Water is running into an open conical tank at the rate of 9 cubic feet per minute. The tank is standing inverted, and has a height of 10 feet and a base diameter of 10 feet. At what rate is the radius of the water in the tank increasing when the radius is 2 feet?

Answer 1

draw a diagram. We have at a depth of h feet, a water surface with a diameter of h feet. So, h=2r

v = 1/3 π r^2 h = 2π/3 r^3

dv/dt = 2πr^2 dr/dt
so we have
9 = (2π)(4) dr/dt

Now you have dr/dt, the rate that the water is rising.

Answer 2

I mean, the rate at which the radius is increasing.