1. (10%) Bᵫ r(t) = (t + sin t cos t)i + (sin

1. (10%) 令曲線由 r(t) = (t + sin t cos t)i + (sin²t)j + (2 cos t)k, t ∈ R, 所定義。

(a) 求曲線上的曲率 κ 。

(b) 求 κ 發生最大值之點, 以及在這些點的單位切向量 (unit tangent) T 及單位主法 向量 (principal unit normal) N。

Sol.

r⁰(t) = (1 + cos 2t)i + sin 2tj + (−2 sin t)k ⇒ T = e₁ = r⁰(t)

|r⁰(t)| = 1 2r⁰(t).

Since

r”(t) = (−2 sin 2t)i + (2 cos 2t)j + (−2 cos t)k,

n = r”(t)− < r”(t), e₁ > e₁ = (−2 sin 2t)i + (2 cos 2t)j + (−2 cos t)k and

|n| = 2√

cos²t + 1, it follows that

N = 1

√cos²t + 1((− sin 2t)i + (cos 2t)j + (− cos t)k) . We have

κ = < e⁰₁, N >

|r⁰(t)| = 1 2

√cos²t + 1.

Therefore,when t = mπ, m ∈ Z, κ has mamimum. Moreover,T = (1, 0, 0), N =

√1

2(0, 1, −1)(n is even) ,N = ^√1

2(0, 1, 1)(n is odd).

2. (18%) 判斷以下級數是絕對收斂 (absolute convergence)、 條件收斂 (conditional con- vergence) 或發散 (divergence)?

(a)

∞

X

n=1

(−1)ⁿ

√n + 1 +√ n (b)

∞

X

n=1

(−1)ⁿ

√n + 1 −√ n (c)

∞

X

n=1

(−1)ⁿ

√n³+ 1 +√ n³ Sol.

(a)(1) lim

n→∞

√1 n

√ 1 n+1+√

n

= 2. Since P∞ n=1

√1

n is divergence, P∞ n=1

√ 1 n+1+√

n is di- vergence.

(2) Since ^√_n+1+^{1 √}_n is decreasing and lim

x→∞

√ 1 n+1+√

n = 0, P∞ n=1

(−1)ⁿ

√n+1+√

n is conver- gence.

Hence P∞ n=1

(−1)ⁿ

√n+1+√

n is conditional convergence.

(b) Since ^√_n+1−^{1 √}_n is increasing, P∞ n=1

(−1)ⁿ

√n³+1+√

n³ is divergence.

(c) lim

n→∞

√1 n3

√ 1 n3+1+

√

n3

= 2. Since P∞ n=1

√1

n³ is convergence, P∞ n=1

√ 1

n³+1+√

n³ is con- vergence.

Hence P∞ n=1

(−1)ⁿ

√ n³+1+√

n³ is absolutely convergence.

3. (10%)

(a) 將 f (x) = tan⁻¹x 表成在 x = 0 的冪級數 (包括一般項及收斂半徑)。

(b) 估計積分 ˆ ¹

4

0

tan⁻¹x

√x dx 之值, 使其誤差 < 1

10⁴(須明確解釋你的估計值之誤差確 實滿足題目中的要求)。

Sol.

Sol: (a) tan⁻¹x = ˆ _x

0

1

1 + t² dt = ˆ _x

0

∞

X

n=0

(−t²)ⁿ dt =

∞

X

n=0

(−1)ⁿ ˆ _x

0

t²ⁿ dt

=

∞

X

n=0

(−1)ⁿ 2n + 1x²ⁿ⁺¹

Let a_n = (−1)ⁿ

2n + 1, by Ratio Test, lim

n→∞|a_n+1

a_n | = 1 ⇒ R = 1

(b) tan⁻¹x

√x = x^1/2−1

3x^5/2+1

5x^9/2−1

7x^13/2+ · · · ˆ ¹

4

0

tan⁻¹x

√x dx = [ 2

3x^3/2− 2

21x^7/2+ 2

55x^11/2− 2

105x^15/2+ · · · ]^1/4

0

= ²₃(¹₂)³− ₂₁²(¹₂)⁷+₅₅² (¹₂)¹¹−₁₀₅² (¹₂)¹⁵+ · · · ( ≡ u₁− u₂+ u₃− u₄+ · · · )

∵ ₅₅² (¹₂)¹¹ < 10⁻⁴ and u_n≥ 0 , u_n& 0 ∀n By The Alternating Series Estimation Theorem, ˆ ¹

4

0

tan⁻¹x

√x dx ≈ 2 3(1

2)³− 2 21(1

2)⁷ = 37

448 , |Error| ≤ 10⁻⁴

4. (10%) 考慮冪級數

∞

X

n=2

1 n ln nxⁿ 。 (a) 求收斂半徑 r 。

(b) 討論在 x = r 及 x = −r 的收斂性。

Sol.

(a)

由比率審斂法 ,

n→∞lim |

1

(n+1) ln(n+1)xⁿ⁺¹

1

(n) ln(n)xⁿ | = |x| < 1 收斂半徑 r =1 .

(b) r=1,

∞

X

n=2

1 n(ln n) 收斂性與

ˆ ∞ 2

1

x(ln x)dx = ln(ln(n))|^∞₂ → ∞ 一致 , 由積分審斂法 , 級數發散 .

r=-1,

5. (10%) 令 z = f (x, y) 為一連續可微函數, 利用代換 x = r cos θ, y = r sin θ 可得一個 恆等關係

(∂z

∂x)²+ (∂z

∂y)² = A(r, θ)(∂z

∂r)²+ B(r, θ)(∂z

∂θ)² 。 試求 A(r, θ) 及 B(r, θ) 。

Sol.

x = r cos θ, y = r sin θ z = f (x, y)







∂z

∂r = ^∂z_∂x^∂x_∂r + ^∂z_∂y^∂y_∂r = ^∂z_∂xcos θ + ^∂z_∂ysin θ

∂z

∂θ = ^∂z_∂x^∂x_∂θ + ^∂z_∂y^∂y_∂θ = ^∂z_∂x(−r sin θ) + ^∂z_∂y(r cos θ) 左右取平方







(^∂z_∂r)² = (^∂z_∂xcos θ + ^∂z_∂ysin θ)². . . [1]

(^∂z_∂θ)² = (^∂z_∂x(−r sin θ) + ^∂z_∂y(r cos θ))². . . [2]

⇒ r²[1] + [2] 左右相加

⇒ r²(^∂z_∂r)²+ (^∂z_∂θ) = r²((^∂z_∂x)²+ (^∂z_∂y)²)

⇒ (^∂z_∂x)²+ (^∂z_∂y)² = (^∂z_∂r)²+ (_r¹2)(^∂z_∂θ)

⇒ A(r, θ) = 1, B(r, θ) = _r¹2 z

6. (10%) 令 z = ye^x+ cosy

x, 求該曲面上之點 (x, y, z) = (1,^π₂,^eπ₂ ) 的切平面及法線方程 式。

7. (10%) 設函數 z = f (x, y) 在 (x, y, z) = (π, π, π) 的附近滿足方程式 sin(x + y) + sin(y + z) + sin(z + x) = 0 。

求函數 z = f (x, y) 在 (π, π) 點, 沿著哪個方向的方向導數 (directional derivative) 最 大? 最大值為何?

8. (10%) 求函數 z = x³− 2x + xy²+ y + 1 的所有臨界點 (critical point), 並判斷它是 極大、 極小或是鞍點 (saddle point)。

Sol.







∂z

∂x = 3x²− 2 + y² = 0

∂z

∂y = 2xy + 1 = 0

(1)

⇒ y = −_2x¹ ⇒ 3x²+_4x¹2 = 2 ⇒ 12x⁴ − 8x²+ 1 = 0 ⇒ (6x²− 1)(2x²− 1) = 0

⇒ x = ±^√1

6, ±^√1

2 ⇒ (x, y) = ±(^√1

6, −

√6

2 ), ±(^√1

2, −^√1

2) are critical points z_xx = 6x, z_yy = 2x, z_xy = 2y ⇒ D = z_xxz_yy− z_xy² = 12x²− 4y²

At ±(^√1

6, −

√6

2 ) ⇒ D = 12 · ¹₆ − 4 ·⁶₄ = 2 − 6 = −4 < 0 are saddle points At (^√1

2, −^√1

2) ⇒ D = 12 ·¹₂− 4 ·²₄ = 6 − 2 = 4 > 0, z_xx = ^√6

2 > 0 is local minimum point

At −(^√1

2, −^√1

2) ⇒ D = 12 · ¹₂ − 4 · ²₄ = 6 − 2 = 4 > 0, z_xx = −^√6

2 < 0 is local maximum point

9. (12%) 平面 x + y +√

2z = 0 與橢球面 (ellipsoid) x² 4 + y²

4 +z²

2 = 1 相交成一橢圓。

(a) 求橢圓上之點與原點的最長距離及最短距離。

(b) 求橢圓面積。

Sol.

Let f (x, y, z) = x²+ y²+ z² g(x, y, z) = x + y +√

2z h(x, y, z) = x²

4 +y² 4 +z²

2 − 1 Then solve the equations











5f = λ 5 g + µ 5 h −(1)

g = 0 −(2)

h = 0 −(3)

¿From (1), we have











2x = λ + µx

2 −(1.1) 2y = λ + µy

2 −(1.2) 2z = λ + µz −(1.3)

⇒











(4 − µ)x = λ (4 − µ)y = λ (2 − µ)z = λ If µ = 4, λ = 0, then z = 0 ⇒ x + y = 0 (∵ g = 0)

substitute y, z with −x, 0 in (3), we have (x, y, z) = ±(√ 2, −√

2, 0) distance =√

4 = 2

If µ 6= 4, then x = y = _4−µ^λ ⇒ z = −√

2x (∵ g = 0) substitute y, z with x, −√

2x in (3), we have (x, y, z) = ±(

r2 3,

r2 3, −

r4 3) distance =

r8 3

So the longest distance = 2, and the shortest distance = r8

3 –(4)

(2) Since the center of the ellipse is at O(0, 0, 0), the semimajor axis = 2, and the semiminor axis =

r8 3 r8