Дифференциальное исчисление функций нескольких переменных. Скляренко В.А - 65 стр.

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6. x =
1
2
(e
u
+ e
v
)
2
, y =
1
2
(e
u
e
v
)
2
, z = u 2 v; u
0
= ln (3), v
0
= 0.
7. x = 2
v
cos u + sin u
, y =
1
cos u + sin u
, z = v (cos u + sin u); u
0
=
1
2
π, v
0
= 1.
8. x =
u + v
u
2
+ v
2
, y =
uv
u
2
+ v
2
, z =
1
u
2
+ v
2
; u
0
= 1, v
0
= 2.
9. x = u ch v, y = u
2
sh v
u
2
, z = u
3
+
sh v ch v
u
2
; u
0
= 1, v
0
= 0.
10. x = v + u
2
ln 3 π, y = 3
u
cos v, z = 3
u
sin v; u
0
= 1, v
0
= π.
11. x = 2
v sin u
1 + v
2
, y = 6
v cos u
1 + v
2
, z =
º
2
4
(1 + v
2
); u
0
=
1
4
π, v
0
= 2.
12. x =
»
2 (u
2
3)(1 + v
2
), y =
»
10 (3 v
2
)(u
2
+ 1), z =
º
u
2
v
2
3;
u
0
= 2, v
0
= 1.
13. x = tg
2
v cos u, y = cos
2
v, z = tg
2
v sin u; u
0
=
1
2
π, v
0
=
1
4
π.
14. x =
2
sh u + cos v
, y =
cos v
sh u + cos v
, z = sin v ch u; u
0
= 0, v
0
= 0.
15. x = 3 u + 3 uv
2
u
3
, y = v
3
3 v 3 u
2
v, z = 3 u
2
3 v
2
; u
0
= 1, v
0
= 1.
16. x = 3 u + ln (v u), y = 3 v ln (v u
2
), z = u
2
+ 2 v + u; u
0
= 1, v
0
= 2.
17. x =
º
u
u v
, y = 2
º
v
u v
, z =
1
3
º
uv; u
0
= 4, v
0
= 1.
18. x = 4 ch u cos v, y = 2 ch u sin v, z = 6 sh u sin v; u
0
= ln 2, v
0
=
1
6
π.
19. x = ue
v
, y = e
v
(cos
2
u + 3 sin u), z = e
v
sin
2
u + 4 cos u
; u
0
= π, v
0
= 0.
20. x = 2 u
2
cos v, y = 4 u
2
sin v, z =
º
2u
2
8
v
2
π
+
π
2
; u
0
= 1, v
0
=
1
4
π.
21. x =
ch u
sin v ch u
, y =
sh u
sin v ch u
, z =
1 + u
sin v ch u
; u
0
= 0, v
0
=
1
6
π.
22. x = u
2
+ ln v, y = v
2
ln u, z = uv; u
0
= 1, v
0
= 1.
23. x = u (2 v u 1), y = v (2 u v 1), z = u
2
+ v
2
; u
0
= 3, v
0
= 1.
24. x =
º
1 + u cos 2 v, y =
º
1 + u sin 2 v, z = 2
º
1 + u + cos u;
u
0
= 0, v
0
=
1
3
π.
25. x =
»
3 (u
2
1)(1 v
2
), y = u (1 + sh v), z = v + u
2
ch v; u
0
= 2, v
0
= 0.
26. x =
e
2 u
1 + v
, y =
ve
u
1 + v
, z =
v
1 + e
u
; u
0
= ln 3, v
0
= 2.
27. x =
»
(u + 2)(v 2), y =
»
(u 2)(v + 2), z =
º
3uv; u
0
= 4, v
0
= 4.
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