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Homework Help: Related Rates Problem that I am getting wrong.

  1. Jun 24, 2011 #1
    1. The problem statement, all variables and given/known data

    Sand is pouring from a pipe at the rate of 16 cubic feet per second. If the falling sand forms a conical pile on the ground whose altitude is always 1/4 the diameter of the base, how fast is the altitude increasing when the pile is 4 feet high?

    2. Relevant equations

    [itex]V = \frac{1}{3}\pi r^{2}h[/itex]

    3. The attempt at a solution

    Firstly, since I am given that the altitude is 1/4 of the diameter, I rearrange the volume formula.

    1/4(2)r = h
    1/2r = h
    r = 2h


    [itex]V = \frac{1}{3}\pi 2h^{2}h[/itex]

    [itex]V = \frac{1}{3}\pi 2h^{3}[/itex]

    Second, I'm given the rate of volume increase as 16, so [itex]\frac{dV}{dt}=16[/itex]

    Differentiation of both sides of the volume equation gives me:

    [itex]16 = 2\pi 2h^{2}\frac{dh}{dt}[/itex]

    So it follows that

    [itex]\frac{dh}{dt}= \frac{16}{2\pi h^{2}}[/itex]

    Solving this for h = 4 gives me ~= .159155

    Which is incorrect.

    Can someone show me where I'm messing this up?

  2. jcsd
  3. Jun 24, 2011 #2


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    Homework Helper

    Here's the problem, I think. You need parentheses around the 2h, like this:
    [tex]V = \frac{1}{3}\pi (2h)^{2}h = \frac{1}{3}\pi 4h^{3} = \frac{4}{3}\pi h^{3}[/tex]

    Ignoring the mistake from earlier, I don't know where that extra 2 came from.

    Now that extra 2 disappeared.

    You really need to take care in typing in the equations.
    Last edited: Jun 24, 2011
  4. Jun 24, 2011 #3
    Check your differentiation of

    [tex]V \, = \, \frac{1}{3} \pi (2) h^{3}[/tex]

    I think you made a small mistake in there. All your work seems ok to that point.

    (P.S. sorry for basically a repeat of what the poster above pointed out. At the time I clicked on "reply" there were no responses posted)
    Last edited: Jun 24, 2011
  5. Jun 26, 2011 #4
    My problem was indeed the volume formula. I needed (2h)(2h).
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