Rising Sand Pile in a Cylinder: Calc. dV/dt & dH/dt

[Physics197]

Homework Statement

Sand is poured into a right circular cylinder of radius ½ m along its axis from above. Once sand completely covers the bottom, a right circular cone is formed on the top.
a. If 0.02 m3 of sand enters the container every minute, how fast is the top of the sand pile rising?
b. How fast is the sand rising along the side of the cylinder?

Homework Equations

dV/dt = 0.02
V of cone = 1/3[pi]r^2h
V of cylinder = [pi]r^2h

The Attempt at a Solution

First of all, would the top of the sand pile and the sand along the side of the cylinder be rising at the same rate?

V = 1/3[pi]r^2x + [pi]r^2y
x = height of cone, y = height of cylinder. therefore x + y = H
Rate of change of H is d(x+y)/dt

Im not even sure if I'm still on the right track.

 

[tiny-tim]

Hi Physics197!

(have a pi: π and try using the X² tag just above the Reply box )

First of all, would the top of the sand pile and the sand along the side of the cylinder be rising at the same rate?

Yes, once the cone is formed on the base (at base height zero), the same shape will be maintained …

any new sand can be taken to trickle down the sides of the cone randomly, with the top angle always the same (and you don't need to know what that is).

x = height of cone, y = height of cylinder. therefore x + y = H
Rate of change of H is d(x+y)/dt]

Yes, but as I said, x is (unknown and) constant.

 

[Physics197]

V = 1/3[pi]r^2x + [pi]r^2y

When solving this, would I just say that the height of the cone (x) would remain constant and when I take the derivative;

dV/dt = [pi]r²dy/dt

and solve for dy/dt.

Then take x+y=H

dy/dt = dH/dt

and say the overall height is changing at the same rate as the height around the sides?

or is there another way to prove this?

Because it seems weird that we actually solve the b) part before the a) part.

Thanks

 

[tiny-tim]

Yes, a) and b) seem to have the same answer.