gravel is being dumped from a conveyor belt at a rate of 15 ft^3/hr and its coarseness is such that it forms a pile in the shape of an inverted right cone whose height is three times its base radius. How fast is the height of the pile increasing when the pile has a height of 10ft

Answer 1

height ------- h
base radius --- h/3

V = (1/3)π r^2 h
= (1/3)π (h^2/9)(h) = (π/27) h^3

dV/dt = (π/9) h^2 dh/dt
for the given data:
15 = π/9 (100)dh/dt
dh/dt = 135/(100π) ft/hr

Answer 2

How did you get "dh/dt = 135/(100π) ft/hr?"

That's the only part I'm confused about.

Thanks for your help btw.

Answer 3

from
15 = π/9 (100)dh/dt

I multiplied both sides by 9/(100π)
which left me with dh/dt on the one side
and 15(9/(100π) ) or 135/(100π) on the other side

Answer 4

oh okay, I ended up getting .43ft/hr. I'm 99% sure that's the correct answer.

Thanks for clearing that up for me, I'm really sucking at related rates, so I'm doing all the problems I can.