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Math Help - Rates of Change

  1. #1
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    Exclamation Rates of Change

    Hi

    How would I do this question:

    Water is running into a conical reservoir, 15 m deep and 8 m in diameter at the top at the rate of 1 cubic m/min. What is the depth when the level is rising 9/32pi m/min?

    Thanks a lot.
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  2. #2
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    Sorry, this "9/32pi m/min" is velocity? Level of velocity?
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  3. #3
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    Quote Originally Posted by Sunyata View Post
    Hi

    How would I do this question:

    Water is running into a conical reservoir, 15 m deep and 8 m in diameter at the top at the rate of 1 cubic m/min. What is the depth when the level is rising 9/32pi m/min?

    Thanks a lot.
    Hi Sunyata,

    V_{cone}=\frac{{\pi}r^2h}{3}

    Write r in terms of h.
    If you split the cone vertically, you have a right-angled triangle,
    height H=15 and width R=4.

    Hence \frac{R}{H}=\frac{4}{15}

    R=\frac{4H}{15}

    As the water flows in, the height and radius of the water vary, but they
    still have the same ratio

    r=\frac{4h}{15}

    Hence

    v_{water}=\frac{\pi}{3}\frac{16h^2}{225}h=\frac{16  {\pi}h^3}{3(225)}

    As the water enters at 1\ m^3 per minute

    \frac{dv}{dt}=1=\frac{d}{dt}\frac{16{\pi}h^3}{3(22  5)}=\frac{16{\pi}}{3(225)}\frac{dh^3}{dt}

    \frac{dh^3}{dt}=\frac{3(225)}{16{\pi}}\ \Rightarrow\ 3h^2\frac{dh}{dt}=\frac{3(225)}{16{\pi}}

    Now substitute your value for \frac{dh}{dt}=\frac{9}{32{\pi}}

    to find "h squared" and take the square root.
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