Gravel is being dumped from a conveyor belt at a rate of 30 ft^3/min. 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?
Vol = (1/3)pi (h^2/4)(h)
= (1/12)pi h^3
d(Vol)/dt = (1/4)pi h^2 dh/dt
30 = (1/12)pi(19^2) dh/dt
dh/dt=30*12/(pi*19^2)
dh/dt=0.317
My homework is online, and it says this is wrong... I tryed typing the whole equation and it still says it's wrong... can someone please check this and tell me what's wrong?
"30 = (1/12)pi(19^2) dh/dt "
Shouldn't it be
30 = (1/4)πh^2 dh/dt?
dV/dt = dV/dh * dh/dt
dh/dt = dV/dt / dV/dh
=30 / (πh^2/4)
I get 0.1058 ft/min.
Check the unit for the answer, as well as the format (decimals? how many? exact expressions?) the online homework systems usually instruct you the format expected.
To find the rate at which the height of the pile is increasing, we can use the formula you mentioned:
d(Vol)/dt = (1/4)pi h^2 dh/dt
However, it seems there is a small mistake in the calculation. Let's correct it:
d(Vol)/dt = (1/4)pi h^2 dh/dt
30 = (1/4)pi(19^2) dh/dt
dh/dt = 30 * 4 / (pi * 19^2)
dh/dt ≈ 0.088 ft/min
So, the correct rate at which the height of the pile is increasing when the pile is 19 feet high is approximately 0.088 ft/min.