Oil is leaking from a pipeline on the surface of a lake and forms an oil slick whose volume increases at a
constant rate of 2000 cubic centimeters per minute. The oil slick takes the form of a right circular cylinder with
both its radius and height changing with time.
At the instant when the radius of the oil slick is 100 centimeters and the height is 0.5 centimeter, the radius is
increasing at a rate of 2.5 centimeters per minute. At this instant, what is the rate of change of the height of the
oil slick with respect to time, in centimeters per minute?
V = πr^2 h
dV/dt = 2πr h dr/dt + πr^2 dh/dt
This is just like your other cylinder problem
To find the rate of change of the height of the oil slick with respect to time, we can use related rates. Let's denote the radius of the oil slick as r and the height as h.
We are given that the volume of the oil slick is increasing at a constant rate of 2000 cubic centimeters per minute. Since the oil slick is in the shape of a right circular cylinder, the volume can be expressed as V = πr^2h.
Differentiating both sides of the equation with respect to time (t) using the chain rule, we get:
dV/dt = 2πrh(dr/dt) + πr^2(dh/dt)
We are given that dr/dt = 2.5 cm/min and we want to find dh/dt when r = 100 cm and h = 0.5 cm.
Substituting the known values into the equation, we have:
2000 = 2π(100)(0.5)(2.5) + π(100)^2(dh/dt)
Simplifying the equation, we get:
2000 = 1250π + 10,000π(dh/dt)
Rearranging the equation to solve for dh/dt, we have:
dh/dt = (2000 - 1250π) / (10,000π)
Now we can calculate the value of dh/dt:
dh/dt = (2000 - 1250π) / (10,000π)
Using an approximate value of π = 3.14, we can calculate the numerical value of dh/dt.