Higher_Mathematics_Part_2

ISBN 978-966-598-316-4

© V. P. Denisiuk, V. G. Demydko,

V. K. Repeta, 2009

ISBN 978-966-598-476-4

© NAU, 2009

INTRODUCTION................................................................................................

4

Module 1

5

FUNCTIONS OF SEVERAL VARIABLES AND COMPLEX NUMBERS .....

Topic 1.

5

Functions of several variables. Limits, and continuous functions ..................

Topic 2.

15

Partial derivatives and differentials of a function of several variables .........

Topic 3.

33

Applications of partial derivatives................................................................

Topic 4.

52

Complex numbers.........................................................................................

Module 2

63

INTEGRAL CALCULUS..................................................................................

Topic 1.

63

Indefinite integral .........................................................................................

Topic 2.

88

Polynomials. The rational functions .............................................................

Topic 3.

99

Integration of rational functions by partial fractions ....................................

Topic 4.

114

Integrals involving powers of trigonometric functions...............................

Topic 5.

128

Integration of irrational functions...............................................................

Topic 6.

144

Definite integrals ........................................................................................

Topic 7.

155

Improper integrals.......................................................................................

Topic 8.

164

Application of the definite integral.............................................................

Module 3

184

DIFFERENTIAL EQUATIONS......................................................................

Topic 1.

184

Differential equations of the first order and the first degree.......................

Topic 2.

205

Differential equations of order higher than the first ...................................

Topic 3.

212

Linear differential equation with constant coefficient ................................

Topic 4.

230

Simultaneous differential equations............................................................

LITERATURE .................................................................................................

242

3

Main concepts

Т.1

y0), or z0 = f (M0), or z = z

M 0 .

To

each

point P in

space we

can

z

assign a unique ordered triple of

z0

numbers, called the coordinate of P. To

do

this (see

fig. 1.1),

from

P

we

P(x0, y0, z0)

construct a line perpendicular to the x,

y-plane, that is, the plane determined by

O

y0

y

the x- and y- axes. Let M be the point

where

the

line

intersects this

plane.

x0

M

From M we construct perpendiculars to

the

x-

and

y-axis at

x0 and

y0

x

Fig. 1.1

respectively. From P a perpendicular to

the

z-axis

is

constructed

which

The distance between two points M(x

, x , …, x

m

,) and

M

0

(x0;

x0;

…; x0 ) is

defined by the formula

1

2

1

2

m

ρ(M , M

0

) = (x −x0 )2

+…+(x −x0

)2 .

1

1

m

m

z

z = f(x, y)

z

zy

y

zx

(x,y,f(x,y))

y y

f(x,y)

Fig. 1.2

x

x

x

The set of all points

Мi, the coordinates of

each

assign

an

inequality

ρ(M , M 0 ) < ε , is called the nearby of point

M 0 . For instance the nearby of

point M0 (x0 , y0 ) is the set all inside points of a circle with radius is ε and

center in M0.

If a line bounds the region D it is called

z

the range of the domain D. The points in

the domain but not on the range are called

2

inside points. This domain is called the

close domain. The domain involving the

points inside the range and on it boundary is

O

called the open domain.

2

y

The domain of a function

u = ln(4 – x2 – y2 – z2)

x

is the set of all points inside a sphere

of

Fig. 1.3

radius 2 and center (0,0,0). The graph is

shown in fig. 1.3.

lim f (x, y) = f (x0 , y0 ) ,

(1.1)

x→ x

0 ,

y→ y0

that is

lim f(M) = f(M0)

M →M0

Note, thet point М approache M 0 by any way in the domain of u.

Let x − x0 =

x , y − y0 =

y . Then formula (1.1) can be represented as

lim [ f (x, y) − f (x0 , y0 )] = 0 or lim u = 0 ,

x→0,

ρ→0

y→0

where Δρ = (

x)2 + ( y)2

is the distance between М and M 0 ,

u is the

2) function is defined at M0 and nearby it, but

lim

f (M ) does not exist;

M →M 0

3) function is defined at all points nearby of M0 and

lim f (M ) exists, but

lim f (M ) ≠ f (M 0 ) .

M →M 0

M →M 0

Properties of the continuous functions of two variables

in the close and bounded region

Assume that z = f (x, y)

is a continuous function in the

Theorem 1

close region D. Then the function

is bounded in D and

Assume that z = f (x, y)

is a continuous function in the

Theorem 2

close region D. There are the points in it, where the

function has the minimum and maximum values.

Assume that z = f (x, y)

is a continuous function in the

Theorem 3

close region D. Then the function is bounded in D and

Т.1

Typical problems

radicand 36 – 4x2 – 9y2 is nonnegative. So, 36 − 4x2 − 9y 2 ≥ 0 .

y

The

range of the

domain is

an ellipse

4x 2 + 9 y 2 = 36

or

in

standard

position

2

x

2

+

y

2

= 1. It

has

a

center at origin and x

9

–3

3 x

4

intercepts are 3 and –3; y intercepts are 2 and

–2. The graph is shown in fig. 1.5. The set of

all points inside an ellipse and on the range of

–2

ellipse are domain of z. For instance, the point

(0, 0) is inside the ellipse because is occur the

Fig. 1.5

inequality 36 – 4 0 – 9 0 > 0. A point (3, 0) is

36 – 4 32 – 9 0 = 0 occur.

on the range of the ellipse because the equality

The range of z(x, y) is [0, 6] .

2. Determine the domain of the function

z =

y − x +

2 + arcsin

x2

+ y2

.

4

Solution. The are two factors acting on the domain

y − x + 2 ≥ 0,

y ≥ x − 2,

x2 + y

or

−1 ≤

2

≤ 1,

2

2

x

+ y

≤ 4.

4

The inequality

y ≥ x − 2

is the

upper

half-plane of

the straight line

y = x + 2 , including it. The inequality

x2 + y 2 ≤ 4 is a circle with the center at

3. Calculate lim (x2 + y2 ) arctg

x

.

x→0,

y

y→0

Solution. A function

arctg

x

is bounded:

−

π

< arctg

x

<

π

.

y

2

y

2

If x → 0 and y → 0 the sum

x2 + y 2 is an infinitely less value. The product