微积分II课后答案详解

∂z ∂x
=
−
Fu' Fu'
+ 2xFv' + 2zFv'
② 两边同时对 y 求偏导�
∂F ∂u + ∂F ∂v = 0 ∂u ∂y ∂v ∂y
F u'
(1
+
z
' y
)
+
Fv'
(2
y
+
2zz
' y
)
=
0
z
' x
=
∂z ∂y
=
−
Fu' Fu'
+ 2 yFv' + 2zFv'
练习 5.5
1. 求二元函数z = x2 − xy + y 2 + 9x − 6 y + 29的值
= 1+
3z 2 x + y2
+
z3
∴u x
+ uy
+ uz
|(1,1,1) =
1 4
+
2 4
+
3 4
=
3 2
x
4�设 z = e y2 ,求证2x ∂z + y ∂z = 0
∂x ∂y
x
x
证明�∵ ∂z = e y2 ⋅ 1 = y−2e y2
∂x
y2
x
x
∂z ∵
= e y2 ⋅ x ⋅ (−2)
dt ∂x dt ∂y dt
=
−
y x2
⋅ et
+
1 x
(−2e 2t
)
=
−
1
− e
e
2t
2t
⋅ et
+ 1 (−2e2t ) t
= − 2e−t
(3) z = x2 − y ,而y = 2x − 3,求 dz
x+ y
dx
解�方法 1�
dz =
d
(
x2
−
2x
+
3 )
=
d
x2 − 2x + 3
(
)
dx dx x2 + 2x − 3 dx 3x − 3
π (
−
π
π ,
+
π
)
6 180 4 180
≈
f
π (
π ,
)+
f x′(−
π
ቤተ መጻሕፍቲ ባይዱ
) + f y′
π
64
180 180
=
1 2
×1+
cos
x
tan
y
|π π
(,) 64
(− π ) 180
+
sin
x sec2
y
|π π
(,) 64
⋅π 180
=0.5023
练习 5.4
1. 求下列函数的导数或偏导数。
�1� z
∂x
∂y
② z = ln(xy),求 ∂z , ∂z
∂x ∂y
解�
∂z
=
1
[ln(
xy)]−
1 2
.
1
.y
=
1
∂x 2
xy 2x ln(xy)
∂z
=
1
[ln(
xy
)]−
1 2
.
1
.x
=
1
∂y 2
xy 2 y ln(xy)
③ z = x ln(x + y),求 ∂ 2z , ∂ 2z
∂x2 ∂x∂y
解� ∂z = ln(x + y) + x. 1
练习 5.1
1.在空间直角坐标系下,下列方程的图形是什么形状?
(1) x2 + 2 y 2 = 4z(椭圆抛物面)
(2) x2 + y 2 = 4z 2 (圆锥面�
(3) x2 + y 2 + z 2 = 1(椭球面�
4 16 9
2.求下列函数的定义域:
(4) x2 + z 2 = 1(圆柱面�
(1) z = − x − y
1
= −2xy−3e y2
∂y
y3
x
x
x
x
∴ 2x ∂z + y ∂z = 2xy−2e y2 + ye y2 ⋅ x ⋅ (−2) 1 = −2xy−3e y2 ⋅ y + 2xy−2e y2 = 0
∂x ∂y
y3
练习 5.3
1� 求下列函数的全微分
�1�求 z = xy 在点�2�3�处�当 ∆x = 0.1与∆y = −0.2时的全增量与全微分
x+ y
x→0
分析�由二元函数极限定义�我们只须找到沿不同路径 p → p0(0,0) 时�所得极限值不同即可。
证明� ① p(x, y) x ( x ≠ 0, y = 0)
f (x, y) = f (x,0) = 1, lim f (x, y) = 1 x →0 y→0
p0 (0, 0)
②当 p(x, y)沿直线y = kx(x ≠ 0)趋于�0�0�时�
f (x, y) = x − kx = 1 − k ≠ 1(k ≠ 0) x + kx 1 + k
综合①②可知函数极限不存在。
练习 5.2
1.求下列函数的偏导数
① z = x3 y − xy3,求 ∂z , ∂z
∂x ∂y
解� ∂z = 3x 2 y − y 3 , ∂z = x3 − 3xy 2
解�
� y≥0 ��x − y ≥ 0
� y≥0
即 �� x ≥ 0
��x 2 ≥ y
函数的定义域为{(x, y) | x ≥ 0, y ≥ 0, x2 ≥ y}
(2)
z
=
3
e
x+ y
+
x− y
解�∵ x − y ≥ 0 �∴函数的定义域为{(x, y) | x − y ≥ 0}
3. 对于函数f (x, y)= x − y ,证明 lim f (x, y)
= e xyz (1 + 2xyz + xyz + x 2 y 2 z 2 ) = e xyz (1 + 3xyz + x 2 y 2 z 2 )
2�设 f (x, y) = e xy2 ,则 lim f (2,1 + ∆y) − f (2,1)
∆y→0
∆y
解� lim f (2,1+ ∆y) − f (2,1) = lim e2(1+∆y)2 ( 0 未定式)
两个一阶偏导数存在�求
∂z ∂z ,
∂x ∂y
解�令 u = x + y + 2, v = x2 + y2 + z2 ,则F (u, v) = 0
① 两边同时对 x 求偏导�
∂F ∂u + ∂F ∂v = 0 ∂u ∂x ∂v ∂x
F u'
(1
+
z
' x
)
+
Fv'
(2
x
+
2
zz
' x
)
=
0
z
' x
=
3
3 x+
4
y
− 1 |(1,1)
( − 0 .0 2 )
=0+ 1 × 0.03 − 1 × 0.02 = 0.005
3
4
(3) sin 290 tan 460
解�令 f (x, y) = sin x tan y
取
x0
=
300 , ∆x
=
−π , 180
y0
=
π 4
, ∆y
=
π 180
则原式=
f
−
y)
⋅2
=
x2 − 2x −1 3(x −1)2
�4� z = x2 y − xy 2 , x = u cosθ , y = u sinθ ,
∂z ∂z ,
∂u ∂θ
解� ∂z = ∂z ∂x + ∂z ∂y
∂u ∂x ∂u ∂y ∂u
= (2xy − y2 ) cosθ + (x2 − 2xy) sinθ , = (2u cosθ ⋅ u sinθ − u2 sin 2θ ) cosθ + (u2 cos2θ − 2u cosθ ⋅ u sinθ ) sinθ
=
u2
ln v,而u
=
x
,v
= 3x
− 2 y.求 ∂z
∂z ,.
y
∂x ∂y
解�
∂z ∂x
=
∂z ∂u
∂u ∂x
+
∂z ∂v
∂v ∂x
=
2u
ln v ⋅
1 y
+
u2 v
⋅3
=
2x y2
ln(3x
−
2y)
+
3
⋅
x2 y2
3x − 2y
= 2x ln(3x − 2 y) + 3x2
y2
(3x − 2 y) y 2
∆y → 0
∆y
∆y→0 ∆y 0
= lim e2(1+∆y)2 (2 + ∆y) ⋅1− 0
∆y → 0
1
= 4e2
3�设 u = ln(1 + x + y 2 + z3 ),在点�1�1�1�处求ux + u y + uz
解� ux
=
1+
x
1 + y2
+
z3
uy
=
1+
2y x + y2
+
z3
uz
解�
�����
z
' x
z
' y
= 2x − y + 9 = 0 = −x + 2y − 6 = 0
�x = −4
� �
y
=1
又∵
A
=
Z
'' xx
=
z
>
0,
B
=
Z
'' xy
= −1,C
=
Z
'' yy
=2
∆ = B 2 − AC = 1 − 4 = −3 < 0
∴ Z |(−4,1) = −1
2.求 二 元 函 数 z = 60x + 12 y − 2 x2 − 2 xy − 5 y2 在 条 件 x + y = 15下 的 极 值
dx
解� 两边同时对 x 求导:
y' cos y + ex − y2 − 2xyy' = 0
y' (cos y − 2xy) = y2 − ex
y' =
y2
cos y − 2xy
3.已知方程 F (x + y + 2, x2 + y2 + z2 ) = 0, 所确定的函数z = f (x, y). 且 F 的
解� F (x, y, z, λ) = 60x + 12 y − 2x 2 − 2xy − 5y 2 + λ(x + y −15)
� Fx' = 60 − 4x − 2 y + λ = 0 ��Fy' = 120 − 2x − 10 y + λ = 0 �� Fλ' = x + y −15 = 0
∂z = ∂z ∂u + ∂z ∂v = 2u ln v ⋅ (− x ) + u2 (−2)
∂y ∂u ∂y ∂v ∂y
y2 v
=
−
2x2 y3
ln(3x
−
2y)
−
2x2 y2 (3x −
2y)
(2) z = y , x = et , y = 1 − e2t , dz
x
dt
解: dz = ∂z dx + ∂z dy
解� dz = ∂z dx + 2 y dy
∂x 1 + x 2 + y 2
dz
(1, 2 )
=
2 1+1+ 4
dx
+ 4 dy 1+1+ 4
=
1 dx 3
+
2 3
dy
�3� u = xy + yz + zx,求du
解� du = ∂u dx + (x + z)dy + (x + y)dz
∂x
= ( y + z)dx + (x + z)dy + (x + y)dz
(2) ln(3 1.03 + 4 0.98 − 1) 解:令 f (x, y) = ln(3 x + 4 y −1)
取 x0 = 1, ∆x = 0.03, y0 = 1, ∆y = −0.02
原式
= f (1 + 0 .0 3,1 − 0 .0 2 )
1
−2
x3
≈
ln( 3 1 +
4 1 − 1) +
∂x
x+ y
∂ 2z ∂x 2
=
∂ ∂x
∂z () ∂x
=
∂ ∂x
(ln(x
+
y) +
x x+
) y
=
1 x+
y
+
x+ y−x (x + y)2
=
x + 2y (x + y)2
∂ 2z ∂x∂y
=
∂
∂z ()
=
∂y ∂x
∂ ∂y
���ln(x
+
y) +
x x+
� y ��
=
1 x+
y
+ 0−x (x + y)2
解�全增量 ∆z = f (2 + 0.1,3 − 0.2) − f (2,3) = 2.1× 2.8 − 6 = −0.12
dz = zx∆x + zy∆y = y∆x + x∆y = 3× 0.1+ 2 × (−0.2) = −0.1
�2�求 z = ln(1 + x2 + y 2 ),当x = 1, y = 2时的全微分
2�计算下列各式的近似值�分析运用公式
f (x0 + ∆x1 y0 + ∆y) ≈ f (x0, y0 ) + f x′ ∆x + f ′y∆y � �1� (10.1)2.03 解�令 f (x, y) = x y , x0 = 10, ∆x = 0.1, y0 = 2, ∆y = 0.03
(10.1)2.03 = f (x0 + ∆x1 y0 + ∆y) ≈ f (x0 , y0 ) + f x′ ∆x + f ′y∆y = 102 + yx y−1 (10,2) ⋅ 0.1 + x y ln x (10,2) ⋅ 0.01 = 100 + 2 + 3ln10 ≈ 108.9
=
y (x + y)2
④ u = exyz , 求 ∂ 3u
∂x∂y∂z
解� ∂u = yzexyz , ∂ 2u = zexyz + yzxzexyz = (z + xyz z )exyz
∂x
∂x∂y
∂ 3u = ∂ ( ∂ 2u ) = ∂ (z + xyz z )2 e xyz = (z + 2xyz)e xyz + (z + xyz 2 )xyexyz ∂x∂y∂z ∂z ∂x∂y ∂z
= 3u 2 sinθ cosθ (cosθ − sinθ ),
∂z = ∂z ∂x + ∂z ∂y ∂θ ∂y ∂θ ∂y ∂u
= (2xy − y 2 )u(− sinθ ) + (x2 − 2xy)u cosθ = − 2u3 sinθ cosθ (sinθ + cosθ ) + u3 (sin 3θ + cos3 ) 2.求下列隐函数的导数或偏导数.
① xy + ln y − ln x = 0,求 dy .
dx
解�两边同时对 x 求导�
y + xy' + 1 y' − 1 = 0, y' (x + 1 ) = 1 − y
yx
yx
1−y
y' = x �
x+ 1 y
y' = y − xy2 x + x2y
② sin y + ex − xy2 = 0, 求 dy
= (2 x − 2)(3 x − 3) − ( x 2 − 2 x + 3) ⋅ 3 (3 x − 3)2
= x2 − 2x −1 3(x −1)2
方法 2� dz = ∂f + ∂f dy
dx ∂x ∂y dx
=
2x(x + y) − (x2 (x + y)2
−
y)
+
−(x + y) − (x2 (x + y)2