Spring 2005 Math 152
Exam 2A: Solutions
c
Mon, 28/Mar
2005,
Art Belmonte
3. (e) The arc length of the curve
r(t) = [x(t), y(t)] = [3t + 1, 4 − t] , 1 ≤ t ≤ 3,
Rbq
Rb
is L = a r0 (t) dt = a (d x/dt)2 + (d y/dt)2 dt
√
R q
R √
= 13 (3)2 + (1)2 dt = 13 10 dt = 2 10 ≈ 6.32 cm.
4. (b) Let f (x) = ln x. We’ll determine K = max f 00 (x),
For specificity, lengths are in centimeters unless stated otherwise.
2≤x≤5
1. (d)
then employ the Trapezoidal error estimate.
• For non-STEPS folks: We have
3
X
m k xk
14
(3) (1) + (5) (−2) + (7) (3)
k=1
=
.
=
x̄ =
3
3+5+7
15
X
mk
1
• Now f 0 (x) = = x −1 and f 00 (x) = −x −2 . Thus
x
1
1
= .
K = max
4
2≤x≤5 x 2
• Therefore,
1
3
K (b − a)3
27
9
4 (5 − 2)
|E T | ≤
.
=
=
=
2
2
4
256
12n
12 (4)
3 (4)
k=1
(Note that this only gives you the x-coordinate of the
center of mass, not the entire center of mass.)
5. (a) The surface area is given by
• For STEPS folks: Let p = [3, 5, 7] be the row vector of
masses and


1 2
r =  −2 5 
3 1
Z
S=
2πr ds =
Z b
a
s
2π x 1 +
s
2
Z 3
dy 2
1
dx =
2π x 1 +
d x.
dx
x
1
X2A / 5: Surface of revolution
X2A / 5: Curve to be rotated about y−axis
2.5
4
be a matrix whose rows are position vectors of the
points. Then m = sum (p) = 3 + 5 + 7 = 15 is the
total mass.
The center of mass is
[x̄, ȳ] =
=
=
y
1
−1
y
5
5
0
0
1
2
3
4
z
x
0
−5
−5
x
6. (c) The improper integral converges to π/4.
Z t
Z ∞
1
1
d x = lim
dx
2
2
t→∞
x +1
1
1 x +1
t
= lim tan−1 x t→∞
1
−1
= lim tan t − tan−1 1
t→∞
π
π
π
− =
2
4
4
=
7. (a) The slope of the curve at a point (x, y) is x 3 y 2 and the
curve passes through (0, 12 ). This gives us a separable
differential equation d y/d x = x 3 y 2 and an initial condition
y (0) = 12 . Solve this initial value problem.
4x
x3
dy
+ 2
y= 2
dx
x +1
x +1
y −2 d y
Then an integrating factor is
Z
4x
2
µ = exp
d
x
=
exp
2
ln
x
+
1
x2 + 1
2 2
= exp ln x + 1
2
2.5
2
2. (a) Put the differential equation in standard linear form.
x2 + 1
1.5
1.5
So the x-coordinate of the center of mass is x̄ = 14
15 .
(Remark: pr represents matrix multiplication, realized
by taking the dot products of rows with columns. This
immediately extends to 3-D center of mass problems in
Calc 3.)
2
3
1
(pr)
m
1
[3 − 10 + 21, 6 + 25 + 7]
15
14 38
,
.
15 15
=
curve
axis of symmetry
3.5
−
1
= −y −1 =
y
−2 =
1
−
=
y
y
.
1
=
=
x3 dx
1 4
4x
+C
C
[Substitute the initial condition.]
1 x4
4
−2
1
2−
1 4
4x
=
4
8 − x4
8. (d) We have
Z ∞
1
0≤
dx
√
x + e10x
0
Z
≤
0
9. (d) With y =
2
3
e−10x d x
t
√
1
converges to a value L ≤
x + e10x
(x
− 1)3/2 ,
Z
=
L
b
a
Z
=
4
1+
r
1+
dy
dx
√
dx
2
= w d y = 35 y d y
dF
=
F
=
P d A = 35 δy (y + 2) d y
Z 5
3
δ
y 2 + 2y d y
5
0
5
+ y2 0
3
125
+
25
= 2400 lb.
(60)
5
3
3δ
5
=
1
Z
4
x 1/2 d x
=
4
2 x 3/2 = 16 − 2 = 14 .
3
1
3
3
3
interior func vals
y
2−1
1
10. (b) The step size is h = b−a
n = 4 = 4 . Partition points
o
n
are 1, 54 , 32 , 74 , 2 . The Trapezoidal Rule gives
X2A / 12: Surface of revolution
curve
axis of symmetry
2
2
1.5
1
1
or T4
=
=
=
1h
2
1
8
1
8
f (1) + 2 f
5
4
+2f
3
2
+2f
7
4
5
3
7
ln 1 + 2 ln + 2 ln + 2 ln + ln 2
4
2
4
5
3
7
ln 2 + 2 ln + 2 ln + 2 ln
4
2
4
+ f (2)
−1
1
2
3
4
5
2
3
4
6
x
x
0
5
−2
y
13. The density ρ is constant. First compute the mass
ZZ
ZZ
m=
ρdA = ρ
d A = ρ A = ρ 14 π (2)2 = ρπ
D
D
then compute x̄, the x-coordinate of the centroid.
ZZ
1
x̄ =
ρx d A
[STEPS folks here; others below.]
m
D
Z 2 Z √4−x 2
1
ρx d y d x
=
ρπ 0 0
√
Z
1 2 y= 4−x 2
x y
d x [ρ is constant]
=
y=0
π 0
Z
Z
1/2
1 b
1 2
2
4−x
x d x [i.e.,
x f (x) d x]
x̄ =
π 0
A a
3/2 2
2
4 − x2
=
− 12 π1
3
0
8
8
=
= 0− −
3π
3π
Triangluar plate submerged in fluid
surface of water
−2
−1
0
y
1
y
3
w
4
5
−2
2
−0.5
11. Here is a diagram of the submerged plate.
2
0
0.5
0
2
3
z
P
X2A / 12: Curve to be rotated about x−axis
2.5
Tn = step size × avg of endpt func vals +
π
4
=
1
=
3/2 2
1 + 4y 2
1
√
√
π
17 17 − 5 5 ≈ 30.85 cm2 .
6
dx
1
=
1 y3
3
R
12. The surface area is S = 2πr ds or
s
2
Z d
Z 2
q
dx
2π y 1 +
dy =
2π y 1 + (2y)2 d y
dy
c
1
Z 2
1/2
= π
1 + 4y 2
2y d y
2
x −1
dA
=
1
10 .
1 ≤ x ≤ 4, the arc length is
s
= δz = δ (y − (−2)) = δ (y + 2)
P
0
e−10x d x
t→∞ 0
t
1 e −10x lim − 10
t→∞
0
1
1 = 1 .
−10t
lim − 10 e
+ 10
t→∞
10
=
∞
Z
∞
dx =
lim
=
Z
Z
1
e10x
0
=
Therefore,
∞
• A slice of the plate is at variable depth z.
R Build up your
integral step-by-step: P, d A, d F, F = d F.
3
• Via similar triangles we have 3/5 = w/y; so w = 35 y.
2
14.
Multiply the SLF by µ and solve.
(a) Factor to see that the differential equation is separable.
dy
dt
dy
dt
=
1 + t − yt − y = (t + 1) − y (t + 1)
=
(t + 1) (1 − y) = − (t + 1) (y − 1)
1
dy
y−1
=
− (t + 1) dt
ln |y − 1|
=
|y − 1|
=
− 12 (t + 1)2 + A
2
2
e A−(t +1) /2 = e A e−(t +1) /2
et/100 y 0 +
e
=
Be
y
=
2
1 + Ce−(t +1) /2
=
xe−2x
=
e−2x y
=
xe−2x
− 12 x − 14 e−2x + C
y
=
− 12 x −
1
4
− 14 + C
C
=
y
=
− 12 x −
+ Ce2x
[initial condition]
1
4
+ 54 e2x
15. Let y = y(t) be the amount of salt in the tank at time t. Since
the tank initially contains pure water, we have y(0) = 0 kg of
salt in the tank at the start. The classical balance law gives
=
=
rate in − rate out
L × 1 kg − 10 L × y kg
10 min
2 L
min
1000 L
=
5−
=
5.
1
100 y
=
5et/100
=
500et/100 + C
=
0 = y (0) =
C =
y =
In the third step we integrated by parts. With u = x and
dv = e−2x d x we have du = d x and v = − 12 e−2x .
Therefore,
Z
Z
1
1 −2x
xe−2x d x = − xe−2x +
e
dx
2
2
1 −2x
1
e
=
− x−
+C
2
4
dy
dt
dy
dt
dy
dt
1
y 0 + 100
y
5et/100
y (10) =
e−2x y 0 − 2e−2x y
0
e−2x y
5
4
=
y
(b) The differential equation is already in standard linear
0
form (SLF):
R y − 2y
= x. An integrating factor is
µ = exp −2 d x = e−2x . Multiply the SLF by µ
and solve.
1 = y (0) =
y
et/100 y
−(t +1)2 /2
± (y − 1)
1 t/100
y
100 e
0
t/100
kg
min
The differential equation is both separable and linear. Let’s
find an integrating factor for the standard
form (SLF)
R linear
1
t/100 .
in the last equation above: µ = exp
100 dt = e
3
500 + Ce−t/100
500 + C [initial condition]
−500
500 − 500e−t/100
500 − 500e−10/100 ≈ 47.58 kg