2 cos θ + 1 2 cos 7

1051微微微甲甲甲01-04班班班期期期末末末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%) Evaluate the integrals.

(a)

ˆ 1

sin x cos²xdx. (b)

ˆ

tan⁻¹(√ x)dx.

Solution:

(a) Method 1.

ˆ 1

sin x cos²xdx =

ˆ sin x sin²x cos²xdx

(Let u = cos x, du = − sin xdx.)

=

ˆ −du

(1 − u²)u² (1%)

=

ˆ du

u²(u − 1)(u + 1)

= ˆ −1

u² + Γ 1/2

u − 1 +−1/2

u + 1du (3%)

= 1 u+1

2ln |u − 1| −1

2ln |u + 1| + C

= sec x +1 2ln

cos x − 1 cos x + 1    

+ C (2%) Method 2.

ˆ 1

sin x cos²xdx = ˆ

csc x sec²xdx

Let u = csc x, dv = sec²dx du = − csc x cot xdx, v = tan x.

!

(1%)

= csc x tan x − ˆ

tan x(− csc x cot x)dx (2%)

= sec x + ˆ

csc xdx

= sec x − ln |csc x + cot x| + C (3%) Method 3.

Let t = tan(x

2). ⇒ sin x = 2t

1 + t², cos x = 1 − t²

1 + t², dx = 2

1 + t²dt. (1%)

ˆ 1

sin x cos²xdx =

ˆ 1

 2t 1+t²

 1−t² 1+t²

2

2 1 + t²dt

=

ˆ (1 + t²)² t(1 − t²)²dt

= ˆ 1

t + 1

(t − 1)² − 1

(t + 1)²dt (3%)

= ln | tan(x

2)| − 1

tan(^x₂) − 1 + 1

tan(^x₂) + 1+ C (2%) Method 4.

ˆ 1

sin x cos²xdx =

ˆ sin²x + cos²x

sin x cos²x dx (1%)

=

ˆ sin x cos²xdx +

ˆ 1

sin xdx

=

ˆ −d(cos x) cos²x +

ˆ

csc xdx (2%)

= 1

cos x− ln |csc x + cot x| + C (3%)

• 有正確寫出 Partial Fraction 但是係數算錯扣 1 分

• 最後答案沒有標示把變數換回x 或 ln沒加絕對值或沒有 +C 扣 1 分 (b) Method 1.

Let u =√

x, du = 1 2√

xdx. (2%) ˆ

tan⁻¹(√ x)dx =

ˆ

2u tan⁻¹udu

= u²tan⁻¹u − ˆ u²

1 + u²du (3%)

= u²tan⁻¹u − ˆ 

1 − 1 1 + u²

 du

= u²tan⁻¹u − u + tan⁻¹u + C

= (x + 1) tan⁻¹(√ x) −√

x + C (1%) Method 2.

Let√

x = tan θ. =⇒ x = tan²θ, dx = 2 tan θ sec²θdθ. (2%) ˆ

tan⁻¹(√ x)dx =

ˆ

θd(tan²θ)

= θ tan²θ − ˆ

tan²θdθ

= θ tan²θ − ˆ

(sec²θ − 1)dθ

= θ tan²θ − tan θ + θ + C (3%)

= (x + 1) tan⁻¹(√ x) −√

x + C (1%) 評分標準：

• 最後答案沒有標示把變數換回x 或答案沒有 +C 扣 1 分

• 微分計算算錯扣 2 分

2. (12%) Evaluate the integrals.

(a) ˆ

xp

8 + 2x − x²dx (b)

ˆ x²− 1 (x²+ 2x + 2)²dx

Solution:

(a) Let x = 1 + 3 sin θ, where −π

2 ≤ θ ≤ π

2. Then dx = 3 cos θdθ. Thus we have ˆ

xp

8 + 2x − x²dx

= ˆ

xp9 − (x − 1)²dx

= ˆ

(1 + 3 sin θ) · 3 cos θ · 3 cos θdθ (2pts)

= ˆ

(9 cos²θ + 27 sin θ cos²θ)dθ

= ˆ 

9 ·1 + cos(2θ)

2 + 27 sin θ cos²θ dθ

= 91 2θ + 1

4sin(2θ)

+ 27 ·−1

3 cos³θ + C (2pts)

= 9

2θ +9

2sin θ cos θ − 9 cos³θ + C

= 9

2sin⁻¹x − 1 3

+1

2(x − 1)p

8 + 2x − x²−1

3(8 + 2x − x²)³² + C (2pts)

(b) Let x = −1 + tan θ, where −π

2 < θ < π

2. Then dx = sec²θdθ. Thus we have ˆ x²− 1

(x²+ 2x + 2)²dx

=

ˆ x²− 1 ((x + 1)²+ 1)²dx

=

ˆ (tan θ − 1)²− 1

sec⁴θ · sec²θdθ (2pts)

= ˆ

(sin²θ − 2 sin θ cos θ)dθ

= 1

2θ −1

4sin(2θ) + cos²θ + C (2pts)

= 1

2tan⁻¹(x + 1) −1

2 · x + 1

x²+ 2x + 2+ 1

x²+ 2x + 2+ C

= 1

2tan⁻¹(x + 1) − x − 1

2(x²+ 2x + 2) + C (2pts)

3. (10%) Find the reduction formula In = ˆ

(ln x)ⁿdx, where n is a positive number.(i.e. write In in terms of I_n−1.) Evaluate the improper integral

ˆ 1 0

(ln x)ⁿdx or explain why it is divergent.

Solution:

Let u = (ln x)ⁿ, dv = dx du = n(ln x)ⁿ⁻¹1

xdx, v = x.

ˆ

(ln x)ⁿdx = x(ln x)ⁿ− n ˆ

(ln x)ⁿ⁻¹dx (2%) Therefore,

In = x(ln x)ⁿ− nIn−1.

ˆ 1 t

(ln x)ⁿdx = x(ln x)ⁿ  

1 t− n

ˆ 1 t

(ln x)ⁿ⁻¹dx

= 1 · (ln 1)ⁿ− t(ln t)ⁿ− n ˆ 1

t

(ln x)ⁿ⁻¹dx

= −t(ln t)ⁿ− n ˆ 1

t

(ln x)ⁿ⁻¹dx.

Claim: The improper integral ˆ 1

0

(ln x)ⁿdx converges for all n ∈ N. Proof. We prove the claim by mathematical induction. First, consider the case n = 1,

ˆ 1 0

(ln x)dx = lim

t→0⁺

ˆ 1 t

(ln x)dx

= lim

t→0⁺−t(ln t) − 1

= lim

t→0⁺−ln t

(¹_t)− 1 L’Hospital’s rule

= lim

t→0⁺−

1 t

−1 t²

− 1

= lim

t→0⁺t − 1 = −1 converges. (2%) Suppose that for n = k − 1, the improper integral

ˆ 1 0

(ln x)^k−1dx converges. Consider the case n = k, we have ˆ 1

0

(ln x)^kdx = lim

t→0⁺

ˆ 1 t

(ln x)^kdx

= lim

t→0⁺



−t(ln t)^k− k ˆ 1

t

(ln x)^k−1dx

 . By L’Hospital’s rule, the limit

t→0lim⁺ ln t

t^−1/k = lim

t→0⁺ 1 t

−1k t^−1k−1 = lim

t→0⁺−kt^k1 = 0 since k is a positive integer implies that 1

k > 0.

Hence, the limit

t→0lim⁺−t(ln t)^k= lim

t→0⁺−(ln t)^k

1 t^1/k

k = lim

t→0⁺−

 ln t t^−1/k

k

= −



t→0lim⁺ ln t t^−1/k

k

= 0, (3%) and the improper integral

ˆ 1 0

(ln x)^kdx = lim

t→0⁺



−t(ln t)^k− k ˆ 1

t

(ln x)^k−1dx



=



t→0lim⁺−t(ln t)^k− k lim

t→0⁺

ˆ 1 t

(ln x)^k−1dx



= −k ˆ 1

0

(ln x)^k−1dx converges.

Therefore, by mathematical induction, we have shown that the improper integral ˆ 1

0

(ln x)ⁿdx converges for all n ∈ N, and it converges to

ˆ 1 0

(ln x)ⁿdx = −n ˆ 1

0

(ln x)ⁿ⁻¹dx = −n[−(n − 1)]

ˆ 1 0

(ln x)ⁿ⁻²dx

= ... = −n[−(n − 1)][−(n − 2)][−(n − 3)]...(−3)(−2) ˆ 1

0

(ln x)dx

= (−1)ⁿn! (3%) 評分標準：

• ˆ 1

0

(ln x)ⁿdx 沒有照定義取右極限 lim

t→0⁺

ˆ 1 t

(ln x)ⁿdx 扣 1 分

• 沒有說明怎麼計算 lim

t→0⁺−t(ln t)ⁿ= 0 的過程扣 2 分

4. (12%) Find the values of a and b for which the improper integral ˆ _∞

0

e^−ax

x^b(1 + x²)dx = ˆ 1

0

e^−ax

x^b(1 + x²)dx + ˆ _∞

1

e^−ax

x^b(1 + x²)dx converges.

Hint: Discuss cases a = 0 and a 6= 0, respectively.

Solution:

(1) (4pt) If a = 0, we have ˆ _∞

0

1

x^b(1 + x²)dx = ˆ 1

0

1

x^b(1 + x²)dx + ˆ _∞

1

1 x^b(1 + x²)dx On the interval [0, 1], since 1

2 ≤ 1

1 + x² ≤ 1, we have 1

2x^b ≤ 1

x^b(1 + x²) ≤ 1 x^b.

Since ˆ 1

0

1

x^bdx is convergent if and only if b < 1 (notice that the integral ˆ 1

0

1

x^b(1 + x²)dx is a definite integral if b ≤ 0, so it is a finite value), we get

ˆ 1 0

1

x^b(1 + x²)dx is convergent if and only if b < 1 by the convergent theorem.

On the interval [1, ∞), since x^b+2≥ x^b, we have 1

2x^b+2 ≤ 1

x^b(1 + x²) ≤ 1 x^b+2. Since the improper integral

ˆ _∞

1

1

x^b+2dx is convergent if and only if b + 2 > 1 (that is, b > −1), ˆ _∞

1

1

x^b(1 + x²)dx is convergent if and only if b > −1.

Therefore, if a = 0 and −1 < b < 1, the improper integral is convergent.

(2) (4pt) If a < 0, since lim

x→∞

e^−ax

x^b(1 + x²) = ∞, the improper intergral is divergent for any b ∈ R.

(3) (4pt) If a > 0, since e^−ax

1 + x² is continuous on [0, 1] by the Extreme Value Theorem, there exist m and M such that m ≤ e^−ax

1 + x² ≤ M, so

m

x^b ≤ e^−ax

x^b(1 + x²) ≤ M x^b. Since

ˆ 1 0

1

x^bdx is convergent if and only if b < 1, we get ˆ 1

0

1

x^b(1 + x²)dx is convergent if and only if b < 1 by the comparison theorem.

On the interval [1, ∞), since lim

x→∞

e^−ax

x^b(1 + x²) = 0 for any b ∈ R, there exists x⁰ such that for all x > x0, we have e^−a2

x^b(1 + x²) < 1.

0 ≤ e^−ax

x^b(1 + x²) = e^−a2x· e^−a2x

x^b(1 + x²)≤ e^−a2x, and

ˆ _∞

1

e^−a2xdx is convergent, ˆ _∞

1

e^−ax

x^b(1 + x²)dx is convergent for all b ∈ R.

Therefore, if a > 0 and b < 1, the improper integral is convergent.

5. (12%) Find the volume of Gulliver’s Tunnel (格列佛隧道), which is half of the solids of revolution obtained by rotating the region bounded by y = 1

1 + e^3x, y = 0, x = −2

3ln 3, and x = ln 2, about the x-axis.

x

−2

3ln 3 ln 2

y = 1

1 + e^3x

Figure 1: Find the volume of Gulliver’s Tunnel.

Solution:

1. The volume by disk method is

Volume = π 2

ˆ ln 2

−²3ln 3

 1

1 + e^3x

2

dx (5pts) .

Let u = e^3x. Then du = 3e^3xdx or dx = 1

3udu (1pt). The upper limit is u = 8 and the lower limit is u = 1

9 (1pt). Thus

Volume = π 6

ˆ 8

1 9

1

u(1 + u)²du = π 6

ˆ 8

1 9

A u + B

1 + u+ C

(1 + u)²du (2pts) . From A(1 + u)²+ Bu(1 + u) + Cu = 1 we determine A = 1, B = −1 and C = −1 and so

Volume = π 6

ˆ 8

1 9

1 u− 1

1 + u − 1

(1 + u)²du = π 6



ln u − ln(1 + u) + 1 1 + u

⁸

1 9

= π 6



ln 8 + ln 9 − ln 9 + ln10 9 +1

9− 9 10



=π 6

 ln80

9 −71 90



(3pts) .

2. The volume by disk method is

Volume = π 2

ˆ ln 2

−²3ln 3

 1

1 + e^3x

2

dx (5pts) .

Let u = 1

1 + e^3x, or x = 1 3ln 1

u− 1



. Then dx = 1

3(u²− u)du (1pt). The upper limit is u = 1

9 and the lower limit is u = 9

10 (1pt). Thus Volume = π

6 ˆ ¹₉

9 10

u

u − 1du = π 6

ˆ ¹₉

9 10

1 + 1

u − 1du (2pts)

= π

6[u + ln |u − 1|]¹⁹⁹

10

= π 6

 1 9− 9

10+ ln8 9− ln 1

10



=π 6

 ln80

9 −71 90



(3pts) .

3. The volume by disk method is

Volume = π 2

ˆ ln 2

−²3ln 3

 1

1 + e^3x

2

dx (5pts) .

Let tan θ = e^32x. Then sec²θ dθ = 3

2e^32xdx, or dx = 2 3

sec²θ

tan θ dθ (1pt). The upper limit is θ = tan⁻¹√ 8 and the lower limit is θ = tan⁻¹ 1

3 (1pt). Thus

Volume = π 2

ˆ tan⁻¹√ 8 tan^{−1 1}₃

1 sec⁴θ

2 3

sec²θ tan θ dθ = π

3

ˆ tan⁻¹√ 8 tan^{−1 1}₃

cos³θ

sin θ dθ (1pts)

= π 3

ˆ tan⁻¹√ 8 tan⁻¹¹₃

1 − sin²θ

sin θ d(sin θ) (2pts) = π 3



ln(sin θ) −1 2sin²θ

tan⁻¹√ 8

tan^{−1 1}₃

(2pts)

= π 3 ln

√8

3 − ln 1

√10−1 2 8 9+1

2 1 10

!

=π 6

 ln80

9 −71 90

 .

4. From y = 1

1 + e^3x we have x =1 3ln 1

y − 1



(1pt). At x = ln 2 , y =1

9 and at x = −2

3ln 3 , y = 9 10 (1pt).

Thus the volume by cylindrical shell method is

Volume =1 2

"

2π ˆ ¹₉

0

y



ln 2 − (−2 3ln 3)



dy + 2π ˆ ₁₀⁹

1 9

y 1 3ln 1

y − 1



− (−2 3ln 3)

 dy

#

(5pts) .

We integrate by parts to get ˆ

y ln 1 y − 1



dy = y² 2 ln 1

y − 1



− ˆ y²

2 1

y²− ydy (3pts)

=y² 2 ln 1

y − 1



−1 2

ˆ

1 + 1

y − 1dy (1pts)

=y² 2 ln 1

y − 1



−y 2−1

2ln |y − 1| (1pts) Thus

Volume = 1 2

"

2π ˆ ¹₉

0

y



ln 2 − (−2 3ln 3)



dy + 2π ˆ ₁₀⁹

1 9

y 1 3ln 1

y − 1



− (−2 3ln 3)

 dy

#

= π

 ln 2 + 2

3ln 3

 ˆ ¹₉

0

y dy +π 3

ˆ ₁₀⁹

1 9

y ln 1 y − 1



dy +2π 3 ln 3

ˆ ₁₀⁹

1 9

y dy

= π

 ln 2 + 2

3ln 3  y² 2

¹9

0

+π 3

 y² 2 ln 1

y − 1



−y 2 −1

2ln |y − 1|

10⁹ 1 9

+2π

3 ln 3 y² 2

10⁹ 1 9

= π 6

 ln80

9 −71 90

 .

6. (12%)

(a) Find the orthogonal trajectories of the family of curves y =p³

x³+ c, where c is an arbitrary constant.

(b) Solve the initial-value problem

y^′+ (tan x)y = sec³x, y(0) = 1.

Solution:

(a) Differentiating y =p³

x³+ c yields dy dx = 1

3 3x²

(x³+ c)²³ = x²

(x³+ c)²³ (1pt) = x² y² (1pt) Alternatively, differentiating y³= x³+ c yields the same result:

3y²dy

dx = 3x² so dy dx =x²

y² (2pts) We want to find a family of curves C : (x, y(x)) such that

dy dx = −y²

x² (2pts) This is a separable equation, we compute

1 y²

dy dx = − 1

x²(1pt) ⇒ ˆ 1

y² dy dxdx =

ˆ

−1

x²dx ⇒ −1 y = 1

x+ C (1pt) . Thus the family of orthogonal trajectories of y = p³

x³+ c is 1

x+1 y = C.

(b) The integrating factor is

e^{´ tan x dx}(2pts) = eln | sec x| (1pt) = sec x ( you can pick any sort of integrating factor) Multiply the integrating factor we get

sec xy^′+ sec x tan x · y = sec⁴x ⇒ d

dx(sec x · y) = sec⁴x Thus

sec x · y = ˆ

sec⁴x dx (1pt)

= sec²x d tan x = ˆ

(tan²x + 1) d tan x = 1

3tan³x + tan x + C (1pt) The initial condition y(0) = 1 implies C = 1 (1pt). Thus the solution is

y = 1

3cos x tan³x + sin x + cos x.

Alternatively, integrating by parts yields ˆ

sec⁴x dx

= sec²x d tan x = sec²x tan x − ˆ

tan x(2 sec²x tan x) dx

= sec²x tan x − 2 ˆ

sec²x(sec²x − 1) dx = sec²x tan x − 2 ˆ

sec⁴x dx + 2 tan x

Thus ˆ

sec⁴x dx = 1

3sec²x tan x +2

3tan x + C (1pt) The initial condition y(0) = 1 implies C = 1 (1pt). Thus the solution is

1 2

7. (12%) Consider the plum flower-like curve (梅花) as Figure 2. It is characterized by the polar equation r = 3

2+ cos 5 2θ

 .

(a) Find the slopes of the tangent lines of the curve at the intersection point P (r, θ) = 3 2,π

5

 .

(b) Set up an integral that represents the length of the whole curve. You don’t need to evaluate the integral.

(c) Find the area of the shaded region.

1 2

3

4

5 6 7

8

9

10

P

polar axis

Figure 2: The plum flower-like curve.

Solution:

-2 -1 0 1 2 3

-2 -1.5 -1 -0.5 0 0.5 1 1.5 2

O

P= [3 2,π

5] = [3 2,11π

5]

Q= [3 2, −π

5] = [3 2,9π

5]

(a) (method 1) Since r = 3

2+ cos 5 2θ



, we have r^′ = −5 2sin 5

2θ

 .(1%) Therefore, dy

dx=

dy dθ dx dθ

= r^′sin θ + r cos θ

r^′cos θ − r sin θ(1%) = −⁵₂sin ⁵₂θ
sin θ +³₂cos θ + cos ⁵₂θ
cos θ

−₂⁵sin ⁵₂θ
cos θ − ³2sin θ − cos ⁵₂θ
sin θ The slopes at P = 3

2,π 5



= 3 2,11π

5

 are:

dy dx    θ=^π₅

(1%) = −⁵₂sin^π₅+³₂cos^π₅

−⁵₂cos^π₅ −³₂sin^π₅ and dy dx    θ=^11π₅

(1%) =

5

2sin^π₅ +³₂cos^π₅

5

2cos^π₅−³₂sin^π₅ (method 2)

x = r cos θ = 3

2cos θ + cos 5 2θ



cos θ

or = 3

2 cos θ + 1 2 cos  7

2 θ

 + 1

2 cos  3 2 θ



y = r sin θ = 3

2sin θ + cos 5 2θ



sin θ

or = 3

2 sin θ + 1 2 sin  7

2 θ



− 1 2 sin  3

2 θ



Therefore, dy dx =

dy dθ dx dθ

=−⁵2sin ⁵₂θ
sin θ +³₂cos θ + cos ⁵₂θ
cos θ

−₂⁵sin ⁵₂θ
cos θ − ³2sin θ − cos ⁵₂θ
sin θ(2%)

or =

3

2

cos θ +

cos

θ
−

cos

θ

−

sin θ −

sin

θ
−

sin

θ

!

The slopes at P = 3 2,π

5



= 3 2,11π

5

 are:

dy dx    _θ=π

5

(1%) = −⁵₂sin^π₅+³₂cos^π₅

−⁵₂cos^π₅ −³₂sin^π₅ and dy dx    _θ=11π

5

(1%) =

5

2sin^π₅ +³₂cos^π₅

5

2cos^π₅−³₂sin^π₅

or dy

dx    

=

3

2

cos

+

cos

−

cos

−

sin

−

sin

−

sin

and dy dx    

=

3

2

cos

−

cos

+

cos

−

sin

+

sin

+

sin

!

(b) L = ˆ 4π

0

pr²+ (r^′)²dθ(2%) = ˆ 4π

0

s

 3

2+ cos 5 2θ

² +



−5 2sin 5

2θ

² dθ

or = 10 ˆ

0

s

 3

2 + cos  5 2 θ



+



− 5 2 sin  5

2 θ



dθ = 2

ˆ

s

 3

2 + cos  5 2 θ



+



− 5 2 sin  5

2 θ



dθ

!

(c) By symmetry, the area will be

Area = 2 ˆ ^π₅

0

1 2

 3

2+ cos 5 2θ

2

dθ − ˆ 2π+_π⁵

2π

1 2

 3

2 + cos 5 2θ

2

dθ

! (3%)

= ˆ ^π₅

0 − ˆ 2π+^π₅

2π

! 9

4+ 3 cos 5 2θ



+ cos² 5 2θ



dθ

= ˆ ^π₅

0 − ˆ 2π+^π₅

2π

! 9

4+ 3 cos 5 2θ



+1 + cos(5θ) 2

 dθ

=  11 4 θ + 6

5sin 5 2θ



(1%) + 1

10sin(5θ)(1%)

     

π 5

0

−      

2π+^π₅

2π

!

= 12 5 (1%)

8. (10%) Let f (x) be a differentiable and increasing function on [a, b], where a > 0. Find a horizontal line y = L that will minimize the function F (L) =

ˆ b

a x|f(x) − L| dx =

ˆ f⁻¹(L)

a x(L − f(x)) dx + ˆ b

f⁻¹(L)x(f (x) − L) dx.

x y

a f⁻¹(L) b

f (x) y = L

Figure 3: Find y = L to minimize the function F (L) = ˆ b

a x|f(x) − L| dx.

Solution:

F (L) = L

ˆ f⁻¹(L)

a xdx −

ˆ f⁻¹(L) a

xf (x)dx + ˆ b

f⁻¹(L)xf (x)dx − L ˆ b

f⁻¹(L)

xdx (1pt)

d

dLF (L) (1pt) =

ˆ f⁻¹(L)

a xdx + L · f⁻¹(L) · d

dLf⁻¹(L)

−f⁻¹(L) · L · d

dLf⁻¹(L) − f⁻¹(L) · L · d

dLf⁻¹(L)

− ˆ b

f⁻¹(L)

xdx + f⁻¹(L) · L · d

dLf⁻¹(L)

= 1 2x²

f⁻¹(L)

a −1

2x²  

b f⁻¹(L)

= 1

2((f⁻¹(L))²− a²) −1

2(b²− (f⁻¹(L))²) = 0 (4pt)

⇒ (f⁻¹(L))²= 1

2(a²+ b²) (2pt)

⇒ f⁻¹(L) = ±

ra²+ b²

2 ( choose

ra²+ b²

2 )

∴f⁻¹(L) =

ra²+ b²

2 i.e. L = f

ra²+ b² 2

! (2pt)

9. (12%)

(a) Suppose that f (t) is piecewise continuous on [0, ∞) and of exponential order. Let F (s) = L{f(t)} be the Laplace transform of f (t). Show that L{f(t − a)U(t − a)} = e^−asF (s), where U(t − a) is the unit step function defined as U(t − a) =

(0, if 0 ≤ t < a 1, if t ≥ a . (b) Express g(t) =

 t, if 0 ≤ t < 1

1, if t ≥ 1 in terms of unit step functions.

(c) Solve the differential equation y^′′+ 4y = g(t), where y(0) = 1 and y^′(0) = 0.

Solution:

(a) (3 points) By the definition of the Laplace transform, we have L{f(t − a)U(t − a)} =

ˆ _∞

0

e^−stf (t − a)U(t − a) dt = ˆ _∞

a

e^−stf (t − a) dt

= ˆ _∞

0

e^−s(u+a)f (u) du = e^−as ˆ _∞

0

e^−stf (t) dt = e^−asF (s), where u = t − a, du = dt.

• Partial credits: 1 point for the definition of the Laplace transform, 1 point for changing the variable.

(b) (2 points) g(t) = t − (t − 1)U(t − 1).

(c) (7 points) We take Laplace transform on both sides of the differential equation and get s²Y (s) − sy(0) − y^′(0) + 4Y (s) = 1

s² − e^−s· 1 s² = 1

s²(1 − e^−s)

⇒ (s²+ 4)Y (s) = s + 1

s²(1 − e^−s) ⇒ Y (s) = s

s²+ 4+ 1

s²(s²+ 4)(1 − e^−s).

Consider the partial fraction 1

s²(s²+ 4) = A s + B

s² +Cs + D

s²+ 4 for some constants A, B, C, and D. We will solve As(s²+ 4) + B(s²+ 4) + (Cs + D)s²= 1, and it implies A = 0, B = 1

4, C = 0, and D = −1

4. So we have Y (s) = s

s²+ 4 + 1 4 · 1

s² −1 4· 1

s²+ 4



(1 − e^−s)

= s

s²+ 2²+ 1 4 · 1

s² −1 8· 2

s²+ 2²



(1 − e^−s).

Hence the solution of the differential equation is y(t) = cos(2t) +1

4t −1

8sin(2t) − 1

4t U(t) −1

8sin(2t) U(t)

  t→t−1

= cos(2t) +1 4t −1

8sin(2t) − 1

4(t − 1) U(t − 1) −1

8sin(2(t − 1)) U(t − 1)

 .

• Grading policy: 3 points for transforming the equation, 4 points for converting back to function in t: 1 point each for t, sine term, cosine term, and shifting operation.

9. (12%) Find the center of mass (¯x, ¯y, ¯z) of the hemisphere S =(x, y, z)|x²+ y²+ z²= 1, x ≥ 0 .

Hint: The center of mass of an object is the average of the coordinates functions. You can consider the hemisphere as a surface of revolution obtained by rotating the curve y =p

1 − x², x ≥ 0, about the x-axis.

x y

z

Figure 4: Find the center of mass (¯x, ¯y, ¯z) of the hemisphere S =(x, y, z)|x²+ y²+ z²= 1, x ≥ 0 .

Solution:

By symmetry, ¯y = ¯z = 0. (2 points)

Viewing the hemisphere as a surface of revolution of a quarter circle y =p

1 − x², x ≥ 0, about the x-axis, the average of the function x over the surface is

¯ x =

´x=1

x=0 x · 2πy ds

´x=1

x=0 2πy ds (3 points) where y =p

1 − x², dy

dx =√−x

1 − x², and ds = r

(1 + dy

dx)²dx = 1

√1 − x²dx (3 points).

The denominator (2 points):

it is just the surface area, which is 1

2(4π · 1²) = 2π by known formula, or by integration ˆ x=1

x=0

2πy ds = 2π ˆ 1

0

p1 − x² 1

√1 − x²dx = 2π.

The numerator (x-weighted surface; 2 points):

ˆ x=1

x=0 x · 2πy ds = 2π ˆ 1

0 x ·p

1 − x² 1

√1 − x²dx = 2π ˆ 1

0

x dx = π.

Therefore, ¯x = π 2π = 1

2, and the center of mass is located at (¯x, ¯y, ¯z) = (1 2, 0, 0).

(Note: if you misunderstood the problem and you correctly found the center of mass of the solid hemisphere x²+ y²+ z²≤ 1, x ≥ 0 to be (3

8, 0, 0), 8 points will be credited.)