Name
Sec
ID
MATH 152 Honors
Final Exam
Sections 201,202
Solutions
1-12
/48
Spring 2006
13
/15+5
P. Yasskin
14
/20
15
/20
Total
/108
Multiple Choice: (4 points each)
1. The temperature along a probe which is π m long is given by T = 75 + sin x for 0 ≤ x ≤ π.
Find the average temperature of the probe.
2
a. 75 + π
CorrectChoice
1
b. 75 − π
c. 75
d. 75π − 1
e. 75π + 2
1
T ave = π
π
∫ 0 (75 + sin x) dx =
2. Compute
1
π 75x − cos x
π
0
1 cos π + 1 cos 0 = 75 + 2
= 75 − π
π
π
1
∫ 0 xe 2x dx.
a. 1 e 2
b.
c.
d.
e.
2
1 e2
4
1 (e 2 + 1)
2
1 (e 2 + 1)
4
1 (3e 2 + 1)
4
u=x
du = dx
1
∫0
xe 2x dx =
CorrectChoice
dv = e 2x dx
v = 1 e 2x
2
x e 2x − 1
2
2
1
∫0
e 2x dx
1
0
=
x e 2x − 1 e 2x
2
4
1
0
=
1 e2 − 1 e2 − − 1
2
4
4
= 1 (e 2 + 1)
4
1
∫1
3. Compute
2x + 2 dx
x(1 + x 2 )
3
a. ln 3 + π
12
b. ln 3
c. ln 3+ π
6
d. ln 3
2
e. ln 3 + π
CorrectChoice
6
2
2x + 2 = A + Bx + C
2x + 2 = A(1 + x 2 ) + (Bx + C)(x) = (A + B)x 2 + Cx + A
x
1 + x2
x(1 + x 2 )
A=2
∫1
3
B = −2
C=2
2x + 2 dx = 3 2 + −2x + 2 dx = 3 2 − 2x + 2 dx
∫1 x
∫1 x 1 + x2 1 + x2
x(1 + x 2 )
1 + x2
= [2 ln|x| − ln|1 + x 2 | + 2 arctan x] 1
3
= 2 ln 3 − ln 4 + 2 arctan 3
− (2 ln 1 − ln 2 + 2 arctan 1)
= ln 3 + π
6
2
4. If y = f(x) is a solution of the differential equation
f(1) = 2, then f ′ (1) =
dy
= x + y 2 satistfing the initial condition
dx
a. 1
b. 2
c. 3
d. 4
e. 5
CorrectChoice
As a solution, y = f(x) satisfies f ′ (x) = x + y 2 = x + f(x) 2 . So f ′ (1) = 1 + f(1) 2 = 1 + 2 2 = 5
∞
5. Compute
∑
n=0
(−1) n π 2n
16 n (2n)!
a. 0
b. 1
2
c. 1
2
CorrectChoice
d. 1
e. −1
∞
∑
n=0
(−1) n x 2n
= sin x
(2n)!
∞
So
∑
n=0
(−1) n π 2n
=
16 n (2n)!
∞
∑
n=0
(−1) n
(2n)!
π
4
2n
= sin π
4
=
1
2
2
6. The region between y = 3x − x 2 and the x-axis is rotated about the y-axis. Find the
volume of the solid swept out.
a. 81 π
5
b. 81 π
10
c. 27 π
2
CorrectChoice
d. 27π
e. 81 π
2
x-integral
3x − x 2 = 0
y-axis
cylinders
3
3
V=
x = 0, 3

r=x
y 2.0
h = 3x − x 2
1.5
∫ 0 2πrh dx = ∫ 0 2πx(3x − x 2 ) dx
3
4
= 2π x 3 − x
4
0
1.0
0.5
= 2π 27 − 81
4
= 27 π
2
0.0
0
1
2
3
x
7. The region between y = 3x − x 2 and the x-axis is rotated about the x-axis. Find the
volume of the solid swept out.
a. 81 π
5
b. 81 π
10
c. 27 π
2
CorrectChoice
d. 27π
e. 81 π
2
x-integral
=
3
∫0
πr 2 dx =
x = 0, 3

y 2.0
r = y = 3x − x 2
disk
x-axis
V=
3x − x 2 = 0
3
∫0
1.5
π(3x − x 2 ) 2 dx
3
1.0
∫ 0 π(x 4 − 6x 3 + 9x 2 ) dx = π
5
5
= π 3 − 3 + 34
5
2
5
4
x − 3x + 3x 3
5
2
= 81π 3 − 3 + 1
5
2
3
0.5
0
= 81 π
10
0.0
0
1
2
3
x
3
8. Find the length of the curve y = ln(cos x) for 0 ≤ x ≤ π .
4
a. ln
2
b. ln
1
2
c.
1
ln
2
d. ln
2 −1
e. ln
2 +1
CorrectChoice
dy
sin x = − tan x
= −cos
x
dx
L = ∫ ds =
π/4
∫0
= ln
dy
dx
1+
2
dx =
2 + 1 − ln(1) = ln
π/4
∫0
1 + tan 2 x dx =
π/4
∫0
sec x dx = ln|sec x + tan x|
π/4
0
2 +1
9. The curve x = t 3 ,
y = t 2 for 0 ≤ t ≤ 1 is rotated about the x-axis. Which integral gives
the area of the surface swept out?
1
a.
∫ 0 πt 2
b.
∫ 0 2πt 3
c.
∫ 0 πt 3
d.
∫ 0 2πt 3
e.
∫ 0 2πt 2
1
1
9t 4 + 4t 2 dt
9t 4 + 4t 2 dt
9t 2 + 4 dt
1
9t 2 + 4 dt
1
9t 2 + 4 dt
dy
= 2t
dt
dx = 3t 2
dt
A = ∫ 2πr ds =
10. Compute
6
∫2
1
∫0
2πy
CorrectChoice
r = y = t2
dx
dt
2
+
dy
dt
2
dt =
1
∫ 0 2πt 2
9t 4 + 4t 2 dt =
1
∫ 0 2πt 3
9t 2 + 4 dt
1
dx
(x − 2) 3/2
a. −2
b. −1
c. 1
d. 2
e. The integral diverges.
6
∫2
1
dx =
(x − 2) 3/2
6
CorrectChoice
∫ 2 (x − 2) −3/2 dx =
−2(x − 2) −1/2
6
2
= −1 + lim+
x2
2
=∞
(x − 2) 1/2
4
11. Compute lim
x0
sin(x) − x
x 2 (e x − e −x )
a. −∞
b. −1
Correct Choice
12
c. 1
6
d. 1
3
e. ∞
lim
x0
sin(x) − x
= lim
x0 2
x 2 (e x − e −x )
x
3
5
x− x + x −⋯ −x
3!
5!
2
3
2
3
1+x+ x + x +⋯ − 1−x+ x − x +⋯
2!
3!
2!
3!
3
5
−x + x −⋯
3!
5!
= lim
3
x0 2
x
x 2x + 2
+ −⋯
3!
2
− 1 + x −⋯
3!
5!
= lim
2
x0
x
2+2
+ −⋯
3!
= −1
12
−1, 1
by its 3 rd degree Maclaurin
2 2
polynomial T 3 (x) = −x − 1 x 2 − 1 x 3 , then the Taylor Remainder Theorem says the error is
3
2
less than
12. If you approximate f(x) = ln(1 − x) on the interval
Taylor Remainder Theorem:
If T n (x) is the n th degree Taylor polynomial for f(x) about x = a
then there is a number c between a and x such that the remainder is
f (n+1) (c)
(x − a) (n+1)
R n (x) =
(n + 1)!
a. 1
b.
c.
d.
e.
4
1
8
1
16
1
32
1
64
CorrectChoice
f(x) = ln(1 − x)
f ′ (x) =
M = max |f (4) (x)| =
x∈[−.5,.5]
−1
1−x
f ′′ (x) =
−1
(1 − x) 2
f ′′′ (x) =
−2
(1 − x) 3
f (4) (x) =
−6
(1 − x) 4
6
= 96
(1 −. 5) 4
For x ∈ − 1 , 1 , |x| ≤ 1
2 2
2
(4)
f (c) 4
|R 3 (x)| =
x ≤ M |x| 4 = 96
4!
4!
24
1
2
4
= 1
4
5
Work Out: (Points indicated. Part credit possible.)
13.
y 10
(15 points) A water trough is 10 ft long
8
and has vertical ends in the shape of the
paraboloid y = x
2
6
up to a height of 9 ft.
4
2
It is filled with water.
Take the density of water as ρg = 60
lb
ft
3
-2
.
0
2
4
6
8
x
Do one of the following two problems. (5 points extra credit if you do both.)
a. Find the work done to pump the water out the top of the trough.
The slice of water at height y with thickness dy has width 2x = 2 y
and length 10.
So its volume is dV = 20 y dy and its weight is dF = ρg dV = 1200 y dy.
It must be lifted a distance h = 9 − y. So the work is
W=
9
∫0
h dF =
9
∫0
9
(9 − y)1200 y dy = 1200 ∫ (9y 1/2 − y 3/2 ) dy = 1200
0
= 1200 3 ⋅ 2 ⋅ 9 3/2 − 2 ⋅ 9
5
5/2
= 1200 3 4 ⋅ 2 − 2 ⋅ 3
5
5
9 ⋅ 2y 3/2
2y 5/2
−
5
3
9
0
= 1200 ⋅ 81 ⋅ 2 ⋅ 2 = 77760
5
b. Find the force of the water on the end of the trough.
The slice of water at height y with thickness dy has width 2x = 2 y .
So its area on the end is dA = 2 y dy.
This slice is a distance h = 9 − y below the surface of the water.
So the pressure is P = ρgh = 60(9 − y). So the force is
F=
9
∫0
P dA =
9
∫0
9
60(9 − y)2 y dy = 120 ∫ (9y 1/2 − y 3/2 ) dy = 120
= 120 3 ⋅ 2 ⋅ 9 3/2 − 2 ⋅ 9
5
0
5/2
= 120 3 4 ⋅ 2 − 2 ⋅ 3
5
5
9 ⋅ 2y 3/2
2y 5/2
−
5
3
9
0
= 120 ⋅ 81 ⋅ 2 ⋅ 2 = 7776
5
6
14. (20 points) Consider the sequence defined recursively by
a1 = 4
a n+1 =
and
a. (4 pt) Write out the first 4 terms and simplify, e.g.
a1 =
a3 =
a2 =
4
2 ⋅ 2 2 = 24 2
8 =2 2
2⋅4 =
a4 =
2a n
8 =2 2
2 ⋅ 24 2 = 28 2
b. (5 pt) Is the sequence increasing or decreasing? Prove it.
The sequence is decreasing. We need to prove a n+1 < a n .
First note a 2 = 2 2 < 4 = a 1 .
Now assume a k+1 < a k . Then 2a k+1 < 2a k and
2a k+1 <
2a k
or a k+2 < a k+1 .
So a n+1 < a n .
c. (5 pt) If it is increasing, find an upper bound.
If it is decreasing, find a lower bound. Prove it.
The sequence is bounded below by 2. We need to prove a n > 2.
First note a 1 = 4 > 2.
Now assume a k > 2. Then a k+1 =
2a k >
2 ⋅ 2 = 2.
So a n > 2.
d. (2 pt) What do (b) and (c) imply?
That the sequence has a limit which is ≥ 2.
e. (4 pt) Find the limit of the sequence.
Method 1: Let L = n∞
lim a n . Then L =
2L
So L 2 = 2L or L 2 − 2L = 0 or L = 0 or L = 2.
By (b) and (c), L = 2.
Method 2: a n = 2 ⋅ 2 1/2
n−1
So n∞
lim a n = n∞
lim 2 ⋅ 2 1/2
n−1
lim 1/2
= 2 ⋅ 2 n∞
n−1
= 2 ⋅ 20 = 2
7
15. (20 points) A brine solution that contains 0. 4 kg of salt per liter flows at a constant rate of 2
liters per minute into a large tank that initially held 100 liters of brine that contains 0. 5 kg
of salt per liter. The solution in the tank is kept well mixed and flows out of the tank at 2 liters
per minute. Let S(t) be the amount of salt on the tank at time t.
a. (8 pt) Set up the differential equation and initial condition for S(t).
dS kg = 0. 4 kg 2 L − S(t)kg 2 L
dt min
100L min
L min
in
S(0) = 0. 5
kg
100L= 50kg
L
out
dS = 0. 8 −. 02S(t)
dt
S(0) = 50
b. (12 pt) How much salt is in the tank after 50 minutes?
Separable:
dS
= dt
0. 8 −. 02S(t)
|0. 8 −. 02S(t)| = e −.02t−.02C
∫
dS
= ∫ dt
0. 8 −. 02S(t)
0. 8 −. 02S(t) = Ae −.02t
−1 ln|0. 8 −. 02S(t)| = t + C
. 02
where A = ±e −.02C
. 02S(t) = 0. 8 − Ae −.02t
S(t) = 0. 8 − A e −.02t = 40 − 50Ae −.02t
. 02
. 02
S = 50 when t = 0:
50 = 40 − 50A
50A = −10
A = −. 2
S(t) = 40 + 10e −.02t
S(50) = 40 + 10e −.02⋅50 = 40 + 10
e ≈ 43. 679
OR
Linear:
dS +. 02S(t) = 0. 8
dt
.02 dt
I = e∫
= e .02t
d (e .02t S) = 0. 8e .02t
e .02t dS +. 02e .02t S(t) = 0. 8e .02t
dt
dt
e .02t S = ∫ 0. 8e .02t dt = 0. 8 e .02t + C = 40e .02t + C
. 02
S = 50 when t = 0:
e .02t S = 40e .02t + 10
50 = 40 + C
C = 10
S = 40 + 10e −.02t
S(50) = 40 + 10e −.02⋅50 = 40 + 10
e ≈ 43. 679
8