1032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. (14%) 求函數 f(x, y) = xy 在曲線 x2+ xy + y2= 1 上之最大值,最小值及其所在之點。
Solution:
使用Lagrange multiplier method:
⎧⎪⎪⎪⎨⎪⎪⎪
⎩
y= λ(2x + y) x= λ(2y + x)
x2+ xy + y2= 1 (4pts)
⇒⎧⎪⎪⎪
⎨⎪⎪⎪⎩ λ=1
3, x= ±
√3 3 , y= ±
√3
λ= −1, x = ±1, y = ∓1 3 (4pts)
將所得各點代入f(x, y)後比較,得到:
f(x, y)在(±
√3 3 ,±
√3
3 )有最大值1
3。(3pts) f(x, y)在(±1, ∓1)有最小值−1。(3pts)
(若過程完全錯誤,這部分不給分。)
2. (12%) 令 f(x, y) = x3− 3λxy + y3其中 λ 為實數,λ≠ 0. 求出 f 之候選點,並判斷其極值性質。(注意:和 λ 之值有關)
Solution:
∇f(x, y) = (3x2− 3λy, 3y2− 3λx) = (0, 0) (4 pts) Now, we solve the following equations,
{ x2− λy = 0 − (1) y2− λx = 0 − (2)
From (1), we get y=x2
λ, put into (2), we have x4
λ2 − λx = 0 Hence, x= 0 or x = λ, so critical point: (0, 0), (λ, λ) (2 pts for each) D(x, y) = ∣ 6x −3λ
−3λ 6y ∣ =36xy− 9λ2 (2 pts)
1. D(0, 0) = −9λ2< 0 → (0,0) is saddle pt. (1 pt) 2. D(λ, λ) = 27λ2> 0 & fxx(λ, λ) = 6λ, then
(i) if λ> 0, f(λ, λ) = −λ3is local min.
(ii) if λ< 0, f(λ, λ) = −λ3is local max. (1 pt)
3. (12%) 計算 ∬
Ω
x2
x2+ y2dA,其中 Ω∶ 1 ≤ x2+ y2≤ 2.
Solution:
Let x= r cos θ, y = r sin θ, then
∫ ∫Ω x2
x2+ y2dA= ∫02π∫
√ 2 1
rcos2θdrdθ= 1 2 ∫
2π 0
cos2θdθ= 1 4 ∫
2π
0 (cos 2θ + 1)dθ =π 2
* The score will be given based on 1. domain of the integral, 2. correctness of the integration. In general 3, 7 or 12 points will be given, some minor adjustments might occur if computational mistakes happened.
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4. (12%) 令 Ω 為直線 y− x = 1, y − x = 2, 2x + y = 0, 2x + y = 2 所圍成的區域。計算 ∬
Ω
(y − x)(2x + y)dA.
Solution:
By change of variables, let:
u= y − x, v = 2x + y x=v− u
3 , y= 2u+ v 3 So the integration region changes from:
y− x = 1 y− x = 2 2x+ y = 0 2x+ y = 2 to:
u= 1 u= 2 v= 0 v= 2 The Jacobian from change of coordinate(x, y) to (u, v) is:
J =
⎡⎢⎢⎢
⎢⎢⎢⎢
⎢⎣
∂x
∂u
∂x
∂v
∂y
∂u
∂y
∂v
⎤⎥⎥⎥
⎥⎥⎥⎥
⎥⎦
=
⎡⎢⎢⎢
⎢⎢⎢⎢
⎢⎣
−1 3
1 3 2 3 1 3
⎤⎥⎥⎥
⎥⎥⎥⎥
⎥⎦
= −1 3
The integration becomes:
∫ ∫ (y − x)(2x + y)dA = ∫02∫12uv∣J∣dudv = 1 3 ∫
2 1
udu∫02vdv
= 1 3∗ (u2
2 )∣21(v2 2 )∣20= 1 評分標準:
1: 變數變換並得到正確的 Jacobian: (3分) 2: 找出在變數變換後正確的積分範圍: (3分) 3: 雙重積分做對: (6分)
在過程中有小錯誤每次一分扣
常見的錯誤像是 Jacobian 沒有加絕對值都是扣一分
另外有同學直接用直角坐標去做,儘管圍城區域有畫對, 但通常積分範圍就開始不正確了, 所以只給1-2分
5. (12%) 設 f(x, y) 為可微函數,且 ∂f
∂x(2, −2) =√ 2,∂f
∂y(2, −2) =√
5。令 x= u − v,y = v − u. 求 ∂f
∂u +∂f
∂v 在 u= 1, v= −1 之值。
Solution:
當(u, v) = (1, −1)時,(x, y) = (2, −2) (1%)
∂f
∂x(2, −2) =√ 2,∂f
∂y(2, −2) =√
5 (3%)
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∂x
∂u = 1,∂x
∂v = −1,∂y
∂u = −1,∂y
∂v = 1 (3%) (∂f
∂u +∂f
∂v) ∣(u,v)=(1,−1)= (∂f
∂x
∂x
∂u+∂f
∂y
∂y
∂u+∂f
∂x
∂x
∂v +∂f
∂y
∂y
∂v) ∣(x,y)=(2,−2) (5%)
=√
2× 1 +√
5× (−1) +√
2× (−1) +√
5× 1 = 0
註1:因為(u, v) = (1, −1)時,(x, y) = (2, −2),之後的偏微分才可以直接帶入√ 2,√
5,所以一定要明確表示這一點
註2:如果不把點帶入,直接用代數,如下:
∂f
∂u+∂f
∂v = ∂f
∂x× 1 +∂f
∂y × (−1) +∂f
∂x× (−1) +∂f
∂y × 1 = 0 就不需要敘述(u, v)與(x, y)之間的關係,可以直接拿到滿分
6. (14%) 令 f(x, y) = excos y+ a sin y,其中 a 為一常數。
(a) (7%) 求曲線 f(x, y) = −1 在點 (0, π) 之切線方程式,以 a 表示之。
(b) (7%) 設方向導數 ∂f
∂⇀u(0, 0) 之最大值發生在⇀u= (3 5,4
5),求 a 之值。
Solution:
(a) 令g(x, y) = excos y+ a sin y + 1 = 0,則:
∇g(x, y) = (excos y,−exsin y+ a cos y)
(3分)
∇g(0, π) = (−1, −a)
(2分) 切線方程為:
0= ∇g(0, π) ⋅ (x − 0, y − π) = −x − ay − aπ
(2分) (b) 因最大值發生的方向為∇f,故⃗u = (3
5,4
5)與∇f平行:
∇f ∥ ⃗u
(5分) 即:
−1−a=
3 5 4 5
a=4 3
(2分)
7. (12%) 計算下列積分值: ∬
Ω
3x2
(x3+ y2)2dA其中 Ω= [0, 1] × [1, 3]。
Solution:
∫13∫01 3x2
(x3+ y2)2dxdy= ∫13(1 y2 − 1
1+ y2)dy
= (−1
y − tan−1y)∣31=2
3− tan−13+π 4
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8. (12%) 計算下列積分值: ∬
R
xey
y dA其中 R∶ 0 ≤ x ≤ 1, x2≤ y ≤ x 所圍成的有限區塊。
Solution:
R= {(x, y) ∣ 0 ≤ x ≤ 1, x2≤ y ≤ x} = {(x, y) ∣ 0 ≤ y ≤ 1, y ≤ x ≤ √y}
∴ ∬Rxey
y dA = ∫01∫
√y
y
xey
y dxdy (7%)
= ∫01ey y ⋅x2
2 ∣
√y
x=y
dy
= ∫01ey y ⋅1
2(y − y2)dy (2%)
= 1
2[∫01eydy− ∫01yeydy] =1
2[ey− (yey− ey)] ∣1
0
= 1
2e− 1 (3%)
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