Calculus (Math 2A)
posted by What is wrong with it? .
Gravel is being dumped from a conveyor belt at a rate of 20 cubic feet per minute. It forms a pile in the shape of a right circular cone whose base diameter and height are always the same. How fast is the height of the pile increasing when the pile is 19 feet high?
Recall that the volume of a right circular cone with height h and radius of the base r is given by v= 1/3pi(r)^2h
okay so this is what i did and got but it isnt right and i don't know why...
20 = pi/4 3(19)^2 * dH/dt
and i solved for dH/dt..and i got 80/1083pi which is 0.0235 but its wrong...why? thanks for the help!

You said the base diameter and height were the same
then if the height is 1
the radius is 9.5 not 19 
diameter = 19, radius = 9.5
I mean 
V= PI/3 (h)(h/2)^2=PI/12 h^3
dV/dt= PI/12 *3h^2 dh/dt
dh/dt= dV/dt*4/(PIh^2)
dh/dt= 20*4/(PI*19^2)=not your answer.
Take a close look at the difference 
Whoops, pay attention to what bobpursley did.