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Math Help - Related Rates Involving A Cone

  1. #1
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    Related Rates Involving A Cone

    A cone is 40 cm in height and 20 cm in diameter at widest point. If it is being filled at a rate of 10pi mL/min, how fast is its height changing when it is half filled?

    So far, I've got that r/h = 1/4, so r = h/4.

    And I've got that dV/dt = pi/48 x 3h^2 x dh/dt

    However, I'm stumped on how to find the height when the cone is half filled.

    Any help is appreciated.
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  2. #2
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    Quote Originally Posted by Shapeshift View Post
    A cone is 40 cm in height and 20 cm in diameter at widest point. If it is being filled at a rate of 10pi mL/min, how fast is its height changing when it is half filled?

    So far, I've got that r/h = 1/4, so r = h/4.

    And I've got that dV/dt = pi/48 x 3h^2 x dh/dt

    However, I'm stumped on how to find the height when the cone is half filled.

    Any help is appreciated.
    volume of full cone ... V = \frac{\pi}{48} \cdot 40^3

    half full ... V = \frac{\pi}{96} \cdot 40^3 = \frac{\pi}{48} h^3

    h^3 = \frac{40^3}{2}

    h = \frac{40}{\sqrt[3]{2}}
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  3. #3
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    Quote Originally Posted by Shapeshift View Post
    A cone is 40 cm in height and 20 cm in diameter at widest point. If it is being filled at a rate of 10pi mL/min, how fast is its height changing when it is half filled?

    So far, I've got that r/h = 1/4, so r = h/4.

    And I've got that dV/dt = pi/48 x 3h^2 x dh/dt

    However, I'm stumped on how to find the height when the cone is half filled.

    Any help is appreciated.
    ..

    half filled means volume is halved.
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