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.

To find the rate of change of the height of the pile (dH/dt), we need to use the volume formula for a right circular cone and the given information.

The volume of the cone is represented by the formula: V = (1/3)πr^2h

Given that the rate of change of the volume (dV/dt) is 20 cubic feet per minute, we can differentiate the volume formula with respect to time to relate the rate of change of volume with the rate of change of height.

dV/dt = (1/3)π(2rh dr/dt + r^2 dh/dt)

Since the question states that the base diameter and the height of the cone are always the same, r = h/2.

Therefore, substituting this relationship into the equation:

20 = (1/3)π((h/2)(dh/dt) + (h/2)^2(dh/dt))

Simplifying the equation further:

20 = (1/3)π((h/2)(dh/dt) + h^2/4(dh/dt))

Now, we can substitute the given height value of 19 into the equation:

20 = (1/3)π((19/2)(dh/dt) + 19^2/4(dh/dt))

20 = (1/3)π((19/2 + 19^2/4)(dh/dt))

Now, let's solve for dh/dt:

dh/dt = 20 / [(1/3)π((19/2 + 19^2/4)]

Evaluating this expression will give you the correct value for dh/dt when the pile is 19 feet high.