rate of change/water poured into cone Water is poured into an inverted cone at a rate of π-units per second. If the radius of the base of the cone is r and its height is 2r, what is the rate at which the depth of the water is changing when the height of the water is 1/2r?
Let h, and R  be the height of the water and the radius of the water surface. Then:

h/2r = R/r ==> R = h/2.

The volume V of the water in the cone is: V = 1/3*pi*R^2*h = 1/12*pi*h^3. 

So: pi = dV/dt = 1/4*pi*h^2*dh/dt = 1/4*pi*(1/(4*r^2))*dh/dt ==> dh/dt = 16r^2
Let h, and R  be the height of the water and the radius of the water surface. Then:

h/2r = R/r ==> R = h/2.

The volume V of the water in the cone is: V = 1/3*pi*R^2*h = 1/12*pi*h^3. 

So: pi = dV/dt = 1/4*pi*h^2*dh/dt = 1/4*pi*(1/(4*r^2))*dh/dt ==> dh/dt = 16r^2