1092 Calculus B 01-03 Final Exam Solution

(1)

1092 Calculus B 01-03 Final Exam Solution

June 24, 2021 1. Consider the differential equation ydy

dt =

√ 1 − y².

(a) (6 pts) If the constant function y(t) = a is a solution of the equation, find the value of a.

(b) (10 pts) Given that y(0) = 1

2, solve y(t) for t near 0.

考慮微分方程式 ydy dt =

√

1 − y²。

(a) (6 pts) 如果常數函數 y(t) = a 是方程式的解，求a 的值。

(b) (10 pts) 給定 y(0) = 1

2，求 y(t)，當t在0附近。

Solution:

(a) Let

√

1 − y²=0. Then y = ±1. Hence

a = ±1 (6%).

(b) Now suppose y ≠ ±1 (1%). Rewrite the equation as y dy

√ 1 − y²

=dt (1%) and integrate on both sides with respect to t, we get

−

√

1 − y²=x + c. (3%).

Since y(0) = 1/2, we get that

c = −

√

1 − (1/2)² = −

√ 3

2 (2%).

Hence

y = ±

√

1 − (x −

√

3/2)² (1%)

=

√

1 − (x −

√

3/2)² (since y(0) =1

2, we should choose +) (2%).

(2)

2. (16 pts) Solve

⎧⎪

⎪⎪

⎨

⎪⎪

⎪⎩

y^′(t) = t(t²−y(t)), y(0) = 3.

(16 pts) 解

⎧⎪

⎪⎪

⎨

⎪⎪

⎪⎩

y^′(t) = t(t²−y(t)), y(0) = 3.

Solution:

y^′+t y =t³ rearrange the eqn I(t) =e^{∫ t dt}=e¹²^t² find IF(5%)

e¹²^t²y^′+e¹²^t²t y =e¹²^t²t³ multiply the eqn by I(t) = e¹²^t² (2%) d (e¹²^t²y)

dt =e¹²^t²t³ rewrite the left-hand side

∫

d (e¹²^t²y)

dt dt =∫ e¹²^t²t³dt integrate both sides (2%) e¹²^t²y =e¹²^t²(t²−2) + c evaluate the integrals (3%)

3 = − 2 + c use y(0) = 3

5 =c find c(2%)

y(t) =t²−2 + 5 e^−t2² the solution(2%)

Page 2 of 7

(3)

3. (8 pts) Suppose that X and Y are independent and E(X) = 1, E(Y ) = 2, Var(X) = 5, Var(Y ) = 7.

Compute E((2X − Y )²).

(8 pts) 已知 X，Y 為兩個獨立的隨機變數，而且 E(X) = 1，E(Y ) = 2， Var(X) = 5， Var(Y ) = 7。

求 E((2X − Y )²)。

Solution:

E((2X`Y )²)

=E(4 X²−4 X Y + Y²) expand (2%)

=4 E(X²) −4 E( X Y ) + E(Y²) E(a X + b Y ) = a E(X) + b E(Y ) (1%)

=4 (Var(X) + (E(X))²) −4 E(X) ⋅ E(Y ) Var(X) = E(X²) − (E(X))² (2%)

+ (Var(Y ) + (E(Y ))²) X, Y idep. ⇒E(X Y ) = E(X) ⋅ E(Y ) (1%)

=4 (5 + 1²) −4 ⋅ 1 ⋅ 2 + (7 + 2²) input the values (2%)

=27

(4)

4. Suppose that X is a random variable with probability density function f_X(t) = ce^−t2+8t²⁰ for some constant c > 0.

(a) (8 pts) Find the constant c such that ∫

∞

−∞

f_X(t)dt = 1. (Hint: ∫

∞

−∞

e^−t²dt =√ π) (b) (8 pts) Find E(X) and Var(X).

(c) (6 pts) Suppose that X₁, ⋯, X_n are independent random variables and X_i ∼X for 1 ≤ i ≤ n.

Use Chebyshev’s inequality to estimate the size of n such that we can derive P ( ∣X₁+ ⋯ +X_n

n −E(X)∣ < 0.05 ) > 0.9.

已知隨機變數 X 的機率密度函數為 fX(t) = ce^−t2+8t²⁰ ，其中 c > 0 是一個常數。

(a) (8 pts) 求常數 c 使得 ∫

∞

−∞

f_X(t)dt = 1. (提示: ∫

∞

−∞

e^−t²dt =√ π) (b) (8 pts) 求 E(X) 和 Var(X).

(c) (6 pts) 假設 X1, ⋯, Xn 為互相獨立的隨機變數而且 Xi ∼X， 1 ≤ i ≤ n。利用 Chebyshev 不等式，估計 n 需要多大，我們才能保證

P ( ∣X₁+ ⋯ +X_n

n −E(X)∣ < 0.05 ) > 0.9.

Solution:

(a) We have

1 = c∫

∞

−∞

e^−(t²^−8t)/20dt

=c∫

∞

−∞

e⁻^{(t−4)2−16}²⁰ dt (3%)

=ce⁴⁵∫

∞

−∞

e⁻^(t−4)2²⁰ dt (1%)

=ce⁴⁵ ⋅

√π ⋅

√

20 (3%).

Hence

c = e^−4/5⋅ 1 2√

5π (1%).

(b) Follows form (a), we have that

X ∼ N(4, 10) (2%).

Hence

E(X) = 4 (3%), Var(X) = 10 (3%).

(c) Let

Z ∶= X1+ ⋯ +Xn

n .

Since Xi∼X for all i and X1, ⋯, Xn are independent, we have E(Z) = E(X) = 4 (1%) Var(Z) = Var(X)

n =

10

n (1%).

Page 4 of 7

(5)

By the Chebyshev inequality, we want to find n so that P (∣Z − E(X)∣ < 0.05) ≥ 1 − Var(Z)

(0.05)² (2%)

=1 − 10/n

(0.05)² (1%)

>0.9, this is equivalent to

0.1 > 10/n

1/400 ⇐⇒n > 40000 (1%).

(6)

5. Let f (t) =

⎧⎪

⎪

⎨

⎪⎪

⎩

c ⋅ te^−t, t > 0 0, t ≤ 0.

(a) (8 pts) Find constant c such that f (t) is a probability density function.

(b) (4 pts) Suppose that X and Y are independent with probability density functions f_X =f_Y =f . Let Z = X + Y . Find the plane region D in the xy-plane such that the distribution function of Z, F_Z(t) = P (X + Y ≤ t), is ∬

D

f_X(x)f_Y(y)dxdy.

(c) (10 pts) Find the probability density function of Z in part (b).

令 f(t) =

⎧⎪

⎪

⎨

⎪⎪

⎩

c ⋅ te^−t, t > 0 0, t ≤ 0.

(a) (8 pts) 求常數 c 使得 f (t) 是一個機率密度函數。

(b) (4 pts) 已知隨機變數 X 和 Y 是獨立的而且它們的機率密度函數為 fX =f_Y =f。令 Z = X+Y 。求 xy 平面的區域D，使得 Z的分配函數， F^Z(t) = P (X +Y ≤ t)，是重積分∬

D

f_X(x)f_Y(y)dxdy。 (c) (10 pts) 求(b)小題中 Z的機率密度函數。

Solution:

(a)

∫

∞

−∞

f (t)dt = c∫

∞ 0

te^−tdt = c (−te^−t∣

∞ 0 + ∫

∞ 0

e^−tdt) = c(0 − e^−t∣

∞ 0

) =c.

(3 pts for applying integration by parts. 3 pts for correct result. Students do not need to use the definition of improper integrals to compute this integration.)

Let c = 1. Then ∫

∞

−∞

f (t)dt = 1 and f (t) is a probability density function.

(2 pts for c = 1.)

(b) FZ(t) = P (X + Y ≤ t) =∬

DfX(x)fY(y)dxdy, where D = {(x, y)∣x + y ≤ t, x ≥ 0, y ≥ 0}.

If t ≤ 0, then D is an empty set or a point {(0, 0)}.

If t > 0, then D is the triangle bounded by x = 0, y = 0, and x + y = t.

(4 pts for D = {(x, y)∣x + y ≤ t, x ≥ 0, y ≥ 0} or D = {(x, y)∣x + y ≤ t}. It is o.k. if students don’t discuss cases t > 0 and t ≤ 0. )

(c) For t ≤ 0, F_Z(t) = 0 For t > 0,

Solution 1:

F_Z(t) =∬

D

xe^−xye^−ydxdy

⎧⎪

⎪⎪

⎪

⎨

⎪⎪

⎩

u = x + y v = y

ÔÔÔÔÔÔÔÔ ∬

Ω

(u − v)e^v−u⋅v ⋅ e^−v∣

∂(x, y)

∂(u, v)∣dudv

= ∫

t 0 ∫

u 0

(u − v)v ⋅ e^−udvdu,

where Ω is the region in the uv-plane bounded by u = t, u = v, and v = 0, and ∣∂(x, y)

∂(u, v)∣ =1.

(1 pt for substitution u = x + y, v = y, or u = x + y, v = x. 1 pt for ∣∂(x, y)

∂(u, v)∣ =1. 1 pt for Ω.

2 pts for the iterated integral.)

Page 6 of 7

(7)

Hence fZ(t) = F_Z^′(t) =∫

t

0 e^−t⋅ (t − v)vdv = 1 6t³e^−t. (2 pts for f_Z(t) = F_Z^′(t) =∫

t 0

e^−t⋅ (t − v)vdv. 3 pts for∫

t 0

e^−t⋅ (t − v)vdv = 1 6t³e^−t.) Solution 2:

f_Z(t) =∫

∞

−∞

f_X(t − s)f_Y(s) ds or ∫

∞

−∞

f_X(s)f_Y(t − s) ds. (3 pts) Hence f_Z(t) =∫

∞

−∞

f_X(t − s)f_Y(s) ds =∫

t 0

(t − s)e^s−tse^−sds (2 pts for the upper and lower limits of the integration 0, t.)

Thus fZ(t) =∫

t

0 s(t − s)e^−tds = 1 6t³e^−t. (2 pt for f_Z(t) =∫

t 0

s(t − s)e^−tds. 3 pts for ∫

t 0

s(t − s)e^−tds = 1 6t³e^−t.)

(8)

6. The number of flaws in a carpet appears to be Poisson distributed at a rate of one every 6m². (a) (8 pts) Find the probability that a 3m by 4m carpet contains more than 2 flaws.

(b) (8 pts) There are two carpets with sizes 2m by 4m and 2m by 5m. Find the probability that these two carpets together contain less than 2 flaws.

假設地毯上的瑕疵個數呈 Poisson 分配，而且平均每 6平方公尺有一個瑕疵。

(a) (8 pts) 求一條 3公尺寬 4公尺長的地毯有超過 2個瑕疵的機率。

(b) (8 pts) 有兩條地毯，各是 2公尺寬 4公尺長和 2公尺寬 5公尺長。求這兩條地毯上總共的瑕疵數少

於 2的機率。

Solution:

(a) The average number of flaws in a 3m×4m carpet is 1 × (3 × 4

6 ) =2. (2 pts)

Let F be the number of flaws in the carpet. Then F is Poisson distributed with E(F ) = 2.

Hence P (F = 0) = e⁻², P (F = 1) = 2¹

1!e⁻² =2e⁻², P (F = 2) = 2²

2!e^−2=2e⁻².

(1 pt for P (F = 0) = e⁻². 1 pt for P (F = 1) = 2e⁻². 1 pt for P (F = 2) = 2e⁻².) Thus P (F > 2) = 1 − P (F ≤ 2) = 1 − P (F = 0) − P (F = 1) − P (F = 2) = 1 − 5 ⋅ e⁻².

(2 pts for P (F > 2) = 1 − P (F ≤ 2) = 1 − P (F = 0) − P (F = 1) − P (F = 2). 1 pt for the final answer.)

(b) Solution 1:

Together the average number of flaws in the two carpets is 1 ×(2 × 4 + 2 × 5)

6 =3. (4 pts) Let F be the number of flaws in the two carpets. Then P (F < 2) = P (F = 0) + P (F = 1) = e⁻³+3e⁻³ =4e⁻³.

(1 pt for P (F < 2) = P (F = 0)+P (F = 1). 1 pt for P (F = 0) = e⁻³. 1 pt for P (F = 1) = 3e⁻³. 1 pt for the final answer.)

Solution 2: Let F₁ be the number of flaws in the 2 × 4 carpet. P (F₁ = 0) = e⁻⁸⁶ and P (F1 =1) = 4

3e⁻⁴³. (2 pts)

Let F₂ be the number of flaws in the 2 × 5 carpet. P (F₂ =0) = e⁻¹⁰⁶ , P (F₂=1) = 5

3e⁻⁵³. (2 pts)

P (F₁+F₂ ≤1) = P (F₁=0, F₂=0) + P (F₁=1, F₂=0) + P (F₁ =0, F₂ =1) (2 pts)

=e⁻⁴³e⁻⁵³ + 4

3e⁻⁴³e⁻⁵³ +e⁻⁴³5

3e⁻⁵³ =4e⁻³. (2 pts)

Page 8 of 7