Related rates calculus problem about oil spill?

Question

We assume that an oil spill is being cleaned up by deploying bacteria that consume the oil at 2 cubic feet per hour. The oil spill itself is modeled in the form of a very thin cylinder as depicted below. Its height h is the thickness of the oil slick. When the thickness of the slick is 0.005 feet, the cylinder is 500 feet in diameter. If the height is decreasing at 0.004 feet per hour, at what rate is the area covered by the slick changing. (That is, the area of the base disc of the cylinder)

Answer 1

Givens

dV/dt=-2
h=.005
r=500
dh/dt=-.004

This is a two part question. First know that

V=πr^2h

Therefore dV/dt=πr^2dh/dt+2πrhdr/dt

-2=π((500)^2*(-.004)+2*500*.005*dr/dt)

dr/dt=199.87 ft/hr

now consider the area of the circle

A=πr^2

dA/dt=2πrdr/dt

dA/dt=2π*500*199.87=627919 sq ft/hr. The bacteria need to eat faster.