微積分甲

982微積分甲^08-13班期中考解答與評分標準

1. (15%) Find the interval of x such that the power series X∞

k=1

x^k

ln (k + 1) converges.

Sol:

(i)

klim→∞

|1/ ln (k + 2)|

|1/ ln (k + 1)| = lim

k→∞

ln (k + 1) ln (k + 2)

= lim

k→∞

1/(k + 1)

1/(k + 2) (l’Hospital’s rule)

= 1.

So the radius of convergence is 1 1 = 1.

(ii) For x = 1,

X∞ k=1

x^k ln (k + 1) =

X∞ k=1

1 ln (k + 1).

Since 1

ln (k + 1) > 1

(k + 1) for each k ∈ N and X∞ k=1

1

(k + 1) diverges, X∞

k=1

1

ln (k + 1) diverges by comparison test.

For x =−1

X∞ k=1

x^k ln (k + 1) =

X∞ k=1

(−1)^k ln (k + 1).

Since 1

ln (k + 1) & 0 as k → ∞, X∞

k=1

(−1)^k

ln (k + 1) converges by the alternating series test.

From (i) and (ii), the interval of convergence is [−1, 1).

評分標準^:

(1) radius of convergence → 9

– Knowing to use ratio test or root test → 3

– Getting some results from one of the tests above but not to the extent of recognizing 1 as the radius of convergence → 4 ∼ 7

(2) Discussion of x = 1 → 3 (3) Discussion of x =−1 → 3

2. (10%) (a) Find the Maclaurin series for f (y) = sin y.

(b) Evaluate Z ^π

2

0

sin (cos x) dx correct to within an error of 0.01.

Sol:

(a)

f (y) = X∞ n=0

f⁽ⁿ⁾(0) n! yⁿ since d^4k

dy^4k sin y = sin y, d^4k+1

dy^4k+1 sin y = cos y, d^4k+2

dy^4k+2 sin y =− sin y, d^4k+3

dy^4k+3 sin y =− cos y we have

f^(4k)(0) = 0, f^(4k+1)(0) = 1, f^(4k+2)(0) = 0, f^(4k+3)(0) =−1, k = 0, 1, 2, 3, · · · (2pts)

f (y) = X∞ n=0

(−1)ⁿy²ⁿ⁺¹

(2n + 1)! (2pts) (b) since the radius of convergence is ∞, and −1 ≤ cos x ≤ 1

⇒ sin(cos x) = cos x − cos³x

3! + cos⁵x

5! − · · · (1pts) Z ^π

2

0

sin(cos x)dx = X∞ n=0

Z ^π

2

0

(−1)ⁿcos²ⁿ⁺¹x (2n + 1)! dx =

X∞ n=0

(−1)ⁿa_n

1 > cos x > 0 for 0 < x < π

2 thus a_n is positive and cos²ⁿ⁻¹x

(2n− 1)! > cos²ⁿ⁺¹x (2n + 1)! ⇒

Z ^π

2

0

cos²ⁿ⁻¹x (2n− 1)!dx >

Z ^π

2

0

cos²ⁿ⁺¹x (2n + 1)!dx 0≤ an≤ π

2(2n + 1)!

⇒ an decreasing to 0, we can apply alternating series test,

a₃ = Z ^π

2

0

cos⁵x 120 dx =

Z ^π

2

0

cos x(1− sin²x)²

120 dx = 1

120(sin x− 2

3sin³x + 1

5sin⁵x)¯¯¯

π 2

0

= 1

225 < 0.01

only have to compute a₀− a1 ,error is less then 0.01 (3pts)

a₀ = Z ^π

2

0

cos xdx = sin x¯¯¯

π 2

0 = 1 a₁ =

Z ^π

2

0

cos³x 6 dx =

Z ^π

2

0

cos x(1− sin²x)

6 dx = (sin x

6 −sin³x 18 )¯¯¯

π 2

0 = 1 9 a₀ − a1 = 8

9 (2pts)

3. (10%) Let z = y + f (x²− y²) and f be a differentiable function in one variable. Find the value of y∂z

∂x + x∂z

∂y when x = a and y = b.

Sol:

∂z

∂x = 0 + 2xf⁰(x²− y²), ∂z

∂y = 1− 2yf⁰(x²− y²) y∂z

∂x + x∂z

∂y = 2xyf⁰(x²− y²) + x− 2xyf⁰(x²− y²) = x So y∂z

∂x + x∂z

∂y

¯¯¯

(a,b) = x¯¯¯

(a,b) = a.

評分標準^:

z 對 ^x作偏微分^{: 3}分 z 對 ^y 作偏微分^{: 3}分

將上述二式併入所求式子^{: 3}分 將 ^{(a, b)} 代入上式得其答案^{: 1}分 其餘錯誤酌量給分

4. (10%) Let f (x) = ln (5− x).

(a) Find the power series representation for f (x) at x = 0 . (b) Find f⁽ⁿ⁾(0).

Sol:

(a) ln(1− t) = −t − t²

2 − ... = − X∞ n=1

tⁿ

n (3 points) and ln(5− x) = ln 5 + ln(1 − x

5) = ln 5− X∞ n=1

(^x₅)ⁿ

n = ln 5− X∞ n=1

xⁿ

n5ⁿ (3 points)

(b) By (a) ⇒ f(0) = ln 5

f⁽ⁿ⁾(0) =−n! · 1

n5ⁿ =−(n− 1)!

5ⁿ , n≥ 1. (4 points) 另一種^:

(a) ln(5− x) =

Z 1

5− tdt

= 1 5

Z 1

1− ₅^tdt

= 1 5

Z X∞

k=0

(t

5)^kdt (2 points)

= hX^∞

k=0

(x

5)^k+1 1 k + 1

i

+ C (2 points)

choose x = 0, then ln 5 = C, so ln(5− x) = ln 5 − X∞ n=1

xⁿ

n5ⁿ (2 points) (b) c_n= f⁽ⁿ⁾(0)

n! (2 points) f⁽ⁿ⁾(0) =−(n− 1)!

5ⁿ , n≥ 1 (2 points) f (0) = ln 5

5. (13%) Let r(t) =ht²,2 3t³, ti.

(a) Find the arc length of r(t) from t = 0 to t = 5.

(b) Find the curvature of r(t) at t = 4.

Sol:

r⁰(t) =< 2t, 2t², 1 >, r⁰⁰(t) =< 2, 4t, 0 >

(in (b) first solution 1%)

(a)

|r⁰(t)| =√

4t²+ 4t⁴+ 1 (3%)

= 2t²+ 1 (2%)

L = Z ₅

0

|r⁰(t)|dt = Z ₅

0

2t²+ 1dt = h2

3t³+ t i5

0

= 250

3 + 5 = 265 3 (1%) (b) r⁰(4) =< 8, 32, 1 >, r⁰⁰(4) =< 2, 16, 0 >

r⁰(4)× r⁰⁰(4) =<−16, 2, 64 >, |r⁰(4)| = 33

|r⁰(4)× r⁰⁰(4)| = 2√

64 + 1 + 1024 = 2√

1089 = 2√

3²· 11² = 66

κ(4) = |r⁰(4)× r⁰⁰(4)|

|r⁰(4)|³ (3%)

= 66 33³ (2%)

= 2

1089 (1%) or

T(t) = 1

2t² + 1 < 2t, 2t², 1 > (1%) T⁰(t) = 2

(2t²+ 1)² < 1− 2t², 2t,−2t > (3%) T⁰(4) = 2

33² <−31, 8, −8 >

|T⁰(4)| = 2 33 κ(4) = |T⁰(4)|

|r⁰(4)| (2%)

= 2

33² = 2

1089 (1%)

6. (15%) The plane x + y + 2z = 2 intersects the paraboloid z = x² + y² in an ellipse. Find the points on this ellipse that are nearest to and farthest from the origin.

Sol:

Method1: Lagrange Multiplier

f (x, y, z) = x²+ y²+ z² (1)

g₁(x, y, z) = x + y + 2z (2)

g₂(x, y, z) = x²+ y²− z (3)

f is to measure distance from the origin. g₁ and g₂ are restrictions.

To find out critical points of f with these restrictions, one needs to solve:

Of = aOg1+Og2

= (2x, 2y, 2z) = a(1, 1, 2) + b(2x, 2y,−1)

Combine with restrictions ,we have 5 unknowns and 5 equations:

2x = a + 2xb (4)

2y = a + 2yb (5)

2z = 2a− b (6)

x + y + 2z = 2 (7)

z = x²+ y² (8)

By (4) and (5) , x = y → (4) and (5)to be

z = 2− 2x

2 = 1− x (9)

z = 2x² (10)

By (9),(10) x = ¹₂ or x =−1 Plug back to (4)-(8),we have two solutions

(x₁, y₁, z₁, a₁, b₁) = (1 2,1

2,1 2,2

3,1

3) (11)

(x2, y2, z2, a2, b2) = (−1, −1, 2,10 3 ,8

3) (12)

Maxia is p

(−1)²+ (−1)²+ 2² =√ 6 Minima is

q

1 2

2+¹₂² +¹₂² = q3

8

Remark: One can choose f (x, y, z) =p

x²+ y²+ z² and do similar calculations to gain critical points.

Score criterion: Eq(4)-(8) +8 ; Solve x,y,z correctly +2 each ; max/min +1

Method2: Lagrange Multiplier (modified)

Since

f (x, y) = x²+ y²+ z(x, y)² = x²+ y²+ (x²+ y²)² and one constrain condition:

g(x, y) = x + y + 2z = x + y + 2(x²+ y²) = 2

then use Lagrange Multiplier

Of(x, y) = aOg(x, y) to solve x,y and z

7. (15%) Find and classify the critical points of the function f (x, y) = (x²+ y²)e^y2−x2. Sol:

fx(x, y) = 2xe^y2−x2(1− x²− y²) f_y(x, y) = 2ye^y2−x2(1 + x²+ y²) (3%) f_x = f_y = 0=⇒(x, y) = (±1, 0), (0, 0) (6 %) f_xx = (2− 10x²− 2y²+ 4x⁴+ 4x²y²)e^y2−x2 f_yy = (2 + 10y²− 2x²+ 4x²y²+ 4y⁴)e^y2−x2 f_yz = 4(−x²− y²)e^y2−x2

(x, y) = (0, 0) =⇒ D > 0, fxx > 0 : local minimum

(x, y) = (±1, 0) =⇒ D < 0: saddle points (6 (Additional error points cause some deduction.) 8. (12%) Let T (x, y, z) = e^{−x2−3y2−9z2}.

(a) Find the directional derivative of T (x, y, z) at P₀ = (2,−1, 2) toward the point (3, −3, 3).

(b) Find the maximum of directional derivative of T (x, y, z) at P₀ = (2,−1, 2).

Sol:

(a) P₀ = (2,−1, 2), P1 = (3,−3, 3)

−−→P₀P₁ = (1,−2, 1) (1%)

v =

−−→P₀P₁ P₀P₁ = 1

√6(1,−2, 1) (2%)

∇T = e^{−x2−3y2−9z2}(−2x, −6y, −18z) (2%)

T_v(2,−1, 2) = v • ∇T (2, −1, 2) (2%)

= −52√

6 e⁻⁴³ (1%) (b) maximim : u//∇T and |u| = 1 (2%)

so take u = 1

√337(−2, 3, −18) T_u = 2e⁻⁴³√

337 (2%)