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Math Help - word problem invovling rate change

  1. #1
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    word problem invovling rate change

    Here's my problem

    Sand falls out of the end of a slurry at the rate of 60 cc/sec. The pile forms a circular cone, the ratio of whose base diameter to height is 3. When the pile is of height 10 cm., at what rate is the height of the pile increasing?

    So far I have V=\frac{1}{3}(\pi)r^2h where r=\frac{2h}{3} but then I can't seem to figure out the rest of the problem. Any help is greatly appreciated.
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  2. #2
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    It is given that the ratio of whose base diameter to height is 3
    So d/h = 3
    2r/h = 3
    r = 3h/2.
    Volume of the cone is
    V = 1/3*π*r^2*h. Substitute r = 3h/2 and find dV/dt to find dh/dt.
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  3. #3
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    Quote Originally Posted by sa-ri-ga-ma View Post
    It is given that the ratio of whose base diameter to height is 3
    So d/h = 3
    2r/h = 3
    r = 3h/2.
    Volume of the cone is
    V = 1/3*π*r^2*h. Substitute r = 3h/2 and find dV/dt to find dh/dt.
    So I'm getting the answer wrong still, maybe you can check my work?

    V=\frac{1}{3}(\pi)(\frac{3h}{2})^2*h <br />
    \\<br />
\\<br />
=\frac{1}{3}(\pi)\frac{9h^2}{4}*h
    <br />
\\<br />
\\<br />
=\frac{9\pi}{12}(h)^3
    <br />
\frac{dV}{dt}=\frac{9\pi}{4}h^2
    <br />
\\<br />
60=\frac{9\pi}{4}(10^2)\frac{dh}{dt}

    I'm getting a negative answering which I know isn't right, Did I setup my calculations right or did I do some mathematical error?
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  4. #4
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    Quote Originally Posted by ascendancy523 View Post
    So I'm getting the answer wrong still, maybe you can check my work?

    V=\frac{1}{3}(\pi)(\frac{3h}{2})^2*h <br />
    \\<br />
\\<br />
=\frac{1}{3}(\pi)\frac{9h^2}{4}*h
    <br />
\\<br />
\\<br />
=\frac{9\pi}{12}(h)^3
    <br />
\frac{dV}{dt}=\frac{9\pi}{4}h^2
    <br />
\\<br />
60=\frac{9\pi}{4}(10^2)\frac{dh}{dt}

    I'm getting a negative answering which I know isn't right, Did I setup my calculations right or did I do some mathematical error?
    Where is the negative answer?
    dh/dt = 60*4/9π*10^2.
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