(10%) Determine the values of x for which the series ∞ X n=2 1 n ln n  1 + 1 x n converges absolutely, converges condition- ally or diverges

1002微微微甲甲甲08-12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%) Determine the values of x for which the series

∞

X

n=2

1 n ln n

 1 + 1

x

ⁿ

converges absolutely, converges condition- ally or diverges.

Solution:

Use Ch.9. theorem11. p.516 ρ = lim

n→∞

1

(n+1) ln(n+1) 1 + ¹_x
n+1 1

n ln(n) 1 + ¹_x
ⁿ      

= lim

n→∞

n ln(n)

(n + 1) ln(n + 1)(1 + 1 x)

=    

1 + 1 x    

=    

x + 1 x

    ρ < 1

x + 1 x

< 1, |x + 1| < |x|

when x < −1

2, converges absolutely x = −1

2, the series =

∞

X

n=2

1

n ln(n)(−1)ⁿ is the alternating series (Ch9. theorem 14. p.521) when x = −1

2, converges conditionally when x > −1

2, diverges

2. (10%) Evaluate lim

x→0

x − tan⁻¹x

e^3x− 1
2x²− 1 + cos 2x .

Solution:

tan⁻¹x = x −x³

3 + ... (2 points) e^3x= 1 + 3x + ... (2 points) cos 2x = 1 −4x²

2! +16x⁴

4! (2 points)

x→0lim

(1 − tan⁻¹x)(e^3x− 1) 2x − 1 + cos 2x = lim

x→0

x⁴+ ...

2

3x⁴+ ... = 3

2(4 points) (There is no partial credit for using L’hospital law)

3. (10%) Find out the value for the series

∞

X

n=0

(ln 2)²ⁿ (2n)! .

Solution:

Since e^x= 1 + x +x²

2! + ... and e^−x= 1 − x +x²

2! + ... (2 points), e^x+ e^−x

2 = 1 + x² 2! +x⁴

4! + ...

1 +(ln 2)²

2! +(ln 2)⁴

4! + ... = e^{ln 2}+ e^{− ln 2}

2 =5

4 (8 points)

4. (10%) Find the Taylor series representation for f (x) = ln(2 + 3x) about x = 1. For what values of x is this representation valid?

Solution:

ln(2 + 3x) = ln(5 + 3(x − 1)) = ln 5

 1 + 3

5(x − 1)



= ln 5 + ln

 1 +3

5(x − 1)



= ln 5 +

∞

X

n=1

(−1)ⁿ⁻¹3

5(x − 1)ⁿ n

(6 points)

−1 <3

5(x − 1) ≤ 1 ⇒ −2

3 < x ≤ 8

3 (4 points)

5. (15%) Consider a curve r(t) = (3t − t³) i + 3t²j + (3t + t³) k, t ∈ R.

(a) Find the unit tangent T(t).

(b) Find the arc length function s(t).

(c) Find the unit normal N(t).

(d) Find the curvature κ(t).

Solution:

γ⁰(t) = (3 − 3t², 6t, 3 + 3t²) (2 pts)

|γ⁰(t)| =p

18 + 18t⁴+ 36t²= 3√

2(t²+ 1) (2 pts)

T (t) = γ⁰(t)

|γ⁰(t)| = 1

√2(1 − t² t²+ 1, 2t

t²+ 1, 1) (2 pts)

s(t) = Z t

0

|γ⁰(u)|du = Z t

0

3√

2(u²+ 1)du = 3√ 2(t³

3 + t) (2 pts)

T⁰(t) =

√2

(t²+ 1)²(−2t, 1 − t², 0) (2 pts)

|T⁰(t)| =

√2 (t²+ 1)²

p4t²+ (1 − t²)²=

√2

(t²+ 1)²(t²+ 1) =

√2

t²+ 1 (2 pts)

N (t) = T⁰(t)

|T⁰(t)| = 1

t²+ 1(−2t, 1 − t², 0) (1 pt)

κ(t) = |T⁰(t)|

|γ⁰(t)| =

√ 2 t²+1

3√

2(t²+ 1) = 1

3(t²+ 1)² (2 pts)

6. (20%) Consider the function

f (x, y) =





 x³

x²+ y² if (x, y) 6= (0, 0) 0 if (x, y) = (0, 0).

(a) Find lim

(x,y)→(0,0)

f (x, y). Is f continuous at (0, 0)?

(b) Find the partial derivatives ∂f

∂x and ∂f

∂y. (c) Is ∂f

∂y continuous at (0, 0)?

(d) Find ∂²f

∂x∂y(0, 0) and ∂²f

∂y∂x(0, 0).

(e) Find the directional derivative of f at (1, 1) in the direction of 5 i + 12 j.

(f) Find the maximal rate of change of f at (1, 1) and the direction in which it occurs.

Solution:

(3 pts)(a)

lim

(x,y)→(0,0)| x³

x²+ y²| ≤ lim

(x,y)→(0,0)|x|| x²

x²+ y²| ≤ lim

(x,y)→(0,0)|x| = 0

⇒ lim

(x,y)→(0,0)

x³

x²+ y² = 0 = f (0, 0) So f is continuous at (0, 0).

(4 pts)(b)

For (x, y) 6= (0, 0),

∂f

∂x(x, y) = x⁴+ 3x²y² (x²+ y²)²,

∂f

∂y(x, y) = −2x³y (x²+ y²)².

And

∂f

∂x(0, 0) = lim

h→0

f (h, 0) − f (0, 0)

h = 1,

∂f

∂y(0, 0) = lim

k→0

f (0, k) − f (0, 0)

h = 0.

(3 pts)(c) Let x = cos θ, y = sin θ. Then

lim

(x,y)→(0,0)

∂f

∂y(x, y) = lim

(x,y)→(0,0)

−2x³y (x²+ y²)²

= lim

r→0

−2r⁴cos³θ sin θ

r⁴ = −2 cos³θ sin θ Hence, ∂f

∂y is not continuous at (0, 0).

(4 pts)(d)

∂²f

∂x∂y(0, 0) = lim

h→0

∂f

∂y(h, 0) −^∂f_∂y(0, 0)

h = 0.

∂²f

∂y∂x(0, 0) = lim

k→0

∂f

∂x(0, k) −^∂f_∂x(0, 0) k

= lim

k→0

−1

k does not exist.

(3 pts) (e)

u = 5 13i +12

13j, ∇f (x, y) = (∂f

∂x,∂f

∂y)

⇒D_uf (1, 1) = ∇f (1, 1) · u = (1,−1 2 ) · ( 5

13,12 13) = −1

13.

(3 pts) (f) maximal rate of change = |∇f (1, 1)| =

√5

2 . The direction vector is parallel to ∇f (1, 1).

7. (15%) Let C be the curve of intersection of the paraboloid z = x²+1 2xy +y²

4 and the circular cylinder x²+ y²= 13.

(a) Find a parametric equation for the tangent line to C at the point (3, 2, 13).

(b) An object is moving along C. If the x-coordinate is increasing at the rate of 4 cm/sec, how fast is the z-coordinate changing at the instant when x = 3 cm and y = 2 cm.

Solution:

(a) [8%]

Let f (x, y, z) = x²+1

2xy +y²

4 − z, g(x, y, z) = x²+ y²− 13 and the curve C = C(s).

∇f (x, y, z) = (2x +1 2y,1

2x +1

2y, −1) ⇒ ∇f (3, 2, 13) = (7,5

2, −1). ...(2 points)

∇g(x, y, z) = (2x, 2x, 0) ⇒ ∇g(3, 2, 13) = (6, 4, 0). ...(2 points)

Since ∇f (3, 2, 13)⊥ the tangent vector C⁰(s)|_(3,2,13) and ∇g(3, 2, 13)⊥ the tangent vector C⁰(s)|_(3,2,13)

⇒ the tangent vector C⁰(s) // ∇f (3, 2, 13) × ∇g(3, 2, 13).

And ∇f (3, 2, 13) × ∇g(3, 2, 13) = (7,5

2, −1) × (6, 4, 0) = (4, −6, 13). ...(2 points) Therefore, the tangent line equation is x − 3

4 =y − 2

−6 = z − 13

13 . ...(2 points)

(b) [7%]

At the point (3, 2, 13), we have the rate dx

dt = 4 cm/sec.

Take d

dt on the equation x²+ y²= 13 ⇒ 2xdx

dt + 2ydy

dt = 0 ...(2 points)

⇒ dy

dt|_(3,2,13)= (−x y

dx

dt)|_(3,2,13) =−3

2 · 4 = −6 cm/sec. ...(1 points) Take d

dt on the equation z = x²+1

2xy + y² 4

⇒ dz

dt = 2xdx dt +1

2 dx dty +1

2xdy dt +y

2 dy

dt ...(2 points)

⇒ dz

dt|(3,2,13)= (2xdx dt +1

2 dx dty +1

2xdy dt +y

2 dy

dt)|_(3,2,13)

= 6 · 4 +1

2 · 4 · 2 +1

2 · 3 · (−6) +2

2 · (−6) = 13 cm/sec. ...(2 points)

8. (10%) Let x, y, u and v be related by the equations

 xyuv = 1

x + y + u + v = 0. Find ∂y

∂x

 u

.

Solution:

令

F (x, y, u, v) = xyuv − 1 = 0, G(x, y, u, v) = x + y + u + v = 0.

法一： 由公式

 ∂y

∂x



u

= −

∂(F,G)

∂(v,x)

∂(F,G)

∂(v,y)

(分子分母各 2 分; v 寫成 u 者以下不給分)

= −    

xyu yuv

1 1

xyu xuv

1 1

(分子分母各 2 分)

= −xyu − yuv xyu − xuv (2分)

= y(x − v)

x(v − y). (由題意知 u 6= 0) 法二： Holding u,

∂F

∂x = 0 ⇒ yuv + x∂y

∂xuv + xyu∂v

∂x = 0, (3分) and

∂G

∂x = 0 ⇒ 1 +∂y

∂x+∂v

∂x = 0. (3分) By Crammer’s Rule,

∂y

∂x =    

−yuv xyu

−1 1

xuv xyu

1 1

(2分)

= y(x − v) x(v − y). (2 分)

小錯誤扣 2 分; 對非 x 的變數做偏微分至多得 4 分。

法三： 由

x + y + u + 1

xyu = 0, (2分) 視 u 為常數, 對 x 偏微分得

1 + ∂y

∂x+ −1

(xyu)² · (yu + x∂y

∂xu) = 0, (完全正確得 6 分)

同乘 x²y²u, 得

x²y²u + x²y²u∂y

∂x− y − x∂y

∂x = 0, 故

∂y

∂x = y − x²y²u

x²y²u − x = y − xy · ¹_v

xy ·¹_v − x = y(v − x) x(y − v). (2 分)

9. (10%) Find and classify the critical points of the function f (x, y) = x⁴+ y⁴− 4xy.

Solution:

令







∂f

∂x = 4x³− 4y = 0 ⇒ y = x³

∂f

∂y = 4y³− 4x = 0 ⇒ x = y³

(2 分) ⇒ x = y³= x⁹

⇒ x(x⁴+ 1)(x²+ 1)(x + 1)(x − 1) = 0

⇒ x = 0, 1, −1.

得 critical points:

(0, 0), (1, 1), (−1, −1). (3分)

設二階測試

D(x, y) =    

12x² −4

−4 12y²    

(2 分), 計算

D(0, 0) =    

0 −4

−4 0    

= −16 < 0 ⇒ (0, 0) is a saddle point. (1分)

D(1, 1) =    

12 −4

−4 12    

= 128 > 0, and f_xx(1, 1) = 12 > 0

⇒ f (1, 1) = −2 is a minimum. (1分)

D(−1, −1) =    

12 −4

−4 12    

= 128 > 0, and fxx(−1, −1) = 12 > 0

⇒ f (−1, −1) = −2 is a minimum. (1分)

10. (10%) If the ellipse x² a² +y²

b² = 1 (a, b > 0) is to enclose the circle x²+ y²= 2y, what values of a and b minimize the area of the ellipse?

Solution:

Since x² a² +y²

b² = 1 enclosed x²+ y² = 2y, and the central of the circle is located at (0, 1). The minimal ellipse that can enclosed the cirlce is tangent to the circle, and the tangent point (x, y) has one possible solution for y variable. Hence, replace y variable into one of the two equation we got

2y − y² a² +y²

b² = 1

Rearrange into quadratic form we have (a²−b²)y²+ 2b²y − a²b²= 0.We require the determiant D =p

b²− 4ac = 0, i.e, b²− a²(b²− a²) = 0. This is our constrain equation.(3 points).Under this constrain, we need to derive the minimal area πab. This come to consider Lagrangian multiplier method.

Then our Lagrangian equation is

L(x, y, λ) = πab + λ[b²− a²(b²− a²)]

(1 point)We take derivative along a and b variable, and the minimal point (a, b) will satisfy following system πb + λ(−2ab²+ 4a³) = 0

πa + λ(2b − 2ba²) = 0 b²= a²(b²− a²)

(3 points)Solve this system we got a =p

3/2, b = 3/√

2, λ = −π/√

3.The minimal area is 3√ 3 2 π.

Remark Constrain equation cost 3 points, and Lagrangian equation for 1 points. Answer and system equations cost 3 points resp.