Calculus
posted by help .
Sand is poured onto a level piece of ground at the rate of 0.25 m^3/min and forms a conical pile whose height is equal to its base diameter. How fast is the height increasing at the instant when the height is 0.50 m? V = 1/3(pi)(r^2)h
I got 3/pie, is that correct?

I got 4/π
let the radius be r, then the height is 2r
V= (1/3)πr^2 h
= (1/3)πr^2(2r)
= (2/3)πr^3
dV/dt = 2π r^2 dr/dt
when 2r=.5
r = .25 and dV/dt = .25
1/4 = 2π(1/16) dr/dt
times 8
2 = π dr/dt
dr/dt = 2/π
d(2r)/dt = d(height)/dt = 4/π 
Thank you!

Where did you get 8 from?