微積分甲

982微積分甲^01-05班期中考解答與評分標準

1. (12%) (a) Evaluate lim

n→∞n(√ⁿ 2− 1).

(b) Find the interval of convergence of the power series p₁(x) =

∑∞ n=1

(√ⁿ

2− 1)ⁿ(x− 1)ⁿ.

(c) Find the interval of convergence of the power series p₂(x) =

∑∞ n=1

(√ⁿ

2− 1)(x − 1)ⁿ. Sol:

(a) (2%) lim

n→∞n(2ⁿ¹ − 1) = lim

x→0⁺

2^x− 1

x = f⁰(0) = ln 2, where f (x) = 2^x. (b) (3%) 1

R = lim

n→∞

n

√

| (√ⁿ

2− 1)(x − 1) |ⁿ=| x − 1 | lim

n→∞ | √ⁿ

2− 1 |= 0 ⇒ R = ∞ the convergence set isR

(c) (7%) radius of convergence (3%)

nlim→∞

n

√

| √ⁿ

2− 1 || x − 1 |ⁿ=| x − 1 | lim

n→∞(√ⁿ

2− 1)ⁿ¹, and

nlim→∞

(√n

2− 1)¹

n = exp (

nlim→∞

ln (√ⁿ 2− 1) n

)

= exp (

nlim→∞

ln[n(√ⁿ

2− 1)] − ln n n

)

= e⁰ = 1.

Thus radius is 1.

At x = 2 (2%): By part (a) and limit comparison test, since lim

n→∞

2ⁿ¹ − 1

1 n

= ln 2 and

∑

n

1 n =∞, p₂(2) =

∑∞ n=1

(√ⁿ

2− 1) is divergent.

At x = 0 (2%): Let an= 2ⁿ¹ − 1, clearly an is decreasing to 0. By alternating series test, p₂(0) =

∑∞ n=1

(−1)ⁿa_n is convergent.

The convergence interval is [0, 2).

2. (9%) Let f (x) = arctan4− x²

4 + x², x∈ R.

(a) Evaluate f⁰(x).

(b) Find the power series representation for f (x) about x = 0. That is, find c₀, c₁, c₂,· · · such that f (x) = c₀+

∑∞ n=1

c_nxⁿ. What is the radius of convergence of this power series?

Sol:

f (x) = arctan4− x²

4 + x² ⇒ f⁰(x) = 1

1 + (⁴_4+x^−x22)² · −2x(4 + x²)− 2x(4 − x²)

(4 + x²)² = −8x

x⁴+ 16 (3%) C₀ = f (0) = π

4 (1%) f (x) = C₀+

∫ _x

0

−t

2 · 1 1 + (₂^t)⁴dt

= C₀+

∫ _x

0

−t 2 ·

∑∞ n=0

(−1)ⁿ· (t 2)⁴ⁿdt

= C₀+

∫ _x

0

∑∞ n=0

(−1)ⁿ⁺¹

2⁴ⁿ⁺¹ t⁴ⁿ⁺¹dt = C₀+

∑∞ n=0

(−1)ⁿ⁺¹

2⁴ⁿ⁺¹(4n + 2)x⁴ⁿ⁺² (2%)

Cn= (−1)ⁿ⁺²⁴

2ⁿ⁻¹n if n = 2 mod 4, Cn= 0 otherwise (2%). Radius of convergence is 2 (1%).

3. (8%) Let g(x) = 1

2x²− 3x + 1. Find g⁽¹⁰⁾(−1), the tenth derivative of g(x) at x = −1.

Sol:

(Method 1.)

g(x) = 1

2x²− 3x + 1 = 1

(2x− 1)(x − 1) = 1

x− 1 − 2

2x− 1 (3%) g⁽¹⁰⁾(x) = (−1)¹⁰× 10! × 1

(x− 1)¹¹ − 2¹⁰× (−1)¹⁰× 10! × 2 (2x− 1)¹¹ g⁽¹⁰⁾(−1) = −10!

2¹¹ + (2

3)¹¹× 10! = 10!((2

3)¹¹− 1

2¹¹) (5%) (Method 2.)

4. (12%) Consider the power series H(x) = x−( 1+1

2 )

x²+ (

1+1 2+1

3 )

x³−( 1+1

2+1 3+1

4 )

x⁴+· · · =

∑∞ k=1

(1 1+ 1

2+· · · + 1 k

)

(−1)^k−1x^k.

(a) Find the interval of convergence of H(x).

(b) Express (1 + x)H(x) as a power series about x = 0. Recognize it as the Maclaurin series of certain function Q(x).

(c) Evaluate 1 2+

( 1 +1

2 )(1

2² )

+ (

1 +1 2+1

3 )( 1

2³ )

+· · · +( 1 +1

2+· · · +1 k

)( 1 2^k

)

+· · · = m.

Sol:

(a) (3%) Let a_k =

k. Since lim

k→∞| ^k

a_k+1| = 1 − 0 = 1, then the radius is 1. Besides, at x = ±1, the k− th term of H(x) satisfies lim

k→∞|ak| =∑^∞

k=1

1 k =∞.

By the test for divergence, H(1) and H(−1) are divergent.

Thus the interval of convergence is (−1, 1).

(b) (5%) Since (1 + x)H(x) =

∑∞ k=1

(−1)^k−1(x)^k

k (2%)

Q(x) = ln(1 + x) by − ln (1 − x) =

∑∞ k=1

x^k

k , x∈ (−1, 1). (3%) (c) (4%) Take x = −1

2 . Then we have m =−H(−1

2 ) = −Q(⁻¹₂ )

1− 2¹₂ =−2 · ln(−1

2 ) = ln 4 5. (15%) Let r(t) =hcos πt

π ,sin πt π ,4

3t³²i, t ≥ 0.

(a) Find the length of the arc 0≤ t ≤ 2 of r(t).

(b) Find T(2), the unit tangent vector when t = 2.

(c) Find κ(2), the curvature when t = 2.

(d) Find N(2) and B(2), the principal unit normal vector and the binormal vector when t = 2, respectively.

Sol:

(a) Since r⁰(t) =< − sin πt, cos πt, 2t¹² > and |r⁰(t)| =√ 4t + 1, s =

∫ ₂

0

√4t + 1dt = 1

6(4t + 1)³²¯¯

¯¯²

0

= 13 3 (3%) (b) T(2) = r⁰(2)

|r⁰(2)| = (0,1 3,2√

2

3 ) (3%) (c) T⁰(t) = <−π cos πt, −π sin πt, t⁻¹² >√

4t + 1− ^2<− sin πt,cos πt,2t√ ^1/2>

4t+1

4t + 1 T⁰(2) = 1

9 {

<−π, 0, 1

√2 >·3 − 2

3 < 0, 1, 2√ 2 >

}

= 1

9 <−3π, −2 3,3√

2 2 − 4√

2 3 >

|T⁰(2)| = 1 9

(

9π²+4 9 + 1

18 )¹

2

κ(2) = |T⁰(2)|

|r⁰(2)| =

√36π²+ 2

54 (3%)

(d) N(2) = T⁰(t)

|T⁰(t)| = 18

√36π²+ 2(−π 2,− 2

27,

√2

54) (3%) B(2) = T(2)× N(2) = 1

√36π²+ 2(√

2,−4π√

2, 2π) (3%) 6. (12%) Evaluate the following limits.

(a) lim

(x,y)→(0,0)

xy

x²+ y² = A, (b) lim

(x,y)→(0,0)

( xy x²+ y²

)¹

x2 = B, (c) lim

(x,y)→(0,1)(1 + x)^x−yx = C.

Sol:

(a) A doesn’t exist

Along the line y = x, lim

x=y→0

xy

x²+ y² = 1 2.

Along either the line y = 0(x 6= 0) or x = 0(y 6= 0), xy

x²+ y² = 0 (using polar coordinate to explain is ok ).

So the limit does not exist. (4 pts)

(b) Both answers (1) and (2) are acceptable (4pts), but you have to explain the reasons.

Answer(1). The question is problematic because ( xy

x²+ y²)^x21 is not defined when xy < 0, and such (x, y) is present in any neighborhood of (0, 0).

Answer(2). B = 0

solution: It follows from 0≤| xy |≤ 1

2(x²+ y²), then 0 ≤ ( | xy |

x²+ y²)^x21 ≤ (1 2)^x21 . Since lim

(x,y)→(0,0)(1

2)^x21 = 0, B exists and equals 0 by Squeeze theorem.

(c) Both answers (1) and (2) are acceptable (4pts), but you have to explain the reasons.

Answer(1). The question is also problematic because (1 + x)^x−yx is not defined in the neighborhood of (0, 1) when x = 0.

Answer(2). C=e⁻¹

lim

(x,y)→(0,1)(1 + x)^x−yx = lim

(x,y)→(0,1)(1 + x)(1 + x)^−yx

= lim

(x,y)→(0,1)(1 + x) lim

(x,y)→(0,1)(1 + x)^−yx = lim

(x,y)→(0,1)exp(−y

x ln(1 + x))

= exp( lim

(x,y)→(0,1)−yln(1 + x)

x ) (by the continuity of exponential function)

= exp( lim

(x,y)→(0,1)(−y) lim

(x,y)→(0,1)

ln(1 + x)

x ) (both limits exist)

= exp(−1 × 1) = e⁻¹ (4 pts)

Many people pass y = 1 into the limit of f (x, y) first and then find lim

x→0f (x,−1) or set y = mx + 1 to find the limit or e.c.t.

But these are wrong ways. In general, lim

(x,y)→(a,b)f (x, y)6= lim

x→alim

y→bf (x, y).

7. (8%) Find the tangent plane at (1, e, e²) on the surface x + ln(y²) + ln(z⁴) = 11.

Sol:

Set f (x, y, z) = x + ln(y²) + ln(z⁴)− 11, 5f(x, y, z) = (1,2 y,4

z),5f(1, e, e²) = (1,2 e, 4

e²) (3 pts) The tangent plane is (x− 1) + 2

e(y− e) + 4

e²(z− e²) = 0 (5 pts)

8. (12%) Suppose that f (x, y) is differentiable at (1, 0), f (1, 0) = 3, and D_u0f (1, 0) = 3

√2 in the direction u₀ = ( 1

√2, 1

√2), while D_v0f (1, 0) = −1√

2 in the direction v₀ = (−1√ 2, 1

√2).

(a) Find the maximum value of the directional derivative D_uf at (x, y) = (1, 0).

(b) Let the space curve r(t) be the intersection of the surfaces z = f (x, y) and z = x + y²+ 2.

Find the parametric equation for the tangent line to r(t) at the point (x, y, z) = (1, 0, 3).

Sol:

(a) Let∇f (1, 0) = (a, b), then we have 1

√2a + 1

√2b = 3

√2 and −1√

2a + 1

√2b = √−1 2. So∇f (1, 0) = (a, b) = (2, 1).

This implies the maximum of the directional derivatives is |∇f (1, 0) | = | (2, 1) | =√ 5, which occurs when the direction is

( 2

√ , 1

√ )

.

(b) Let F (x, y, z) = f (x, y)− z and G(x, y, z) = x + y²+ 2− z.

∇F |(1,0,3)= (2, 1,−1), ∇G|(1,0,3) = (1, 2y,−1) |(1,0,3) = (1, 0,−1)

∇F |(1,0,3)× ∇G|(1,0,3) = (2, 1,−1) × (1, 0, −1) = (−1, 1, −1) So the tangent line is x− 1

−1 = y− 0

1 = z− 3

−1 . 評分標準^:

(a) (1) 列出方程組^{: 3}分 ⁽²⁾ 方程組解出答案^{: 2}分 ⁽³⁾ 寫出方向導數最大值為何^{: 1}分 ⁽⁴⁾ 其餘 錯誤酌量給分

(b) (1) 算出兩個區面的梯度函數^: 各²分 ⁽²⁾ 求出切線方向向量^{: 1}分 ⁽³⁾ 寫出切線方程式^{: 1} 分 ⁽⁴⁾ 其餘錯誤酌量給分

9. (12%) Let U = x³y, and x, y, and t satisfy







x⁵+ y = t, x²+ y³ = t².

(∗)

Around t =−1 and (x, y) = (−1, 0), by Implicit Function Theorem, the relationship (∗) defines two differentiable functions x = x(t) and y = y(t).

(a) Evaluate dx

dt and dy

dt at t = −1 and (x, y) = (−1, 0). (Hint. Differentiate (∗) and derive a system of equations of dx

dt and dy dt.) (b) Evaluate dU

dt at t = −1 and (x, y) = (−1, 0).

Sol:

(a) Differentiating (*) with respect to t, we have

5x⁴dx dt +dy

dt = 1, (3分⁾

2xdx

dt + 3y²dy

dt = 2t. (3分⁾

dx

dt = 3y²− 2t 15x⁴y²− 2x

¯¯¯¯

t=−1,(x,y)=(−1,0)

= −1

−1 = 1, dy

dt = 10x⁴t− 2x 15x⁴y²− 2x

¯¯¯¯

t=−1,(x,y)=(−1,0)

= −10 + 2

2 =−4.

(若前面做對 ^,則到此做對可累計各得 ⁴分 ^, 此小題滿分⁸ 分⁾ Or, at t =−1, (x, y) = (−1, 0),

5dx dt +dy

dt = 1, (3分⁾

−2dx

dt =−2. (3分⁾

Hence dx

dy = 1,dy

dt =−4. (若前面做對 ^,則到此做對可累計各得 ⁴ 分^, 此小題滿分⁸ 分⁾ (b)

U = x³y, dU

dt = ∂U

∂x dx

dt +∂U

∂y dy

dt, (得 ²分⁾

= 3x²ydx

dt + x³dy

dt. (到此做對 ^, 累計得 ³分⁾

At t =−1, (x, y) = (−1, 0), dU

dt = (−1)³· (−4) = 4. (到此做對 ^, 累計得⁴ 分⁾