Quiz #11

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1
151 WebCalc Fall 2002-copyright Joe Kahlig
Quiz #11
MATH 151 Section
Name:
October 11, 2002
This is due at the start of class on Thursday.
1. A metal rod has the shape of a right circular cylinder. As it is being heated, its length is
increasing at a rate of 0.005cm/min and its diameter is increasing at 0.002cm/min. At what
rate is the volume changing when the rod has length 40cm and diameter 3cm?
The volume of this rod is given by the formula V = πr 2 l. Since the diameter is given in the
problem and d = 2r or d2 = r, we can rewrite the volume as V = π4 d2 l. Given in the problem
dl
is dt
= 0.005, dd
dt = .002, l = 40, and d = 3. Now take a derivative and plug in the known
values.
dV
π
dd π 2 dl
= dl ∗
+ d
dt
2
dt
4 dt
dV
π
π
= ∗ 3 ∗ 40 ∗ .002 + ∗ 9 ∗ .005
dt
2
4
dV
= 0.41233 cm/min
dt
2. A person flying a kite holds the string 5 ft above the ground level, and the string is released at
a rate of 2ft/sec as the kite moves horizontally at an altitude of 105ft (see figure). Assuming
there is no sag in the string, find the rate at which the kite is moving when 125 ft of string
has been released.
Answer:
Given in the problem is x = 100, z = 125,
dz
dx
dt = 2, and dt = 0 (since the kite is only
moving horizontally).
Using the formula x2 + y 2 = z 2 and
105 feet
solving for y gives y = 75. Now take
a derivative and plug in the known values.
2x dx
dt
+
2y dy
dt
=
z
x
y
5 feet
2z dz
dt
dy
2 ∗ 125 ∗ 2
=
= 3.3333f t/sec
dt
2 ∗ 75
3. Water is leaking out of an inverted conical tank at a rate of .01m3 /min. The tank has height
of 8m and the diameter at the top is 6m. Find the rate of change of the radius (located at
the water level of the tank) when the height of the tank is 3m. V = 13 πr 2 h
From the problem we know dV
dt = −0.01 and h = 3
The other information is labeled on the picture. The
relationship between the radius and the height is given
by: hr = 83 or h = 83 r
V = 13 πr 2 h = 13 πr 2 38 r = 89 πr 3
8
dV
2 dr
dr = 3 πr dt
note: if h = 3 then r =
2
9
8
−0.01 = π
3
8
Solving for
dr
dt
9
8
dr
dt
gives
dr
dt
= −.0009431m/min
6m
3m
r
8m
h
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