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Math Help - Related Rate problem

  1. #1
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    Related Rate problem

    Gravel is being dumped from a conveyor belt at a rate of 11 m^3/min The coarseness is such that it forms a pile in the shape of a cone whose base diameter and height are in the ratio of 5 to 2 . How fast is the height of the pile increasing when the pile is 6 metres high?

    <br />
\displaystyle v=\frac{1}{3}\pi*r^2h<br />

    <br />
\displaystyle v=\frac{1}{3}\pi(\frac{5}{4}h)^2h<br />

    <br />
\displaystyle \frac{dv}{dt}=\frac{\pi}{3}(\frac{5}{4}*h)^2+h(\fr  ac{\pi}{3}*2(\frac{5}{2}*h*\frac{5}{4}*\frac{dh}{d  t}))<br />

    \displaystyle \frac{dv}{dt}=11

    <br />
\displaystyle \frac{dh}{dt}=\frac{11-\frac{\pi}{3}(\frac{5}{4}*h)^2}{\frac{11}{3}*h(\fr  ac{10}{4}*h)*\frac{5}{4}}<br />

    When I finally plug in 6 for h, I get −.0587184655608
    which appears to be wrong!

    What am I doing wrong here?
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  2. #2
    MHF Contributor Unknown008's Avatar
    Joined
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    Mauritius
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    <br />
\displaystyle \frac{dv}{dt}=\frac{\pi}{3}(\frac{5}{4}*h)^2+h(\fr  ac{\pi}{3}*2(\frac{5}{2}*h*\frac{5}{4}*\frac{dh}{d  t}))

    This is not dv/dt, this is dv/dh as you differentiated V with respect to h.

    And you needn't apply the product rule here...

    V=\dfrac{1}{3}\pi(\frac{5}{4}h)^2h = \dfrac{25}{48} \pi h^3

    \dfrac{dV}{dh} = \dfrac{25}{16} \pi h^2

    Then use the chain rule:

    \dfrac{dV}{dt} = \dfrac{dV}{dh} \times \dfrac{dh}{dt}

    And solve for dh/dt at h = 6
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