Related rates question involving volume of cone

  • #1

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?
 

Answers and Replies

  • #2
SteamKing
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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.
 
  • #3
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
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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
LCKurtz
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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.
 

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