eu = e2u2ln2v+v2e2u(4u2ln v v + 2ve2u) 解法2 f (x(u, v), y(u, v

1002微微微乙乙乙01-05班班班期期期中中中考考考解解解答答答和和和評評評分分分標標標準準準 1. (10%) 令 f(x, y) = e^2x2+y2, x(u, v) = u ln v, y(u, v) = ve^u. 求 ∂f

∂v。 Solution:

解法1

∂f

∂v = ∂f

∂x · ∂x

∂v +∂f

∂y · ∂y

∂v

= (4xe^2x2+y2) ·u

v + (2ye^2x2+y2) · e^u

= e^{2u2ln2v+v2e2u}(4u²ln v

v + 2ve^2u) 解法2

f (x(u, v), y(u, v)) = e^{2u2ln2v+v2e2u}

∂f

∂v = e^{2u2ln2v+v2e2u}· (2u²· 2 ln v · 1

v + 2ve^2u)

F評分標準

(1)解法一第一步寫對得4分，第二步四項每項1.5分，x(u,v)、y(u,v)沒代回式子扣兩分 (2)解法二f(x(u, v), y(u, v))寫對得3分，微分第一項寫對得3分，後兩項每項2分

2. (10%) 設 f (x, y) = (x²+ 1)²y − xy³。

(a) 求 f(x, y) 在點 (1, 2) 沿 (2, 1) 的方向導數。

(b) 在點 (1, 2)處，f(x, y) 沿哪一方向的方向導數最大。

Solution:

(a) Since ∇f (1, 2) = (4x(x²+ 1) − y³, (x²+ 1)²− 3xy²)|_(1,2) = (8, −8) (2%) and ~u = ( 2

√5,−1

√5) (1%),

D_~u = ∇f (1, 2) · ( 2

√5,−1

√5) = 8

√5 (2%) (b) 由課本定理可知，沿(1, −1)方向∇f(1, 2)會有最大值。 (5%)

3. (10%) 求 e^(x2−y2)+ x sin z = 1在 (1, 1, 0) 的切平面方程式。

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Solution:

Find the tangent plane equation of

e^(x2−y2)+ x sin z = 1 at (1, 1, 0) Let f = e^(x2−y2)+ x sin z − 1

then

∇f = (2xe^(x2−y2)+ sin z, −2ye^(x2−y2), x cos z)

∇f |_(1,1,0) = (2, −2, 1) (8 points) Hence the equation is

2x − 2y + z = 0 (2 points)

4. (15%) 使用 Lagrange 乘子法求 f(x, y) = x²y 在 x²+ 2y² = 6 限制條件下的最大值與最小值。

Solution:

令

f (x, y) = x²y 和 g(x, y) = x² + 2y²− 6, 則由 Lagrange multiplier method:

 Of(x, y) = λOg(x, y),

g(x, y) = 0, (3 分) 可得







2xy = λ · 2x, x² = λ · 4y, x²+ 2y²− 6 = 0.

(3分)

由 2xy = λ · 2x ⇒ x = 0 or λ = y.

case 1: x = 0. 此時代入 x²+ 2y² − 6 = 0 得 y = ±√

3. (3 分) case 2: λ = y. 此時代入 x² = λ · 4y 得 x² = 4y². 故解

 x² = 4y²,

x²+ 2y²− 6 = 0, 得 (x, y) = (2, 1), (2, −1), (−2, 1), (−2, −1). (4 分) 綜合上述, 計算

f (0,√

3) = 0, f (0, −√

3) = 0, f (2, 1) = 4, f (2, −1) = −4, f (−2, 1) = 4, f (−2, −1) = −4, 得最大值為 4, 最小值為 −4. (2 分)

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5. (15%) 求f(x, y) = x⁴+ y⁴− 4xy + 1的極大值、極小值與鞍點。(假如存在的話)

Solution:

1^◦ 找出候選點。(5%)

令 5f(x, y) = (fx, f_y) = (4x³− 4y, 4y³− 4x) = (0, 0) , 則得到

 4x³− 4y = 0

4y³− 4x = 0 =⇒ x³ = y

y³ = x =⇒ x⁹ = x 若 x = 0 ,則 (0, 0) 為候選點。

若 x 6= 0 ,則 x⁸ = 1 ⇒ x⁸− 1 = 0

⇒ (x⁴+ 1)(x²+ 1)(x + 1)(x − 1) = 0 ⇒ x = 1 or x = −1 (只需要實數解) 故 (1, 1), (−1, −1) 為候選點。

2^◦ 計算 D(x, y)。(4%)

D(x, y) =    

f_xx f_xy f_yx f_yy

=    

12x² −4

−4 12y²    

= 144x²y²− 16

3^◦ 使用二階測試判斷各個候選點。(6%)

1. 在點(0, 0)

D(0, 0) < 0 =⇒ (0, 0) 為鞍點。

2. 在點(1, 1)

D(1, 1) > 0 & f_xx(1, 1) > 0 =⇒ (1, 1) 為極小點，極小值為 f(1, 1) = −1 。 3. 在點(−1, −1)

D(−1, −1) > 0 & f_xx(−1, −1) > 0 =⇒ (−1, −1) 為極小點，極小值為 f(−1, −1) =

−1。

6. (10%) 求 Z 1

0

Z 1 x

cos(y²)dydx。

Solution:

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Z 1 0

Z 1 x

cos(y²) · dydx

= Z 1

0

Z y 0

cos(y²) · dxdy ( Fubini’s : + 4pts )

= Z 1

0

cos(y²) Z y

0

1dx · dy

= Z 1

0

cos(y²) · y · dy ( Integration of Monomials : + 2pts )

= Z 1

0

cos(y²)

2 · d(y²)

= sin(y²) 2

y=1

y=0

( Integration of cos(y²)y : +2 pts )

= 1

2 sin(1²) − sin(0²)
= 1

2sin(1) ( Computation of sin(x) : +2 pts )

7. (15%) 求 Z Z

Ω

e^(x2+y2)2 dA, 其中 Ω : 1 ≤ x² + y² ≤ 2 且 y ≥ 0。

Solution:

Z π 0

Z

√ 2 1

re^r22 drdθ (5 points) = π(e − e¹²) (10 points)

8. (15%) 利用變數變換計算Z 1 0

Z 1−x 0

(x + y)³²(y − x)²dydx。

Solution:

Assume u = x + y and v = y − x ⇒ x = u − v

2 y = u + v

2 (4%)

and compute the Jacobian J = 1

2 (1%)

Now we have to consider the range of integration. In x-y coordinate the range is an area which is enclosed by x = 0, y = 0, and y = 1 − x. Because we use the change of variables, we should find the corresponding integral range.

x = 0 ⇒ u − v

2 = 0 ⇒ v = u

y = 0 ⇒ u + v

2 = 0 ⇒ v = −u

y = 1 − x ⇒ u = 1 (3%)

So by above argument we obtain the range of integration in u-v coordinate, we rewrite the integral

Z 1 0

Z u

−u

u³²v²1

2dvdu = Z 1

0

1

6u³²v³|^u_−udu = 1 3

Z 1 0

u⁹²du = 2 33

(7%)

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