1022微
微乙 01-05班
班期 中 考 解 答 和 評 分 標 準
1. (12%) 求過曲面 y − x = 4 arctan (xz) 上點 (1, 1, 0) 的切平面方程式。
Solution:
令 f (x, y, z) = y − x − 4 tan−1 (xz)
∇f (x, y, z) = (−1 − 4
z
x
, 1, − 4
)
2
1 + (xz)
1 + (xz)2
∇f (1, 1, 0) = (−1, 1, −4)
(2%)
∇f (1, 1, 0) · (x − 1, y − 1, z − 0) = 0
2. (12%) 設在點 (x, y) 之溫度為 T (x, y) = 100e−x
2
−3y 2
(6%)
(4%)
, 此處 T 以攝氏 ◦ C 計，x, y 以 m(公尺) 計，
(a) 在點 (1, −1) 處，往那個方向增加最快？
(b) 此最快增加率（以 ◦ C/m 計）為何？
∂T
(c) 求 方向導數 * (1, −1)* 3 4 .
∂u
u=( 5 , 5 )
Solution:
Since ∇T = T (−2x, −6y) (3pt) ⇒ ∇T (1, −1) = 200e−4 (−1, 3) (3pt)
(a) in the direction: ∇T (1, −1) = 200e−4 (−1, 3) (2pt)
√
(b) fastest increasing rate: | ∇T (1, −1) |= 200e−4 10 (2pt)
3 4
(c) directional derivative: ∇T (1, −1) · ( , ) = 360e−4 (2pt)
5 5
ˆ
3. (12%) 求
0
1
ˆ
1
1
x6
1
dy
dx.
1 + y7
Solution:
ˆ
0
1
ˆ
1
1
x6
ˆ 1
1
dy
dx
=
1 + y7
0
ˆ 1
=
0
ˆ
0
y6
!
1
dx dy
1 + y7
y6
dy
1 + y7
ˆ
1 1 1
dy 7
7 0 1 + y7
1
= ln(1 + y 7 ) |10
7
1
= ln 2
7
(4 points)
(2 points)
=
4. (12%) Ω 由 2x + 3y = 0, 3x + y = 0 與 x − 2y = 1 所圍成的區域. 計算
ˆ ˆ
I=
(x − 2y)3/2 (3x + y)1/2 dA.
Ω
Page 1 of 4
(4 points)
(2 points)
Solution:
Use changing variable, let
⇒
then Jacobian is
u =
v =


 x
=

 y
=
x − 2y
3x + y
(2%)
−3u + v
7
u + 2v
7
3
−
J = | 17
7
1
| =
7
1
7
2
7
(1%)
(4%)
¨
Therefore
I
(x − 2y)3/2 (3x + y)1/2 dA (4%)
=
ˆ
=
=
=
=
Ω
1ˆ u
1
u3/2 v 1/2 · dvdu
7
0
0
u
ˆ 1
2 3 3
1
u 2 v 2 du
7 0 3
0
ˆ 1
2
u3 du
21 0
1
(1%)
42
By the way, you can also write
ˆ
1
ˆ
1
u3/2 v 1/2 ·
I=
0
1
5. (15%) 決定 z = ye− 2 (x
2
+y 2 )
v
1
1
dudv =
7
42
的極值候選點，並決定其極值性質。
Solution:
1
Let f (x, y) = ye− 2 (x
2
+y 2 )
fx
fy
fxx
fyy
fxy
=
=
=
=
=
1
2
2
−xye− 2 (x +y )
2
2
1
(1 − y 2 )e− 2 (x +y )
2
2
1
(x2 y − y)e− 2 (x +y )
2
2
1
(y 3 − 3y)e− 2 (x +y )
2
2
1
(xy 2 − x)e− 2 (x +y ) = fyx
(1%)
(1%)
(1%)
(1%)
(1%)
If we solve fx = fy = 0, which equal to solve:
xy
=
0
2
=
0
1−y
and thus the critical points are (0, 1) and (0, −1).
Now let
f
D(x, y) = xx
fyx
D(0, 1) =
1
2
> 0, fxx (0, 1) = −e 2 < 0
e
D(0, −1) =
1
2
> 0, fxx (0, 1) = e 2 > 0
e
(4%)
fxy fyy ⇒ (0, 1) is a local maximum
⇒ (0, −1) is a local minimum
6. (10%) 設 f (x, y) = x3 y 5 − x2 y − y 3 , x = x(u, v), y = y(u, v).
Page 2 of 4
(6%)
已知 x(1, 3) = 2, y(1, 3) = 1
1
∂x
1
∂x
(1, 3) = ,
(1, 3) =
∂u
5
∂v
2
∂y
1
∂y
1
(1, 3) =
,
(1, 3) =
∂u
11
∂v
4
∂f
∂f
求
及
在 (u, v) = (1, 3) 的值.
∂u
∂v
且
Solution:
We know that
∂f
∂u
∂f
∂v
∂f ∂x ∂f ∂y
+
∂x ∂u ∂y ∂u
∂f ∂x ∂f ∂y
+
∂x ∂v
∂y ∂v
=
=
(3pts),
(3pts).
Moreover,
∂f
∂x
∂f
∂y
=
3x2 y 5 − 2xy
=
5x3 y 4 − x2 − 3y 2
(1pt),
(1pt).
Thus, when (u, v) = (1, 3),
∂f
|(u,v)=(1,3)
∂u
∂f
|(u,v)=(1,3)
∂v
7. (12%) 求
¨ p
=
=
1
+ 33 ×
5
1
8 × + 33 ×
2
8×
1
23
=
11
5
1
49
=
4
4
(1pt),
(1pt).
x2 + y 2 dA, Ω 為心形線 r = 1 − cos θ 的內部。
Ω
Solution:
Use polar coordinate, let x = r cos θ, y = r sin θ, then we obtain
¨ p
ˆ
x2 + y 2 dA =
Ω
2π
0
ˆ
1−cos θ
r2 drθ
(4pt)
(1 − cos θ)3
dθ
3
(1pt)
0
2π
=
0
ˆ
ˆ
2π
1
cos3 θ
− cos θ + cos2 θ −
dθ
3
3
0
2π
ˆ
θ
2 + sin2 θ 1 2π
= − sin θ −
(1 − sin2 θ) d sin θ
−3
3
4
0
0
2π
5π 1
sin3 θ
=
−
sin θ −
(1pt)
3
3
3
0
5π
(1pt)
=
3
=
Page 3 of 4
(1pt,1pt,2pt,1pt)
轉完極坐標若錯誤(積分函數不是 r2 ) 得 0 分. 轉對而範圍寫錯得 2 分
8. (15%) 求 f (x, y) = x2 + xy + y 2 在橢圓上 x2 + 2xy + 2y 2 = 1 之極值。
Solution:
Let f (x, y) = x2 + xy + y 2 and g(x, y) = x2 + 2xy + 2y 2 = 1
By Lagrange theorem,∇f (x, y) = λ∇g(x, y)[2pts]


1

= λ(2x + 2y) . . . (1)
 2x + y
 λ=
2
x + 2y
= λ(2x + 4y) . . . (2) [3pts] ⇒
3
 2

2
 λ=
x + 2xy + 2y − 1 = 0
. . . (3)
2
√
1
3
The maximum of f (x, y) is f (∓ 2, ± √ ) = .[2pts]
2
2
1
1
The minimum of f (x, y) is f (0, ± √ ) = .
[2pts]
2
2
Page 4 of 4
,x = 0
√
,x = ± 2
1
, y = ±√
2
1 [6pts]
, y = ∓√
2