Related rates question involving volume of cone

1. Nov 21, 2014

needingtoknow

1. The problem statement, all variables and given/known data

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?

2. Nov 21, 2014

SteamKing

Staff Emeritus
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.

3. Nov 21, 2014

needingtoknow

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.

4. Nov 21, 2014

Staff: 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.

5. Nov 21, 2014

LCKurtz

The volume of a cone is $V=\frac 1 3 \pi r^2 h$. You must avoid putting in the instantaneous values before you differentiate to get the related rates equation.