA')
720.
.
7,0.
.
x
d(/
8
)
=
when
717. For *
.
.
16
713. cf(l
^5 = 2^Ax, A5 = 2^ Ax +
722.
"J 256
//
d/-0.009.
^=^-0.0436.
0.0698.
~
1
'
709.
2 2 ft-4-.
'
v
726.
.
-j;
2
4o
1
4
" 707.
9
694 a > 0; b) 2e>al 695 - a)
777 4a
/"T^?.
c)
= 0.
= ;=i 0.00037.
723.
I
2xe~**dx. 727. \nxdx. 728. x
y*xe 735.
738. 740.
~dx.
IJ
737. a) 0.485; b) 0.965; c) 1.2; d)
0.045; e)
5"=^ 2. 25;
8
739. 565_cm ^^4.13; ^70^=8.38; 200=5=5. 85. 741. a) 5; /70=^4.13; J/ ^/10=^2.16; .
"""
d) 0.9.
742.
1.0019.
743. 0.57.
744. 2.03.
748.
~(^)
sin,ln, + fc).
753.
2 .
-.754.
(d **' iy
(1
750.
~ + 0. 025^0
^)*/i
~
.
^640=^ 25 b)
749.
3.2"
sin
3.
l.l;c)0.93; *( dx ? .
(1
(dx)*. 752.
751.
81.
X2 )
-e
/a
__ Answers
408
e
755.
xcosa
sin(*sina + na)-(d*)
758.
No. The point
763.
(2,
+
765.
4).
2(
3U'
C<8,<1.
0
0<6<
c*
770.
s
sin
-
~
error
^
/
x
x
( _ = H
l
\
x \ (1 --
/
I
2!L_ 8
= 77:.
get:
f
x \
,
2
-f
where
|2
where
J,,
= 8 *, 2
?-6*,
n
775.
=
both cases g
We
Solution.
6jc;
have
factors in
powers
of
#,
e
in
__L
^
2
^ + -_x -- x.
1
1
I
,
1
^1-f-)
i
;
40
Expanding both
.
J
.
= 0.
~ sin
x-^- +
sin g 2 ,
-L .
&
_i_ *
(
[
we
*=
1
than =7 5!
i
762.
x=(x- l)-i (*-l)
-
A
b)'
;
*
less
is
exist.
(
j_ a
In
not
function.
769. sin
1.
~ + ^-
3
The
773.
768.
0<0<
x = x-
the
of
does
(2)
/'
=i+x + J+|J. + ...+ ^lL.+5jeS
772. Error: a)
1.
b)g = -~.
1
1
since
No,
discontinuity
|= +6 (x- 1),
^ = 8,*, 0<9 <1;
where
757.
.
a
is
=^;
a)
where
1)8>
x = ~x
/l
I
;
y
.
-^~^1H //T"
,
y ---
-x \
8 1
x
J y*
^~9~2-
Tne n
expanding
x
x
powers 778
oo
788.
1.
1
we
,
779.
~.
788.
same polynomial
get the
780. 3.
1
n
for
802
of
789.
793. 0.
~
781. 790*.
1.
795. -1
-
804.
1
Find
lini
,
-^
u ~>0
791. a.
0.
796.
805.
.
where S
~
2
hh
783. 792.
797.
.
806.
oo
1.
IT
.
sin a)
(a
807.
>
n
799.
1.
1;
a
n
for
800.
808.
1.
the exact
is
.
784. 0. 785.
oo.
for
1
^77.
810.
1.
expression
for
=
801.
e*.
e
e
R2 = -~-
x x =^rlH---r-n~-2
\2
e
S
a
782. 5.
o
803.
e
the
l;
1.
Hint. area
3"^ of
the segment (R
is
the radius of the corresponding circle).
Chapter
(00,
III
812. ( increases; (2, oo), decreases. oo, 2), decreases; 813. 814. increases. (2, oo), increases. oo), 0) and (2, oo), increases; (0, 2), decreases 815. 2) and (2, oo), decreases. 816. 1), 817. increases; (1, oo), decreases. 2), 8) and (8, oo), decreases. 818. (0, 1), decreases; (1, oo), increases. 819. ( and (1, oo), in00, 1)
811.
2),
(00, (00, (00,
creases;
creases;
(
1,
f
,
1),
decreases
oo
820.
increases. 822. j,
(
(00,
(00,
(2,
00,
(2,
oo),
0),
increases
increases. 823.
821.
(
(0, oo,2),
],
de-
decreases;
409
Answers
(0,
824.
increases.
oo),
(2,
oo), increases
decreases; (1,
1),
and
00, a)
(
(a,
=
*=-
when
j
0)
and
828.
No
(00,
825.
decreases.'
oo),
827. ymax
.
=
1296 when x = 6. extremum. 830. t/m in = when x=Q; t/ mln =0 when x= 12; t/max when x= 1; t/ mfn =^ 0.05 when 0.76 when x=5=0.23; i/ max = 831. t/min^ 2 833. #max = 832. No extremum. x=^1.43. No extremum when x = 2. -
when *=0;
|/
= 3/3 when x = i/
840.
[/
n
=2
x=
when 837.
j/
2,
=
845.
y mln
when x = 2 value
is
m=
when x = Af
=
75-
^= 12
=
for x
l
fjkJ:^
1
c f/
= ~;// min ==0
x
min
for
when x =
1;
j- (fc^O,
for x
2,
1,
is
is
(2fe
+
r ".
(fe-^0,
1) JT
when x =
844.
f/
m?n
-.
-l when 4
when x = 0;
=
No extremum.
M = 7rwhen
value,
= 5.
//max--2
m^=0 when
852.
1; M = 27 when x = 3. m = 1579 when x =
Each
side
whose
x
863>
-f-
1)
= l; M=JI 854.
-j-
when
m-
a)
;
6
M = 3745
when
be equal to
--
10;
864. The
Isosceles.
865.
m
850.
\.
The
side
The altitude must be
866.
.
x
terms must
of the
Smallest
849.
when x = (2fc
m=~
851.
must be twice the other
of
tlit
half
the
-g-
altitude
-^L y3
the radius of the given sphere.
R
mln
-7--
Altitude of the cylinder,
where
f/
=6
= --
x=\
square with side
a
cut-out square must be equal to 867. That
x
848.
...).
b)
861.
4.
<7
side adjoining the wall
868.
846.
= g when x=l. x~ greatest
= 10; M = 5
2,
when
842. y mln
= /2
s/"3 when
y mta =
839.
max
when x = 2K
K^3"
x=1
when
1.
=
y min
r/
max =
*,
The rectangle must be
base.
= 0.
x
t/
5
when
y min=l
Ji;
835.
836.
^==^
9rr
^ n; ^/max^^cos
l when 853. /n=s when x=l; M=-^2o6 when x = 5;
862.
when
fcjT
p=
when x = 3.2.
yg
l
x== _l.
856.
=9
r
x=
x=12.
#max
= 3/3"
min
ymln ==0when
2
and x
f/
(/
=
847.
834.
= V"3 when x = 2^3; *=1; ma x = when x =
w ^n
t/max^^-
=2
max
841.
...).
x
^L; yO
*=12
Scos^when
843.
when
m ax-= 5 when
1,
x
i
mm -0 when
838.
=
m
is
;
to
equal
radius 869.
the radius of the given sphere.
of
the
its
Altitude
diameter
base of
R ]/
the
of the base
ro
,
where
cylinder,
870. Altitude of the cone,
i<
__
RV'2 --
410
__ Answers
4
where
R
is
the radius of the given sphere.
where
R
is
the radius of the given
cone
-jr-r,
3
where
r
871. Altitude of the cone,
Radius
872.
sphere.
876.
.
=
The altitude
of the cylindrical
+ *yo = 1. ^* XQ -^-
878.
879.
The
the vertices of the rectangle which
/
3 \ -j J.
1
881.
~Y=.,
(
arc cos
4-
885.
a)
and
t-y-.
P min = y%aqQ. spheres,
the
impact
with
m*V +w
b),=
=;
If
a
=
n=
-
"I/
(if
r
this
^
is
battery
,
number
(
the physical
points ^"3")
points
3),
oo,
6)
2,
J,
.
*.
^_
884.
.
886.
not an integer or
is
meaning
M,
M lj2
concave up;
...);
oo,
-4r
m
,=
obtained
is
as
as close as possible to the
concave
2),
0),
(
(4/e
+3)
0)
1/~3,
(
(|/*
0)
and
-^, (4
Af 2 ( 6,
down;
0(0,
+ 5)~
l)it,
2&Ji),
f(2fe+l)y, concave down(fe=0,
sas of the points of inflection are equal to
xkn.
0).
,
895.
|-V
and (1^3,
points of inflection,
concave up; ((2k
not a divisor of
(2,
oo),
892.
(6, --|Vo(0,
^3), concave up;
(0,
is
of the solution
must be
of
inflection
of
P
^L
inflection. (00, oo), concave up. 12), point concave down, ( 3, oo), concave up; no points of inflection. and (0, 6), concave up; ( 6, 0) and (6, oo), concave down;
inflection
and
-jr-
1,
M (2,
oo,
of
I/ ^M-
a divisor of TV). Since the internal resistance
is
=
(
2
(-~-a;
of
l
follows: the internal resistance of the battery 2 external resistance. 889. y h. 891. ( o 893. 894.
coordinates
\f~Mm. Hint. For a completely elastic impact of two imparted to the stationary sphere of mass m after sphere_of mass m 2 moving with velocity v is equal to v
concave up;
where
887.
2
the
6J/~2,
8
equal to the greatest of the numbers
AM=a
883.
7V,we take the closest integer which of
is
The
d
velocity
888.
.
^
880.
that
zero;
= \l*
a^Tand
on the parabola
lie
The angle
882.
arc tan
of the ellipse.
the crossthe sector
is,
must be
part
sides of the rectangle are
a and b are the respective semiaxes
/HI
jr, that angle of
the vessel should be in the shape of a hemisphere. 877. h
is,
R,
of the base of the
the radius of the base of the given cylinder. 873. That
is
whose altitude is twice the diameter of the sphere. 874. q> section of the channel is a semicircle. 875. The central
/~2
-^
oo),
896.
concave
down
J
oV
(00,
concave down;
(&
=
897. (2/m,
1,
898.
i2, [0, \
...);
^
the abscis),
V&J
concave
Answers
[7^,
down; ;
899.
oo
V/>
concave
M
up;
0),
concave
up;
900.
(00,
3)
_JL
^=
f
(0,
oo),
and
(1,
)
W)
(Y*
J
oo,
(
inflection.
Y
411
a point of inflection.
is
concave down; 0(0, 0) concave up; (3,
a point
is
oo),
of
concave
1),
= 901. of inflection are M ( 3, T and M ( = = = 0. 902. *=1, x^=3, = x= 903. 904. x. 2, y= *, 905.0 = right 907. jc-= = x, right. 906. = left, left, y=-x, left, = 2, left, y-=2x 2, right. 909. 0^2 910. jt-=0. = *, right 908. 911. * = 0, 0=1. 912. i/^O. 913. *-=!. 0=1, left, = 0, right. = x n, left; y = + right. 915. y a. 916. ma x = when x 0; 914. 4 when x = 2\ point of inflection, Af,(l, ~~ 2 917 ^/max^ when min = r = 0; points of inflection Ai l|f when jf=V 3; m ~) == 4 when x = 918. y min = Q when je=l, point of inflection, M, (0, 2). m ax 919- f/max^ 8 wnen ^ = 2, m n = when = 2; point of inflection^ M (0, 4). when A'^0; of inflection 920. M lt2 (Y5, 0) and points //min^ down; points
A'
1,
2
l
J.
j
1.
iy
1,
1,
l,
y
JC
JT,
)-
r=:
in
1
-
jc
1,
f
1;
//
Jt
i
!
\,
totes, x
x
923.
924. (farm
x
0=-xl. ~
1,
= 0.
= 0.
=
~
when
1
\
Ai
J2/"3,
lf
of
point
= --J;
x
Ai(5, \
of
0(0,
inflection, jc^=
*---
1;
0min
and /^(4,
B
(8,
A
(
and
4) 3,
Q
V
940.
+
938.
= 0; m n= x s
2
x
asymptotes, x
x
0,
=4
=
when x =
V%
0max=
935.
2.
when jg=
^nax^
936f
~YH max
=
4
when x =
of
points
= 0.
when
Asymptotes,
941.
x
= 3. jc=
1;
1;
4;
max j
ym n
=y
\
942.
2
and
max
\\
4r J 0. y
=l when
x-^4;
929.
Point
0.
--97 930. nnx = 4 when max =
'16 ,
931.
= Q and =2
f/
= 3v 933.
;c
0(0,
/I
0).
932.
(8,
934.
A
(0,
End-point,
A(Y$* 0), 0(0. 0) inflection, M (1^3 -f 2 f^J,
point of
M
l
(Q
x t
m = M 1>2 (1,
= 0,
points
of
and M f (1, (when x = 0)
0);
1)
1
in
~~
m
i
n
=2
= 0,
2)
and
4)
End-points,
when
1
-
inflection,
asymptote, 939. max -=2
K 2); asymptote, 4 wh'en x==4; point of inflection, 0(0, ~~ ^ 3 / 4 4 when x when x 2, 0mi n
inflection,
when x =
J
'
and
= x
asymptotes, r
asymptote,
0);
M lt (\ \ x--2 and
= 0.
and
2
and
937. Points of inflection,
0).
asymptote,
943.
0);
l
V M, t(l, when
2
=2
/
x~\\
vvnen
m in^=
0);
x.
=4
928.
2 V 2 when 2) are end-points; ma x are end-points. Point of inflection,
B(Y$*
0=^
0);
asymptotes,
-;
=
=-
=
points of inflection, 0(0, 0)
asymptotes, x
77-]; y
T2); asymptote O. asymptote, AT
1/2,
when x--0; asymptotes,
asymptote,
8
when
when x-=2; asymp-
inflection,
m jx ==l whenx=l;
-^)'.
inflection,
M(
of
points
2
926. # ma x-~
^2
mm
inflection,
= 0,
*
min
(l,
f/
of
/y
inflection M, lt 4 when i=l;
of 1;
point
wnen 0max~'Qo
925.
asymptote,
and
Points
922.
when jc=l;
when x^O;
2
^max^
-
when x
^
max
=3
927. y m n
921
-
=
when 944.
x
= 0;
0min^
= 4;
=K
asymptote,
T7= 1/2
0.
0);
when
x=2.
Answers
412
J/max
1/~JT
=
TT= when V 2 -o-
M
of inflection,
of
M, x
3a,
(
= 4;
949.
6
=2
rnax
of
point
/min=
953.
=g
min
f/
y
=
2 2,
I
of
= 0;
M
when x = g;
*-
as
1+0
x
-- ,0
jc
points
=
j n+
=
2Jfeji;
t/
(
~+
fare cos
On
the
*=n; M, 4 nal
2fen
(
[0,
(fe
= 0,
= 0, # =
1;
n + kn,
-j-
~^-J+2fejx,
m in
=
2 20, 2it],
ji,
[
...);
i/
max
=
1
1
=
when
2A5Ji
(fc
t/
max
=
Odd
of
/
.
when
of
x=
V
Asymptote, x oo).
2;
points
defined on
\
max
the
= 2,
...);
function
with
1,
Periodic
inflection,
~;
q
=
V$
M k (kn,
M
/
l
2
ml n=0.71
f
when * =
86;
^=0.37); J
e
when and
0)
961. Periodic function with period
j
1;
956.
m in =
(0.57,
periodic function with* period
when x = 0;
^=
2n. y m n
inflection,
*
x=
= 0,
/
39
# = 0.
"""472)-
1
x (as
period
j Ji-f2/5Ji;
,
-y
not
is
1
asymptote, A:-
1^0.63;
with
960.
points
yg )^l5j.
962.
jc
the function
*=
when
when ^ = 0; points 0.95).
when
max
and
(-rr==
~V,
,
(e
t/
^e ^ 7
=
*
M
= 0.
2
when
74
e
2,
ji],
-
J.
=
Oj.
_
1,
950.
1,
y m{n =
j+
wn en y
asymptote,
);
2
f\ e
function
f
interval i/
#
inflection,
2
+
^3
y mln ==
2ji.
Ai
inflection,
of
948. f/max 1111162
g
f lf 2
= -F ^0-54
max = ^2" when A:=
Mk
of inflection,
= 0.
asymptotes, x=*l. (when x-+ oo) and y
,
o
period
max
Points
'
~\
M
inflection,
959. Periodic
.
point
1.33);
=
Asymptotes,
when
Nk
of
when x=l;
inflection,
(limiting end-point). 955.
M
interval
x=
f/
inflection, 1>2 (1.89, 957. Asymptotes, r/ Q.
958.
of
point
point
point
asymptotes,
point of
0)
when x = 6;
/2 max =
t/max^
0.70);
0(0,
3 = -0-7=
947.
M
\
-n-l,
J
,
951.
^14.39,
954.
D
*/
M^J
^ = -4=.,
wnen
x-*0.
946.
points of inflection,
(e"'*
i
3,
[
^
/ 8 ^ 2 i/"~ ^2
when ^ = 0.
=
m
asymptote, y
;
)
inflection,
>:
j-
a,
2 (
/
Af,
= Q.
asymptote, y
;
)
M
and
)
\
-j-
when * = 0;
=
-^
of
/
w min
when
y-+Q
y
/iooy
inflection,
952.
945.
j^r=V, asymptote, x = 2
when
jc=l;
when
x=l
/12,
points
//
/min
5-
points of inflection,
I
V
M
inflection,
3
asymptotes,
;
J ')
Q
f
=
\
(3 3,
*
*.
2;i.
2
2xc.
when
0.13) and
On
inter-
i/max=l when
__ Answers
=
*
Ji
T
^nin
'
=
When * =
1
=l
M
2
(1.21,
when
= 2n;
M,(2.36,
0.86);
A4 e (5.50, 0).
*
of
points
with
function
when * = 3
asymptotes, x=-r-Ji
Even
965.
= c
^+
Mk
inflection,
+ ^Jt kn,
^min
(
skill:?, function
A:
= arccos
^zr
;
(/
=
lO^ with m
n
i
M
;
J^
(jt^arcsin
1
On
2it.
x
the
when x^Ji;
0(0,
x
0);
M k (kn, function,
//
and
Odd
/
m in =
n + 1 +2^n when
*-*+oo).
inflection (centre of 974. Odd
Oj
;
Even
966.
-
when
^min=--F-
n]
t/
(/j
ma x
M!
t/max^
f
-y
,
Oj;
-i/g)-
0,
2,
1,
57
1
Odd
969.
when
...). A:
=
function.
1
t/
JI
A;
(/
+-fr
,
+ oo). Point of inflection, + ?JT when max = -^-
when
972.
(/
m
JC==1;
symmetry)
|
of
points
/JJT;
2,
1,
(as
(
l
inflection,
1
2fcrc);
1
n
=
when x =
^max=
(0, Ji);
"
1
-^-^
--
1
asymptotes,
...).
(as
971.
x-*
Even and
oo)
(node); asymptote, y
Whe "
X=S ~" I;
y = x + 2n
(left)
=
P oint
and
l.
J
y=x
= 1.856 when -= M of 0, inflection, + (when asymptotes, point -^y and # = * In2. (as ^-^+00). 975. Asymptotes, # = oo) and y = -^
(right).
l;
function.
t/
m in=1.285 when f
;
j
^-^
(1,
function.
[0,
^= when
O
2k 4-
=^.^1
#==
q
oo)
970.
j JI+/JJT.
vvhen
0.34).
= -J + 1 asymptotes, x^-y -^ (6 = 0, = 0; asymptotes, = t/mln^ when x
+ /en;
-j-
1,
= -
M
of
points
interval
inflection,
57)
1
t
(
n;
...);
1
[0,
fejt)
(fcjt,
'
asymptotes, x\.
2,
1,
the
;
of
points
(2
graph
= 0,
^-)
,
when
\
inflection,
interval
function. Points of inflection, M* Even function. End-points, 4, 2 83, Af J(1.54 (cusp); points of inflection, lf of
0.86);
(- (-/I)-=
-.(/S-i/S)' Odd points
of
-=.
= arccos
967. 968.
Limiting
(4.35,
5
= 0whenx-n;t/ mln =
8
when
=
max
0.86);
period
On
when
1
Ai^O.36,
asymptotes, *
...);
2n
wnen ^==0; points
period
= -F=
(fc
=
#min
V~2 ~ ym n = L
2n.
+ 2foi
2,
1,
period
t/
arccos-^;
= arccos --7=)'
(arc periodic
with
when ^ =
rr^s
Af t
= 0,
(fc
-y- J
5
M
0.86);
period
ji
5
f
inflection,
964. Periodic function with
function
periodic
*=
wnen
l
M 4 (3.51,
0);
Periodic
963.
= ^- 7
ymax
;
413
JC=1;
t/
t/
max
ji
Answers
414 976.
j/
*=1;
m n =1.32 when i
2jt.
period
t/
=
min
asymptote, x
when
x=^-ji
-g-
= 0.
+ 2&JT;
t/
...); points of inflection,
2,
1,
+
arc sin
and
^- +
2;ri
i/^ + 2/m, UlA
M^ Urcsin
2
wnen * =
max ^=e
/
= 0,
(fc
with
977. Periodic function
2
e
I
g
978. (2fe+l)n,e 4(0, 1) End-points, ^ B(\, 4.81). Point of inflection, M(0.28, 1.74). 979. Points of inflection, 4.81 (as *-*-|-oo). 0.21 (as x -+ (0.5, 1.59); asymptotes, # oo) and r/ 980. The domain of definition of the function is the set of intervals (2kn t where fc Periodic function with period 2ji. ... 0, 1, 2, 2/jJt-f-ji),
A^ and
I
=
M
=
t/
when * =
=
max
The
981.
domain
of
Points
-^-
+
=
(/e
r/=1.57;
4
lf
2,
1,
2/jJi
i/-^ 1
y mln
x
=
as
-*
x
Domain
of
(limiting
period
=
+ kn.
i/
min
Limiting
asymptotes,
...);
definition,
increasing
<~
2/jn
]x
|
y mm =l when
2jr
984.
Asymptote,
985.
End-points,
^ 0.69
-
2ji.
period
monotonic
0;
with
986.
987.
>
end-point).
when ^^0.
^-^ +
2,
1,
definition,
with
function
Periodic
x
kn.
of intervals
= 0,
asymptotes,
...);
x
asymptotes,
...);
the set
function
Periodic
1.57
1
(fe
983.
as
-+
0)
(2fcji,
0.
2,
p0.37; y
integer.
x
1,
57);
an
is
of
...).
= 0,
,(1.31,
*=
Mk
asymptote,
(fc~0,
is
2,
1,
Domain
982.
2fejt.
0,
(&
definition
k
inflection,
function;
x
2&Ji
of
where
(2fc+-jrjJi
x=
+
when
4 (+0,
end-point,
Q);
i
t/max^ 6
e
^ 1-44
when # = e=^2.72; asymptote,
(/
=
of
point
!;
inflection,
when / = (0 = 3); y mm = 1 M, (0.58, 0.12) and M 2 (4 35,1 40). 988. Jc inln = when / = (x^3) 989. To obtain the graph it is sufficient to vary / from to 2jt. a (cusp) when ^min^ awhen^-ji(// = 0); ^max = awhen / = (#-0); min 1
1
1
t/
*
=
+-2-
.
3T
^
inflection
t
when
f
=
fl
(cusp)
when
<
7--
V'Se \
when
*
=
8 ,
-y
U = 0);
points of inflection
7jT
,
-j-
*=
1
I/" 2,
i.e.,
to=
^)l
^max =
XJ.
f
j
i.e.,
=
44
,
r wnen
m in=
Jf
+
ma x 5jl
r4
,
4
1
990.
/
3jT
= -j-
.
when
(x;
= 0);
-TT=;
V
1
i/
= 2x
when
t
/
_ /"^"t
asymptotes, jc=0 and y
(cusp); asymptote,
wn ^n
-*
= 0.991.
+ oo.
992.
=
l(jK:
= e);
and when
points of
t=
j
xmin =l i/
m in =
andymln =i when
*
= 0.
Answers
$93. ds
=
d
u
cosa =
dx,
415
;
y
<*>
where
= x sin
tanh
a
998. ds
-.
997.
.
= 2asin~rff;
cos
ds
a
sin--;
cosa=
a
cos/; sin
sin
a
=a
sin/. 1000. ds
"(p;cosp=
cos
dx:
Z
Z
d^;
= cosh
cos
1002.
a=
-~-
.
999. ds
=
cosp
=
Z
V^l +(p
-
ds^
995.
.
tit
I/
(X
~3asin ^ cos/
62
Ka 2
c=-
2
d
ds==
dcp;
tp*
cos*
1003.
.
l
= an2z
1
zzz
K^36.
1006.
4^
/(-=
1010.
^"3^2"
1007.
1012.
.
both
at
1005.
.
=
"
vertices.
&2
; '
ds==
KB =
~
1011. (
^).,0,3.
(-!"*..
,=
.
a sin 2t
1016.
.
.
,
!023.
. |
^a
-ilfl,
(
2
3)
1019.
-8.
-^-
a cos
3
1017.
.
-pK
2
= ~(X
8
p)
)
R
and
sin
p
= cos 29.
K^ [
-f-
,
R^-
= |a/|.
1018.
1020.
-
^
1026.
1009.
1014.
-^-
f k2
d(p;
.
3 1015.
1004.
cos-|-.
.
=
/?
=
(2,2).
l025 -
-
(semicubical parabola). 1027.
_ 8
^c
8 ,
where
c
2
=-a 2
6
2 .
Chapter IV In the answers of this section the arbitrary additive constant
ted
for the
sake of brevity. 1031.
y
aV.
1032.
f
/
2x 3
+ 4* + 3*. 2
C
is
1033.
omit-
+
416
Answers
1037.
'-+--
--' a 2 x---a'x
1038.
J.
j/
<*
''---~. '*'---
JC .
3x 2
x
3x*
fi
1040.
'
13
4//2
1042.
+ 1043.3=
-
In -4= 2/10
1044.
= sec
2
-x;
b)
x
.
1045.
ln(x+
2
/4 + x
1048*.
tanhx. Hint. Put tanh z
b) x
1;
/"lO
-^~ln (x+ /x ^).
1047. arc sin
x-cothx.
^j~
1050.
a
'
_i
I
x
a
J Solution.
^
numerator
the
|.
^ 1063.
2
1065.
--^-.
|
cotx
1052.
we
1057.
c--l|.
Inlx
1059.
|.
~~'
a
3*- 2
]/
_Jiy(a_-
1062.
y.
1066.
1
^=
2 ).
In
/7 + 2
+ b + x V ab
+ V7 + 8x
1064. 2
1
~. .
j
fll
,,
.
-L arc tanx
=
get
1054.
.
1
I06li
:ln
In
1049. a)
.
Solution, f
In (2
1073.
2/2
* 1
.
1
1067.
Put
denominator,
~+
(x+W
/x +l.
Hint.
Solution.
.
1056.
1060.
"
r
1058.
the
by
-I' + T
j
.
sin-
r-j
.
Dividing
.
arc
c
1053.
1055.
x.
a
=
tan
arc
1046.
).
x=l
1051. a in
.
2
tanx
a)
{
s
+
1009.
/
1068.
x
1072.
/"2
arc
-~=
tan
arc sin x
/
Answers
1075.
.
5
^ In (a * + 2
1079.
y
V^=
31n|x +
2
1077.
2 ft
=l n 5
-
2
1078.
1080. -i arc sin
^.
^*
1). 1076-.
(
-ln|jt
arc tan
)-}-
417
^-
2
4
l
1081.
.
are tan 1082.
4
1083.
1|.
^(arcsmx)
8
V
2
1086.
\-e~
a
l
1090.
.
2 1092.
-
1084.
.
,/
-H-
r?
2
,
3 Ina i
1095.
In
1097.
|
e*
1|
i
+ 3)
In (2-
3 In 2 1102.
~
1098.
.
-
~
1109.
Xcos(logA:)
Hint. See
arc sin
HOI.
x
__
!. Hint. Put in
1109
lilt.
sin
2
jc=
~ tan
sin
(fljc
+
cos 2).
tan(x
).
1117.
-^-cos(l
^'.1119.
Xln
sin
-
.
1122. 5 In
sin
I
4-
1125.
In
]
1120.
. |
x
-^ cot x
ln|sinx|.
1121.
21n|cos/"x|.
1123.
1.
tan*
1118.
r).
ln|cosx|.
a
+
ft
1
1
2
--
1110.
tan 1
lOx
1112.
ft).
a In
1116.
+ ^)-
In
1108.
/"x.
~(1
008(0
o
ax 1113.
4
arc tan (a*).
jJ
1104.
e*
^-cos2ax. 1107. 2
1106.
hint
'
!n~5" 1100.
_^L_J. 1103.
2sni -^r.
1096.
.
4
Hint.
-Jr V
x
~
1099.
(a
r
1105.
e
2
1126.
^.
|sin
1124.
K
(a
2
ft)X
^
1127.
1
1129.
-
1128.
4a
-
1130.
l
o
sin
1132.
1131.
4
-^-tan -^.
U33.
-| 5
1134 .
_3^_lJL.
14-1900
M55.1
U 36.
tan
(
j).
+ 2au.) l(ln tanf |
|
23 Answers
418
4- In 3a
|
ocot3x|
6
tanh
1140. In
1144. In
I
-r-
sinhxl
-
ln(x
1152.
.
lr
1155.
1153.
.
In
tan x
|
fls n
n49
+
|
-
1
1147.
.
arc tan
J/
~
1150.
In
ln|x*-4x+
1
Incoshx.
1143.
ln|tanhx|.
1146.
"-
1148.
Xarc tan
1142.
*/(5=x^.
1145.
1139.- +-
.
D
2arctane*.
1141.
.
.
1138.
.
x
_. -------
1
1137.
.
| )-
]/ 1151.
.
x
-
.
sec
1156.
.
1)
'
1157
.
*. 1158.
arc sin (x 2 ).
1159.
.
&
a
in
1160.
taniw-x.
a
/(^ Vx
2 In cos
1165.
|
1172.
e
sln
1168.
.
1169.
1170.
^
-l^arcsin^A
1173.
1175.
.
\n(e*
+
^21n
1171.
.
tan
1167.
tan^-l.
|
K.
=.
-i-ln
1166.
1
x
a f
1181.
/"l <40 -
1187.
-
.
2 jc
.
- 1=r: 7
1182.
1185.
arc sin
ln(secx+
.
arc tan
/tanx\ I == }
.
1183.
-2cot2x. 1
2
l/~sec x-i- 1).
1186.
1
.-
2).
= In
-p=.
+ 2arctanx.
ln|x|
1174.
Ve*x
tan
jr
ln|tanox|.
arc cos -^
1184.
V 5 -f
sin
2x
sin2x .
Hint.
2
v
= x
r
J
1189.
s inh(x
.
1190.
-Latanh*.
1191. a)
-~=arccos^
when x
> ^2;
b)
In (1
+*-*);
Answers
(5*'-3)';
c)
-
d)
2arctan \e*
1195.
.
V 2x4-1
aarccos
2
arccos
1204.
if
,
x
> 0,
and arccos
In*
1201.
-.
-~ V2~x -^ V2-x
1202.
.
1196.
1.
Putx =
Hint.
.
TT arc sin x.
ln(sinx+- J/"l+ sin**)-
e)
1
1194. In
419
In
21n Inx-f |
--
|
2 .
(
~J
x<0*)
if
Hint.
'
Put
x
= ~-
substitution x
=
be used in place
may
~ x arc
1216.
sin x-}-
V
x2
1
1217.
^r~-
1214.
.
^T^Y-
the
of
^
(9x
2
f
In (1 -|-x
x--^ *
xcosx.
sinx 12 18.
x arc tan
3*
st
1215.
Note. The
substitution.
trigonometric
1212.
x.
.
^ o
+
igf y
^
of undeter-
xV*dx = (4x 2 + Bx + C) e*x
= (Ax + Bx -f C) 3e + (2Ax -f B)e9X 2
3Jf
.
* and Cancelling oute equating the coefficients of identical powers of 3
122 B=
=
whence ^
;
-^-;
=Q
(x)e
ax
where P
,
polynomial
Hint.
*) is
).
after differentiation,
X 2e* x
a
2
6x-f- 2). Solution. In place
by parts we can use the following method
of repeated integration mined coefficients:
or,
xlnx
1211.
a|.
1213.
1206.
.
good
See
rt
of degree
Problem
c=
2f
= 3C + In
the
x,
we
get:
,
C
6eneral
^ orrn
\
p n to e ax dx =
is the given polynomial of degree n and Q n (x) is n with undetermined coefficients 1219. e~* (x 2 -f-5}
(x)
X
1218*.
1220.
-3e
T
"(x-h9x
2
+ 54x+ 162).
Hint.
See
Henceforward, in similar cases we shall sometimes give an answer thai for only a part of the domain of the integrand.
Answers
420
Problem
cos 2x
mined
-
1221.
1218*.
Hint. It
where P n (x) polynomials
px dx
(x) cos
= Q n (x) cos PX + R n (x) sin
X
arc sinx
+ In
|
sin
3
-|-
x
|
1236.
x8
f
Inx
l
O^2
x
-^
(x+ V^l+x
Ay2 ^Ar
x
)
tan ~^ 2
1
arc sin x
-j 4
2,
2
2
e* (sin x
1232.
X
xcotx-f
1230.
.
'
2
1228.
.
x
.
sinx
-
cos x)
x2
V
x
ft
,
+2 V ,
-
get
+ -^ In (1 + x
rt
2
2
1246.
1+
- -- -- ,
In
cos 2x
I
x
1
]/T=Ixx
x
""
xcos
.
2
^
4
'
x2 x
1
1
1
\
and
+2 V
x
+
.
1248
'
(2 In x)
+
'
2xsin (2 In x)
10
a=x
1
du = dx
2
X
x (arc sinx) 2
1244.
).
xtan2x
._
1247.
V
X 2
^
V
In x 2 In x ~^-^--
^ arc tan 3x-^ + ^o In(9x +l). 1243. ~-^X
1242.
.
/ cos2x -2sin2x
1240.
1
x.
fy-*
2
x
1
2
1238.
-1).
Inf-^-x. +X
1245.
2x.
1237. 1,
1239.
x arc tan x
_
,
x2
-/m -_
_
+ -3x.
1).
arCMn *
Xarcsinx Xarcsin
1234.
(x*+
[ln(lnx)-l].lnx.
X (arc tan x) 2
we
1225.
i
.
1250.
1218*).
"
cos (In x)J.
X
x In
1229.
x*.
1238.
1241.
(x) are
ln *
x
arc tan x
1227.
.
1231.
.
and R n Problem
(x)
(see
ij(,
2
l/T
-|-
undeter-
of
|
Qn
2xlnx + 2x.
xln x
1224.
4
/"xlnx
2
method
the
apply
the given polynomial of degree rc, and degree n with undetermined coefficients
is
of
^-Inx--^. y o
1226.
1222.
form
2
1223.
2x
also advisable to
is
coefficients in the
Pn
sin
4^8
=
*
C
Whence
.
^ ( x ~r
x dx
and
x*dx
d
=
2
2(x +l)
1)
125K Hint.
xy a
8
^x* + -s-
a=-
^
Utilize
arc sin
.
the
1
identity
Solution. Put
j..
x;
we have f }/aa
a ~2
2
~
1252.
= Ka x and -x'dx = x ^a -^-
\
r
xa
-
f -/-^2 \
J
Ka
=. ,f x 2 dx+a2 l
dx
,
J
r
x
whence
2
-
-;
a*
x 2 ].
^a2
Answers
Consequently, 2
y~a*x
\
~
l257
-
=
-
Fn
arctan
x*
2 -fa arc
Problem
1258
TTT
2
-ln
1259.
.
2
~.
sin
1252*.
Problem 1252*. 1255.
Hint. See
.
Va
See
Hint.
.
arc sin -| o 2
dx = x
2
421
4A
^~-
.
/ f ~
ln( *
a
2
1260. x
2).
-
y
arc tan
7=-.
y
7
Larcsin^P^. ^
,262.
x-f31n(x
1261.
l
2
arc tan
7
1263. arcsin(2jc
1).
2
n
(jc
z
3
--T
~
1256.
2t
arc tan (x
(x
-~
1254.
arc tan
T
'
~ V& +**
1253.
+
(A:3).
1264. In
1/2
2
1265.
-2 V -x-x* -9 arc sin
1266.
4jc-f5.
1
=^
.
1
In
r
1267.
'71
5/"1 1268. In
.
?=. 1270. arc
arc sin
1269.
x arc sin
1271.
.
"*
,273.
9 arc sin
-
(
+ 2*
2 In
1272.
+
sin -
y$
+5).
1
1274.
8
2x4-1 -
arc tan
1276.
o
o
1278.-ln|cosx + 2+ ~'
y\
1279.
1281.
jc
In 2 x
4 In x
+ 31n|x
1283. In
l
2 arc sin
l!L~.
1280.
o 3|
.
31n|A;
2|
5*
+ In
1284.
^
1282.
*
2
(x-4)'
- -b
x \
+a
I
.1285.
1+*
(*-!)' 1286.
2
30
27 1288.
1290.
11
^
4-j 1289.
=
2U-3)
-^-r .
49
U
5)
49 (x
+ 2)^4*
x~5 x-f-2
x
l
1292.
1293.
ln|
x-3|-
In
U~l + [
In
X
Answers
422
X arctan(* + 2). x*-x V
x
,
arc tan
2
+
In
1300.
+ji-i*-
_|arctanx- TT^-jT -^
1305.
x
1304.
1
2
5=41.
l
f
arc tanx.
In (x
5
x
1310.
y ln|* +H 7
ln|x|
1
1
o
2
FT
^/(a* +
6)']
1315.
56
^
X -6
-3
In
In
1324.
(A:
1 |
2 arc tan
-J-
~
x7
1)
.
tan x
_ J_ J
'
1
Sx^Sx*
x
1
2 arc tan
1317.
-^=.
1322.
I)
7
arc +1)-^o
.
^3
Xarctan]/|-.
=il.
3 In
7
7(jc
3
(l+/l.
3^/1-6 ^/"x 1320.
l)"
(A:
._-
1
s
4(x
9(jc_l)>
= (*
1
arc tan
1312.
1
1313.
1
1 . i
.
Put
Hint.
.
1
1
*+l
tan (x-^1).
4
.
1311.
2
1306.
2
X[2
1303.
4r (8
+ 21n
-
|
1
'+4^
arc tan
1298.
1299.
.
1302.
x
arc tan x
,,., 1297.
In
1295.
.
1
K"-\
+ arctan(*+l). -
= arc tan
+
In
1294.
-
1316. 1318.
1319.
+ J/l + 6 arc tan |
*/l.
arc tan
^1-x.
1323.
-
(jf_2)+-I
In
/
where
423
Answers
=7+l).
1327.
-
*-li
.
-+i
1329.
-arc sin -^
1330.
1331.
.
4/7=1
-* + l).
-retan
.
*
-2
-cos.,.
(*
1340.
1334.
where
.
1337.
1332.
+1)
-
,336.
.
2
1 .
1338.
sin
x
1339.
g-sin'x.
-cosx+-jtos>
x
1342.
*J^-**. |841.4eo.|-^|.
>-.-.+ --^=
cot*
i
_ 7 i_sin6x. 144
1347.
__-. 35
,849.
-1
21
._. ._
, 344 .
3 In
|
tan x
cot
jc
Jt
1348.
.
tan x
tan + -^ o
+2lnt.n.
|
+ To tan
x
5
x
1351.
^.
1350.
8
1353.
.
COS 2
"[
ln
,357.
3 cos*
*+i
t,n|
_S2*lf_in|sinx|.
.
1354.
-
1358.
4
cot'
x
sin 4 A:
In
8
8siu*x
+ cot x + x.
1361.
.
1362.
xlB
-1 J/55?x + -| J/cSS*-^ J/cw
SrRj{"^ .
-cosx.
,366.
.369.
-
arc tan
^' .
.
1367.
,370.
11
-tan! --f
1359.
-
!. 1363. 2 J^twix. 1364.
where I
" iain
-+3-.
-
1365 '
.368.
lE_^M+
J-J^
cos-
137 ,.
!!H +
Answers
424
5*
sin
sin
7*
^ Jo
,
tan
-!-
1374.
+
[
tan
In -f *
1373.
1376
--
2
tan|
cos 4* --5- cos 2*. ^ cos 6* -^ z4 o lo
1372.
.
'
* In
1377.
1378. arc tan (l \
.
-
tan
We
+
3 cos x Solution. Whence 4- P (2 sin* -{-3 cos*)' 12 5 .,, Q
a= !3' r
I
*
iiC
-f-
arc tan
1381.
5>
f
have
lO
-^^
fraction by ccs 2 x
,
= j2^_
<.
In ||*--^ lo id
+
2 cos x
a
===
+ 2p = 2 p3sinx + 2crs*
= 3,
3^
(2 sin x
3a
and, ,
2^77+3^7^
J rfx
Si
J
1379.
/
3 sin x
put 2u
We
P== -!3-
+ tan 4V ^
+
2sin *
|
+
+
3 cos x) consequently,
12
==
13
5 2
ln
.
|
1380.
In |cosx
situc(.
1*3
Hint. Divide the numerator and denominator of the
)
-=- arctan
1382.
.
See
Hint.
.
V*
V
Problem
See
Hint.
Problem 1384.
1381.
1381.
4 In x
1
1387.
V2
2
/" 2 -f
sin
Vi
sni2*'
~~"
/\
r~~~
/"
t388.
4-ln^^^.
arc ian
1 .
1
Use the identity
sin*
2
sin*)
sin
In
cos*
-.
r cosh * !
1
1396.
.
,
2coth2*.
.
12 or
-j=. O
arc tan
/
3
1394.
5)
.
+m+ 1401.
I
r
-JT-
In (cosh
1397.
V
If
sinh*
*
*)-^^ 2
1399. arc tan (tanh*). 1400.-~=.arctan
* iA-?\l
(f V
Hint.
2
'
* ,
.
cos*
* *
-j-sin
cnth 8 *
coth*
1398. *
identity
+ 21n '
~
x
tan
1393.
tanh
the
arc tan
x+cos*
1-j-smx
^>
1395.
*
1390.
sin*
3
Sm 1392.
Use
Hint.
1
1
sit
(2
=
1389.
2
sinh * ^ *
sinh 2x r *
*
^^
5 ,
A
.,
..
..
...
Hint. Use the identity
J
,
cosh*
-^ (sinh* %
-J*
cosh*). '
1402.
-7=1 In (/" 2 cosh* 4-
^2
^ cosh 2*).
425
Answers
1403.
~ ^2 + x + In (x + 2
1404.
.
H06.
1405. -
J/"^IT~2
1407.
|~
_
1408.
3+ Kx
8ln| x
2
In
6x
[
2x
+ +2 1
1410.
7|.
|
x
+ yV
^ VV
1409.
.
In
r
^
17)
F
4|*
6x
7
I
JT
97
**
I
1413.
--Uarctan. 2
-54-~']
1416.
:
1419
1423.
4-
+
4 sin 4x4-c< c s4x:\
sin
_~Z + o
._.
142
*! a1)
V>
2*
+2
|arctan(2x+l). 1436.
,.
.
V\+x*)2 V\
arCCOS (5x "~ 2)
/.^-x'.-
2x+10-9...
4 arc tan (x .434.
n cosxsin -'x
M
l428 '
1430.
...
1432. In
-
~
X
.
arcta " +,3 2?
tanx.
e*
~
1425.
X
sin2x
-
xln'(x+
1424.
-
1426
-TT (2
1422. x
lnk*-l|4--O
-25.v'-3.
Y*
1415.
1418.
2
'
2
-
1417.
1421.
1412
L=ln
1414.
1-x
2
1411.
.
1).
1431.
1).
_+
1433.
j In
.
1437.
=
1435. 2 ,
.
arc tan
In
x'
n
1
Answers
426
l-x 2 2
^
'"
'
2x
.
arc tan
1439 l439
x
-
1
1
-^r
1440.
.
3 1442.
l"\
In
/2X-45VJ/(2jc)
1443.
.
/
J/5^x
(
x 2 4T,'
4 In
I)
^lVr>_ 9 _ arc sin (8*
(8x--l)
1458.
----
arc tan
,r-
where
,
2
cott-
1462. *
cos5<
1463.
3cos5x tan
T^ In
~sin2x.
1467.
X arc tan
.
1
+
+2
-
In
+ secjc|-i-cosecjc. 2
A.n|\ -^ ). >
ax
y 2
1473.
In
|
tan *
'
+
r-^-^.
2
ax).
1477.
1475.
^uu-1).
*
-
~ (cos
2
x
|~cot
2
*-
cos 2
6)
t^ +
tanS *
1470. arc tan (2 tan
.
an
2
1472.
^"tan
x tan 1478
x+\).
T -r=X 1
-^= xarc j/
+ 2+ -
tan x
|
,465.
.
-= arc tan
1469.
hi
In
1
ln| tanjc
Xarctanl
X In (sin
tan 2
^
.
1)
= 1461.
2
J/JT^t).
1
1453.
1471.
+
(1
1450.
.
1452.
1466.
.
J/x+1
-1
1448.
X arc sin
--5-^
1444.
.
2
2
1446.
5
tan
3
2
3*+--
V
\
?=?
|A3
tan
.
1479.
y
2
^74. J_ x a
In |cos 3x
le-.
/
/
|
.
1476.
fin |O=5-
427
Answers
1480. 1483. .n
.
J481.
^-?
+ ln
|
sinhxl
-
1485.
.
4- 1" cosh 2x. 1487. 5
1486.
x.
\
gL-^ + lln |e*
1488.
.
r
|l+cot*|-co tje .
1489.
2|.
*coth*4
i-arctan^^-
-1 */(?
1490.
x^-'+fir 1494.
In
1496.
(coslnx
lcos5*~
-
t499
'
+ sinlnx).
|sin5*V
1497
1498.
f
x2
2) arc
tan
.
1 [(^
y x sin 5x + 3x cos 5x + -|
^^r
'
1500.
In (2 -' ,
Chapter V 1501.
vjg~.
1502.
a.
fc
x=\
Hint. Divide the interval from tervals so that the abscissas of
metric
progression:
sina
Problem
See
Hint.
xl
x =^l,
the
=x q
L
in
=
2)
a
at?
1512.
-f- cos x
x
1,2, sum
-. x
5-^ = 2 sinh*.
**
1516.
gral
.
in o
/i
cos
_
1513. *
=
1517.
.
.
.,
xn
= x$ n
Hint.
1
)
-
'
'
n
.
Solution.
1\
+ ~~T" J ,
'
formula 3da
1)
=
1511.
1518.
1
the
1508.
.
1,2,3, ...). 1514. In
sin x.
In.
1506.
.
Utilize
(
/
156.
the x-axis into subindivision should form a geo-
cos n-}--^ a ^~ 2 V 2; J
1510.
1/1,2, ^r-^V^=+ 7T +
X on the interval
of the function /(A:)
1505.
on
cos*.
1
In*.
1509.
x b q*,
_l
2sin
=5
points of
x2
t
1504.
3.
to x
1507.
1505.
+ sin2o+...+sin/io =
- r -;
1503.
2.
The
be
sum as
4
mtc -
lim
Therefore,
[0,1].
.
o
.
may
~~
1515.
n
sn
-*oo
i
-^ 2
-
1519. In 2. Solution.
.
--o-+."H-- \ may
M\ 1+1arH [
The sum
i+Ji n
1
+
J
n /
s" n
= -i- H n+1
/i
\-^ + +2
. . .
H /I
be regarded as the integral
+M sum
=
of
428
__ Answers
=
the function /(x)
form
the
t520.
p 1526.
xk
=l+ n
= arc tan
tanh
; 1
1552.
1555.
-=.
1556.
=
-
1524
"T 4
^2- =
f J
1
-
1525
^r o
-1=
-4o
-
arc tan 2
~.
1533.
.
|o
15 38.
~.
1543. sinh
^-r
if
,
1557.
Diverges.
= -i-
p^l. p>l;
~
2ji.
(*
1549.
}
if
1547. Di-
p
:
f(x)dx+{ f(x)dx where
/
t
(jc)
= x^-
.
1564.
8
l + ~ln3 3
1
x)?"
(1
n.
Diverges.
1565. 3 1569. Converges. 1570. 1566. Diverges 1567. Converges 1568. Diverges verges. 1571. Converges. 1572. Diverges 1573. Converges. 1574. Hint. B(p
1563.
~. 2
1554.
1559.
T
.
e J
1550.
Diverges.
diverges, 1558.
1546. 2.
=
cosh
1
2 \
+ -|-sinh
Diverges.
t560. ,-!-. 1561. Diverges. 1562. -i k Ina
1
-
~.
1534.
1539.
In 2.
4
^-
if
diverges,
1553.
1.
tang
1545.
.
In 2.
H-*
1529. arc tan 3
1
-^-
p
if
,
1537.
1542. arc
(In 2)
p
1551. Diverges.
1532.
^.
%-+\-. 4 o
^7= + ir. 9^36
1541.
1548.
1531.
.
1523
-
35~-321n3.
ln-|
=
o
oo
1528.
.
1536.
tanh(ln3)
verges.
j
sn oo
5
V
1
O
lim
Therefore,
= 334-
152
1522.
.
In
1530.
.
~ In
1544.
2 ..... n).
!,
where the division points have
[0,1]
n ->
1527.
.
1540. 0.
=
o
-|- 1
-jln-|
1535.
on the interval
(&
~
1521.
.
j-7
--
--
1 ;
4
since
lim
/(x)x
1
Con%
q)=
-^=l
1
o
and lim
j
^)
(1
"^/(x)=
1,
both integrals converge when
p<\ and
1
q
1
-> i
that
is,
when p>0 and
-x p
~l
Hint. ?
1575.
(/>0.
(p)=\f (x) dx+(
f
where
(x) dx,
1
e~*. The
first
integral converges
when p>0,
when p
the second
is
IL 2
1577. 2 /~~2
arbitrary. 1576. No.
("
In
2
/7rff.
3
1578
00
r
1580.
1584.
C^f1+' f !?"**#. 1 J
2~
,
1585.
2
1589. 4
n.
1581. x
-^
1586.
1/5 1590.
~ 1" o
= (6
a)/+a. 1582.4 -^
2 1^1
H2.
1591. In
.
4-^
^-|y
1587.
^
.
21n3. 1583.8
1~~. 4 1592.
1588.
~ + T^
4
^JT. 21^3 /""S-
1593
-
3
nr o
Answer^
1594.
--
-Z
1599.
.
2
1
I
600
^rr^ o
1601.
1-
-
429
+ l^. 4-(^ 2
1 6(>2.
1603.
1.
00
1604.
z
.
.
2
1605.
z\
I
b z-
606
r(p+l)=C
Solution.
-
we put *P
the formula of integration by parts,
~ du = px. p l
e~ x dx
U,
v=
dx,
xPe~*
dx.
= dv.
Applying
Whence
e~*
and
If p is a natural number, then, applying formula account that
we
p times and taking into
(*)
get:
"
1>35
(2fe
2-4-6
2 ,
n=
ii
2k-\-
number;
an odd number
is
1
even
an
is
246
2k
1-3-5 ... (2k /
t608
.
W
^7 V
...
9
128 _ "
1609.
-
.
/
'
3l5
_ 63jt 10
~"
^,
^B
'
512 Hint.
'-Hr^
Put
1610. a) Plus; b) minus; c) plus Hint. Sketch the graph of the integrand for values of the argument on the interval of integration 1611. a) First; b) second; c)
1612.
first.
1617.
Hint. 1624.
~
2
1618.
The integrand 1.
1613.a.
1625. --
|-
10
4O
1639. ab[2 1641.
2a2
m
1636.
2
In 3.
~.
.
In
1642.
(2+
2n + -
and
1631.
.
na8
1640.
.
-g-
.
1643.
6n--|
.
1620.
t638
-
1632. -~-p
2 .
.
s
= 32
1626.
4~
.
1633.
4-i
.
1623
%< l <-^-12.
0
J
= 2(coshl -1). ^+~~2 6
n
/~~3)).
2
-^a
naa
~~4-O
1637.
15n.
1644.
See Appendix VI, Fig. 23. 1647. a2 f 2 1648.
^jt
of the sign of the function.
1630.
O
2 arc sin -i.
1616.
.
-|
1619.
.
Take account
Hint.
1635. 4.
V"l
1615.
increases monotonically. 1621.
1627. 2. 1628. In 2. 1629. 1634.
~.
1614.
1049.
Hint. See Appendix VI, Fig. 27.
yln3.
+ ~-V
^ji-i|-?
1645.
Hint. See
and
1.
1646. 3jra2 .
Hint.
Appendix VI, Fig. 24.
~ n+ p
.
1650.
Answers
430
1651. 3na2
2
1655.
-|na
Hint. See
.
^
Pass
Hint.
1664.11)^2.
VhT^tf.
1667.
l/T+ln(l+/n
1673.
1674. 2a
8
1687.
1691. 1696.
b)
(5
t;^
= ~; &
^(15 2
6jia;
,705.
+4
c- 2 ).
av
16 In 2).
2
^. 1710.
~a o
s .
2
O
na 2 /"p^.
8
V5
1)
2n6 2 -f
_
(eccentricity
-
of
Ao "P
1
ellipse).
8 .
~
1695.
1U 1701. a
10
1703.
-na.
1707.
}.
nabh (
1704.
4U
1708.
+ -^oC /)
1
.
nabc.
1713. 5
1715.
1717.
Do
1719.~rta 2 1720. .
^tfx*. 8
~nh*a.
1699.
no*.
1690.
-
1694.
.
1688. -1
.
lUi)
1712.
nln. +
+ 4) = ^(2 + sinh2). z 1722.
*.
i[5^T-8].
=
torus.
^^. z
na'.
1721. 4n*ab Hint. Here. y b external surface of a torus; taking the
face of a
Jia
\
n(5-^T) +
17l8.^(e*-ff 4
1711.
=
je
0]. 1682.
^
1685.
In 3).
10
1706.
+l).
1676.
.
V2" + ln ( 12"+
1693. 1698.
2
1678. I6a. 1679.
-
1689.
.
J
,
1
1716.
2
1702.
+$ +fl 6).
DOT -!;
1714.
2n a.
1697.
l
+ *(x* ^ 4-JW &
2
(9n -16).
c)
-Ji
= 2n. 1692.^^. o
5
1709.
1688.
[
1672. ~-(e
fc=ln
4- [4 +
1684 .
ln(2+VT)
+
^^
1680. 8a. 1681. 2a
JQ!.
, 68 3.
^(10/161). a= 1.
sinh 2
cosh q
ln~
1675.
Fig. 29. 1677.
.
.
1665.
2
V'"?11!). 1671.
+
/T".
Appendix VI,
See Appen-
See Appendix VI,
Hint.
coordinates.
polar
4
Hint. See
~.
1659.
the
the lo P
1668.
1670. In (e
l-f-lln-|.
aln.
to
For
-
See Appendix VI, Fig. 22.
1656. 8jtV. Hint.
Fig. 28. .
Hint
fl2 -
0<+oo
1658. a2
.
-
1654.
.
Hint. Utilize the formula
1666.
1669.
6na2
1653.
Appendix VI,
1657.
Fig. 30.
VI,
+ 2ab).
varies within the limits
t
parameter
dix
1652. n(b*
.
-^(e1
Taking the plus sign, we get the minus sign, we get the internal sur-
arc sine; 2) 2na 1723.
a)
o
2
+
-\
81|-^ In
,
where
~~~8
b) 16n
2
a 2 ; c)
~ a
Answers
~jia 2
1724.
1725.
.
2
2ita
(2-
2
J=y = y.
M x =M K =-~a
1730.
a 24-sinh2
x
;
1729.
= y = ~a.
-
1737< *~=Jia
^^SB-
I736t
^ss-
2
<
_
2na 2
1731.
1732. x
.
4
.
^=|-
;
M x =- V M X =M K = ^-
1727.
.
M & = ^.
;
asina
._00
na 2
1726.
/2~).
M = ~-
1728.
43 1
a
= 4a
,__
1738>
-
Divide the hemisphere into elementary spherical slices of area da by horizontal planes. We have da 2jiadz, where dz is the altitude of a slice.
tion.
=
a
az dz
2ft f
Whence
z=
2nfl
tance of
-j-
=
o
= z
Ct
Due
-?r.
2
from
altitude
symmetry, x = y = Q.
to
vertex
the
the
of
cone.
At
1739.
a
dis-
Partition the
Solution.
cone into elements by planes parallel to the base. The mass of an elemen2 tary layer (slice) is dm,YftQ dz, wnere Y * s * ne density, z is the distance
=
the
of
from
plane
cutting
vertex
the
of
the
cone,
Q
= -J-Z.
Whence
h Jt
z
\
n
;
=
=
3
3
/
1740.
h.
f
0;
0;
+-g-
_^
\
a )-
Due
Solution.
to
symmetry.
7
we partition the hemisphere into elementary #~=0. To determine layers (slices) by planes parallel to the horizontal plane. The mass of such an elementary layer dm ^nr^dz, where Y s tne density, a is the distance 2 2 is 2 the of the cutting plane from the base of the hemisphere,
5T-=
*
r= ^a
a
2
z
[a
radius of a cross-section.
11 "33
1742. I a
1745.
= -z-ab*\
/=yjt
concentric
the
moment
s
/?J).
The
circles.
of inertia 7
have:
z
)
zdz o
=
3 =-Q-a. 1741. /==jta
11
4""
I b =z-^a
(#J
We
2
4
b.
1743.
7=-/?& 8
.
1744.
15
Solution.
We
mass
each
of
= 2nCrdr =
partition
such
y n(/?*
/
= JW& "4 --
the
8
r
a
ring
element
;
IW b
into
= -T na*b. 4
elementary
dm = y2n.rdr
#J);( Y ==1). 1746.
/
.
=
and
^nR*Hy.
Solution. We partition the cone into elementary cylindrical tubes parallel the axis of the cone. The volume of each such elementary tube is dV 2nrhdr, whe r e r is the radius of the tube (the distance to the axis of
to
the
=
cone),
h
=H
[
\
s-
)
is
the altitude of the tube; then the
moment
of
Answers
432
/=Y
inertia ia
cone. 1747.
/
2jttf (
=
2
Ma 2
-
5
cylindrical
1
volume dV
*
dr
^lr^
the
Then the moment
where y
is
radius of
= |- Ma
lows that y
of inertia I
_
2
V--=2n 2 a 2 b\
1748.
.
= 4nay
I/
I
S--=4n 2 ab.
4 f/^-rr
9
b) x==(/
= T7-n.
sen
that
1U
so
centre of
1750. a) x^=0,
the
the
circle; b)
^=4o
cone obtained from rotating
where b
is
the base, h
rem, the same volume of
trom
gravity
is
V
1749.
= ~^/?
2
//
2
a2
=A
na'v,
a)
fol-
ita'y, it
o
7-=//~=4 o
Solution.
The volume
about of
its
the
base,
is
x
\\hence
base.
of the solid
is
triangle.
where x
-z-b'i.
and the origin
diameter
the
1753.
x=
The elementary
Hint.
a;
the
=
double
a
--
equal to V
rft/i
o
By the Guldm
distance
of
the
is
force
of
(force
,
centre
tlu>
2
H -
D flt,=
2
theo-
1751.
v
vt' ^
t
.
2
i
-lnl+--.
1752.
-
The coordinate axes are cho-
Hint.
the altitude
the
1
"
2
71
with
a triangle
2Jtx
I/ f
r
r
o
coincides
jc-axis
/
4
M=
o
---
elementary
L./-*dr
1
of the
An elementary
h = 2a
a tube,
and since the mass
e density of the sphere,
t
densit y
into
sphere
a is its altitude.
the
is
Y
the given diameter.
is
the
is
r
where
'
partition
which
axis of
2nrhdr, where
=
We
Solution.
.
the
tubes,
r
^)
\
-f
1754.
is
gravity)
the
equal to
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
weight
= ynR
1757.
of
2
water in
dx,
A=ryR
2
H
2 .
l^&
1759.
A = ynR*H.
on
mass
a
m
is
1760.
equal to
of the earth. Since for r
1758.
^=
F= k
=R
A ---^ R*TM
^
tn0h
j-
where
,
A=
When/i = oo we have A
f
The
7
kgm.
force
acting
the distance from the centre
r is
F=mg + R
m^-
Solution.
/lo.-mg/?.
;
we have
sought-for work will have the form
^Q 79-10* -0 79- 10
it
follows that
kM=gR
2
The
.
h.
\
J
kr-dr *
= mgR.
1761.
kmM 1.8-10
-
{ -^-
.
,
)
\KK-\-nJ
4
ergs.
=
Solution.
__ Answers
The
7
force of interaction of charges
is /
433
= ^i
work
dynes. Consequently, the
-J-=S Jd.x
= *o*i(/ --- \ = 1.8-10 X X / isothermal process, pw = p 1
1
*' 4
)
*
4=800 nln2
1762.
ergs.
kgm. Solution. For an
2
\
tV The work performed
in the
expansion of a gas
v\
from volume v
volume v
to
A=
is
l
\
J
p di>=p
In
i>
^
A
1763.
.
UQ
an
For
Solution.
adiabatic
the Poisson
process,
i>i
*^1.4,
holds
A=
Hence
true.
=5=
15,000
kgm.
^0
( J
law
pv
k
= pj>^
where
t
k
P
-Ms -^dvk k v
I
4
4=~jiu.Pa.
1764.
the pressure
Solution.
on unit area
a
If
of the
the base
the radius of
is
support p
=
p z
The
.
3ta a
by
frictional
.
forces
Therefore,
on
a
of
revolution
complete
A=^x
work
the complete
force
a
-^~rdr. The work per-
is
one
in
ring
frictional
2uP
ring of width dr, at a distance r from the centre,
formed
them
of a shaft,
f
r*
dr
=
is
JifiPa,
a
Solution. j M# = ^~da where dK=^L2
1765.
2
.
t
from
the
axis
of
The da
rotation,
kinetic
an element
is
the
is
Q
of
a
of
energy
particle
area, r
surface
of
the disk
the distance of
is
= -^j.
Q
density,
it
Thus,.
R
Whence 1767.
K = ^/?
2
co
2
= 2.3-10
K-^J,-*-^ 8
kgm. Hint. The amount
to the reserve of kinetic energy. 1768.
1770.
P = abyxh.
1772. 633
Igm
1771.
P = ^-_
1773. 99.8 cal.
p = -^-. o
(the vertical
1774.
M^~
1769.
,
of
766
work required
p^fo +
component gf
* = | X M*W.
.
fe
^fr)
b
.
1776.^-
__ 1777 15-1900
Solution.
n
-
Q==
cm. 1775.
^^-^ 2
3
10>r
(k
is
the
a
Q= (v2nrdr = .
^n
equat ,
directed upwards).
is
a
gravitational constant).
is
a6
2 f( a
8
UJ A Hlnt
-
r*)rdr
Answers
434
along the large lower side of the rectangle, and the y-axis,
perpendicular to
ua it
in
middle.
the
S=\J
Solution.
1778.
a
on the other hand,
dv,
7^=a, at
vt t'a
whence
1780.
=
dt
dv,
M*=:
\
and
acceleration time
the
consequently,
(xt)ktdt + Ax=j(l'x*).
/=l- = S.
is
1781.
Q =0.12 TRI*9
S=
-
Hint.
cal.
Use the Joule-Lenz law. Chapter VI
V=
1782.
2 (i/
x 2 )*.
1783.
(x
+
/"4z
t/)
Oje//
1786.
/<*,
x2 )
= \+x-x
replace
~
by
x.
_,
Then .
;
It
z
_
1=^;
1789.
f (x,
y)
in the
form
=
T
4
^ Ri
Solution.
o~~^
= x~\+V"y. * === (a
In
Hint. 2
~|-
1
)
and,
Solution.
the
f
J
x
2
Then
and
y.
+1
and
+ y~u,
1790. / (a)
*=1+/(VT
/(w)=:w
When *=1
a)
+
= l/V
Designate
v, x
identity
hence,
-
z
f (x)
/(-=M= I/
remains to name the arguments u and
then
1788.
.
(xy) 2
UV
-
^=
=
.
Hint. Represent the given function
/?
1787.
2
-}-3
_...
-f. j^S.
2
2
we
1791.
-}-2a.
have
the "
1)
put
/(
=
identity
/
Single circle
with centre ai origin,
=
^
=
*; c) halfincluding m the circle (x y 1); b) bisector of quadrantal angle # #==0 (x y>Q)\ d) strip contained plane located above the straight line * lines ( between the straight lines (/= 1, including these e) a 1 and /= 1, includsqua're formed by the segments of the straight lines l
+
+
K#<1);
#=
(Kx^l,
(x^y^x
<
=
+
y=
t
*<
>
x
=
,
+
_ __ Answers
=
435
~y > 0); ,k) the entire j/-plane; 1) the entire *t/-plane, x*(x* parabola y with the exception of the coordinate origin; m) that part of the plane located above the parabola y* x and to the right of the (/-axis, including the points of the t/-axis and excluding the points of the parabola (*:^0, y > V x)\ 0; o) the n) the entire place except points of the straight lines *=1 and t/ x2 + y 2 n (2k + 1 ) ( = 0, 1, 2, ...). family of concentric circles 2n k 1793. a) First octant (including boundary); b) First, Third, Sixth and Eighth octants (excluding the boundary); c) a cube bounded by the planes x= 1, 1, including its faces; d) a sphere of radius 1 with centre y~ 1 and z
+
=
=
<
<
1794. a) a plane; the level lines are at the origin, including its surface 0; b) a paraboloid of revostraight lines parallel to the straight line *-f */ lution; the level lines are concentric circles with centre at the origin; a hyperbolic hyperbolas; paraboloid; the level lines are equilateral c) d) second-order cone; the level lines are equilateral hyperbolas; e) a parabolic 0; t/rf- 1 cylinder, the generatrices of which are parallel to the straight line x the level lines are parallel lines; f) the lateral surface of a quadrangular pyramid; the level lines are the outlines of squares; g)_level lines are parabz C ]fx i) the level Ifnes olas y^-Cx h) the level lines are parabolas y 2 2 2*. 1795. a) Parabolas are the circles C (* (C y ) 0); b) hyper2 2 C 2 d) straight lines y ax-{-C; bolas xy^C(\ C 1); c) circles jt -f*/ lines y-=Cx(x^Q). 1796. a) Planes parallel to the plane c) straight 0, x-\-y-\-z^=Q\ b) concentric spheres with centre at origin; c) for u one-sheet hyperboloids of revolution about the z-axis; for u 0, two-sheet hyperboloids of revolution about 2 the same axis; both families of surfaces 2 za are divided by the cone * 4-r/ 1797. (u 0). b) 0;c) 2; a) 0; k not limit does limit e does not exist. Hint. In Item(b) exist; f) d) e) coordinates In Items and to consider the variation of x (f), (e) pass polar kx and show that the given expression and y along the straight lines y may tend to different limits, depending an the choice of k. 1798. Continuous. 1799. a) Discontinuity at je 0; 0, y b) all points of the straight line of line of is the circle x c) (line y discontinuity); discontinuity 2 the tines of the coordinate axes. l; d) discontinuity are
=
+
\
;
=
+
|<
>
y^Cx*
=
=
;
>
<
=
=
\
=
1800 is
=
Hint. Putting y
=y
l
continuous everywhere,
when
=
f/ 1
q^M^O.
^0,
2v u (y) = is
since
we
for
get the function (?,(*)
yl
^
discontinuous at the point
=
the denominator * 2
when
Similarly,
jt
From
everywhere continuous.
is
2
function z
=^ const,
(0, 0)
= *, = const,
,
2
-|-f/
the
^0, and function
the set of variables x, y, the
since there
is
no limz. Indeed, X
=
which
-*
r cos cp,f/ r sin
evident that then z -* sin 1801.
~^3(jc 2
1803.
^=
1805
dx
ay),
^ = 3(#
^
dy
= (
x*
+
z
3
l*'
dy
2
=
Answers
436
.-s,^i/ ox
xy*
x
x*
dz
V2x'-2y*
yx*
cos^.
x
dy
*
ln
,810.
ox
x
dz
V2x*-2y*
oz 4813.
1815.
-- -
1820.
/;(1,2,
= 4,
0)
1821. r.
5^-.
1828.
-.
tan p
x-=zxtf2*"
-5-=x2*"ln2,
-r-=i/2*'ln2,
tan Y =-~- 1829
=
1827.
cp (x).
2-^
tana = 4, tanp=co,
1)
^=y *
-
= arc tan ^ +
1826. z
^(1,2,0) = !
= ~,
fy (\, 2, 0)
1,
1814. r
.
1'
If^T'
to see that the function is equal to zero over the entire x-axis and the entire i/-axs, and take advantage of the definition of partial derivatives. 1831. A/ Be convinced that f'x (0, 0) ^(0, 0) 0. 4Ax+ A(/ 2Ax 2 2 4dx &\ -f 2AxAi/-t-Ax Ar/; a) A/ b) A/ df dy\ d/=- 0.062. df 2 1833. dz'^3{x*y)dx 3(y*x)dy. 1834. dz 2xy*dx 3x*y dy. 1835. de
Check
+
2 \**
I
v
i. 2
(xy
dxx
=
=
y dy). 1836. dz
= sin 2xdx
+
+
=
=
+
sin 2(/di/.
1837. dz
=
/
y.
18401 dz
2
=
= +
=
= 0.
1838. dz
1841. d0
=-
=
2
-
x
(xdx
d
+ ydy). dx
1839.
^=
1842.
.
*
d/(l,
1)
=
dx
sin
x 1843.
du =
1845.
du=xy +
t/z
dx
+/x dy + xy dz.
y^. (5dz
3dx
y
1846.
1844.
;
4di/).
1848. d/
live to inner dimensions).
r/dx
:74
= 0.062
1850.
=
=
r
(x
dx
+
y dy
+ z dz).
zdx+\-xzdy +
+
~
dw-
du
-r-
+ xdt/--dz.
x
1847. d/ (3, 4,
cm; A/==0.065 cm. 1849. 75
cm. Hint. Put the differential
cm 3
5)-
(rela-
of the area
o
the sector equal to zero and find the differential of the radius from that. 1851. a) 1.00; b) 4.998, c) 0.273. 1853. Accurate to 4 metres (more exactly,
of
4.25 m).
1854.
n ag ~Ef.
1855.
da==-^
77
(dy cos
a
dxsia
a).
1856.
^=
Answers
-sin*
In sin*). 1861.
d*
-.
=_.; x'-l-y
= 0.
z)[^(x,
i/.
y)
1865.
l867 -
S=^
-j)
+
dx
=W: *
dx
ty'v (x,
^l-
.l.
_ 221 66
+ /^(*,
1862.
dx -,--,. 1+x 1
1
= 2x/,
1863.
1864
437
( *-
The
1873.
y)
>+<
"
increases
perimeter
at
a rate of 2 m/sec, the area increases at a rate of70m*/sec. 1874.
20/52 V5 km/hr.
1875.
_
- 9 ^3
1876.
1877.
.
1878.
1.
J^..
1879.
O
A
68
c) (7,2,1).
1884.
cosa=s-r-, o
cos p
cp
-^QQOQ-r ^=83 37.
9/-3/
=
j (5/-3/). 1886. 6/-f 3/ +
1885.
COSY =
,
1891.
I7 1/1
9 = -7=. y 10
= 1894.^=0. 1895.^ ^ xd^ ^x
.
r
r
dy dz A
= mn;
^
0)
v (0,
and
tiation
_
the
,,
irp-yi+^i [~2 that For x ticular,
fj,,(0,
2 f/
cos
(0.
0)
2
/y
1
= /i(n
and
for
0)=
of
1
+ yvj
(when
any 1.
1899.
2jc sin (xy).
(jry)
definition
4\
grad a|=6;
1889. tan (p^= 8. 944;
,897.
^yi^
|
^-;= ^'
'
1898.
1887.
3
1888. cos
-O-.
^ ^=
lfiol
2A?.
(/,
1902. Hint.
1).
a
partial
A'
+
/^ (0,
Similarly,
1 J/
y.
MO.
)
Whence
find that
0)
verify that
f",IJC
=
and
-
.
= m (m
Using the rules
derivative,
?4 0)'
y)^ we
(0,
y
r
of
differen-
f' x (x,
1;
in par-
0) = 1.
1903.
,
"i-i-
-g^r--/ii\". 1904.
-if
/("'
0)
+ 4*0/
Jll ,
(M,
y)4-*
2
/
w (,
y) =3
consequently,
fxlt (Q, y)=
\Q,
1);
u).
o);
438
_
Answers
1905.
= q>W + t915. + * (y). 1916. d*z=e*y x (*, /)=*cp d*u = 2(xdy dz + ydzdx + zdx dy). 1917. + x dy)* + 2dx dy]. Y* X 1919. dz = d*z = 4
1914. u(*.
X
I(y
1>(
!/)
((/)
dx
1918.
=
(f
= a M (a, o) x x 1921 dz = (ye*f'v + eVf"uu + <2ye +yf'nv + y*e* w dx* + + 2 (&f'u + e*f'v + xe*yfuu +e x + y +xy f + ye* x fo dx dy + 1922. d'z = e* (cos y dx* + (xeyfu + x*e*yfuu + 2xe*+yfuv + e**Q dy\ 1923. d'z = 3 sin y dx dy 3 cos y dx dy + sin y dy # cos x d* 1924. d/(l, 2) = 0; d f(l, 2) = 3sinxdx*dy 3cosy dxdy + xsiny dy*. 2
2
1920. d z .
f'
)
(\
2
2
)
tlv
8
9
-
).
2
z
d 2/
1925.
+ C.
1926. xy
^?+p + C.
1931.
1927.
y ln(^ +
2
2
1929.
a=
1932.
1,
|/
0)=
0,
(0,
+ 2arc
)
6=
z
1,
^,=
1938.
1.
Hint.
Xdx+Ydy.
expression
2
condition
1939.
f^f'y
of
1940.
.
+C
'
l933
total xy
C.
-
JC
2 //2
+ y + 2 +C f
1937.
the
~+
1930.
2z-j-C. 1935.
t/z
the
+ C.
tan
+ 2^ + 3x2 + ^ 1936. + + + c.
Write
1928.
o
=
8 1934. x
.
~ + sin x + C.
x't/
for the
differential
a=f(z)d2 + C.
=
1941.
a
=
-^y
;ri= dx2
a2
^I=-T-I. dx* a*y*
2 a5^1; y*
of straight equation of a pair r
dx 2
=
1946 1943
1945
.
=
=3
I
The equation M defining// &
^= ^
ln
-1;
or
x^ *
1944
*%2
=8
is
the
= ^ dx y _
or
*
1
-8.
.
axy'
dz_x <9jc
1943.
lines.
(\-y)* dy
^42.
2
xf/
2
'
%~~
dx
(axy)*
dz_Gy
yz z
'
dx 2 2
3
3xz2 (xt/
z
'
a )
^^
dz_zsinx
x
'
dx 2
cosy
,
x2
dz
'
dx
cosx
t/sinz*
d/y
'
_
Answers d*z
^
d*z :
dx*~
a*b*z>
:
dxdy
*;*;
1954. d2
439
1953.
:
dy*''
=
dx 2
2
dw; *
i/
-^^dr/ 2*
f .
1955. d2
= 0;
d 2 2=
).
Tc(^ IO
1961.
a
-r-
=
+
2 rf l/
l956
)-
T"
r-^
1;
=
dy);
z
=00;
dxfy
dz
-
11111
"^"*
= 0.
T~9
=
1964.
dw
=
-
u
:^; 1965.
=-
,966. a)7
dz
c cos y
1967. ,-
.
sin
|).
1971.
a)
-20; pp^-.
sina
1974.
/
q,^. 1969.
,
=F;(r,
cos
q>
9)^.
- +
+ ^O.
b)
g-0.
~=0.
1975.
tt
cp)
cos
dz 1968.
dx 1970.
-
1972.
~-. z== 0.
1976.
^
-
Answers
440 3*
1983.
=
+ 4j/-H2z
/ a 2 -f &2
2
169
= 0.
+ 4y + 6z =
1985. x
21
1986.
the tangent planes are the points (1, 1. 0), -f and (2, 0, 0), to the j/z-plane. parallel to the xz-plane; at the points (0, 0, 0) There are no points on the surface at which the tangent plane is parallel to c
Ai
1987
_
l
the x#-plane. 1991.
-^
* Projection on the
on the xi/-plane:
1994. Projection
.
._w_ )=0
=
(
<
/z-plane:
_ ~n
1.
3y*
on
Projection
I^Q
2
the
xz-plane:
= |
3x 2
\
4-z
|
A 1=0. ,
2
Hint. The line of tangency of the surface
with the cylin-
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 2 ax + 2bxy 1996. /(* k) /t, y cy*-{- 2(ax + by)h + 2(b* + cy) k i-ah 2 2 1997. (xH-2) /)=! /(*, 2(jc-f2)(t/~l)-f3((/ -f 26/ifc-K/z 1)2,
der
i/)
= 2/i + fc-M + 2/i/e-f/i
-_1)H2(*-1)
(2
/(^
2001.
z)-f
(/,
,
+
2
A/(x,
1998.
+ =
=
+
+
2[/i(x~t/
*.
1999.
^
r
+
i
2005
>
[(jt
-
+ =
2
I)
4-
y
+ k,
I)
(//
z+/)
z) .
+
1)
2
f(x
+ (y
+
2003.
Dl
Un?^
a)
.
= (x + H,
/ (x, y, z)
2000.
-z) + k(y-x
2004.
1).
+
+
(*/-!)--(
+ x,+.2002.
-!)(!/-
2
b)
+
Hint. Apply Taylor's formula for 2006. a) 1.0081; b)J).902. r in the neighbourhood of the point (1,1); the functions: a) f(x,y)=\ x y* in the neighbourhood of the point (2,1). 2007. z= 1 -}-2(x~-l) b) f (x, y) 2 when x=l, 10 ( X -l)(y-l)-3 (y- !)'+.. 2008. z mm -(t/-l)~-8(x~l) 2010. z min -= 2011. z max =108 1 whenx=l, 0=0. i/=0 2009. NocxVcmum. (3 n z_4M)p2j
yy
=
=
+
= 3,t/ = 2.20l2. z min = There
is
8
when x =
no extremum
Y^ y =
for
1^2
x=j/ = 0.
when
and
2013.
2 max
^
at the
=
^
jc
=
= = 0. i/
1"\ at points
2017.
min
=
T^ 2015.
y==
,
a"d
""Ff
zm i n
of the circle
4 3-
a
and*=
X^-TTL, t/---^
points
when
b
a
.
.
=
*2
when x =
y==
'~'Vf'
x=0 = 0;
wne "
21
+y = 2
-j
,
1.
/
2016. zmax
=
,
2 14 '
F?'
nonrigorous
= 1^3 whenx=
z=l.
3
ab
b
* ==
=
at
V
3 the points
x
2018.
w min
;?Tnax
"=1
maximum 1,
=
y
=4
1.
when
=Y
J/1' z== ^ 2019. The equation defines two functions, of which one has a maximum (z m ax 8) when *=1, f/=2; the other has a minimum 2jwhenx 1,0 = 2, at points of the circle (x 1) 2 + (t/ -f 2 2 -^ 25, (Zimn =
JC
=
functions has a boundary ext-emum (z 3). Hint. tions mentioned in the answer are explicitly defined by the
eacn
of
these
;
The
func-
equalities
__ Answers
e=3
441
2 1^25 (x\)* (f/-f2) and consequently exist only inside and on 2 2 the boundary of thecircle (x 25, at the points of which both I) -|- (l/ 2> functions assume the value 2 3. This value is the least for the first function and is the greatest for the second. 2020. One of the functions defined by the the other has a 2, 1, r/ equation has a maximum (*max"-~~~ 2) for x minimum (2 min 1) for x 2, both functions have a boundary extremum 1, y
=
+
=
=
=
=
at the points of the curve
*=#=-. 2023.
=5
2022. 2 max
2
s
12*
4i/
x=l,
for
=
*
for
imta
4jt
= 2;
2 min
2024.
max
f/
~. ^~.
,
+ 16r/
.
33=0. 5
----
9
x
for
=
=
(f 2,
1)
f
z
=
j/
r
f
fl
2)
=2
x==
for
/"T
^^ T
I/
-IT;
b) greatest
= 0,
2030.
2).
=
1,
x
= 0.
2=
j/
value
1.
y
=
2
ll m,
,
m 2 -f m,
i>
-rp 3
=
y
of
~T
the
ellipsoid.
2046
2
Major
-
=
for
=
/""2"
j
value
= =
x
for
^
for
V
^
smallest
y-^0;
1,
Y
t/
(in-
x
j/2V,
44 a
+
2b
^2V*
s=s3i
j/
= 6,
^
f
.
,
-|- J
The dimensions
2 Q43.
of
the
c
= = 26+2V,
2a
Isosceles
2041. Jgaa
.
2c
2044. *
2036.
2
"T^' where V 3
V
axis,
+
b
--
>/a. 2039. Ai
\/a
>
~p,~/>, and
7^-* 3
,
J/2V,
\/a
\/a
2Q42
Y
axes
*=
for
1
x==
for
y=l; ,
1
*2a
parallelepiped are
TFT
= 0,
y-~Q (boundary minimum). = 2, y = 1 (boundary maximum); smallest (boun(internal minimum) and for x = 0, y =
\
Cube. 2038. a
1a -i-
2
z== "~
2032. Greatest value 2
are 2040. Sides of the triangle fe
v
2031. a) Greatest value
value
=
2034. Cube. 2035.
dary minimum).
27
for
2
smallest
;
t/=
2 for x
triangle. 2037.
2.
f T)'--(f =3 Greatest value
3
=
?
2025. r/
a)
ternal maximum); smallest value 13 for x 2033. Greatest value 2
value
r
I/
f T)=
2,
(1,
"3
for x
= 5 + *:i, /
/^ \
#--2
1,
4
b) smallest value 2
y
f
I)' (41,
(2,
=
for
=-j
= 2, 2 = = 2, 2= 2, "max^ 9 *=1, = 2 = 0; Wmin^C for x = t/=0 2^C. *= at the points .6Mor x = 2, y=4, 2=6. 2028. u max -4 1,
2026. MMIX -=fl 2 2.4 2027. w max
x
for
2 max
-j/
=
2021.
minor
b,
a,
*
and
= --
axis, 26
= 2.
c
are
2045.
Hint.
the
x=
semi-
-
The square
,
of
the distance of the point (x,y) of the ellipse from its centre (coordinate origin) 2 2 The problem reduces to finding the extremum of the function is equal to x -f-r/ 2 2 9. 2047. The radius of the base of the cylinder 5t/ x*-\-y* provided 5* Bxy .
+
+
=
Answers
442
~
is
|/2 + -~=,
R
the altitude
must connect the point
sphere. 2048. The channel
/
with
the
^ 14
2049.
1^2730.
the ray passes a
a
cos
the ray J
of
the
of
point
sin
p
BM=^
,
of the function /(a,
a
=
a=p.
function
/^/ji/^
2052.
7 2 , / 8)
f (/
must
lie
ft
which
at
between A, and B t ;
to finding * the
-
1
01
t/2
:
-5t\2
AI
:
~ provided that a tan COS p
-5-
Find
Hint.
.
the
minimum
a + b tanp = c.
minimum
the
of
Kl
= /J/?; + /*# +
that provided second kind
/J/? t
2
The isolated point
M,
is
B^M = b tan p. The duration of motion
-- --
UjCOS 2051.
length
j
The problem reduces r
.
-5
y 2 cosp
|J)
its
;
into the other,
A,M =a tan a,
-
1
a, cos
,
)
Uj
-- --
is
(
"T
7 V^L -
5 \
1 1
Hint. Obviously, the point
medium
one
the radius of the
is
the parabola (-o-
of
-^-
EE^^!
2050.
frm
line
straight
R
--^=, where
|/2
+/ +/
/!
2
8
-/.
2055. Tacnode (0, 0). 2056. Isolated point (0, 0). 2057. Node (0, 0). 2058. Cusp of first kind (0, 0). 2059. Node (0,0). 2060. Node (0, 0). 2061. Origin is isolated point it if a>6; is a cusp of the first kind if a and a node if a b. 6, 2062. If among the quantities a, b, and c, none are equal, then the curve does not have any singular points. If a b
2053.
2054.
(0, 0).
of
Cusp
(0, 0).
=
<
a
of the 2
-f-
^
/3
kind. 2063.
first
=
y=x.
/ ''.
2067.
whose equations,
2068.
if
z
2064.
2
A
+
2066. **/
pair of conjugate equilateral hyperbolas,
x#^=-^-S. the axes of symmetry
coordinate axes, have the form
y^R.
2065.
y ^2px.
of
the
ellipses
g
xy~
~
2069. a)
are
taken
as
the
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; b) the discriminant curve y Q is the locus of cusps and is not an enof the family ;c) the discriminant curve y velope; d) the discriminant curve decomposes into the straight lines: * (locus is
=
of nodes)
2073.
x^ a (envelope).
and
1/~3V
1).
2074.42.
2079. a) Straight line; b)
2082. 4/
2
(*
for
f:=0,
A;
= 2cos_^,
0=1/^13
z?=-2y+3ft,
;
f/
for
y=~ *8
2075.
parabola;
+l). 2083. ^ = 3 cos
= 4/ 2084. + 2ycos/-l-3Ar;
2070.
#=4
c) ellipse;
/
= 2sin/, any
w=
/;
2i;
for
XQ
e
+Z
Q.
= 3/
^^-
t
^
11
2080.
= Q, v--=4j, w=.
(screw-line);
-2/cos
2072.
2077.
d) hyperbola.
(ellipse); for
w=
i ^
2076.
5.
sin
2071. 7
.
2v 2
2ysin
=
t\
2
u=
3/; fo
2/sinH
w2tor
any
^
Answers
x=cosacoso);
2085.
a
co/ sin
of
sin
uy
+
=
a>
2
r
2 2088. 0"|/ a
.
screw
=
(*
-y-
= --
-
1
/
[(sin
,
2
/
a cost
are
2090. T
.
sin
t
a
cos
*
=
+
-
(*
x + 2z
COS(T,
0./J*.
^
acos/ ,
^
sin
/
cos p
=
==
+ 4r/-f-12z
114
2096.
=
p
=
(normal
M
2
|,
(4, .
ing
(tangent); j
-_ cipal normal);
,
=
:
2V
y
_
z
=y+ =
plane);
(tangent); x
I
(tangent);
u
.
,
*
(principal normal);
-, 1
z=
/2
of
= sin/; plane);
x
(tangent);
+
plane).
(pri "~ t
cosv,
= 0.
\
^ilx*
(binormal);
.
/
^
cosp 2 =
1
=-=;
cos|J 1
=_ 2 ^
/4
.
b
(oscu^ating
2
TT~
2
z
^-~j-
2
1
cosines
(osculating
~T?T
= _
^.=^l = i^l
2097.
= cos/;
=
8
6r/-f-z
^
cos Y
;
u
_
= __t>r
2
2
12*
plane);
= j
=^T
*"T
=
...
=
f
=
(rectifying plane). 2095.
v
(binormal)
a
directi on
The a cos
=
=
2093.
asint
normal)'
a =--===
k]\
p
/\
- zbt
.
VI
/;
cos(v,z)=0.
^r&cos/r = y
.
.
^ sin
=
+
-;
"; P^=. /5
/105 ..
v
1^3^
=
2)
=
5
a>/
cos t)J
The direction cosines of the principal normal are cosa, 1 z 2094. 2* 0. cos YI (normal plane); y
=
art
cof
+ *);
+
t
(sin
^ cos
t
cos a sin
cousin a cos
the angular speed of rotation of the
is
(o*
[(cos
^--
smt
tangent
2aa>u sin
.,
y-
t
cos
i
\ *> (tangent); /
b
x -acost
i
v= a cos CD/
(circle); 2
co /
|;
,
^"
;
co
|
where w = -
2
+ cosO /+(sin
-
the
/i
v
;
T= --
2091.
fc).
2092. t
cos
+
V aV + u
2089.
.
= u=K
CD*
2086.
.
= sincof w +
sin acoscof; z
y cos
cofc
443
jc
"""3"'
+ = (osculat= + 2- = z--2 i/
2
i/
-g
2098. a)
(normal plane); b) __
(tangent); x (tangent);
+ # + 4z-10 =
2_>^3 2|/*2^=0.
x
+
r/
2
a)
6 2 )ejz
-=
c)
^|
1
Z
2/*3z =
2101.
c) b*x\x a*yly + (a ~ = ing plane);
(normal plane);
2
O
2
2
(a
6 2 ).
^
3
2
==
~~A""J: f 3
+ y = 0. 2100. x = 18 = 0; 6y + 2z 2102. 6.r8/z + 3 = (osculat-
(normal plane); 2099. x 9x 0; # z 9 b)
4*
=a6
f
=^"' 2
(P rinci P al normal);
Answers
444
* +6 J 3y
=o
+
(
binormal )
(osculating
TSBS
;
= 0.
\i
-
fl>
2107.
when
M .
/=!,
2
= =u
1
'
>
(principal normal);
}
/+ *
v== /
>
J7*
t
K
K= 99
/C=4-1/T7 / r 14
w,
2jc
-
+
=*
= 0,
/"To
2106
_
2108. a)
When i
w w =2;
;
V~i\ b)
a)
2112.
.
plane);
|7^=i
/
19z~27
2111.
#
*=0
bx
(binormal). 2103.
= -^=.,
oy x
2,
=2
wn
= 0,
/~TQ
I/
~
.
*
'
"^ j^
Chapter VII 9 2113.
2120.
x=l;
9^
44. O -. O
x
2114.
= ^r
*=-!;
2125.
*
/
= 0; t/=y25
= 2.
J*Jf(*. f/
1
;
=
2117. 50.4. 2118.
y
6;
x = 0;
x
It 00 y)d(/.
= 2.
= 3.
2122.
2126.
2129.
2
2
.
=x
2
=
2
r/
2JC
JC
+3
2Af
72
f (x,
12 '
2I32
'
2
;
/ (x,
2130. 1
2119.2.4.
;
21
00
X
f(x,y)dy.
t/
^^
J dj/ J
(
12
J/^
dy
^2
ix = JdxJ/(*. 00
10
00 42
2131.
i/)rfx
*
t
1
y
y\
2127.
11 2128.
x=2
1;
TT/7*
44
2116.
\JL
4
= 3.
~.
2115.
L**
2121. ^
9
7t
ln~
,/T y T
t/
= + 9; A:
Answers
2133.
dx
j
Vl -
t
f (x,
j
""
i/)d
d*
y)dy =
f(x,
-K 4 -*
+J
d
"
dx
*,
j
~
a
r
f
dy
\
J ~
2
\
1
dx
r dy
\
j
-
-V*--tp
y
c/
~yr^p
vr~^y~*
V* -X*
2
y)dx +
f(x,
J
* -
2134.
/(*,*/)
J
x*
-
i
+j
445
Kl+
2
JC
2
/(Jt,
-
-
f(x,y)dy=
J V* -n*
,
!/)
^+
*
U. y)d^+
f
f
dy
-vihT
a
I-A:
i
-
2135.
1+ Kl
(x,
r e) \
J
\)
-
+ 2aa r
dy
\
J
K
48
2136.
dx. 2137.
_
'
l dlj
jc
a
-
I
1
f (x,
\
y)
dy=
*j
i/
-
a
f(*>y)** +
\
f (x t
%)
-
1
1
81
dy
f (x,
y)
dx
+
_ 1
y*
dy
\
*
JL
12
2138
o^
-
ao
Ji
Vu
_
Vx
i
22 f (x, y)
dt/
2
o
20
*
l
1
ax oo
2
w "
a
y) dx\ d)
dx
\
-V i^Kr~7^
/a
/(AT,
rfy
o
1
1
Jf
f
=
-4f/ a
2
-
y) dy
Vci* - v*
a
i-#
i
/ (*,
_o^
r~; - x?
dx
a)
f(x,y)dy-
j
-^rn?
!
]*y
Va?
J
-
y*
f(x.y)dx.
d[/
/
(x t y) dx*
t/)
d
444
__ Answers
x
*=0
bx
(binormal). 2103.
(osculating
plane);
=
\
'
>
__ Q
(principal normal);
v,y.
,
30
+
= 0.
192-27
2107.
9100 2109.
-
21
2108. a)
.
- n -K-e
.1 a)
fl
flfj*
n
-
Q
when
OC
2113.
4~.
2120.
~,
1*
2114.
2121.
= ^-
#=3
2123.
2125
t/
y
2115.
= x\
= 0'
y\
/=10
r,
1^*25
x*'
11
t/
=
1
y
= 0;
y = 4.
= 0*
x
=3
2122.
TT/7.^
2124.
y)d
,
u
.
2119.2.4.
2 ;c
;
t/
= * + 9;
= ~\
=x
z'
y
= 2x;
u
21 = /U. 00 dr
y) dy.
2-0
oo
2130.
y)dy.
=
y
2126
V)
2129.
^y /
ZX + 9
2
f
dx
/ (x.
2X
1
72
12
y_
y = 2.
6;
X
J
10
00 42
= 2,
1
oo
*
t
K
X
1
1
= 0,
12 */(*' 00 jc
*/(*.
7=
-nM b) P R-Q-
.
2127.
2128.
;
Chapter VII Q 2116. -~ 2117. 50.4. 2118.
x=2
1;
4
= 3.
x
jc
~ jf
Ing.
/
,
K=
^=
/=!,
O
jc==l;
When
2112.
-
i^i
^=2;
K~=-
>^2; b)
a)
f(x,y)dx.
y
i
r
r 2131.
\
dy
\
/(^,
i/)^+
\
^
f(x,y)dx=
\
\
dx
-i
\
f(x
t
y)dy+
-*
12 /.
2132.
^
dx Ja
2JC
f (x,
y)dy=\dy
\
f (x t
y)dx.
Answers
445
1
2133
.
4-
J<*x i
J " ~~ *
dx
(x.
dx
f (dx
\
C
c/
y) dy
=
dy
j
+J "
y) dy
/ (x,
J
y) dy
/(*,
y)dy =
+
f
dx
\
/(x, y)
dx+
\
C
dy
dy
f(x,
j
rr
J
f (x,
f(x,
y)
y)dx +
j
t/
i/
+J
dx
Vl + *
2
f (x,
dx
y)
/ (x,
J
V9 -X*
2
\
dx
V&~^
/ j -vr^*>
2134.
+J
dy
/ (x, y)
j
r
r
dy
\
J
/(x,
\
J
\y f (x,
j
2135.
y)dx.
a)
- x*
dx
b)
y)
dy
=
dy
\
1
/ (x,
a
[/
^
48
2136.
2138.
f dy
j
dy
/ (x, y)
'
!L
- x*
f(x>
dx. 2137.
+j
a
a
ta
\
dy
o
2(j
dy
\
J
*
2a
SI
T
<
dx
-
1
*. ffMir.
Ji
y)
-
a
12
f (x,
Vx d*
y) dx c)
*
JL
j
_
2a
2
S
\
y) dxt
y) dx; d)
ax oo
+ 2u
f (x.
Jdy J
i
f (Xt
-1
2
=
-
Ki~^
-
dy
y)
f (x,
-t/'
* ~a
P
'a
Va*
a f (x>
j
_L.
EHi 2
J
dx
f (x,
y) dx
+
\ 2
dy
\
JL 3
/ (x,
y) dx.
/ (x,
y) dXr
Answers
446
+ dy(f(x,y)dx J
( J
2139.
o
2 fl a
.
a
20 za
P
P
/(*, */)dx
j
+ jd*/ o
o
2141.
yf1
J
10J
f(x,y)dx +
J
2d
20, 2a
di/
j o
a+Vtf^H*
_
T
11-*
8
f(x,y)* + jdxj
dx
VT
V
2
1
^
o
2 i/
2~~
p
[dy
f(x,y)dx.
\
c/
Va*
a
_ ]/ aG a2 _ y* yl
r
2140
dy
\
J
V8
a
a_
a
a
a
a
2
00
VT
i
f(x,
dx
2142.
y)*.
VT^lfi
^I*
VR*-y*
2
/(A:,
2143.
t/)efy.
/ (x,
J j 00
f(*,y)dx.
dy
1
2
arc sin
jt
2144.
fdf/
arc ^in
o
2149.
2153.
-p
/(A:,
f/)dx. 2145.
2146.
.
.
~ a.
2147.
2148.^-
-^
-g-
y 15jt
1.
2150.
6.
y
f
In 2
2151.
8
a)
dx
2154.
.
2152.
;
-|
xr/dt/
=
^
b)
16 ;
c)
2
.
1"
_
2155.
.
o
i
2JT/?
//
=
/
1
Hint,
2156.
C
|-Ji/?'.
j
//
dx
== dr/
/?(1-COS/)
*Jl
\
o
o
(S)
=
C dx
cosO^
/?(!
\
ydy> where
the
last integral
is
obtained from
o
o
D4
the preceding one
2159.
by
flf
the
substitution
+i--
216
x=R
(*
sin t).
2157.
1
2158. --
.
-
o
Jt
1
^
T"
sin q>
^T
no T [(p
\
J
r/(rcos9, rsin(p)dr.
2161.
\
J o
cos
(p
.
Answers jn
i
sin
\
sin
*Ji
i
T
sin (p
a /COS
4
2164.
\
\
(tan
f
r/ (r cos
r sin q>)
cp,
r dr.
+
dr
5Jt
a /COS
\
dq>\
2
a cos
2165.
f J
f J
2171.
~nab.
r/ (r cos
dcp
r2
Hint.
-sin(pdr=-^.
2166.
The Jacobian
= abr.
I
is
1.
= u(\ u
as
v
1
2172.
^
v)
and
cfu
\
The
0);
the
when
=
c,
/(w
\
of u:
when x
limits of integration are
wy,
uv}udu.
Jacobian x=r-0,
Limits
.
I
is
uv^au
(1
t;),
\
/
2173.
.
2~y dv
Of
\
whence
=
i
fr \
-TT
variation
of
the
define
whence of
a
=
since
v:
1;
= ^-^
+a
;
for
y
= ftx
we
find
dw
"1
f
(2
'
"~9~)^
u
Hint '
^
ablf? 2 L\/i
-r]arctan^-r-
k2 J
ter
chan S e
^
variables, the equa-
^
tions of the sides of the square will be u
2174.
= u. We
^
B
4-
We
Solution.
u(l~u)=0,
1
i
^
-
o
y--^uv\
functions
follows that
wr^axjt
u=r-r-5
dr.
i-
+p q
(since
q>)
g
P i
2167
4-*-
3
x
r sin
2
limits
2(J)
T
""T"
have
\
\
dq>
2
\
d
+
/(tanp)dcp
cs a
n r dr
\
/ (tan q>) dcp
\
\
o
n_ 4
4-
oo
2163.
r/(rcosq>, rsinq>)dr.
(f
cos 2
4
\
d
sin
,
t_
4
2162.
447
bli
+ -Ahk
.
\
= v;
Solution.
w
+ u = 2;
u
u
The equation
= 2; of
u the
=
w.
curve
448
__ Answers
r4 **/*
2
[
rj-cos*(p-- rrsin
4ht upper
follows that
we
r= y
limit,
cos 2
p-
to the
sin
pDue
to
2
sin 2
T^
cp
^2
cp
limit
whence
0;
the
entire
of
first
\
quadrant:
dxdy = 4
\
.
a)
4-g-;
-;
a)
j
^dy 2
b)
dx ~
+
~
1
jJia
2184.
.
Vy
+ ^dy
j
a *-
2177
T5.
2180.
2183.
2
6.
2188.
1(6 o
1188.
,.96.
2202.
2205.
2209. Hint.
2214.
*.
;
^dx 2179
flZ -
-
a-- + -.
dy.
^
*
w
-
Hlnt
-
-
2182.
Change the variables
x2y = u,
I(R o
a )ln-.
2187.
a).
2193.
A.
abrdr.
~.
2194.
^.
d
2195. ~-
.
4
2197.
iia'(a
J.
2,98.
o
2206.
(
2
.19..^. 2200.^. 220,.^.
^"2"~ 1)-
2207.
nafrc.
2^'
|-
IB (!--). 2210. Change
~^
2203.
P).
^.
4(m
2l78
-
u=di/(l-^^=:C^r(l-x)dy.
the 1
ft)/?
2217. 8aarcstn
3 .
2218.
-~ a
-i O
2204.
(6
J~
2211.
L
yT
xt/
2
= u, ~=f.
-1^(3 /"T o
to polar coordinates. 2221.
o
=
1).
.
2212.
2
|-jw
\(
(2
L
~-
2213.
2219.
(
V
"2
1).
3
^a
.
2
Hint. Integrate in
.
'
2208.
5).
ft
variables
2215.
.
.
Q
Paas
-
a)(p
IX
^-^-
dx\ b)
Hint.
lOjt.
sin'cp
aVa*~x*
2181.
2185.
ourselves
_
\
o
_j>
Vy
i
1175.
- b* -
dy
\
(S)
cos'
^
it
real,
quadrantal angle
confining
fa*
y
be
integration relative
of
integral,
ak
the
first
region
^
tan
must
r
for the
of the
arc
fo
and
be
r will
for
Since
.
q>
symmetry
we can compute
axes,
lower
the
J,
2 rr cos
q>
ak have fan
whence
cp
2 2 /"a 6
KT-1).
+6 c +c a 2 2
2
the j/z-plane. 2216.
8a 2
.
1+5LV
2220.
1
I
3na2
.
4
2
2 .
Hint.
Hint. Pass to
__ 2222.
coordinates.
polar
-5-
2223. 8a arc tan
8aa
Hint. Pass to polar coordinates.
and
a
a
r
C
.
fl
T T
V2 L
-
i
Integrate by parts, and then
ya
9 2
x'
=?
9 2
+c
a
8fl
arc sin
\
J
\r
^^
variable
the
change
2 Vb* + c*-a }/a + c
b
^-
,
a
C
ady
Hint.a=\J dx\J
-^5
the answer. 2224
449
a1
/.__
2
Answers
2
sin
^
In
Iransform
t\
'-=)
Hint.
.
^
Pass to polar coordinates -
H=0.
2228
/x
2231.
2226.
m
^ ^=
=4
2232.
-
2229 /t
a)
"'
^J; lz
o
^=
'
L.
2225.
!
2227. *==*?,"" o (4
.
J*
*-;lr-o.
-
-
(D*-d4 );
b)
A
=
a4
2234.
.
|-
-3-
/=- f
Hint.
a*.
2230.
/^ Voj?
a
2233.
;
JT)
^
f
-K^I 16 In
2235. x
=
is
(/
29 ~
.
d=
to
equal
V of the straight line.
The distance
Hint.
-JL and 2
2236. 7
=
is
by means
found
5
^fca
of the point (x, y)
[7
/~T+3
In
from the straight lint
of the
normal equation
(}^"2~+l)],
where k
is
the
proportionality factor. Hint. Placing the coordinate origin at the vertex, the distance from which is proportional to the density of the lamina, we direct the coordinate axes alonjj the sides of the square. The moment of inertia is determined relative to the x-axis Passing to polar coordinates, we have jt_
_jt_
a sec
4
/x =
oo I
d
2238. /o
=
\
^--
a cosec
2
(jp
kr
(r sin q>)
2
r
dr
+
\
dq>
n_
~
jia
4 .
VR*~^7*
a
2242. a
-1
=
Hint. For the variables of integration take
( dx )
-*
dr 2237. /
Jg
Jifl
4 .
T 2239.
y (see Problem 2156). 2240.
2241.
(r sin q>)V
d
i
R
kr
\
H
\
J
dy
x \
J
/ (*,
0, x)di
/
and
Answers
450
VT=T*
i
2243.
^1 -*'-/, dy
J
/(*,
J
-y
-i
1
2244.
2248 .lln21
4n# 15
i
5
_^
-
u
,
2254.
-
2258.
-
.
=f-
2253..
2252. -jutfc.
2257.
2249.
|.
-~a2
2255.
/?.
??fl%. 9
2259. 10
(
2256.
.
~ Jia
2260.
?c
|
|-r 3
Selut ion.
.
v
4
2acos(p
C
C dx
f
40
dz
= 2fdq>
f
Q
L
L 20 COS
2
r'dr
=
"25"
00
-T^^T-
2
3
f(2acoscp)*
1
^
1
2na 8
nnot
,
226h
Hint.Pass
3
19
2263.
4).
=0;
t/
4
the cylinder the jq/-plane
_ =0;
^(a + b +
2265.
2266.
c).
_
=0, 2=0.
Introduce
Hint.
a.
2269.
~ iz
coordinates.
a2
2
^(6c
6 2 ).
2
(3a
+
coordinates.
spherical
2
4/i ).
Hint.
For
the axis of
we t^ke the z-axis, for the plane of the base of the cylinder, The moment of inertia is computed about the x-axis. After the
passing to cylindrical coordinates,
rdydrdz
to cylindrical
2
2=-^D -
*=~, o
2268.
nabc.
2264.
_
_ x
Pass
-^-jt.
~(3jt
2267.
Hint.
2262.
coordinates.
to spherical
from
the
x-axis
is
equal
to
square of the distance of an element r z sin 2
+z
q>
2 .
2
2270.
(2/z
+ 3a
2 ).
Hint. The base of the cone is taken for the xr/-plane, the axis of the cone, for the 2-axis. The moment of inertia is computed about the x-axis. Passing to cylindrical coordinates, we have for points of the surface of the cone: r
= j-
2);
(/
and the square
of
the distance of
the element
2 2271. 2jtfcQ/i (1 the x-axis is equal to r 2 sin cp-f z 2 proportionality factor and Q is the density. Solution. .
cos a),
rdydrdz from where
The vertex
k
is
the
cone is taken for the coordinate origin and its axis is the z-axis. If we introduce spherical coordinates, the equation of the lateral surface of the cone will be a,
\|)r=--
and
the
equation
of
the
plane
of the
base will be
2t
From
the
follows that the resulting
r=-sin
-. ip
stress is directed along the
symmetry The mass of an element of volume dm = p/ 2 cos dcp dtydr, where Q the density. The component of attraction, along the z-axis, by this element it
z-axis. is
of the
of unit
tj)
mass lying at the point
is
equal to
j
sin
ip
=
Q
sin
i|?
cos
ty dtp dq>
dr.
Answers
451
L-a The resulting attraction
oo
equal to
is
h cosec
2
27i
\
\
dq>
\f
Q sin
\
d\|?
\f
cos
dr.
\|>
o
2272. Solution. We introduce cylindrical coordinates (Q, cp, z) with origin at the centre of the sphere and with the z-axis passing through a material point whose mass we assume equal to m. We denote by % the distance of 2 be the disfrom the centre of the sphere. Let r= >^Q 2 -h(| z) tance from the element of volume dv to the mass m. The attractive force of the element of volume dv of the sphere and the material point m is directed
this point
along
and
r
equal to
numerically
is
kym
density of the sphere and dv qd^dQdz jection of this force on the z-axis is ._,
dF
/\
kmydv
=
y=
where
,
Q
the
The
pro-
-^
... dQ
z
H
f
kmy *^~
is
la
the element of volume.
is
=
c s (rz)
^
2
dz.
d
Whence
R 4
But since -x-ynR'
M,
follows that
it
F=
kMm
(p>
2275. a)
p
2273.
for
w
P
p
>
a; c)
(p
> Q);
Hint. Differentiate L. 2277.1. 3
n2
p
dye~ x
y*e~*y
-^-^ A !'?
d)
P~1~P
.
> 0)
(p
7
00
2276.
\
J
5
---
b)
0);
P
?.
*5
.
C-^f/= J
2278.
twice.
P
ln~. a
o
2279. arc tan
--
arc
tan
2282. arc cot
2280.
n (l+a). -^l 2,
2284. -1
2285. -5.-
.
24
AW
AW
2283.
4--
1.
p
polar coordinates.
from S
=
\
\
In
-^
2288.
.
~ o
coordinate
the
consider 7
2287.
V
x*
n
-^ 4a-
2286.
V
(
'
a2
1
its
that
e-neighbourhood,
+ y*dxdy, where the eliminated region
is
at
=
JJ o6o \d(p
Whence
\ r
the
Passing to polar
origin.
271
1
Inr
lim 7 8 e->o
dr=
--=.
round the straight
\
we have
coordinates,
1
JL^
-5-
In r
le^JJ rfcp~2jt(-j~-Q-lnen
\
rrfr
\ *
-T-). * /
^
e
-5L. 2290. Converges for 2 line
is,
a circle of
(S 6 )
radius e with centre
Pass to
Hint.
.
1).
2289. Converges. Solution. Eliminate
with
together
origin
.
2281.
(/
=*
with
a
a>
narrow
1.
strip
2291. Converges. Hint. Sur-
and
put
I
- =a
i
^
J.' (S)
K(* K v
2 I/) y/
Answers
452
i-
.
2292.
t
2
2297.
l-2-^! arc
2301.
1).
~ll
)
|-[(I+4Ji
2298.
.
^
tan
M*
2302. 2jra
2
^(lOVTO
2303.
.
/7~
a*V 2. 2300.^(56
2299.
.
Hint.
1).
ftl
be interpreted geometrically as the area of a cylindrical sur-
may
y)ds
.
d
flu \
5m
fot
Converges
c face with generatrix parallel to the z-axis, with base, the contour of integra tion, and with altitudes equal to the values "of the integrand. Therefore, o
/
S=
xds, where
\
C
the
is
OA
arc
parabola t/=--x 8
of the
e (0, 0)
and
a V~3. 2305. 2
(4,6). 2304.
V
V
V^ + u
2306.
2308.
2ita
2
2
r
V
f ji
fl
a)
+ 4jTu
^aTT^
"
2
-5-;
o
b) 0;
c)
2309.
Use the parametric
To
;
a
2
b
,/
V
(a
.
J
+&
2
equations of a
fl
(-q
.o a
2311.
cases 4. 2314.
all
2315.
circle.
.
)
/
V
2jta
2 .
30
)
4;e)4. 2313. In
d)
2307.
2310.40-
.
2
h*\
a
2
fl
"*"
4-|rln
10
4 2312.
2
l/~a 2
^
arc sin
r
b*-\
[
that connects the
a *b
/ points
2
-r-a& o
2
2316.
.
Hint.
2rc.
2
sin 2.
xa
2317.
2318.
0.
a) 8;
12;
b)
c)
2;
d)
A
;
2
e)
\n(x
+ y);
//a
+ J{
ty (y)
dy-
2319.
2
2322.
a) 62;
b)
+ ln2;
c) -L
1;
4
+
1
d)
Y
f 9(jc)rfjc-f J
f)
2.
2320.
y\
Vl-fft e*~ y (x + .
c)
2325.
1
fi-
y)
+ C\
d) ln|Jt
+ j^"M //,
2328.
JJ
x2
a)
8 -
+ |/| +
2326
4o
+ 3xy
-
a>
2329
-
2
2f/
C.
+ C;
b)
2na(a +
2323.
~ 20
b>
0^ 233
^r-^
1;
-
c)5 4--
z
x8
x y-{-x 2 2 n/? cos b). 2324.
a
K
=
d) 0.
2;
2331
-
2327. /
2332
-
-
d
a ) Ol
(S) 2/iJi.
Hint In Case (b), Green's formula is used in the region between the b) contour C and a circle of sufficiently small radius with centre at the coordinate origin 2333. Solution. If we consider that the direction of the tangent coincides with that of positive circulation of the contour, then cos(X,n)==
= cos(y, 0=/. as
S
is
hence,
(X, n)
ds= J
ds =
the area bounded by the contour C. 2335.
not applicable. 2336. nab.
2337. -| Jia
o
2 .
Hint. Green's formula
4.
2338. 6jia
2334. 2S, where
2 .
2339.
-|a *
2 .
Hint.
is
Put
__ Answers
45$
for 2341. n(R + r) (/? + 2r); 6n y=tx where t is a parameter. 2340. ~. R r Hint. The equation of an epicycloid is of the form x = (R -f r) cos f r ~rcos^JL. *, y = (/?-f-r)sin/ rsin^J^/, where / is the angle of turn of 2
%
radius
the
2342. Ji(/? is
cycloid
Problem 2345.
V c)
a
of
r)(#
~ nR
2r),
drawn
circle
stationary 2
for
2
6
-^-(a
2
where k
),
work,
mgz,
/
potential,
mg(z
=
a
is
2 2 );
l
L(jc
of
the
+ + *)
work, 4r (#
f
2346.
= -ii-,
/
potential,
epicycloid 2344. mg(z,
FR.
factor,
proportionality
b)
tangency. the hypo-
of
corresponding 2343.
r
of
point
The equation
Hint.
'=-7-
obtained from the equation 2341) by replacing r by
the
to
2
r
(see z 2 ).
Potential,
a) -
work, 2
2347.
).
-|jta*.
2348.
^
!
2353.
2354.
a.
/J*"
/i*.
51)
V
10 (5
^
2355. a)
0;
b)
-{(
(cos
a
J
+ cos p
j
2356. 0. 2357.
.
na\
2358.
4;i.
2359.
=
2360.
a*.
(V)
2365. 3a4
2366.
-^ 2
2378. grad(rr)
/'(/)
the vector
=---
COS
(
f
2385. a)
>r)
/
^=0
;
r ot
r) t
2387.
2388. div grad
2392.
2a)/i
U= 2
a)
5^
=
for
j~Ji/?
(/(r)<7) ,
where
2
57
2
z
.
= lgrad(y|
= ^-^cxr. is
The
3y
3Jfe;
~
c)
;
b)
div
rot (/r)
=
a=^
2380.
/ (r)
+
~= /'
(r).
.
-^-;
2386. div tf=-0;
c)
div
rotv = 2co,
where
unit vector parallel to the axis oi rotation.
a
2 );
when a = 6 = c. 2383.
div(r^)-=~,
b)
+ 2//
points except the origin.
= c 2 2376. grad (/ (>1)=9/ x = y = z. 2377. a) ~; b) 2r ,
2382.-?-.
IJ_r.
^+^+^
//(3/?
2
the level surfaces are planes perpendicular to
7^,
/i
2371. Spheres; cylinders.
2363.
.
= xy\
= c;
divr=3, rotr=0;
= LW( Ct co=co/f
2379.
c.
5
+/=c *
d)
Jia
O
2373. Circles, x 2
2372. Cones.
~
2367.
.
;
rot
~ n/?
2
//
grad 2 (
flux is equal to
U = 0.
+ 2//
2 ).
2391.
2393.
4jtm. Hint.
2
3;i/? //.
divF=Oat
When
all
calculating
454
__ Answers
the
flux,
use
theorem.
the Ostrogradsky-Gauss
2394.
_ ~~TtD
2395.
2n*h*.
.
r
2396.
/=f
rf(r) dr. 2397. .
2398.
.
.
-L.
.
1
2402.
24 8
potential;
U=xyz + C',
b)
2400. Yes.
Chapter 2401
No
a)
.
2403.
^
VIII
2404.
.
.
2405. -
.
24 06.
-
2416. Diverges. 2417. Converges. 2418. Diverges. 2419. Diverges. 2420. Diverges. 2421. Diverges. 2422. Diverges. 2423. Diverges. 2424. Diverges. 2425. Converges. 2426. Converges. 2427. Converges. 2428. Converges. 2429. Converges. 2430. Converges. 2431. Converges. 2432. Converges. 2433. Converges. 2434. Diverges. 2435. Diverges. 2436. Converges. 2437. Diverges. 2438. Converges. 2439. Converges. 2440. Converges. 2441. Diverges. 2442. Converges. 2443. Converges. 2444. Converges. 2445. Converges. 2446. Converges. 2447. Converges. 2448. Converges. 2449. Converges. 2450. Diverges. 2451. Converges. 2452/Di2455. Diverges. 2453. Converges. 2454. Diverges. 2456. Converges. verges. 2457. Diverges. 2458. Converges. 2459. Diverges. 2460. Converges. 2461. Di2462. Converges. 2463. Diverges. 2464. Converges. 2465. Converges. verges. 2466. Converges. 2467. Diverges.
k- > !
2468. Diverges. Hint,
2470. Con-
1
verges conditionally. 2471. Converges conditionally. 2472. Converges absolutely 2473. Diverges. 2474. Converges conditionally. 2475. Converges absolutely. 2477. Converges absolutely. 2478. Converges 2476. Converges conditionally. absolutely. 2479. Diverges. 2480. Converges absolutely. 2481. Converges conditionally. 2482. Converges absolutely. 2484. a) Diverges; b) converges absolutely; c) diverges; d) converges conditionally. Hint. In examples (a) and (d) CO
consider the series
2
( a 2k-i
+ a zk)
anc
^
in
examples
and
(b)
(c)
investigate
2486
Converges
fe=i 00
separately the series
2a k=i
00
2k-\ and
2 a^' k=i
2485<
Diver ^ es
-
-
absolutely. 2487. Converges absolutely. 2488. Converges conditionally. 2489. Diverges. 2490. Converges absolutely. 2491. Converges absolutely. 2492. ConCD
verges absolutely. 2493. Yes. 2494. No. 2495.
*
1
7.9*1
f9n_iv
^T
-
3/2
;
converges. 2496.
n=l converges.
2497. Diverges.
2499.
Converges. 2500. Converges.
fi=i
Hint. The remainder of the series
a
geometric
progression
may
exceeding
be evaluated this
by means
remainder:
Rn
'
of the
an
"9*
sum ,
of
+
__
Answers
455
--
-'' 2504.
m + "-=~
-f
2505
-
For tne
<.<.
iven series
it
is
Solution.
to find the
easy
exact value of the remainder:
Solution.
We
multiply
-:
by
*"+ 2
/
1
\
M+ "
Whence we obtain is
nv nv .
=n
/
1
\ zn
VTj
/
.
v n+
m
* .
A
1
"**,
. '
I
I
= / + 16 T5
4 /
+\
i
r
}_
16
From
this
the series
we
find the
above value
Rn
of
Putting /i=0, we find the sum of
2507. 2;
2506. 99; 999.
S-=(pY-
.
3;
5.
2508.5=1.
Hint.
when x < 0; S-=0 when > 0, S= 2511. Converges diverges for x< Converges absolutely for x> for 0
^~ x = Q. 2510.
an
2509.5-1 when
x
1
1.
1,
x.
-
.
.
.
,
oo, since from cosn*->0 xs^O, and cos n.x does not tend to zero as n thus, the necessary condition for converwould follow that cos 2n.v -* when 2/m < x < gence is violated when x<0. 2516. Converges absolutely
for
1
it
;
it diverges. 1, 2, ...); at the remaining points <(2fc + l)ji(fc=sO 1 for x ^ 0. 2519. x 2518. absolutely Converges Diverges everywhere. f
2517.
> x< 2523. x < 4,
2520.
x>3, x
x>l,x<-l.
2521. 2524.
x^l,
x
-Kx<
i-,
2522.
x^5j,
y
Hint.
1.
.
For these values
*
00
1
of x,
both the series
x k and
the series
^
converge>
When '^l^
1
__
456
and
when
x
|
\
^
=-
the
,
Kx<0,
2525.
Answers
term of the series does not tend to zero
general
0
l
2526.
\
2528.
1
2532.
4
2536. 2540.
2530. !<*
~ <*<
2538.
o
.
.
-5-
2542.
^
6
\
2541.
2
2527.
K V oo
2533. 2537.
_ 1
e
The diver-
1
x 1 of the series for is obvious (it is interesting, however, to note that the divergence of the series at the end-points of the interval of converthe aid of the necessary condition is detected not only with 1 gence
gence
|
|
x=
by means
of convergence, but also
d'Alembert
of the
<
When|x|
test).
1
we
have
=
lim
x
n\
n
nIn
lim
n
-
|(rt+l);t
|
oo
lim n-* oo
1
x
,
|
readily obtained by means of 1'Hospital's rule). 2543. l<;x^l Hint. Using the d'Alembert test, it is possible not only to find the interval of convergence, but also to investigate the convergence of the given series at the extremities of the interval of convergence. 2544. Hint. Using the Cauchy test, it is possible not only to find the interval of convergence, but also to investigate the convergence of the given series at the extremities of the interval of convergence. 2C45. 2 x 4. 2548. 2. 2547. 2549. 2546. (this equality
is
l
< <
2
2550. x
=
2554.
3
3
e
3
2558.
0
7
2551.
3
2555.
3.
2559.
1
2
4
l
1
2552.
.
1
1
-j^^ ~ \
series
2561. 2566.
diverges,
lim n -+
oo
e
.
|
(~-l< x
-ln(l-A:)
ll n
lf(|xl)
2579. .
arc
2577.
tan x
sum 2586.
^ farctanx--
In
x of
3.
the series
2587.
ax
x
3
\^}
x
|<
(|xl<1)
(\x\<
1).
2585.
2580. .
-
I
z
z
|
I
<
1
(-
In(l-f-x) 1)..
2563.
1
2570.
_
-
2583
^---^
.
Hint. Consider the
x*
^-+-=
=l+
(|
2
2560.
1
|
2582>
2384.
^
ye 1<*<3 2562. l
2576.
2578.
since
\
...
(see
^
Problem
_ 00
2579)
for
2688. sin
jc=~^=: x
+
-
'
=
Answers
xsm
/
cos (jc-|-a)=cos a
2589.
457 **
gating
the
1
1
)""
+ --^5S A
,
*
.
-j-
~2
Hint.
When
investi-
7*
use
remainder,
theorem
the
on
integrating
a
power
series
2592
r
(
_ l\"-
on-i
n=o oo
< *<
e**=
2595.
oo
1
+V
^
oo< x <
,
oo
2596.
n=i
(_-oo
oo
2599.
.
(-!)"
2600.
2608.
oo)
oo
(-3
rn-
T (-0" H
04/1-S y2n
2609.
(-OO
* 2611. 2-f
22^11
+3
2610. 8
~
-|.._ +
2601.
(2rz) |
.
**
2s 32 -
...
2591
o.
+(-o^ 3Z ^Z + TTi-"
x*
^ cos a + %r sma+'-rr cos a
a
-
2!
"*"
2W
-
-4-
3"
""
1
f
^ (=
l
W JC
(~oo
1
I)""
vn
458
Answers
*
<-<
V
1
2612.
g
/
1
+V
f|
-
(l+2- n )
I
l)
(
+ o7rFT
*"
(
'
~ 2 < * < 2)
'
2614.
).
2615. ln2
1
Ti
2616.
(-.i
V(-.l) x =o
(-D n+I
2018.
(I*KD.
'"
2622.
e(
T+S-;2624.
219.
c+
1-^4-...).
262S
-(^+75+^+---)
J
-
x
!+
2623.
+
+...
+ x* + jx>+... 2626. Hint. Proceed= &sin
ing from the parametric equations of i/ pute the length of the ellipse and expand the expression obtained in a series 2 of 2628. *' e. 2x* 5x 2= 78 59 (x-H) 14(x of powers 2629. / (je 5x 8 4* 2 4) ( oo
+ + =
+ +
< <
;
2630.
Ar
n -1
V(
l)
^"^
(0
2631.
n=i
V
00
2632.
(
V
1)" (*
(
1)"
(Q
< x < 2).
AI=O 00
(n
+
6
(x
1)
2).
+ \) n
(
2
< x < 0).
2633.
2634. /l=0
(,<. 1.3(^-4)' r 4-6 2
L3.5(*-4) 4.6-8
2"
2J(-D"
n=i
n=i
(0
U
X
2637.
4-i
"T---TI
JL
(2 ttll)!
2636.2+^-1^ 1.3.5...(2n-3)(x-4) 4.6-8...2n
O^K
00 )-
""^
*""
2638.
~+
Answers
Make
Hint.
the substitution
=
.
45J>
and
t
expand
\RI<<1.
2641.
2642.
T
-
\nx
|
in
of
powers
1
R \<
.
t.
~
2643.
p r ve that
exceed 0.001, it is necessary to evaluate the remainder by means of a geometric progression that exceeds this remainder. 2644. Two terms, that is, 7
X2 1
TT.
x*
Two
2645.
terms,
Z
2647. 99; 999. 2648.
*J<
0.7^68.
2659.
1
|
-"
r
_
/y
_
\2 .
I
0.005. 2650. 2.087. 2651.
<0
0621
2656.
+ V (-l)JfZ^L.
2660.
|
R \<
x |<0 39; \x\
0.608
2655.
2646. Eight terms, i.e.,
.
o
2649. 4.8
U|< 0.22. 2652.
0.39;
2fi54.
1.92
i.e.,
1
~-
x
~~ 03 Z
2657.
Z
0.2505
V-L jL*
!
2658.
0.026.
i
/
(2/0'
00
2661.
V n=
1
(
i
JO
2662.
a
1-1-2
V
x)";
(//
l-^-y +
J
ry
i/1
<
j
Hint.
1
x^
2663.
geometric progression
Hinl.
lJC
= (l-.^)(l-y).
yW
|T^^=- + !
i
I
^
(_
2664.
~~
K//<1).
Hint, arc tan -j
i
2665.
/ (Jt
+
2
\-2bh-\-ck
.
/i,
#+
2666.
A;)
~^-=arc tan x + arc
tan (/(for
A^y
= ax* + 2bxy + cy* + 2 (a* + fy) + 2 (bx + cy) /i
f(\+h,
2-\-k)
f(\
t
2)
=9/i
21/r -f3/i
-2^.2667.
2669.
l+x+-
+
-+...
n\
\x"< 0.69; -^ 0.4931.
\ o o
(-00
Y
18 2653.
1-f
2670.
i
Answers
460
%-l n=o
(n)=^=%.
^ + 4^(-l)";S(
2873.
n)
= n. 2674.
n=i
X
[^-
L2a
y
x ^4
V tr^! + <*a* +
(fl
n
cos n*
sin
n*) 1
rt
l>fl
l)
(
|tlyl
?
m nf
a*
S (+
;
Ji)
= cosh an.
2 Sln 2675.
a
is
nonintegral; sin ax
if
a
is
an integer;
gn
n
J
X
5(n) = 0.
n=i 2676. Ji
2a
I
L
S(
integer;
2678.
Jl )
+ jL* y _i )B (
if
or
n=i
= cosan.
a
is
nonintegral; cosa*
t\r
J 2677.
= :
CO
2680.
V
tin
.
sin
"-
^
if
a
is
an
Answers
2689
461
slnn/t .
/
2691
i_
.
2694.
4n-l
Solution.
2 P --
H
\
1)
2a
= -~
f /(*) cos 2/i* dx
/
(*
-jj-
W
cos 2/ix d
Ji
/ (x)
cos 2nx dx.
we make
If
the
x
substitution t**-=
in
the
first
Jl integral and
identity (n = 0,
t
=x
second, then, taking advantage of the assumed
in the
z
/(-^+< 1,^2,
)
= -/
will
<
(^
)
=
&,
f
(y +
/ (^) sin
=/
/
)
2nx dx =
f
2697.
~. T
sin
Sin
-|-
Case
as in
(\^
(1),
with
account taken
^ leads to the equalities 6 2II
t
of
=
the assumed
(/i=l,
2, ...).
'
/cos sinh
o 2n =0
n
t
T
The same substitution Identity
that
be seen
...);
n 2)
readily
~
J
/
[t /IJTJC
n
> (-1)"
2698.
cos
sin-
4
_8_
where
1 ^
nx
-16\\-1)"-'
-~,
(2'?
1^1
_-!? ,(2'H-l)Jix o
-il
2702.
b) 1)2
n=i
_ A V ^ J 2 1+ 1
n=o
S
!)
n*
2
9
2'mx
.
1
rs
2
mx
Answers
462
Chapter IX 2704. Yes. 2705. No. 2706. Yes. 2707. Yes. 2708. Yes. 2709.
no.
Yes;
a) b) = 0. 2716. = 0. 2715. 2xy' = Q. 2717. = = x 2xyy'. 2720. xyy (xy*+ 1)= y' = y. 2719. 3y = 0. 2724. / 2721. w=-xr/'ln~. 2722. 2xy" + y'=Q. 2723. 2725. #" 3f/V -^0. ^"'^O. 2728. (1 +y' )y"' + = 0. 2726. = 0. 2727. 2X = cos*. 2732. r/== 2730. 2731. x = 25. 2729. y = xe ^^.r^Sg-^ + g^ 4g 2X 2738. 2.593 (exact value y = 2739. 4.780 [exact o = 3(e 1)]. 2740. 0.946 (exact value y=\). 2741. 1.826 (exact value value 2742. cot //-tan x + C. 2743. ^= y = 0. 2744.
Yes.
2710.
/
2714. / 0. 2718.
xi/'
;u/'
2{/
2
r/
f
2
1.
(/"
2r/
i/'
I/
2
2
2//'
f/"
t/
2
2
.
//
/
e).
).
i/
2
2
..
V
= lnCx
2
2745.
.
2
= a + -^-. 2746. tan y^=C (1 -~
-
-
2748. 2e
y
=1^6 (l+e x 7|
= C. ment
.
2
2760.
2762. if
8x 5x
2752. 2754.
=
*=2px. 2761.
t/
x
;
= 0. 2747. = C sin i/
jc.
2751.
y=l.
2750.
.
|
or
hyperbola #
+^
2 J/
= ax
2
Hint.
.
to
respect
x,
=
2758
.
^2
2 -
I/
By
hypothesis
we
get
a
The
Hint.
seg-
= C. 2759. = r/
differential
equation.
-*.
y=V 4
^2
*
+ 21n-
^
Family
of circles x
2
2764.
.
Pencil of lines y^=kx. 2765. Fa-
+ y = C 2766. Family of hyperbolas x y = C. =~ ~ 2769. + (y b) = b 2768. = xln~. x x & z
mily of similar ellipses 2x* 2767.
8
)
+ 2y+l=2tan(4x + C). 2753. + 10y + C = 31n lOx 5y + 6| 2755.
f/^Cx
t/
with
twice
x
9
equal to
is
e
i+^srrf-j.
2749.
).
2757. Straight line
Differentiating
2763.
= C.
of the tangent
= Ce a
;
+ if
1
2
2
2
.
2
2
.
f/
.
r/
X .
x
= Ce y
.
2771.
+ ln|(/|=C. 2775. y
=
xl
C)
2773. y=5
x. C,
(x
2776.
2778.
2
2 t/
=C
2
2 ;
^-; (x
+y
ln|4jc +
8t/
(x
*
2)
= 0.
2 //
= 4;
2774.
= (xy + 3). + 5| + 8j/-4;t = C. !)
(;
i/
(x*
=
jc.
2772.
+ y*)* (x + y)*C.
2777. 2779.
x 2 =l-2[/.
__ Answers
463
2780. Paraboloid of revolution. Solution. By virtue of symmetry tUe soughtfor mirror is a surface of revolution. The coordinate origin is located in the source of light; the x-axis is the direction of the pencil of rays. If a tangent at any point (x, y) of the curve, generated by the desired surface being cut by the xi/-plane, forms with the x-axis an angle q>, and the segment connecttan 2q> (x, y) forms an angle a, then tan a ing the origin with the point
M
=
-
y
=
= y The desired differential equation is 2= = 2Cx-\-C*. The plane section is a para2jq/' and its solution is y* The desired surface is a paraboloid of revolution. 2781. (x y) 2 C*/ = 0. Hint. Use the fact that the area x 2 )-=Cx x = C(2# + C). 2783. But tana
.
.
tan 2 q>
1
=
=
M
tancp ^
;
x
j/'.
*/*/'
bola. 2782.
2
2
2
.
(2(/
X is
to
equal
~--* 4 +
y dx.
\
2
2787. x
.
t>x and ~.
respect
2
2
2793.
r/
= Cx
2784. y
Y
x
_
In
|x
+ + cos y = C. 2
1
= Q/
f/
= xln-.
2789.
.
x (y
2
+C
+ C)
x(
2
2C|/)
=x
8
2
2808.
2799. x
= 01n
;
is
lnU|
.
=
with
*
2790.
#
=
singular
2
.v
1/
= C.
is
no
singular 2 /
2801.
2812.
= 0.
2810.
2813.
= 0.
2815.
integral.
1
General
integral
"o"""^~ r"^
General
x
integral
1/^3
1
2816.
C2 +
2
(A
HO singular integral. 2814. General integral there
1.
x*' =
2806.
~ + ~==C.
2809.
2
integral
+-=
2803.
v
2 singular integral x
^2Cx;
t/
linear
.
2800.
.
(xsiii[/-j-f/cost/~ sin^)c =tC.
= 0;
there
~+C J=0;
x
ry2.|-C
2
=2
2811.
c.
is
y=^+
2 arc tan-
//
e
2786.
.
2 2794. x ^=
= C.
-
~
ab
2
2
+ arc sin x)
= 0.
?807.
= Cx + x
The equation
Hint,
!/
2788. x
2785.
|.
f/^-^cosxi
-~-
sin x. 2817.
A
- + C,
Singular 2821.
In
solution:
r
V p
2
t/
= 0.
2820.
+ + arctan~ = C, 2
i/
4y = x
x^\n
y
1
^
+p
1 ,
In |p
p .
Singular
x|
= C+
solution:
.
y=e
Answers
464
2822.
2823
g-C*+;
y=2*.
L=4
l* =
nlPL-zIl'sin P + C '2825.<
from which x
equation
= *
.
y=xu\
j/
2829.
= Cx + C; */
= Cx +
The
Hint.
,
defined as a function of p
is
2827.
+ y*=\.
and the family of geneous, equation;
2824.
homogeneous. 2826.
is
no singular solution 2
-g
;
/
= 4*.
2830.
differential
2828.
JO/-C
A
2831.
circle
=
+
astroid x*" a z i>. 2833. a) Homoy*i> uv\ c) linear in y\ y uv, d) Bernoulli's with variables separable; f) Clairaut's equation; reduce
tangents. 2832. b) linear in x\ x its
The
=
=
y = uv\_e) to y^xy' V^y^l g) Lagrange's equation; differentiate with respect to x\ h) Bernoulli's equation; y = uv\ i) leads to equation with variable? separable; u--x-}-y\ j) Lagrange's equation; differentiate with respect to x\ k) Bernoulli's equation in r, x=uv\ 1) exact differential equation; m) linear; y = uv\ n) Bernoulli's equation;
+ = Q/ 4
2 2835. x
(/
1
y
2836.
.
= uv.
/=i~.
lnC; singular solution, solution,
y = ~.
2840.
.
3//
2842. y
=
2834. a)
i/
2837.
=
In
sin-^
xi/(C-~
f- (x+l)
.
jc|
In
1
+ C;
x)=
x^y-
b)
2838.
1.
y=-Cx+ VaC\ singular 2841. ***-^ -arc tan y~
2839.
+ L~1J = C. ln
= x(l+Ce*).
|
2843.
y=
1
Ar= ).
2844.
X -f2847.
x = Ce
sIn
^-2a(l+sini/).
2848.
1
i + Sx + y + ln [(x-3)
10 1
y-
8 1 |
]
= C.
2
-= In Cx.
2849. 2 arc tan
2852.
l- + r
cos*. + 4o
-l.
2860.
2863. y
=
X
ln|x|
2855.
xy
2858. x
= C.
= C(y
2850.
2853.
1).
(/
**=! ---\-Ce
= * arc sin (Cx).
2856. x
y .
2851. x 3
2854. y i ^Ce" ix
= C^-- (sin y + cos
= CeV-
=-Ce^
^
2859.
/).
+
y sin
5
2857.
2
jc+
py-
-0. 2
y x
ij
2879. f/-0.
+
y^Cx + -~,
2873.
i/
= ~^/2?.
+ 4# ^Q/ 2876. 0=-=* 2880. 2881. y = (sin x + cos x). i/^-^= = 2. 2883. a) y x; b) Cx, where C
^C.
9
2875.
2
p
2
3
.
1.
-i-
2874.
2877. r/-=x. 2878.
(2x
f
2
x'-f-x
+ 2x + l).
*/
= 2.
/
2882.
=
y
2x is arbitrary; the point (0,0) t/ 2 singular point of the differential equation. 2884. a) y 2 x\ h) # 2px; 2 2 2 2 2 (0,0) is a singular point. 2885. a) (x C) 4-(/ -=C b) no solution; c) x -=x; t/ ---=e-*
is
=
a
=
+
;
2891.
2-=0
.
.
8 i/
.
e~ x
--1
-C
f .v
+
.
A:
that
the
area
is
to
equal
:c
y dx
\
J
and
arc
the
length,
f
to
x=^^--f
Cf/.
2
2897.
//
H^l/ 2
|
-^^
/
o
<>
2896.
^
a singular point. 2886. // --e". 2887. y^(V~2a Kx) 2 2888. f/ 2 2889. r-=Ce^. Hint. Pass to polar coordinates. 2890. 3y 2 2 2 2892. x 2 -|- (// 2893. // 2 [- 16x^0. 2894. 6) -_6 Hyperbola r-^/fq) 2 2 2 --C or circle x i-j/ 2895. t/--i- (e-v e- x ). Hint. Use tha fact
is
(0,0)
- 4C (C
-j-
a
x).
2898. Hint. Use the fact that tha
resultant of the force of s>ravit\ and the centrifugal force is normal to the surface. 'lakmg the r/-axih as the nxis of rotation and denoting by co the angular velocity "of rotation, \\ e get for the plane axial cross-section of the desired surface
the differential equation g-'-'==o) 2 x.
2899. p_^ e --*
Hint,
The
prts-
sure at each level of a vertical column of air may be considered as due solely to the pressure of the upper-lying layers Use the law of Boyle-Mar otte, according to \\hich the density is proportional to the pressure. The sought-for differential
equation
.-.kw-dx.
2901.
is
s=/>-j--a; ~ /
one hour.
2904.
kpdh. 2900.
dp
3
will decay in 100 years. Hint.
=^35.2
dQ==
sec.
kQ
Equation
Hint. Equation n(/i f
dfi.
Q
=
Qo
2905.
K
m^ = mgkv
~=
:
t
*f~V
a)e~
Q-=Q (4-)
T
as
;v=
y
2910.
ds
Equation ki
In
2903.
.
the initial quantity
of
~V udf.
=n
2908< u "~^
2909. 18.1 kg. Hint. Equation
4 2?6
^ =kQ.
2fc)d/i
(y)"*
tionality factor). Hint. Equation
16-1900
rpm.
Hint.
T^a + (T Q
2902.
V
(o^lOO(-g-J
s^klw.
/
2906.
2907.
"^
'
n^-
.
^
is
__
t
=^=
Hint,
a
g~tanh ft I/
=
Q
~)
[(/? sin
wf
.
Answers
466
1
+ L~ = E sinco*.
Hint. Equation Ri
].
2912.
x
+C
z.
2914.
#=<:,
+ {;, In
2915.
= C ed?
e
y
2916.
l
2918.
(x
C,)
w
=
i/
=
/=
2913.
.
2911.
r/
= aln : 2 ;y
= C.
+C
in~|
2021.
2923.
2
(singular solu-
2925. y
tion).
= C x(x
/
l
8
2929.
=C x=C
2935.
2930.
1).
sin(C,
=x+
x^^
2933.
-
/
2y
2
4x z
2934.
=\.
y=^x
2
2941.
.
2944.
^
-'e-r
2947.
x=
2950.
2954. w
--e"^ 2
= + Cj + D
x2
+
n-
(^i
2951.
2
No
solution.
2932.
.
~ 2939
^8
-
-2)
(i/-|
*
2948.
x.
2
2
-
//-C, //
In
2938.
y
=
"--*
-
-
2943.
.
//
2028. y
.
//=-*-]-!.
2945.
1-e'
t/^sec .
x=
2942.
t/-
= Cx
Jt^C,
2937.
ln|
2940.
8
2931. y
l.
.
2936.
.
l
2926.
(singular solution).
y=
2927.
.
-
r/-^=
= ~- + C
-
y
2946
-
1 -
e*
/y
-"
2949. 2952. y
= ex
.
2953.
//-2
In
|
--
A' |
2
+C
o-Cj^+l)*
C 2 )A: + C 2
2.
Singular solution,
y^
4-
2959.
(x
(x.
.
Singular solution,
1
^ ;
r/
= C.
2955.
3
+C.
2956.
2957. -singular solution, t/=-
2960. Catenary,
x
2 )
r/
= 2at/~ a
=a
2 .
.
2958.
cosh
Cycloid, x
.
x
Circles.
Circle,
=a
(t
2
z
(x sin /).
y = a (1
C^-\-kC\
C,)
= a2
.
cos
2961. /).
<7
O. 2963. Parabola. 2964.
^-e""* +
2962.
e
"^
-0.
Parabola,
__ Answers
4- C*
x.
-
Xcosh The
differential
=g(sina
2
/
=
-fC 2 lnx.
2973. y the
= A + Bx
2
the substitution
2967. In
.
No, b) yes>
+
.
2972.
= C,x-f
f/
Hint. Particular so-
.
AT
/2
---
method
the
By
.
x3
C ^~-\-A, C z
find:
x
\"*/
= */,.
//
O
v
we
-}
(
y< 2968. a)
= --!- 4 * + ^L
2974.
-{-x\
homogeneous equation r/j.v,
variation of parameters
B
Use
Hint.
k
=
3y"+4i/~ 2//-0
y'"
(C,sm x-}-C 2 cosx).
^
~
2966. s
2969. a) t/'-{-y Q\ b) y" 2y' -f f/-=0; 3jt 5x f 2x 8 2971. //-= 2970. j/
no, g) no, h) yes
f)
mg
m-r^ at*
Hint*
E(l uation of
'
cos a)
u,
300 d~x ~A~Z~
Hint. Equation of motion,
2x//'4-2f/-0, d)
2965
s^^- (sin a
motion,
of
^' ^\T)'
Hint. Equation of motion,
).
d) yes, e) no,
yes, A-
l
y
tn/
6.45 seconds.
c)
g
r
= 77
^22
= a.
constant horizontal tension, and
a
is
Law
jiicosa).
\
c)
equation
I/
Xlncoshf/
H
H
where
|-C 2t
467
2975.
-\-B
l
y
the
of
= A-\-
= +
2X
Ccos x-Hn|sec x-f-tan ^I-f-sinx In cos v| xcosx. 2976. f/ C,e -h C 2 sin x. 2977. //^-C,e" 3 * {-C 2 e 3V 2978. // -=C t -|- C 2 e x 2979. */ C, cos x v 2980. //--e' (CicosA'-l-C 2 Sin A-) 2981. y=~ e~** (C t cos 3v-|-C 2 sin 3x) 2982. y -|
sin x-\ 8
H-C'jC
.
~
-
2983. i/-.--e- (C,e r
(C^i-Cx)^-* *
- CV*
-r
A
C,e"
.//-4tf 2992.
x
//=-
C
sin
^4cos2v
+
/J
y(A\ -\-B\i-C}
4- xe
av
^-i-3A 2998.
-I
//
--.
C 2 sin
C,c*
7 'v
I-
C2
x
sin x.
t>
+
(/;
C a x)
2995. [/^(C.-j
2xJ
nx
2
e
+2
-{-
--
z
f)
3001. y
,
16*
cos 2x 4-
C 2 sin
2x)
+
//
= a cosh
+ +
+
2
.
4 cos 2* h e)
e
x
;<
x \-F)x
'
+ 4x
(2x
a x)
2
-f-
3003.
!,
2996. y
3).
P~ X -[-{-
= ^ (C
ic
<>**.
~ xex
.
3000. y
= C, cos x-f-
2 -p-
o
(3 sin 2x-|-cos 2.v). 3002.
i/
=
== (C 1
3005.
sm2x). 1
sin 2x.
f/-
0,
sin
+ C, sin
i/=l. 2991.
;
s
x e
x
A;
= C,+C e- x + ~x + ^r.(2cos2x X
J-
+
a)
= C,e* + C e~ 2X .
y
C, cos
- C,
x
3004.
C2
-[-
>
ft
=
i/"TT
/ I
= sin 2x. 2990.
= (C, -f C
/
6
J^^fex
If
xe** (Ax* Bx C); b) fl sin x), d) g* (^ cos x xc* [(Av -j- fiv C) cos 2x+ (>x*-f
+1
2999. y
1
x
ax
2997.
.
r/
2994.
);
2984.
).
j/-=C, cos
+ CvV
2v
sin
+ C^~* rl
2
2986. //=.g
)
2988. f/-=e~*. 2989.
.
2993. c)
s
X sin
-|-C 2 e
+ e lv
/y-0
-|-/Jsin2x;
2
(C t e
]
0,
---x 2
* 2
(/=*
<
k
if
;
2985.
=
|
*
3006.
//
cos 2x -f
-^-
(sin
x
(/=
+ sin 2x).
__ Answers
468
$007.
1)
x
=C
A
1
cosco/
+ C 1 sinci)f + -j
-^slnp/;
2)
= Cj cos w/ + C, sin
x
2
$/
2
1
2
3010.
~
y^CfV + CStx**. 3009. ^ = 2X = **(<:! + C x + x 3011. = C + C e + -|x. yA-e = ^ + 4-( 3cos2x + sin2x 3013. y o
-f.cosorf. 3008.
2Jf
).
t/
3012.
2
1
CD/
)-
~ = (CjCos3jc + C sln3A;)gx + ~ = C, + C e * ~ (cos* + 3 sin*) 3018. + C x + *V w + ^4-^. 1U o
X*~* + |-e*.
= (C
3017. y
2
cosx.
+C
2
ie(4x + l)-^-^ + j.
3021. y sin
= C 6" * + C 1
1
2x
(3 sin 2*
-|-
2xcosx). 3024.
i/
3x)+~(2x
(2
2
x)e
8X .
= C e~~ x
+ 2cos2x).
2x
^. (sin
3023.
.
f
8;f
I)e
j(3x
= C, + C c
.
IJC
1
y
the
(2x
1
x -f
cos x
+
-j-
*/
(5, cos x
t/^C, cos 3x-
3028. y
+ x)^^ +
sin
cosines to the
of
product
=
ex
-x
2jce*-~ x
1
sin
3022.
3026. t/^
= C e-* + C ^-
y = Cj cos x + C 2
|/
x)e*. 3025.
2
J
3027. y
Transform
Hint.
-f r^sinSx.
IJC
a
+ 2 cos2x) + -j
3029.
8030.
3020.
= C e x H-C e- x + ~ (% +
X
2
*/
1
3019. y
= C, cos 2x
y
8
1
_^_*. xslnx
3016.
x
x
<
-^-
sum
of cosines.
+ C ex +xe*sin x + e* cos x. 3032. = C, cos x + C = tan ~ + --- 3033 + cos;cln cot x x x 3035. 3034. y = (C + C x)ex + xe \n\x\. x t/=^(C + C x)e-~ + xe^ -= 3036. 3037. y = Cj cos x C sin ^ + x sin x + cos x In cos x C, cos x + = + C sinx *cos* + sinxlnjsin*|. 3038. a) y C e* + Cf-* + (e* + e~*)x V 1/t Xarctan^; b) = C,^ +e*'. +Cf-* 3040._EquatIon of motion, 3031. y
*
2
l
t/
-
-
2
|
\Ti
2
l
-f-
s
I
2
l
2
/
|.
(/
|
2
/
.
l
*
-
2# -= -5
sin30/
t/
60
-~^
g^sin
900
"^7 cm
sec.
| .
.
ff
x-
.
.
t
rec koned
X
'
3041.
from the position of
4 k (x g is the distance of l), where x -jjt"=*4 the point of rest of the load from the initial point of suspension of the spring,
rest of the load,
I
is
the length of
then
the spring at rest;
+ xy
therefore, k (*
/)
= 4,
hence, -i^L*
8
k(x
y),
where *-4, ^ = 981 cm/sec
1 .
3042.
m
d*x.
dt*
= k(bx)k(b + x)
Answers
and x = ccos
=
) J
6~ In (6+ 1^35). 3044.2a)r= ^35). 3044. ~=r=gs; -gs;=: j/ y -In
3043. 6
.
t
2
e~ wi Hint. The differential equation of motion + e-"*); b) r = (e = o)V. 3045. y = C, + C,e* + C e 12X 3046. y = C, + C
.
(e'tf
2 is
y
t
f
469
)
2,(H
^ or
.
3
C 4 e" x 2 3049. 3048. i/ -= C, C 2 x C,e* 2 C 2 sin x) +e~ x (C 3 cos x -\3050. // c_e* (C, cos x C 4 x) sin 2x 3051 y (C, (C, -j- C 2 x) cos 2x ^ / i/T
+
+
=
.
3065. w--_C e" x 1
3066.
//
= Cj
4-
+C
2
y=-e~
+
/
x
2
-\-e
(
sin
3068. y -- (C,
2
3073.
3075. 3077.
j/
C2
In x)
C, cos (2 In x)
3070. y
3071.
-I-
= C,JC-hC
2
~ +C
x 2 |-C 3 x 3
r/^Qx-H-
.
2 .
3074.
.
x
-1
sec x
|/-j
cos
V
v
t
3069.
.
8 /
+ cos x In i/-j
j
xH-
\
I/"Q-
~ \ 4
p^sin ^r- x
cos v
|
tan x sin x |
+ x sin x.
\ )
2 / /3 y C,x + -%
2.
-|-x
3
.
sin (2 In x). 4
3072.
/=- C,-|-C 2 (3jcH-2)-
f/^Q cos
(In x)-f
C 2 sin
/3
.
(In x).
= (^+l)MC + C ln(jc-H)] f (x-}-l) = C x + C x +-o * 3076 2 = x(lnx + ln x). 3078. y = C, cos x + C sin x, z = C cos x C sinr x y = e~ (C cosx + C 2
3
i/
3
-
2
1
!/
1
1
2
2
f/
i
3079.
-^* C 4 sin x)
+
c,sinjcH-g
c(;sx-l
C 2 cos x-\-C 3 -
3067.
V
+
+
3
2
3080. y
l
=
(C,
Cj
Qx) e
2
t
Answers
470
3081. x
=C
L
I/-,-
f
2 e
l
8
(^C
3082.
\
I/"Q~
cos^* + C sin-^
2
x^CV' = = C + C + 2sinx, z= 2C, 2C x)
tj,
3083. y 3084.
3085.
/
2
1
2
2
e~ x 2x(3 + 4e-*), = 10^~8e ~^ + 6/-l; y=
0=14(1 3086. x
b) In
+ = arc
2
2
f/
dz
-
hence,
r
y% ..
2 r
2
/
Vx +y 2
we
^2
2
the
find
xdx -\-udy
v=
.
==C 2
.
,
,
Z |f
.
c)
;
x
+ (/-|-z
..
..
whence dx + dy+d*
2
+ +z 2
2
-= dx
we have
-- Q and,
1
.
,
In (* v
=
J/
9 2
+ ,
-ryw
Hint.
6.
dz
cf/y
y_ 2 z _ x consequently, x + + z = C >
jc
]/
-f
:
x __y
9v 2 ) /
p
2
we have
of derivative proportions,
//
homo-
2
In
integral
,
A;
10.
Hint. Integrating the
Whence lnz^=
.
of derivative proportions, properties r r t
&+ 12/+
first
....
0,
f x (5 + 4^-
*"*)
(1
%&* + &** +
Then, using the properties
y dy /
+i/
x
=
+ C,.
= -; xdx r=
+C
--
geneous equation
= arcxtan
tan
9
z
)
8t
]^x
C 2 (2A:+ 1) 3 sin x 2 cos *. z^Q-fCj,*) e-* + 5x 9;
A;
.
+ ln C, and, ,
,
2
r
the
Applying
= dx+dt/-\-dz
-
;
Q
x dx 1
.
Similarly,
-
-
x dx -\- if dy ~\- z dz n z(x y) 2 C 2 Thus, the integral curves are the circles x + // + z C ,;t 2 -f// 2 -}-z 2 Co 4-z From the initial conditions, Jt=l (/=!, z 2, we will have C l = 0, C 2 = 6.
y(zx) .
1
f
3089.
f
/==
z= 3090. y
C 1
1
*2
+
2C.A;
~^ +
+ -| (3 In * + In x2
f-
1).
= C^ V
Solution.
mjj-kv x m~-^kvy \
mg
for the initial conditions:
when
Answers x
0,
r
0,
I/Q
*
+ mg = (kv
kv v
,
^
sinTr
t,
Vm
2-2*
3093. [/-
x2
,
3097.
X2 ^ z
+ *% x
r/
-j
1
^~
x'
3101. //m
1
_|
~~t o
4 t
4
9
lions:
3104.
.
u(0,
-
//
-.2
JT
A
-]-~t o
7
;
;
3105. u
= -4-
"(^.
2
--qi~~ x
the
:
(x,
JfH-~
.
9 -
^ ie
'
method
for
l
conver ^ es for
serics
undetermined
of '
"-"
coefficients.
series converges for
oo
undetermined coefficients.
of
SCneS
the
..
.}
.
3103.
I
0- -0,
(/.
M (x, 0) -=
cos /m) sin
(1
^
Conver 6 es
^-
A
sin
sin
x-
3102.
M-^cos^sin^.
for
a
'l<-
v l
2
l~oT*
~r-
Hint. Use the condi-
I
,
-^-
Hint.
.
Use
the
conditions:
/
/
u(x,
~y
cos
~T"
sin
~T~
"
Hint
n-\
du
^
undetermined coefficients.
of
9
sin
1)
v
O-o,
^~
=
,'+...
J
tt<"/l 2
= 0.
l
2ig+ '"'
^""
!
0----0. "
w(0,
*
4
+ r
!
!
x
-}-
method
Hint. Use the 2
8
Hint. Use the
.
y
equations of motion:
...' the series converges
r
Use the method x
u~
^=-,1,
X4
77
Hint.
-
+
=
k
= acos
3092. x
.
differential
.\ e 2
Q
+
X3
2
3099. //--I
3100. r/--'
y=(y +
3094.
.
+ o4 + + pr-r Z*3
'
3098. f/-^x
The
^ ^^ 1^ .-.I^+-
,
3096. /t
/
+ mg)e m
Hint.
vx
.
mu 2
yl + lx +
3095.
s\na
1.
-\
a-
Q
we obtain
Integrating,
Q ft
f
*
sina.
i>
,
^
'"
= -~
UV
cosa,
i;
Xo
471
'
^ se
~
llie
conf^ tlons:
forO
0)
3106. u
=--t A n cos n=o
^~ r ^
}
"" sin
n **
^"^
,
where
the
coefficients
/!=
Answers
472
(2
,*. f
"+ 1)JtJf dx.
Hint. Use the conditions
^-sin .
400 7 >u===
\^
_2^
1
,, -
3
(1
<,.a)*
-o.
0=0.
to,
v
nnx
.
cos rui) sin
Hint. Use the" conditions: u
(0,
ioo a
e j-rrr-
/)-=0,
u (100,
Chapter
= 0,
u
(x, 0)
= 0.01
x(100
x).
X
<1
<1
3108. a) <1"; <0.0023/ mm; <0.26/ gm; <0.0016/ . c) b) 3109. 0.0005; <1.45/ 0.005; <0.16/ c) b) a) <0.05; <0.021/ ; 8 since the number lies between 47,877 3110. a) two decimals; 48-10* or 49- 10 and 48,845; b) two decimals; 15; c) one decimal; 6*10 2 For practice1 -purposes 2 3111. a) 29.5; there is sense in writing the result in the form (5.90.1)- 10 2 or 18.5 1.643.2. 3112. 10 18470.01; 84.2; b) c) the result of a) c) b) subtraction does not have any correct decimals, since the difference is equal with a possible absolute error of one hundredth. to one hundredth 2 Hint. Use the formula for increase in area of a square. 3113*. 1.80.3 cm 2 19.90.1 3115. 3114. 0.30.1. 30.00.2; 43.70.1; c) b) a) 3116. a) 1.12950.0002; b) 0.1200.006; c) the quotient may vary between 48 and 62. Hence, not a single decimal place in the quotient may be considered certain. 3117. 0.480. The last digit may vary by unity. 3118. a) 0.1729; 2 3 s 3120. a) 1.648; b) 4. 025 0.001; b) 277- 10 ; c) 2. 3119. (2.050.01)- 10 cm s cm 2 Absolute error, 65 cm 2 Relative error, c) 9.0060.003. 3121. 4.01- 10 0.2 cm; sina 0.440.01, a-2615' 0.16/ 3122. The side is equal to 13.8 3125. The length of the pendulum 35'. 3123. 270.1. 3124. 0.27 ampere should be measured to within 0.3 cm; take the numbers ;t and q to three decimals (on the principle of equal effects). 3126. Measure the radii and the generatrix with relative error 1/300. Take the number n to three decimal places (on the principle of equal effects). 3127. Measure the quantity / to within and s to within 0.7/ (on the principle of equal effects). 0.2/ 3128. ;
;
<
<
;
.
,
.
.
;
.
m
.
.
.
,
+
=
.
.
Answers
473
J129.
3130.
the
4 Hint. Compute the first live values of y and, after obtaining A {/o^24, repeat number 24 throughout the column of fourth diilerences. After this the
remaining part of the table from right to left).
is
tilled
in
by the operation
of
addition (moving
Answers
474
b) 0. 229; 0.399; 0.491; 0.664. 3132.
3131. a) 0.211; 0.389; 0.490; 0.660;
=~
1822;
^
x* x*. 3134. y x4 0.1993; 0.2165; 0.2334; 0.2503. 3133. \+x oe QC JC x2 20 for x^5.2. Hint. When computing 8; J/^ 22 fr x 5.5; ^T T2 x2 20 take 11. 3135. The interpolating polynomial is y A; for 10x4- 1; 0=1 when x 0. 3136. 158 kg! (approximately). 3137. a) 0(0.5)-= 1,
+ +
=
=
+
+
=
0(2)=11;
= -,
b) 0(0.5)
J/t 2
)^
3
3138
-
I-
-
325
1.01.
3139.
3140. 019. 3143. 0.31 and 4 0.25; 2.11. 3141. 2.09. 3142. 2 45 and 1.86; 3144. 2.506. 3145. 0.02. 3146. 024. 3147. 1 27 3148. 35; 1 53 1.88; 3149. 1.84. 3150. 1.31 and 0.67. 3151. 7.13. 3152. 0.165. 3153. 1.73 and 0. 3154. 1.72. 3155. 138 3156. x 0.56 0.83; 0.83; i/= 056; 1 1997 3160. By the trapezoi3157. *=1.67; 22. 3158. 4 493. 3159. dal formula, 11.625; by Simpson's formula, 11 417. 3161. 995; 1; 0.005; 3164. 0.79. 3163. 069 1.3-10- s A^O.005. 3162. 0.3068; A 0.5/ 3166. 0.28. 3167. 0.10. 3168. 1 61. 3170. 0.09. 3169. 1.85 3165. 0.84. 3171. 0.67. 3172. 0.75. 3173. 0.79. 3174. 4.93. 3175. 1 29. Hint. Make use of the parametric equation of the ellipse x cost, 0-= 0.6222 sin/<:nd trans-
=
=
x-
0=1
=
;
.
_JT
2
foim
is
the
formula
of the arc length to the
the eccentricity of the ellipse. 3176. y l (x)
x7
xn
x 15 -
63 v3
o?9
3177
59535-
Qy2
T+T""* -2.
+1: 3178.
z
'
^
x2
W=~ T~
iW^ - 2 3x
1
= x, (x)
f/ 2
=x
\
3
o
'
e 2 cos 2 /-d/,
}/~l
3
,
y2
(x)
xs
X
o
3x 2
*T T
'
W-|
=x (x)
|,
2x 2
where
7
= xT + x^
uo
e
x3
,
y3
(x)
To +
~ J X3
yl
22
=
= 3.36.
form
+ 3x-2,
Zl
(*)=-i
^(x)-x~ + ^. =
3180. 3181. z (1)==2. 72 0.80. 3.72; 0(2) 0(1) 3 15. 0-1.80. 31S3. 3.15. 3184. 0.14. 3185. 0(0.5) -3 15; z (0 5)0.18. 3187. 1.16. 3188.0 87. 3189. x (n) -3.58; 0(0.5)^0.55; z (0 5) 3190. x' (ji) -=0.79. 429+ 1739 cos x 1037 sin x 6321 cos 2x -f- 1263 sin 2x 3191. 96cos x-j-2. 14 sin x 1242cos3x 33sm3x. 1.68 cos 2x 0. 53 sin 2x 1.13 cos 3x 0.04 sin 3x. 3192. 0.960 0.851 cos x 0.915 sin x -|608 sin x [-f-0. 542cos2x 4-0. 620sin2x -[-0.271 cos 3x -|-0. 100 sin 3x. 3193. a) 0.414 cos x 4- 0.1 11 cos 2x |-0.056cos 3x. 4- 0.076 sin 2x4- 0.022 sin 3x; b) 0. 338
3179. 3182. 3186.
0(1)
=
6491
+
+
+
+
+
+
APPENDIX
I.
II.
Greek Alphabet
Some Constants
476
Appendix
III.
Inverse Quantities, Powers, Roots, Logarithms
Appendix
477
Continued
478
Appendix IV.
Trigonometric
Functions
Appendix V.
Exponential, Hyperbolic and Trigonometric Functions
479
480
Appendix
VI.
Some Curves
(for
Reference)
-f
1.
-/ 4.
2.
Parabola,
Cubic parabola,
1
Graph
3.
Rectangular hyperbola,
-/
of a fractional
5.
The witch
of Agnesi,
function,
6.
Parabola (upper branch),
7.
Cubic parabola,
481
Appendix
Semicubical
8/7
parabola, 2
8a.
=x
^ *
or
Neile's p arabola, 2 (
y-^-c'
x^t*
or
_
9.
:
Sine curve and cosine curve,
y
10.
--sin
.v
and
//
=
Tangent curve and cotangent curve, and / cot#.
=
Appendix
482
11.
Graphs
of the functions
//-=secx and
y=arc
f/
sin x-
y^arc cos
x
Ji
x
'
A *
arc cos
\ >
y arcsinx
Graphs of the inverse trigonometric functions # = arc sin x and y~ arc cos x.
x
483
Appendix
T
13.
Graphs
of the
//
= arccot#.
.
14.
Graphs f/
of the exponential
= g*
and y
= e~*.
x
cot
x
"V \^^ arccot
inverse trigonometric functions
//=-arc tan x and
arc cot
functions
-
486
Appendix
\
25.
Bernoulli's lemniscate,
24. Strophoid, *
u2 y
= ATs^-M
ax !
.
27.
Hypocycloid i
=
\ y or
26. Cycloid,
x
= a(f
y
a
(1
A
(astroid),
a cos 3
x
fl
sin 8
2
2
S
8
-f/y
/,
/
=a v
.
sin*),
cos
28. Cardioid,
= a(l+coscp).
/)
29.
Evolvent (involute)
of
the circle
sin*)*
{;::
487
Appendix
31.
^30. Spiral of Archimedes,
32.
Logarithmic r
= *'?.
33.
spiral,
Hyperbolic spiral, a
Three-leafed rose, r a sinSip.
=
34. Four-leafed rose, r asiti2(p.
=
INDEX
B Absolute error 367 Absolute value
number
of a real
Bending point 84 11
Absolutely convergent series 296, 297 Acceleration vector 236 Adams' formula 390 Adams' method 389, 390, 392 Agnesi
Witch
of 18,
Bernoulli's equation 333 Bernoulli's lemniscate Beta-function 146, 150
155,
Binormal 238
Boundary conditions 363 Branch of a hyperbola 20, 480 Broken-line method Euler's 326
156,480
Algebraic functions 48
Angle between two surfaces, 219 Angle of contingence 102, 243 Angle of contingence of second kind 243 Antiderivative 140, 141 generalized 143 Approximate numbers 367 addition of 368 division of 368 multiplication of 368 powers of 368 4 roots of 368 subtraction of 368
Approximation successive 377, 385
Arc length Arc length Archimedes
of a curve of a space
158-161
curve 234
spiral of 20, 65, 66, 105, 487 Area in polar coordinates 155, 256 Area in rectangular coordinates 153, 256 Area of a plane region 256 Area of a surface 166-168, 259
Argument
11
Astroid 20, 63,
105,
Asymptote 93 horizontal 94 inclined 94 right horizontal 93 right inclined 93 left left
vertical
93
486
Cardioid 20, 105, 486 Catenary 104, 105, 484 Catenoid 168 Cauchy's integral test 295 Cauchy's test 293, 295 Cauchy's theorem 75, 326 Cavalieri's "lemon" 165 Centre of curvature 103 Change of variable 211-217 in a definite integral 146 in a double integral 252-254 in an indefinite integral 113
Characteristic equation 356 Characteristic points 96 Chebyshev's conditions 127
Chord method 376 Circle 20, of of
104
convergence 306 curvature 103
osculating 103 Circulation of a vector 289 Cissoid 232 of
Diodes
18,
485
Clairaut's equation 339 Closed interval 11 Coefficients Fourier 318, 393, 394 Comparison test 143, 293, 294 Composite function 12, 49
486
Index
Concave down 91 Concave up 91
probability 19, 484 sine 481 tangent 481
Concavity
Cusp 230 Cycloid 105, 106, 486
direction of 91
Conchoid 232 Condition Lipschitz
489
385
Conditions
boundary 363 Chebyshev's 127 Dirichlet 318, 319 initial 323, 363
D'Alembert's test 295 Decreasing function 83 Definite integral 138 Del 288
Conditional extremum 223-225 Conditionally (not absolutely) convergent series 296 Contingence angle of 102, 243 Continuity of functions 36 Continuous function 36 proper lies of 38
Dependent variable
Convergence circle of 306
interval of 305 radius of 305
region of 304
uniform 306 Convergent improper 270 Convergent series 293 Coordinates
integral
of centre of gravity 170 generalized polar 255 Correct decimal places in sense 367 Correct decimal places in a narrow sense 367 Cosine curve 481 Cotangent curve 481
a
143,
broad
Coupling equation 223 Critical point of the second kind 92 points 84 Cubic parabola 17, 105, 234, 480 Curl of a vector field 288 Critical
Curvature centre of 103 circle of 103 of a
curve 102, 242
radius of 102
second 243
Curve cosine 481
cotangent 481 discriminant 232, 234 Gaussian 92 integral 322 logarithmic 484
11
Derivative 43 left-hand 44 logarithmic 55 nth 67 right-hand 44 second 66 Derivative of a function in a given direction 193 Derivative of functions represented parametrically 57 Derivative of an implicit function 57 Derivative of an inverse function 57 Derivative of the second order 66 Derivatives of higher orders 66-69 one-sided 43 table of 47 Descartes folium of 20, 21, 232, 485
Determinant functional 264
Determining coefficients first method of 122 second method of 122 Diagonal table 389 Difference of two convergent
series
298
Differential of an arc 101, 234 first-order 71
higher-order 198 principal properties of 72 second 198 second-order 72 total,
integration
of
202-204
Differential equation 322 homogeneous linear 349 inhomogeneous linear 349 Differential equations first-order 324 forming 329 higher-order 345
linear 349, 351
490
Index
Differential equations of higher powers first-order
337
Differentials
method
343 and higher orders 72 Differentiating a composite function 47 Differentiation 43 of implicit functions 205-208 of
of third
tabular 46 Diocles cissoid of 18, 485
Direction of concavity 91 Direction field 325 Dirichlet conditions 318, 319 function 40 series 295, 296
theorem 318 Discontinuity 37 of the first kind 37 infinite 38 removable 37 of the second kind 38 Discontinuous function 270 Discriminant 222 Dicriminant curve 232, 234 Divergence of a vector field 288 Divergent improper integral 143, 270 Divergent series 293, 294
Domain Domain
11
of definition 11
Double in
integral 246 curvilinear coordinates
in polar coordinates 252 in rectangular coordinates
253 246
coupling 223 differential 322 Euler's 357 exact differential 335 first-order
differential
324
homogeneous 330, 351, 356 homogeneous linear differential 332, 349
inhomogeneous 349, 351, 356 Lagrange's 339 Laplace's 289, 291 linear 332 of a normal 60, 218 of a tangent 60 of a tangent plane 218 with variables separable Equivalent functions 33 Error absolute 367 limiting absolute 367 limiting relative 367 relative 367 Euler integral 146 Euler-Poisson integral 272 Euler's broken-line method Euler's equation 357 Even function 13 Evolute of a curve 103 Evolvent of a circle 486 Evolvent of a curve 104 Exact differential equation Exponential functions 49,
Exkemal point Extremum
327,
328
326
335 55,
483
84
conditional 223-225 of a function 83, 83, 222
Double point 230 Factor
Elimination
method
of
359
Ellipse 18, 20, 104, 485
Energy kinetic 174
Envelope equations of 232 of a family of plane curves 232 Epicycloid 283
Equal
effects
principle of 369
Equation Bernoulli's 333 characteristic 356 Clairaut's 339
integrating 335 Field direction field 325 nonstationary scalar potential vector 289 scalar 288 solenoidal vector 289 Field (cont) stationary scalar or vector 288 Field theory 288-292 First-order differential First-order differential
Flow
lines
Flux of Folium
288
vector 288
71
equations 324
288
a vector field of
or vector
288
Descartes 20, 21, 232, 485
Index
Force lines 288
491
logarithmic 49 transcendental, integration of 135 trigonometric 48 trigonometric, integrating 128, 129 Fundamental system of solutions 349
Form Lagrange's 311
Formula Adams' 390 Green's 276, 281, 282 Lagrange's 145 Lagrange's interpolation 374 Leibniz 67 Maclaurin's77, 220 Newton-Leibniz 140, 141, 275 Newton's interpolation 372 Ostrogradsky-Gauss 286-288 parabolic 382 Simpson's 382-384 Stokes' 285, 286, 289 Taylor's 77, 220 trapezoidal 382
Gamma-function
Formulas reduction 130, 135 Fourier- coefficients 318, Fourier series 318, 319 Four -leafed rose 487
146,
150
Gaussian curve 92 General integral 322 General solution 359 General solution (of an equation) 323 General term 294 Generalized antiderivative 143 Generalized 255 polar coordinates Geometric progression 293, 294 Gradient of a field 288 Gradient of a function 194, 195 of a function 12 Greatest value 85, 225, 227 Green's formula 276, 281, Guldin's theorems 171
Graph 393,
394
282
Fraction
H
proper rational 121
Function 11 composite 12, 49 continuous 36
Hamiltonian operator 288
Harmonic
continuous, properties decreasing 83 Dinchlet 40 discontinuous 270
of
38
even 13 of a
Homogeneous
function 12
implicit 12 increasing 83 Lagrange 223, 224
multiple-valued periodic 14 single-valued vector 235
11
11
Functional determinant 264 Functional series 304 Functions algebraic 48 equivalent 33 exponential 49, 55, 483 hyperbolic 49, 484 hyperbolic, integration of 133 inverse
series 294, 296, 297 Higher-order differential 198 Higher-order differential equations 345 Higher-order partial derivative 197 Holograph of a vector 235 Homogeneous equations 330, 351, 356
12
Functions (cont) inverse circular 48 inverse hyperbolic 49 inverse trigonometric 482, 483 linearly dependent 349 linearly independent 349
linear
differential
equation 332, 349
Hyperbola
17, 18, 20,
485
rectangular 480 Hyperbolic functions 49, 484 integration of 133
Hyperbolic spiral 20, 105, 487 Hyperbolic substitutions 114, 116, 133 Hypocycloid 283, 486 I
Implicit function 12
Improper integral convergent 270 divergent 270 Improper multiple integrals 269, 270 Incomplete Fourier series 318, 319 Increasing function 83 Increment of an argument 42 Increment of a function 42 Independent variable 11 Indeterminate forms evaluating 78 79 t
492
Index
Infinite discontinuities 38 Infinitely large quantities 33 Infinitely small quantities 33 Infinites 33
Interpolation formula Lagrange's 374 Newton's 372 Interval of calculations 382 closed 11
Infinitesimals 33 of higher order 33 of order n 33 of the same order 33 Inflection
of of
Interval (cont)
points of 91
open
Inhomogeneous equation 349, 351, 356 Inhomogeneous linear differential equation 349
363 322 convergent improper 143 definite 138 divergent improper 143 double 246 Euler 146 Euler-Poisson 272 general 322 improper multiple 269, 270 line 273-278 particular 322 probability 144 singular 337 surface 284-286 triple 262 Integral curve 322 Integral sum 138 Integrating factor 335
Initial conditions 323,
Integral
Integration basic rules of 107 under the differential sign direct 107 by parts 116, 117, 149 path of 273, 274, 280
1 1
interval 372 circular functions 48 functions 12
hyperbolic functions 49 interpolation 373 trigonometric functions 482,
483
Jacobian 253, 264
Kinetic energy 174
109
of differential equation of power series 361, 362 Integration of functions
Integration
by means
differential
Integration of total differentials 202-
204 Integration of transcendental functions 135
Interpolation of functions 372-374 inverse 373
table Inverse Inverse Inverse Inverse Inverse
Involute of a circle 20, 106, 486 Involute of a curve 104 Isoclines 325 Isolated point 230 , Iterative method 377, 378, 380
region of 246-248 by substitution 1 13
numerical 382, 383 Integration of ordinary equation numerical 384-393
convergence 305 monotonicity 83
Lagrange's equation 339 Lagrange's form 311 Lagrange's formula 145 Lagrange's function 223, 224 Lagrange's interpolation formula 374 Lagrange's theorem 75 Laplace equation 289, 291 Laplace transformation 271 Laplacian operator 289
Lamina coordinates of the centre of gravity of a, 261 mass and static moments of a 260 moments of inertia of a 261 Least value 85 Left-hand derivative 44 Left horizontal asymptote 94 Left inclined asymptote 94 Leibniz rule 67, 269 Leibniz test 296, 297
linear 13, 372
Lemniscate 20, 105, 232 Bernoulli's 155, 486 Level surfaces 288
quadratic 372
L'Hospital-Bernoulli
rule
78-82
493
Index of
385,
Limit of a sequence 22 Limiting absolute error 367 Limiting relative error 367 Limits
Minimum Mixed
of inertia 169 static 168
straight 17, 20 Line integral
Monotonicity
application of 276, 283 of the first type 273, 274, 277, 278 Line integral of the second type 274, 275, 278-281 Linear differential equations 349, 351 Linear equation 332 Linear interpolation 372 of a fa nation 13 Linearly dependent functions 349 Linearly independent functions 349 Lines flow 288 vector 288 Lipschitz condition 385
21,
105,
point of a function 151
Mean-value theorems 75, Mean rate of change 42 Method 1
150
389, 390, 392
chord method 376 of differentials 343 of elimination 359
Method
nth derivative 67
Nnbla 288 Napier's number 28 Natural trihedron 238 Necessary condition for for
convergence an extremum
106,
Newton-Leibniz formula 140, 141, 275 Newton's interpolation formula 372 Newton's method 377, 379 Newton's serpentine 18 Niele's parabola 18, 234, 481
Maclaurin's formula 77, 220 Maclaurin's series 31 1, 313 Maximum of a function 84, 222
Adams
N
trident of 18
derivative 55 functions 49
M
Mean value
11
Multiplicities root 121
Newton
curve 484
spiral 20,
intervals of 83 Multiple-valued function
293 Necessary condition 222
force 288
Maximum
point 84
partial derivative 197
Moment
one-sided 22
Line
Logarithmic Logarithmic Logarithmic Logarithmic 487
successive
approximation 381, 389 of tangents 377 of undetermined coefficients 121, 351 of variation of parameters 332, 349, 352 Minimum of a function 84, 222
Pascal's 158 Limit of a function 22 Limit on the left 22 Limit on the right 22
(cont)
Euler's broken-line 326 iterative 377, 378, 380 Milne's 386, 387, 390 Newton's 377, 379 Ostrogradsky 123, 125 Picard's 384, 385 reduction 123
Runge-Kutta 385-387, 390
Node 230 Nonstationary scalar or vector Normal 217 to a curve 60 equations of 218 principal 238 Normal plane 238
field
288
Number Napier's 28 real
11
Number
series 293 Numerical integration of functions 382, 383 Numerical integration of ordinary
differential
equations 384-393
One-sided derivatives 43 One-sided limits 22
Open
interval 11
494
Index
Operator
Hamiltonian 288 Laplacian 289 Order of smallness 35 Orthagonal surfaces 219 Orthagonal trajectories 328 Osculating circle 103 Osculating plane 238 Ostrogradsky-Gauss formula 286-288 Ostrogradsky-Gauss theorem 291 Ostrogradsky method 123, 125
Parabola 17, 20, 104, cubic 17, 105, 234
105,
Niele's 18, 234, 481 safety 234 semicubical 18, 20, 234,
480, 485
481
Parabolic formula 382
critical 84 stationary 222, 225 Polar subnormal 61 Polar subtangent 61 Potential (of a field) 289 Potential vector field 289
Power
series
305
Principal normal 238 Principle of equal effects 369 Runge 383, 386 of superposition of solutions Probability curve 19, 484 Probability integral 144
353
Product of two convergent series 298 Progression geometric 293, 294 Proper rational fraction 121 Proportionate parts rule of 376
Parameters variation of 332, 349, 352 Parametric representation of a function 207
Quadratic interpolation 372 Quadratic trinomial 118, 119, Quantity
Partial derivative
hirheg-order 197 "mixed" 197 second 197
infinitely large 33 infinitely small 33
Partial sum 293 Particular integral 322 Particular solution 339 Pascal's lima^on 158 Path of integration 273, 274, 280
Period of a function 14 Periodic function 14 Picard's method 384, 385
Plane normal 238 osculating 238 rectifying 238 tangent 217 Point bending 84 the second (of discontinuity 37 double 230 extremal 84
critical of
of inflection 91
isolated 230
maximum minimum
84 84 singular 230 stationary 196 of
tangency 217
Points characteristic 96
123
kind)
92
Radius of convergence 305 Radius of curvature 102, 243 Radius of second curvature 243 Radius of torsion 243 Rate of change of a function 43 mean 42 Ratio (of a geometric progression) 294 Real numbers 11 Rectangular hyperbola 480 Rectifying plane 238 Reduction formulas 130, 135, 150 Reduction method 123 Region of convergence 304 Region of integration 246-248 Relative error 367
Remainder 31 Remainder of a series 293, 304 Remainder term 311 Removable discontinuity 37 1
Right-hand derivative 44 Right horizontal asymptote 93 Right inclined asymptote 93 Rolle's theorem 75 Root multiplicities 121
495
Index
four-leafed 487 three-leafed 20, 487 Rotation (of a vector field) 288
Solenoidal vector field 289 Solution (of an equation) 322 general 323, 359 particular 339
Rule
Spiral
Rose
Leibniz 67, 269 1'Hospital-Bernoulli 78-82 of proportionate parts 376
method 385-387, principle 383, 386
Runge-Kutta
Runge
of
390
Safety parabola 234 Scalar field 288
Subnormal
Scheme twelve-ordinate 393-395
Second curvature 243 Second derivative 66 Second deferential 198 Second-ordeP differential 72 Second partial derivative 197 Segment of the normal 61 Segment of the polar normal 61 Segment of the polar tangent 61 Segment of a straight line 20 Segment of the tangent 61 Semicircle 20 Semicubical parabola
Archimedes 20, 65, 66,
18, 20, 1M4, -181
Series
convergent 296, 297 with complex terms 297 conditionally (not absolutely) convergent 296 convergent 293 absolutely
Scries (cont) Dirichlet 295, 296 divergent 293, 294 Fourier 318, 319 functional 304
harmonic 294, 296, 297 incomplete Fourier 318, 319
487
61
polar 61 Substitutions
hyperbolic 114, 116, 133 trigonometric 114, 115, 133
Subtangent 61 polar 61
Successive
method
approximation 377, 384, 385, 389
385
of
Sufficient conditions (for an
extremum)
222
Sum integral 138 partial 293 of a series 293, 304 of two convergent series 298 Superposition of solutions principle of 353 Surface integral of the first type 284 Surface integral of the second type 284 Surface integrals 284-286 Surfaces
level
288
orthogonal 219
Table
Maclaurin's 311, 313 number series 293 operations on 297 power 305 Taylor's 311, 313 Serpentine
diagonal table 389 of standard integrals 107 Table interval 372 Tabular differentiation 46 Tacnode 230
Newton's 18 Simpson's formula 382-384
point "of 217 Tangent 238 Tangent curve 481 Tangent plane 217 equation of 218 Tangents method of 377 Taylor's formula 77, 220
Sine curve 481 Single-valued function 11 Singular integral 337 Singular point 230 Slope (of a tangent) 43 Smallest value 225, 227
105,
hyperbolic 20, 105, 487 logarithmic 20, 21, 105, 106, 487 Static moment 168 Stationary point 196, 222, 225 Stationary scalar or vector field 288 Stokes' formula 285, 286, 289 Straight line 17, 20 Strophoid 157, 232, 234, 486
Tangency
Index
496 Taylor's series 311, 313
Term general 294 remainder 311 Test d' Alembert's 295 Cauchy's 293, 295 Cauchy's integral 295 comparison 143, 293, 294 Leibniz 296, 297
Weierstrass' 306
computing volumes by means evaluating a 265
of
268
in rectangular coordinates 262 Trochoid 157 Twelve-ordinate scheme 393-395
U Undetermined coefficients method of 121, 351 Uniform convergence 306
Theorem Cauchy's
75, 326
Dirichlet's 318
Theorem
(cont)
Lagrange's 75 Ostrogradsky-Gauss 291 Rolle's 75
Theorems Guldin's 171
mean-value Theory
75, 150
288-292 Three-leafed rose 20, 487 Torsion 243 field
Tractrix 161 Trajectories
orthogonal 328 Transcendental functions integration of 135 Transformation Laplace 271 Trapezoidal formula 382 Trident of Newton 18 Trigonometric functions 48 integrating 128, 129 Trigonometric substitutions 114, 115, 133 Trihedron natural 238 Trinomial quadratic 118, 119, 123 Triple integral 262 applications of 265, 268 change of variables in 263
Value greatest 85, 225, 227 least
85
mean
(of a function) 151, 252 smallest 225, 227
Variable
dependent 11 independent 11 Variables separable an equation with 327, 328 Variation of parameters 332, 349, 352 Vector acceleration 236 of binomial 238
normal 238 tangent line 238 velocity 236 Vector field 288 Vector function 235 Vector lines 288 Velocity vector 236 Vertex of a curve 104 Vertical asymptote 93 Vertices of a curve 104 of principal
of
Volume Volume
of a cylindroid 258 of solids 161-166
W 1
Weierstrass test 306 Witch of Agnesi 18, 156, 480
Work
of a force
174, 276,
277