Math 1100-6
November 10, 2008
Page 1
Math 1100: Assignment #5 Solutions
Due: Friday Nov 7, 2008 @ 10:30am
No late assignments will be accepted
Please complete the following questions for the assignment. Please be mindful of “good
presentation” criteria outlined in the syllabus as you’re preparing to hand it in.
§11.4
9. Given
x2 + y 2 = z 2
(x, y, z) = (3, 4, 5),
dx
dz
= 10,
=2
dt
dt
dy
We want to find
at the above coordinates. To this end, we (implicitly) differentiate
dt
the main equation to get the following:
dx
dy
+y
dt
dt
dy
3 · 10 + 4
dt
dy
dt
x
→
15. We are given the volume is V = x3 and that
dV
dt
→
⇒
= z
dz
dt
= 5·2
=
−20
= −5
4
dV
dx
= 64. We want to find
when x = 6in.
dt
dt
dx
dt
dx
64 = 3 · 36
dt
16
dx
=
≈ 0.592 in/s
dt
27
= 3x2
dx
16. Given
= 8ft/min, we want to find the rate of change of the surface area (S = 6x2 )
dt
when the edges are 24ft long.
→
dS
dt
dS
dt
= 12x
dx
dt
= 12 · 24 · 8 = 2304 ft2 /min
√
dp
dx
20. Given p = 40 + 100 2x + 9 and
= $1/month, we want to find
when x = 20.
dt
dt
100 dx
dp
= √
dt
2x + 9 dt
100 dx
→
1 = √
49 dt
dx
7
=
dt
100
Math 1100-6
November 10, 2008
Page 2
Therefore, supply is changing at a rate of 0.07 units/month.
32. Let L be the length of the cable and x be the distance of the boat to the dock. We know
dx
dL
that
= −3ft/min and we want to find
when x = 12ft. Note that L is related to x
dt
dt
by
L2 = x2 + 52
√
At x = 12ft:
L =
169 = 13
dL
dx
→
L
= x
dt
dt
dx
L dL
⇒
=
dt
x dt
13
=
(−3) = −3.25ft/min
12
So the boat is approaching the dock at -3.25 ft/min when x = 12ft.
38. The volume of a cylinder is V = πr2 h and we are given that the radius is 18in
dV
dh
(CONSTANT), while
= 200in3 /min. We want to find
when h = 30in.
dt
dt
dh
dV
= πr2
dt
dt
dh
50
→
=
dt
81π
§11.5
8. We are given:
p
(p + 1) q + 1 = 1000
a) To find the elasticity when p = 39, we first need to find q...
p
40 q + 1 = 1000
p
q + 1 = 25
→
q = 624
To find the elasticity, we must compute
dq
:
dp
p
p + 1 dq
(1) q + 1 + √
2 q + 1 dp
20 dq
→
25 +
25 dp
dq
⇒
dp
= 0
= 0
= −
Therefore, the elasticity is
p dq
q dp
39 125
=
·
624 4
≈ 1.95
η = −
625
125
=−
20
4
Math 1100-6
November 10, 2008
Page 3
b) This is elastic because η > 1.
c) An increase in price will decrease revenue.
9. We are given:
p =
1
ln
2
µ
5000 − q
q+1
¶
=
1
[ln(5000 − q) − ln(q + 1)]
2
a) To find the elasticity when p = 3.71 and q = 2 we must compute
dq
1
, so:
=
dp
dp
dq
·
¸
1
1
1
=
−
−
2
5000 − q q + 1
5001
= −
2(5000 − q)(q + 1)
5001
= −
2(4998)(3)
dp
dq
dp
dq
Therefore, the elasticity is
p dq
q dp
3.71 2(4998)(3)
·
=
2
5001
≈ 11.123
η = −
b) This is elastic because |η| > 1.
10. We are given:
q =
5000
−1
1 + e2p
To find the elasticity when p = 1 and q = 595 we must compute
→
dq
dp
dq
dp
10000 e2p
1 + e2p
1000 e2
= −
1 + e2
= −
Therefore, the elasticity is
p dq
q dp
10000 e2
=
595(1 + e2 )
≈ 14.80331
η = −
dq
.
dp
dq
. Note that
dp
Math 1100-6
November 10, 2008
Page 4
13. We are given:
Demand F’n: p = 30 − q
Supply F’n: p = 6 + 2q
Let t be the tax on an individual item. We want to find t = t(q) by finding the new
equilibrium between supply and demand, i.e.
Set 6 + 2q + t = 30 − q
t = 24 − 3q
With this equation, the tax revenues are:
= tq = 24q − 3q 2
T
dT
= 0 = 24 − 6q
dq
→
q = 4
Set
Therefore, the tax required is t = 12.
16. We are given:
Demand F’n: p = 2100 − 3q
Supply F’n: p = 300 + 1.5q
Let t be the tax on an individual item. We want to find t = t(q) by finding the new
equilibrium between supply and demand, i.e.
Set 300 + 1.5q + t = 2100 − 3q
t = 1800 − 4.5q
With this equation, the tax revenues are:
T
= tq = 1800q − 4.5q 2
dT
= 0 = 1800 − 9q
dq
→
q = 200
Set
Therefore, the tax required is t = 900 and the tax revenue is T = 180, 000.
20. We are given:
Demand F’n: p = 1000 − q
1 2
Supply F’n: p =
q + 2.5q + 920
30
Let t be the tax on an individual item. We want to find t = t(q) by finding the new
equilibrium between supply and demand, i.e.
Set
1 2
q + 2.5q + 920 + t = 1000 − q
30
t = 80 − 3.5q −
1 2
q
30
Math 1100-6
November 10, 2008
Page 5
With this equation, the tax revenues are:
= tq = 80q − 3.5q 2 −
T
1 3
q
30
dT
1
= 0 = 80 − 7q − q 2
dq
10
√
−70 ± 702 + 4 · 800
→
q =
2
= 10
Note: We require q ≥ 0
Set
Therefore, the tax required is t = $41.67 and the tax revenue is T = $416.67.
§12.1
9.
Z
(33 + x13 ) dx = 27x +
12.
Z
16.
2
3 5
(8 + x 3 ) dx = 8x + x 3 + C
5
Z
(17 +
22.
2
5
26.
Z ³
x14
+C
14
Z
√
2 5
x3 ) dx = 17x + x 2 + C
5
3
4 1
x− 2 dx = − x− 2 + C
5
1
3x8 + 4x−8 − 5x− 5
´
dx =
42.
1 9 4 −7 25 4
x − x + x5 + C
3
7
4
Z
R =
M R dx
= −0.003x2 + 36x
R(75) = $2683.13
45.
dx
dt
→
=
x(t) =
x(52) =
1 3
t4
600
1 7
t4
1050
7
1
52 4 ≈ 0.959
1050
Math 1100-6
November 10, 2008
Page 6
49.
dB
dt
= −2.1136t + 8.2593
a) The trust fund will start to decrease after t =
8.2593
2.1136
= 3.9077 years.
b) Integrating to solve for B(t), we get:
B(t) = −1.0568t2 + 8.2593t + C
Because B(0) = $74.07billion, we set C = 74.07, so
B(t) = −1.0568t2 + 8.2593t + 74.07
c) There will be no more money in the account when B(t) = 0, i.e. when
→
0 = −1.0568t2 + 8.2593t + 74.07
√
−8.2593 ± 8.25932 + 4 · 1.0568 · 74.07
t =
−2 · 1.0568
= 13.147 years
There will no longer be any money in the account after approx 13.147 years (i.e. early in
2013).