Higher_Mathematics_Part_2

∂u =

1

y ,

∂u

=

1 x

,

∂u

= − 1

xy ,

∂x

2z x

∂y

2z y

∂z

z 2

a =

1

i +

1

j − 2k

,

a = −

1

i −

1

j − k ;

1

2

2

2

2

2

1

1

1

1

−

+

−

− 2 (−1)

2

2

2

1

cos ϕ =

2

=

.

1

+

1

+ 4

1

+

1

+ 1

3

Since

4

4

4

4

1 .

ϕ = arccos

3

5. Examine z = x2 − 2y3 − 2x + 6y for relative extrema.

Solution. If we set

∂z

∂z

= 2x − 2 and

= −6y 2 + 6 ,

∂x

∂y

Then according to (2.11) we will have

2x − 2 = 0,

2

+ 6 = 0.

− 6y

This gives x1,2 = 1 and y1,2 = ± 1. Thus M1 (1; 1)

and M 2 (1; − 1) are the

∂ 2 z

= 2

,

∂ 2 z

= 0 ,

∂

2 z

= −12 y .

∂x2

∂x∂y

∂y 2

At M1 (1; 1) ,

= 2 (−12) − 0 = −24 < 0 .

A = 2 , B = 0 ,

C = −12 ,

Since

< 0 , there is no relative extremum at M1 (1; 1) .

At M 2 (1; − 1) ,

A = 2 , B = 0 , C = 12 ,

= 24 > 0 .

Since

> 0 and A > 0 , there is a relative minimum at M 2 (1; − 1) . The

function is zmin

= z(1; −1) = −3 .

6. Examine

z = x3 + xy 2 + x2 y for relative extrema.

41

umin = u(M1) = 0, 4 .

z = 3x + 2y3

8. Find the relative extrema of

subject to the constraint that x

and y must satisfy

x2

+ y2 = 1, x ≥ 0.

4

Solution. Function z can be described parametrically:

π

π

x = 2 cos t , y = sin t , t −

;

. Then we get

2

2

z(t) = 6 cos t + 2sin3 t .

To examine z(t) for relative extrema:

z′(t) = −6sin t + 6sin 2 t cost = 3sin t(sin 2t − 2) ;

sin t(sin 2t − 2) = 0, sin t = 0,

t = πn, n Z .

π

π

′

For t −

;

we have t = 0

. It is the only critical point. If t <

0, then z >

2

2

z(t) has a relative maximum at

t = 0 . We

0 and if t > 0, then z′ < 0. Hence

calculate xmax = 2 cos 0 = 2 , ymax = sin 0 = 0 and zmax = 6 .

Thus the function z = 3x + 2 y3 , subject to the constraint

x2

+ y 2

= 1, x ≥ 0 ,

has a relative maximum at (2; 0) .

4

9. Find

the

relative extrema of z = 8 − 2x − 4y subject to the

constraint

∂L

= −2

+ 2λx =

0,

∂x

λx = 1,

x

= 1/ λ,

∂L

= −4

+ 4λy =

0,

λy = 1,

= 1/ λ,

∂y

y

2

+ 2 y

2

= 12;

2

= 1/ 4,

∂L

= x2

+ 2y2 − 12

=

x

λ

∂λ

0;

43

z = x2 + x(x − 1) + 2(x − 1)2 = 4x2 − 5x + 2 .

Since x goes from 0 to 1 on КС, the minimum of z on КС is at M2(

5

; -

3 ).

The last, on КА: y = − x − 1, x [−1; 0] . Thus, if (x, y) is on AB,

8

8

z = x2 + x(− x − 1) + 2(− x − 1)2 = 2x2 + 3x + 2 .

Since x goes from –– 1 to 0 on КА, the minimum of z on КА is at M3(-

3

; -

1 ).

4

4

max z(x, y) = z(1; 2) = 11 ;

min z(x, y) = z(0; 0) = 0 .

D

D

Self-test and class assignments

Т.3

Find the equations of the tangent plane and normal line to the given

surface z(x, y)

at the given point М.

1.

z = x2 + 3y 2

, M (2,1) .

2.

sin 2 x + sin 2

y + cos2 z = 2 , M (π / 2, π / 4, π / 4) .

6. Compute the directional derivative of

z = ln(x2 + y 2 )

at

M (2; 1) in the

direction from M to N (6; 4) .

7. Compute the directional derivative of

z = x2 − 3xy + 2y 2

at

M (1; 1)

in

the direction of its gradient.

8. Find a gradient of function u = ze x− y+ 2z at point M (1; 0; 0) .

9. Find the points where a gradient of function z = ln(x − y −1 )

is equal to

i + j .

10. Find an angle between the gradients of a function

z = arctg(x / y)

at

points M1 (1; 1) and M 2 (−1; − 1) .

45

11.

z = x2

− 4xy + 5y 2 − 2x + 10y .

12.

z = x3 + 3xy 2 − 15x − 12y .

13.

z = x2

+ xy + y 2 − 4 ln x − 10 ln y .

14.

z = 2x4 + xy3 .

15.

z = x

1 + y + y 1 + x .

16.

z = e x2 − y (5 − 2x + y) .

Find the relative extrema

of

the

given functions u(x, y, z) subject to the

given constraint f(x, y) = 0.

17.

u = x2 − y 2 ,

if x + 2y − 6 = 0 .

18.

u = xyz ,

if

x + y + z = 6 .

19.

u = 4x2 + y 2 ,

if

1

+

2

− 4 = 0 .

x

y

20.

u = xy 2 z 2 ,

if

x + 4y − 2z − 10 = 0 .

21.

u = 16 − 10x − 24y ,

if

x2 + y 2

− 169 = 0 .

22.

u = 4x2 + 2y 2 + z 2 ,

if

x2 + y 2

+ z 2 − 1 = 0 .

23.

z = x 2 − xy + y 2 − 3x + 3y , D = {(x, y)

x ≥ 0, y ≤ 0, x − y ≤ 3} .

24.

z = x3 + y3 − 3x − 9 y + 9 , D = {(x, y)

x

+

y

≤ 3} .

25.

z = 3x2 + 3y 2 − x − y , D = {(x, y)

x ≤ 5, y ≥ 0, y ≤ x − 1}.

26.

z = 2x2 y − x3 y − x2 y 2 , D = {(x, y)

x ≥ 0, y ≥ 0, x + y ≤ 6} .

27.

z = 4 − 2x2 − y 2 , D = {(x, y) y ≥ 0, y ≤ 1 − x2 } .

1.

4x + 6y − z − 7 = 0 ,

x − 2

=

y − 1

=

z − 7

.

2. y − z = 0 ,

x − π / 2

=

y − π / 4

=

4

6

− 1

0

1

=

z − π / 4

.

3.

y + z + 2 = 0 ,

x − 2

=

y + 1

=

z + 1

.

4.

M1(1;0;0), x − y − 2z − 1 = 0,

−1

0

3

3

x − 1

=

y

=

z

;

M2 (2;0;0), x + 2y + 4z − 2 = 0,

x − 1

=

y

=

z

. 5. 5 або –6. 6. 22 / 25 .

1

2

1

2

4

−1

7.

2.

8.

gradu = ek .

9. (2; 1), (0; ––1). 10. 180° .

11. (–5;–3) is a point of the

46

π

π

3.1.28.

z = arctg(cos x +sin y) ,

M

;

.

3

6

3.1.29. z = 3 x2 + 3xy − 4x + 3y − 1,

M (1; 2) .

3.2.1.

u = arctg( x / y) , v = ln(x 2 + y ) , M (1;1) .

3.2.2.

u = 3x2 + 4y2 + z2 , v = 4 2x + 5y + 3z , M (1;1; 3) .

3.2.14.

u = x2 + y 2

(x2 + z 2 )−1/ 2 , M1 (3; 4; − 4) .

3.2.15.

u = ln( xy +

yz ) , M1 (1; 4;1) .

3.2.23. u =

x3

+ xy 2 + z 2 , a = 4i + 4 j − k .

3

3.2.28. u = arcsin

x

, M (1;1; 2) .

yz

3.3.1. z = 4x +

y 2

+ 2y + y 2 .

3.3.2.

z = ln(xy) +

1

+

2

.

x

x

y

3.3.3. z = x2 + 8xy + 20y2 − 20y .

3.3.4. z = 3 + 6x − x2 − xy − y2 .

49