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Math Help - Rates Of Change

  1. #1
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    Rates Of Change

    Water is poured into an upside down open cone at a constant rate of 24cm^3 per second. Depth Of water at any time t seconds is hcm. What is the rate of increase of the area of the surface area of the surface S of the liquid when the depth is 16cm.

    Surface Area S is the circukar area inside the cone. the radius of the base is 30cm and the heightof the cone is 30cm

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  2. #2
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    Quote Originally Posted by MC Squidge View Post
    Water is poured into an upside down open cone at a constant rate of 24cm^3 per second. Depth Of water at any time t seconds is hcm. What is the rate of increase of the area of the surface area of the surface S of the liquid when the depth is 16cm.

    Surface Area S is the circukar area inside the cone. the radius of the base is 30cm and the heightof the cone is 30cm

    As the volume of water is:

    V(t)=\frac{1}{3} \pi (h(t))^3

    we have:

    \frac{dV}{dt}=\pi (h(t))^2 \frac{dh}{dt}=24

    so:

    \frac{dh}{dt}=\frac{24}{\pi h^2}

    At any time we have:

    S(t)=\pi r^2=\pi h^2

    so:

     <br />
\frac{dS}{dt}= 2 \pi h \frac{dh}{dt}=\frac{48}{h}<br />

    RonL
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