X k=1 lnk k diverges(1 points if the above process is correcet) method(2) for p=1 ∞ X k=1 lnk k after k = 3 the series with nonnegative terms and ∞ X k=3 lnk k &gt

(1)

1012微微微甲甲甲07-11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. (12 points) Test the following two series

∞

X

k=1

ln k

k^p , where p = 1 and p = 3/2, for convergence.

Solution:

(a) method(1) for p = 1

∞

X

k=1

lnk k let u = lnx and du = 1

xdx ˆ ∞

1

lnk

k dk = lim

b→∞

ˆ lnb 0

lnu

u du = lim

b→∞

1

2(lnb)²→ ∞(4points) By Integral test(1 points)

∞

X

k=1

lnk k diverges(1 points if the above process is correcet)

method(2) for p=1

∞

X

k=1

lnk k after k = 3 the series with nonnegative terms and

∞

X

k=3

lnk k >

∞

X

k=3

1 k

and

∞

X

k=3

1

kdiverges (p-series with p = 1)(4 points) By Basic comprasion test (1 point)

∞

X

k=1

lnk k diverges(1 points if the above process is correcet) (b) method(1)

for p = 3 2

∞

X

k=1

lnk k³² let u = lnx and du = 1

xdx dv = dx

x³² and v = −2x⁻¹² ˆ ∞

1

lnx x³² = lim

b→∞[(−2lnx

√x )|^b₁+ 2 ˆ b

1

dx

x³²] = 0 + lim

b→∞2(−2)x⁻¹² |^b₁= 4 < ∞(4points) By Integral test(1 point)

∞

X

k=1

lnk k³²

(2)

converges(1 points if the above process is correcet)

method(2) for p = 3

2

∞

X

k=1

lnk k³²

lim

b→∞

lnk k³2

k⁵⁴ = lim

k→∞

lnk

k¹⁴ = 0(L⁰HospitalRule) because

∞

X

k=1

1

k⁵⁴ converge(a p-seies with p = 5

4 > 1) (4 points) By limit of comparison theorem(1 point)

∞

X

k=1

lnk k³² converges (1 point if process is correcet)

2. (12 points) Determine whether the series converge or diverge.

(a)

∞

X

k=1

(√ k −√

k − 1)^2k.

(b)

∞

X

k=1

(k!)² (5k)!.

Solution:

2-(a) Let ak= (

√ k −√

k − 1)^2k lim

k→∞(ak)¹^k

= lim

k→∞(√ k −√

k − 1)²

= lim

k→∞

1

√k +√ k − 1

²

= 0 < 1

∴

∞

X

k=1

a_k converges by root test.

2-(b) Let ak=

∞

X

k=1

(k!)² (5k)!

lim

k→∞

a_k+1 ak

= lim

k→∞

((k+1)!)² (5(k+1))!

(k!)² (5k)!

= lim

k→∞

((k + 1)!)² (5k + 5)! ·(5k)!

(k!)²

= lim

k→∞

(k + 1)!) (k!)

2

· (5k)!

(5k + 5)!

= lim

k→∞(k + 1)²· 1

(5k + 5)(5k + 4)(5k + 3)(5k + 2)(5k + 1)

= 0 < 1

∴

∞

X

k=1

ak converges by ratio test.

test使用正確: 2 points

(3)

極限算對: 2 points

註明收斂條件後寫出正確答案: 2 points

3. (12 points) Find the interval of convergence of the series

∞

X

k=1

lnk + 1 k

x^k.

Solution:

a_k= ln(k + 1 k ) 1

ρ = lim

k→∞

ak+1

a_k = 1.(you should use L‘H thm) and then you must check the endpoint.

x = 1, Σ^∞_k=1ln(k + 1

k ) = Σ^∞_k=1ln(k + 1) − ln(k) = lim

k→∞ln(k + 1) = ∞.

x = −1,Σ^∞_k=1ln(k + 1 k )(−1)^k lim

k→∞a_k = 0 da_k

dk < 0, and a₁> 0 ⇒ a_k decrease.

=⇒ [−1, 1) is convergence interval.

4. (12 points)

(a) Find the Taylor series for f (x) = (x²+ x + 1)√

x + 1 at x = 0 up to the third power of x.

(b) Let f (x) = ln

r1 + x²

1 − x². Find f⁽¹⁰⁾(0).

Solution:

(a) Find the Taylor series for f (x) = (x²+ x + 1)√

x + 1 at x = 0 up to third power of x Solution:

sincep

(1 + x) = 1 + 1 2x −1

8x²+ 1

16x³− ...

so f (x) = (x²+ x + 1)(1 + 1 2x − 1

8x²+ 1

16x³− ...) = 1 +3 2x + 11

8 x²+ 7

16x³+ ...

(each coefficient of power of x , 2 points) (b) Let f (x) = ln

r1 + x²

1 − x² Find f⁽¹⁰⁾(0) Solution:

since f (x) = 1

2(ln(1 + x²) − ln(1 − x²))(2 points) also ln(1 + x²) = x²−1

2x⁴+1 3x⁶−1

4x⁸+1

5x¹⁰...(1 points) and ln(1 − x²) = −x²−1

2x⁴−1 3x⁶−1

4x⁸−1

5x¹⁰...(1 points) compare the coefficient of power x¹⁰ we have f⁽¹⁰⁾(0) = 10!

5 (2 points for exactly right answer and right expension)

5. (10 points) Find the curvature κ(t) of the curve r(t) = e^ti + e^−tj +√ 2tk.

Solution:

κ(t) = ||T⁰(t)||

s⁰(t) =||T⁰(t)||

||r⁰(t)|| (5%) r⁰(t) = (e^t, −e^−t, √

2), ||r⁰(t)|| =p

e^2t+ e^−2t+ 2 = e^t+ e^−t (2%)

T(t) = 1

e^t+ e^−t(e^t, −e^−t, √

2) = 1

1 + e^−2t, −1 1 + e^2t,

√ 2 e^t+ e^−t

!

(4)

T⁰(t) = 2e^−2t

(1 + e^−2t)², 2e^2t (1 + e^2t)²,

√

2(e^−t− e^t) (e^t+ e^−t)²

!

= 2

(e^t+ e^−t)², 2 (e^t+ e^−t)²,

√

2(e^−t− e^t) (e^t+ e^−t)²

!

||T⁰(t)|| = p4 + 4 + 2(e^2t+ e^−2t− 2)

(e^t+ e^−t)² = p2(e^2t+ e^−2t+ 2) (e^t+ e^−t)² =

√ 2

(e^t+ e^−t) (2%) κ(t) =

√2

(e^t+ e^−t)² (1%) sol:

κ(t) = ||r⁰(t) × r⁰⁰(t)||

||r⁰(t)||^2/3 (5%) r⁰(t) = (e^t, −e^−t, √

2), r⁰⁰(t) = (e^t, e^−t, 0) (2%) r⁰(t) × r⁰⁰(t) = (−√

2e^−t, √

2e^t, 2), ||r⁰(t) × r⁰⁰(t)|| =√

2(e^t+ e^−t) (2%)

κ(t) =

√2

(e^t+ e^−t)² (1%)

6. (15 points) Let f (x, y) =





 x²y

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 derivative ∂f

∂x at (x, y) = (0, 0) and at (x, y) 6= (0, 0).

(c) Is ∂f

∂x continuous at (0, 0)?

Solution:

(a) By polar coordinate,(let x = r cos θ, y = r sin θ)

|g(r, θ)| = |f (x, y)| = |f (r cos θ, r sin θ)| = |r³(cos²θ sin²θ

r² | = |r(cos²θ sin²θ)| ≤ r.

Hence,

| lim

(x,y)→(0,0)

f (x, y)| = | lim

r→0, θ:any angle

g(r, θ)| ≤ lim

r→0r = 0 ⇒ lim

(x,y)→(0,0)

f (x, y) = 0 = f (0, 0)...(5pts)

So f (x, y) is continuous at (0, 0).

Note: If you didn’t not emphasis that the angle θ is arbitrary, you only get the credits at most 4 points.

(b)

fx(0, 0) = lim

h→0

f (h, 0) − f (0, 0)

h = lim

h→0

0 − 0

h = 0...(3pts) For(x, y) 6= (0, 0)

fx(x, y) = y(2xy³)

(x²+ y²)²...(2pts)

(c) Since

y=x,x→0lim fx(x, y) = lim

y=x,x→0

y(2xy³)

(x²+ y²)² = lim

x→0

2x⁴ 4x⁴ = 1

2 6= 0 = fx(0, 0),

the function fx(x, y) is not continuous at (0, 0)...(5pts)

7. (12 points) Let u = u(x, y) be a function of rectangular coordinates x, y. Then u can be expressed in polar coordinates r, θ with x = r cos θ, y = r sin θ. Express ∂u

∂x and ∂u

∂y in terms of r, θ, ∂u

∂r and ∂u

∂θ.

(5)

Solution:

Sol. (I)

r(x, y) =p

x²+ y² (1%)

θ(x, y) = tan⁻¹(y

x) (1%)

∂r

∂x =1 2

2x

px²+ y² = cos θ

∂θ

∂x = 1

1 + (^y_x)²(−y

x²) =− sin θ r

∂r

∂y =1 2

2y

px²+ y² = sin θ

∂θ

∂y = 1 1 + (^y_x)²(1

x) =cos θ

r (1.5% each)

∂u

∂x = ∂u

∂r

∂x+∂u

∂θ

∂x (chain rule: 2%)

= ∂u

∂rcos θ −∂u

∂θ sin θ

r

∂u

∂y = ∂u

∂r

∂y +∂u

∂θ

∂y (chain rule: 2%)

= ∂u

∂rsin θ +∂u

∂θ cos θ

r

Sol. (II)

x(r, θ) = r cos θ y(r, θ) = r sin θ

∂x

∂r = cos θ, ∂y

∂r = sin θ

∂x

∂θ = −r sin θ, ∂y

∂θ = r cos θ (1% each)











∂u

∂r = ∂u

∂x

∂r+∂u

∂y

∂r

∂u

∂θ = ∂u

∂x

∂θ +∂u

∂y

∂θ (chain rule: 2% each)

⇒











∂u

∂r = ∂u

∂xcos θ +∂u

∂y sin θ

∂u

∂θ = ∂u

∂x(−r sin θ) +∂u

∂y(r cos θ) Solve the system of equations for ∂u

∂x and ∂u

∂y (say, by substitution or by Cramer’s rule), we obtain











∂u

∂x = ∂u

∂rcos θ −∂u

∂θ sin θ

r

∂u

∂y = ∂u

∂rsin θ +∂u

∂θ cos θ

r (4%)

(Note: if you write the chain rule as in Sol. (I), you only get half the score for calculating the partial derivatives listed in Sol. (II), and vice versa.)

(6)

8. (15 points) Let f (x, y) = xe^y+ cos(xy).

(a) Find the direction (a unit vector u) in which f (x, y) increases most rapidly at (2, 0) (that is, f_u⁰(2, 0) is maximal).

(b) Find the direction in which f (x, y) decreases most rapidly at (2, 0).

(c) What are the directions of zero change in f at (2, 0).

Solution:

(a) The direction which f (x, y) increases most rapidly at (2, 0) is ∇f (x, y) = (e^y− y sin (xy) , xe^y− x sin (xy)) where x = 2, y = 0. That is ∇f (2, 0)

k∇f (2, 0)k = (1, 2)

√5 .

(b) decreasing most rapidly is the inverse direction of ∇f (x, y) . That is −∇f (2, 0)

k∇f (2, 0)k =− (1, 2)

√5 .

(c) The direction of zero change in f (x, y) is those vector v satisfying hv, ∇f (x, y)i = 0. That is (2, −1)

√

5 and

−(2, 1)

√5 .

Grading:

The computation of ∇f (x, y) has 3 points. The rest of each question has 4 points. Those who write the correct concept but calculation is wrong or insert wrong parameter will gain only 2 point in each questions.