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Math Help - Related rates problem

  1. #1
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    Related rates problem

    First, I want to say I am terrible with word problems... here is the problem:
    Sand falling from a hopper at a rate of 10pi ft^3/sec forms a conical pile whose radius is always equal to half its height. How fast is the radius of the pile increasing when the radius is 5ft.

    I believe the equation is V = (1/3)*pi*r^2*h
    here is what I know:

    the derivative of the equation is: (2/3)pi*rh

    r = (1/2)h = 5
    h= 2r = 10

    To plug all this in (2/3)pi (5)(10)

    = (100/3)pi

    =104.7198 ft^2/ sec

    Is this right?
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  2. #2
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    You would be correct if h were a constant.

    However, h is a function dependent on r. The correct derivative is therefore

    \frac{d}{dt}\left(\frac{1}{3}\pi r^2 h\right)=\frac{d}{dt}\left(\frac{2}{3}\pi r^3\right)=2\pi r^2\frac{dr}{dt}.

    Hope this helps.
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  3. #3
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    That being said, then it would be:
    (2/3)pi(5^2)

    = (50/3)pi

    which turns out to be 52.3599 ft/sec?
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  4. #4
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    Not quite. We are given that

    \frac{dV}{dt}=2\pi r^2\frac{dr}{dt}=10\pi.

    When r=5, this becomes

    2\pi\cdot5^2\cdot\frac{dr}{dt}=10\pi.
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  5. #5
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    I think I see what you are trying to say now. Get dr/dt on one side so it will be:

    10pi/50pi = .2 ft/sec?
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  6. #6
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    That is correct.
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