Related rates question involving volume of cone

[needingtoknow]

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]

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.

 

[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.

 

[mfb]

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]

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.

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.