# Related rates question involving volume of cone

## Homework Statement

Sand falls from a conveyor belt at the rate of 10m^3/min onto the top of a conical pile. The height of the pile is always 3/8ths of the base diameter.

How fast is the radius changing when the pile is 4 m high?

3. The Attempt at a Solution

V = pir^2 (4/3) -- volume of a cone

dv/dt = (4pi/3)(2r)(dr/dt)
10 = (4pi/3)(2)(16/3)(dr/dt)
10 = (128pi/9)(dr/dt)
dr/dt = 0.224

The answer is 0.1119 m/sec. What am I doing wrong?

SteamKing
Staff Emeritus
Homework Helper

## Homework Statement

Sand falls from a conveyor belt at the rate of 10m^3/min onto the top of a conical pile. The height of the pile is always 3/8ths of the base diameter.

How fast is the radius changing when the pile is 4 m high?

3. The Attempt at a Solution

V = pir^2 (4/3) -- volume of a cone

You might want to confirm this formula for the volume of a cone. Remember, volume has units of L3, and there are units of L2 here. There is also no accounting for the height of the cone in this formula.

The formula is correct I just mislabelled it. I meant to say that V = pir^2 (4/3) -- volume of a cone when h = 4.

mfb
Mentor
Might be right (I did not check the prefactor), but the height is not constant. It changes together with r. You cannot consider one change and ignore the other one.

LCKurtz