Asked by Anmy
Here's one we haven't worked with before: A circular oil slick of uniform thickness is caused by a spill of 1 cubic meter of oil. The thickness of the oil is decreasing at the rate of 0.1 cm/hr as the slick spreads. (Note: 1 cm = 0.01 m.) At what rate is the radius of the slick increasing when the radius is 8 meters? (You can think of this oil slick as a very flat cylinder; its volume is given by V = pi(r^2)h, where r is the radius and h is the height of this cylinder.)
Answers
Answered by
oobleck
if the thickness is h, then
dh/dt = -0.1
v = πr^2 h = 10,000 cm^3 so r^2 = 10000/(πh)
dv/dt = 2πrh dr/dt + πr^2 dh/dt
since v is constant, dv/dt=0, and so
πr^2 dh/dt = -2πrh dr/dt
r dh/dt = -2h dr/dt
when r=8,
8(-0.1) = -2(100/√(8π)) dr/dt
dr/dt = -√(2π)/125 cm/hr
dh/dt = -0.1
v = πr^2 h = 10,000 cm^3 so r^2 = 10000/(πh)
dv/dt = 2πrh dr/dt + πr^2 dh/dt
since v is constant, dv/dt=0, and so
πr^2 dh/dt = -2πrh dr/dt
r dh/dt = -2h dr/dt
when r=8,
8(-0.1) = -2(100/√(8π)) dr/dt
dr/dt = -√(2π)/125 cm/hr
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