Higher_Mathematics_Part_2

2.1.25. а) z = 3x4 y5 + 5x − 2 y4 + 2 ;

b) z =

ln(x − y)

.

ln(x3 + y3 )

2.1.26. а) z = 4x−2 y−3 − 5x − 4y5 + 8 ;

b) z =

ln ln(x ctg y)

.

x2

2.1.27. а) z = 9x−2 y6 + 3x4 − 2y + 3 ;

b) z =

sin 2 x − cos2 y .

2.1.28. а) z = − x2 y−3 + 8y3 + 6x −1 ;

b) z =

tg2 x + ctg2

y .

2.1.29. а) z = 3x3 y5 − 6xy2 + 2x − 7 ;

b) z = sin(ln(2x + 3y)) .

2.1.30. а) z = 4x3 y7 + 4xy2 + 4y − 6 ;

b) z = ln cos(4x − 3y) .

2.2. Find the differential dz for the given functions z(x, y)

at point М.

2.2.1.

x + 2 y + z + ez = 0 , M (1; − 1; 0) .

2.2.2.

x2 − y + z + ln z = 0 , M (0; 1; 1) .

2.2.3. sin(x + y) + sin( y − z) = 1 , M (0; π / 2; π / 2) .

2.2.4. sin(x + y + z) + z = 1 , M (π / 4; π / 4; 0) .

2.2.5. e x+ z + e y+ z = 2 , M (0; 0; 0) .

2.2.6. 2xy + yz + z3 = 0 , M (−1; 1; 1) .

2.2.7.

xyz + e z = x , M (1; 2; 0) .

2.2.8.

x − y − z = tg z , M (2; 2; 0) .

2.2.9. e xyz = z , M (0; 1; 1) .

2.2.10.

ln z + ln x = y − 1, M (1; 1;1) .

2.2.11. xyz = ln z , M (0; 1; 1) .

2.2.12.

x + 4y − z = e z , M (−3; 1; 0) .

2.2.13. sin z + cos(x − y + z) = 1 , M (π / 2; π / 2; 0) .

2.2.14. 3 x + 3 y + 3 z = 4z , M (8; 1; 1) .

2.2.15. tg x + tg y + tg z = 2 cos z , M (π / 4; π / 4; 0) .

2.2.16.

x2 − y 2 − z 2 + xz + 4x + 5 = 0 , M (−2; 1; 0) .

2.2.17.

x2 + y 2 − z 2 + xz + 4 y − 4 = 0 ,

M (1; 1; 2) .

2.2.18. 2x2 − y 2 + 2z 2 + xy + xz − 3 = 0 ,

M (1; 2;1) .

2.2.19. 4y2 − z2 + 4xy − xz + 4z − 10 = 0 ,

M (1; − 2;1) .

2.2.20.

x2 − 2y 2 + z 2 + xz − 4y − 13 = 0 ,

M (3; 1; 2) .

31

2.2.21.

y2 + z2 + x − 3z = 0 , M (1; 1;1) .

2.2.22.

y 2 + z 2 − xy − z( y + x) = 0 ,

M (1; 2;1) .

2.2.23. 2(x4 + y 4 + z 4 ) − 3xyz = 0 ,

M (1/ 2;1/ 2; 1/ 2) .

2.2.24. sin 2 x + sin 2 y + sin 2 z − 1 = 0 , M (π / 4; 0; π / 4) .

2.2.25. ex + e y

+ ez

= 3ex+ y+ z , M (0; 0; 0) .

2.2.26. esin x + esin y

+ ecos z − 3 = 0 ,

M (0; 0; π / 2) .

2.2.27.

xy + yz

+ zx

− 12 = 0 , M (3;

2; 1) .

2.2.28.

x3 + y3 + z3 + xyz − 4 = 0 , M (1; 1; 1) .

2.2.29.

x +

y +

z + xyz − 8 = 0 ,

M (1;1; 4) .

2.2.30. 2x + 2 y + 2z + 1 = 2x+ y+ z ,

M (0; 1; 2) .

2.3. Use the differential to estimate the given expressions.

2.3.1. (0,97)4 (0,95)−2 .

2.3.2. (0, 94)2 (0, 98)−3 .

2.3.3.

(3,1)2 + (3,9)2 .

2.3.4. 3 (1,94)3 + (4,06)2 + 3 .

2.3.5. sin 5° cos80° .

2.3.6. sin 6° tg 48° .

2.3.7.

sin 8° + 4 cos 5° .

2.3.8.

(5, 85)2 + (8,1)2 .

2.3.9.

(4, 75)2 + (11, 8)2 .

2.3.10.

arctg

1, 08

.

1, 04

2.3.11. tg 42° tg 50° .

2.3.12. sin 27° cos 4° .

2.3.13.

(2,94)3 − (1,2)2 − 1 .

2.3.14.

4 (6, 9)2 + (4, 8)2 + 7 .

2.3.15.

sin 85° + 3cos 6° .

2.3.16. e0,15 (3,05)−2 .

2.3.17. sin 6° cos 8° .

2.3.18. e0,1 sin 85° .

2.3.19.

arctg

1,1

.

2.3.20. arctg(0, 97 1, 08) .

0, 96

2.3.21.

4 (2,9)2 + (2,05)2 + 3 .

2.3.22. 5 (2,9)4 − (3,86)3 + 15 .

2.3.23. ((1,86)3 + (0,94)3 )2 .

2.3.24. 3 (2,05)4 − (3,04)2 + 1 .

2.3.25. (1,04)3 (0,96)−1/ 3 .

2.3.26. ctg 40° ctg 47° .

2.3.27. ctg 41° + ctg 51° .

2.3.28. 5 (4,92)2 + (2,94)2 − 2 .

32

Т.3

Main concepts

3.1. Tangent plane and normal to a surface

Consider a surface that is a level surface of a function u = F(x, y, z). Let

M0(x0, y0, z0) be a point on this surface where the vector F =

∂F(M 0 )

i +

∂F(M0 )

∂F(M0 )

∂x

+

j +

k ≠ 0. The tangent plane to the surface at the point M0

∂y

∂z

Table list 1.1

The equations

The equations of a tangent plane

of a surface

F(x, y, z) = 0

∂F(M0 )

(x − x0 ) +

∂F(M0 )

( y − y0 ) +

∂F(M0 )

(z − z0 ) = 0

∂x

∂z

∂y

z = f (x, y)

z − z

0

=

∂f (x0 , y0 )

(x − x )

+

∂f (x0 , y0 )

( y − y )

∂x

0

∂y

0

Table list 1.2

The equations

The equations of a normal line

of a surface

F(x, y, z) = 0

x − x0

=

y − y0

=

z − z0

∂F(M0 )

∂F(M0 ) ∂F(M0 )

∂x

∂y

∂z

z = f (x, y)

x − x0

=

y − y0

=

z − z0

∂f (x0 , y0 )

∂f (x0 , y0 )

−1

∂x

∂y

33

z

Let a function u = f (x, y, z) be defined and

М1

continuous a nearby point M(x0, y0, z0). There is

М γ

l

a unit vector l

= {cosα; cosβ; cosγ} at М (α, β

β

and γ are angles with x-, y- and z-axes respec-

α

tively). A point M1(x0 +

x, y0 + y, z0 + z) lies

О

y

on the straight line which passes though M in the

direction

of

the vector

l (fig. 1.12).

The

x

Fig. 1.12

distance

l

between M and M1 by the

Pythagorean theorem,

l = ( x)2 + ( y)2 + ( z)2 .

Directional derivative of a function u(x, y, z). The derivative of u at

M(x0, y0, z0) in the direction of the unit vector l (it is denoted

∂u

) is

∂l

∂u

= lim

u =

lim

u(M1 )−u(M )

∂l

l→0

l

l→0

l

∂u

∂u

Theorem

If u(x, y, z) has continuous partial derivatives

∂x ,

∂y

∂u

and

, then the directional derivative of u at M(x0, y0, z0) in the direction of the

∂z

unit vector l = {cosα; cosβ; cosγ} is

,

∂u =

∂u cos α +

∂u cos β + ∂u cos γ

(1.10)

∂l

∂x

∂y

∂z

where cosα =

l

x

, cos β=

ly

, cos γ

=

l

z

are directional cosines of the

| l

|

| l

|

| l |

vector l , | l |=

lx2 +ly2 +lz2

is the magnitude of l .

The gradient of u(x, y, z). The vector

∂u(x0 , y0 ,z0 )i + ∂u(x0 , y0 ,z0 )

j + ∂u(x0 , y0 ,z0 ) k

∂x

∂y

∂z

is the gradient of u at M(x0, y0, z0) and is denoted grad u(M) or u

34

The number M is called the maximum (or global

z

maximum) of f over a set D in the plane if it is the largest

value of f(x, y) in D. A relative maximum (or local

maximum) of f occurs at a point (x0, y0) in D if there is a

circle around (x0, y0) such that f(x0, y0) is the maximum

value of f(x, y) for all points (x, y) within the circle.

О

y

Minimum and relative (or local) minimum are defined

similarly.

x

The function z = x2 + y 2 in fig. 1.13 has a relative

Fig. 1.13

minimum when x= y = 0, which corresponds to a low

point on the surface. In other words, the expression x2 + y 2

is positive number

relative maximum at (0; 0) and zmax = 2 . The change of z

z(0;0) < 0 .

Let us look closely at the surface above a point

z

(x0, y0) where a relative maximum of f occurs. Assume

2

that f is defined for all points within some circle

around (x0, y0) and possesses partial derivatives at (x0,

y0). Let L1 be the line y = y0 in the xy plane; let L2 be

the line x = x0 in the xy plane. (See fig. 1.15). Assume,

О

y

for convenience, that the values of f are positive)

2

Let C2 be the curve in the surface directly above

x

the line L1. Let C1 be the curve in the surface directly

Fig. 1.14

y0

y

x0

L2

x

L1

Fig. 1.15

Let f be defined on a domain that includes the point

Theorem 1

(x0, y0) and all points within some circle whose center is

∂f

exist at (x0, y0), then both these partial derivatives are 0 at (x0, y0); that is,

∂y

f (x0 , y0 )

= 0,

x

f (x0 , y0 )

= 0.

y

In other words, since

dz(x0 , y0 ) = 0 .

(1.11)

Let

M 0 (x0 , y0 )

be

a

critical

point

of the

Theorem 2

function z = f (x, y) . Assume

that the partial derivatives

∂f

,

∂f

,

∂ 2 f

,

∂ 2

f

and

∂ 2

f

are continuous at and near M0. Let

∂x

∂y

∂x2

∂x∂y

∂y2

A =

∂ 2 f (x

0

, y

0

)

,

B =

∂ 2

f

(x

0

,

y

0

)

, C =

∂ 2 f (x

0

, y

0

)

,

= AC

− B2 .

∂x2

∂x∂y

∂y 2

φ(x, y) = 0.

( 1.12 )

A function

z = f (M )

subject to the constraint has the relative extrema

at M 0

when x and y must satisfy (1.12 ), if a nearby point M 0 , exists such

that for any point M (x, y) ( M ≠ M 0 ) in this nearby on line L an inequality

f (M ) > f (M 0 )

( f (M ) < f (M 0 ) )is provided

For example, the relative extrema of

z = x2 + y 2

z

subject to the constraint that

x + y = 1 is at M 0 (1/ 2;1/ 2)

and the total relative extrema is at O(0; 0)

(fig. 1.16 ).

We can transform z, which is a function of two

variables, into a function of one variable so that now the

function

reflects

constraint

x + y = 1 .

Solving this

equality for y, we get y = 1 − x , which when substituted

О

1 у

for y in z = x2 + y 2 gives

х

1

М0

z = x

2

+ (1 − x)

2

=

2x

2

− 2x + 1 .

Fig. 1.16

37

L(x, y, λ) = f (x, y) + λϕ(x, y) .

(1.13 )

Т.3

Typical problems

1. Find the equations of the

tangent

plane and

normal line to a surface

z = x 2 y + xy3

at point (2, 1).

= 22 1 + 2 13 = 6.

Solution. We have

x0

= 2 ,

y0 = 1. First compute z0

Evaluating

∂z

and

∂z

at (2, 1):

∂x

∂y

∂z

=

2xy + y

3

,

∂z

= x

2

+ 3xy

2

;

∂z(2;1)

=

5 ;

∂z(2;1)

=10 .

∂x

∂y

∂x

∂y

F = x3 + y3 + z3 + xy 2 z

3 − 4 ;

∂F

= 3x2 + y 2 z3 ,

∂x

∂F

= 3y 2 + 2xyz3 ;

∂F

= 3z 2 + 3xy2 z 2 ;

∂y

∂z

∂F(M 0 )

= 4 ;

∂F(M

0 )

= 5

;

∂F(M 0 )

= 6 .

∂x

∂y

∂z

An equation of the tangent plane is

4(x − 1) + 5( y − 1) + 6(z −1) = 0 , or 4x + 5y + 6z − 15 = 0 ,

and an equation of the normal line is

y −1

x − 1

=

=

z −1

.

4

5

6

3.

Calculate the

derivative

of

a function u = x2 y + xy 2 z + z3

at

point

M (0;1; − 1)

in the direction of the vector l =−2i + j +2k .

Solution. We find the directional cosines of the vector l

:

l ={−2, 1, 2}, | l |= 4 +1+4 = 3 ,

cos α = −

2

, cos β =

1

, cos γ =

2

.

3

3

3

Then we calculate the partial derivatives of a given function at M (0;1;−1) :

∂u(M ) = 2xy + y 2 z

M = −1 ;

∂u(M )

= x2 + 2xyz

M = 0 ;

∂x

∂u(M )

∂y

= xy2

+ 3z 2

M = 3 .

Then, by (2.10),

∂u

∂z

2

1

2

8