GRADING CRITERIA

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1062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. (13%) 求 ∬

Ω

(x − y)²⁰dA，其中 Ω 是由 2x + y = 0, 2x + y = 1, x + 2y = 0 及 x + 2y = 1 所圍成的平行四邊形。

Solution:

(2pt) Let x + 2y = u, x − y = v that is x = 1

3(u + 2v), y = 1

3(u − v) ...(2 pt)

∣ J ∣ = ∣ RR RR RR RR RR RR RR R

∂x

∂u

∂x

∂y ∂v

∂u

∂y

∂v RR RR RR RR RR RR RR R

∣ = ∣ RR RR RR RR RR RR RR R 1 3

2 1 3 3

−1 3

RR RR RR RR RR RR RR R

∣ = 1

3...(2 pt)

Note that the new region Ω^′ will enclosed by:

2x + y = 0 → u − v = 0 2x + y = 1 → u − v = 1 x + 2y = 0 → u = 0 x + 2y = 1 → u = 1 So we have

Ω^′=

⎧⎪

⎪

⎨

⎪⎪

⎩

u − 1 ≤ v ≤ u 0 ≤ u ≤ 1 ...(1 pt)

∬Ω

(x − y)²⁰dA =∬

Ω^′v²⁰1

3dA = 1 3∫

1 0 ∫

u

u−1v²⁰dvdu = ...(1 pt)

= 1 3∫

1 0

[ 1

21v²¹]^v=u_v=u−1du = 1 63∫

1

0 u²¹− (u − 1)²¹du...(2 pt)

= 1 63{[

u²²

22]^u=1_u=0− [

(u − 1)²²

22 ]^u=1_u=0}...(2pt)

= 1 63( 1

22+ 1

22) = 1

693...(1 pt)

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2. (12%) 計算二重積分

∬Ωe^−(x²^+y²⁾dA, Ω = {(x, y) ∶ 1 ≤ x²+y²≤2, y ≥ 0, y ≥ x ≥ −y}.

Solution:

...(2pt) Let x = r cos θ, y = r sin θ...(2 pt)

∣ J ∣ = ∣ RR RR RR RR RR RR RR R

∂x

∂r

∂x

∂y ∂θ

∂r

∂y

∂θ RR RR RR RR RR RR RR R

∣ = ∣ ∣cos θ −r sin θ

sin θ r cos θ∣ ∣ =r...(2 pt)

y = x → θ = π 4 y = −x → θ = 3π

4 So we have:

Ω =

⎧⎪

⎪⎪

⎨

⎪⎪

⎪⎩ 1 ≤ r ≤

√ 2 π

4 ≤u ≤ 3π

4 ...(1 pt)

∬Ω

e^−(x²^+y²⁾dA =∬

Ω

e^−r²rdrdθ =∫

3π 4

π 4

∫

√ 2 1

e^−r²rdrdθ...(1 pt)

= ∫

3π 4

π 4

[

−1 2 e^−r²]^r=

√2 r=1 dθ =∫

3π 4

π 4

1 2e⁻¹−

1

2e⁻²dθ...(2 pt)

= 1

2(e⁻¹−e⁻²) π 2 =

π

4(e⁻¹−e⁻²)...(2 pt)

3. (13%) 求曲線 10x²+12xy + 5y² =14上和原點 (0, 0) 相距之最近點及最遠點。

Solution:

assume f (x) =

√ x²+y² using Lagrange equation

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⎧⎪

⎪⎪

⎨

⎪⎪

⎪⎩

2x = λ(20x + 12y) 2y = λ(12x + 10y) 10x²+12xy + 5y² =14

⇒ x y =

10x + 6y

6x + 5y ⇒6x²−5xy − 6y²=0 (1)

⇒ (3x + 2y)(2x − 3y) = 0

condtition 1: 3x + 2y = 0 40

9 y²−8y²+5y² =14 y = ±3

√ 14 13 x = ∓2

√ 14 13 f (∓2

√ 14 13, ±3

√ 14 13) =

√

14 (max)

condtition 2: 2x − 3y = 0 45

2 y²+18y²+5y² =14 y = ± 2

√ 13 x = ± 3

√ 13 f (± 3

√

13, ± 2

√

13) =1 (min)

score: Find f(x) gets 3 pts. Using Lagrange equation gets 3 pts. Didin’t find the answer of Lagrange equation loses 2 pts separately.(There are two lines or λ you will get.) Answer is 1 pt separately.

4. (12%) 設 f (x, y) 是雙變數 x, y 之函數。令 g(t) = f(1 + 2t, 3 + 4t). 求 g^′(0) 及 g^′′(0)。(用 ∂f

∂x(1, 3),

∂f

∂y(1, 3), ∂²f

∂x²(1, 3), ∂²f

∂x∂y(1, 3), ∂²f

∂y²(1, 3) 表出)

Solution:

令 x = 1 + 2t、y = 3 + 4t。

由鏈鎖律得到

g^′(t) = ∂f

∂x dx

dt +∂f

∂y dy

dt =2∂f

∂x(x, y) + 4∂f

∂y(x, y)

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再次微分得到

g^′′(t) = 2 [ ∂

∂x(∂f

∂x)dx dt + ∂

∂y(∂f

∂x)dy

dt] +4 [ ∂

∂x(∂f

∂y)dx dt + ∂

∂y(∂f

∂y)dy dt]

=4∂²f

∂x²(x, y) + 16 ∂²f

∂x∂y(x, y) + 16∂²f

∂y²(x, y) 將 t = 0 代入得到

g^′(0) = 2∂f

∂x(1, 3) + 4∂f

∂y(1, 3) g^′′(0) = 4∂²f

∂x²(1, 3) + 16 ∂²f

∂x∂y(1, 3) + 16∂²f

∂y²(1, 3)

評評評分分分標標標準準準各階微分，列出鏈鎖律公式得 2 分，微分正確得 2 分，代入 t = 0 算出正確答案得 2 分。

5. (12%) 求過曲面 xe^yz+ln(y³z²) =tan⁻¹( x

z) 上點 (0, 1, 1) 的切面方程式。

Solution:

令 f(x, y, z) = xe^yz+ln(y³x²) −tan⁻¹( x z)。偏微分得到

f_x(x, y, z) = e^yz−

1 z

1 + (^x_z)

2

f_y(x, y, z) = xze^yz+ 3y²z²

y³z² f_z(x, y, z) = xye^yz+2y³z

y³z² −

−_z^x₂ 1 + (^x_z)² 代入 (0, 1, 1) 得到

f_x(0, 1, 1) = e − 1, f_y(0, 1, 1) = 3, f_z(0, 1, 1) = 2 因此切面為

(e − 1)(x − 0) + 3(y − 1) + 2(z − 1) = 0 化簡得到

(e − 1)x + 3y + 2z = 5

評評評分分分標標標準準準偏導數正確各得 2 分，代入正確各得 1 分，列出切面公式得 2 分，切面正確得 1 分。

6. (13%) 設 f (x, y) = (x²

4 +y²−1)

2

+x²y² (a) 求在點 (2, 1) 的梯度 ∇f。

(b) 求在點 (2, 1) 沿著向量 (3, 1) 之方向的方向導數。

(c) 在點 (2, 1)，函數 f(x, y) 增加最快的方向為何(以單位向量表示)？並求沿該方向之方向導數。

(d) 求通過點 (2, 1) 之等高線在點 (2, 1) 處之切線方程式。

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Solution:

(a)

In order to fine the gradient of f at (2, 1), we need to find ∂f

∂x and ∂f

∂y first.

∇f (2, 1) = (∂f

∂x, ∂f

∂y)∣

(2,1)

= (2(x²

4 +y²−1)x

2 +2xy², 2(x²

4 +y²−1)2y + 2x²y)∣

(2,1)

= (6, 12) (b)

Directional derivative of f along (3, 1) at (2, 1) is the inner product of ∇f (2, 1) and Ð→

u where Ð→

u is unit vector parallel to (3, 1)

∂f

∂Ð→

u = ∇f (2, 1) ⋅Ð→

u = (6, 12) ⋅ ( 3

√

10, 1

√

10) =3

√ 10 (c)

From the definition of gradient, we know the direction that will make f increasing most rapidly is parallel to the gradient of f . Hence the answer to the first question is the unit vector parallel to (6, 12) which is ( 1

√ 5, 2

√ 5)

The directional derivative of f along ( 1

√ 5, 2

√

5) at (2, 1) is (6, 12) ⋅ ( 1

√ 5, 2

√ 5) =6

√ 5

(d)

Since the normal vector of tangent line of contour line at (2, 1) is ∇f (2, 1), its equation is 6(x − 2) + 12(y − 1) = 0

GRADING CRITERIA

(a) There are

3

points in this subproblem. Find ∂f

∂x and ∂f

∂y correctly get

1

point respectively, compute the final answer correctly get

1

point

(b) There are

3

points in this subproblem. Find unit vector correctly get

1

point, know how to do inner product get

1

point, answer correct get

1

point.

(c) There are

4

points in this subproblem. Find the direction in the form of unit vector get

2

point, find directional derivative correctly get

2

points. You’ll loss 1 point with each computation error.

(d) There are

3

points in this subproblem. Find the normal vector of tangent line get

2

points, answer correct get

1

point.

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7. (13%) 找出函數 f (x, y) = e^y²^−x²(y²+x²) 的候選點，並決定它是局部極大值，局部極小值或是鞍點。

Solution:

f (x, y) = e^y²^−x²(y²+x²)

(2分)fx= −2x ⋅ e^y²^−x²(y²+x²) +e^y²^−x²⋅2x = e^y²^−x²2x(1 − x²−y²) (2分)fy =2y ⋅ e^y²^−x²(y²+x²) +e^y²^−x²⋅2y = e^y²^−x²2y(1 + x²+y²) 要算候選點,即為求∇f(x, y) = (fx, f_y) = (0, 0)之(x,y)解,

(3分)若fx =0, 則x = 0 或 x²+y² =1; 若fy =0, 由於1 + x²+y² >1 ≠ 0, 勢必y = 0. 綜合上面取交集,得到候選點有三點:(0, 0), (1, 0),及(−1, 0).

候選點可以使用D(x, y) = fxxf_yy−f_xy² 判斷,因此我們需要計算二次偏微分:

(1分)fxx= −2x ⋅ f_x+e^y²^−x²{2(1 − x²−y²) +2x(−2x)}

=e^y²^−x²(4x⁴+4x²y²−10x²−2y²+2)

(1分)fyy=2y ⋅ f_y+e^y²^−x²{2(1 + x²+y²) +2y(2y)}

=e^y²^−x²(4y⁴+4x²y²+2x²+10y²+2)

(1分)fxy =f_yx=2y ⋅ f_x+e^y²^−x²{2x(−2y)} = e^y²^−x²(−4xy³−4x³y)

(3分)把候選點們代入,得到D(0, 0) = 2 ∗ 2 − 0 ∗ 0 = 4 > 0且fxx(0, 0) = 2 > 0; D(1, 0) = D(−1, 0) =

−4e⁻¹∗4e⁻¹−0 ∗ 0 = −16e⁻²<0.

故在(0, 0)發生局部極小值,而(±1, 0)皆為鞍點.

1.偏微分有計算錯誤扣一分,幾乎沒積出形式不給分.

2.候選點如果三個都有寫出來但是還多寫不管幾個都只扣一分;如過有漏寫幾個,僅扣少寫幾個的分數.

3.如果沒算出來候選點,卻有寫如何判別且全部正確,1分.

4.D(x,y)的每個點的計算錯誤,頂多扣一分.

8. (12%) 求 ∬

Ω

sin x

x dA，其中 Ω 為 x 軸和 y = x, x = 1 所圍區域。

Solution:

(3分)把邊界畫出來,灰色部分為所圍出的區域.

(3分)因此根據Fubini的定理,所求積分可以寫為∫ ¹

0 ∫

x 0

sin x

x dydx或∫

1 0 ∫

1 y

sin x x dxdy.

(3分)後者的積分無法寫出來,因此我們用前者,所求=∫

1 0

sin x x ∗xdx (3分)= − cos x]¹0 =1 − cos 1.

1.圖若沒畫,有寫出正確的範圍不扣分;圖畫錯一個地方扣一分.

2.兩個不同方向的積分有寫出來其中一個都給3分,錯一個積分範圍扣1分,寫反扣1分.

3.最後答案一個+-號寫錯扣1,積分結果錯扣2分,差太多沒分.