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  1. #1
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    Question Help...

    The surface area of a cube is changing at a rate of 8 cm^2/s. How fast is the volume changing when the surface area is 60cm^2?

    How would i find the length and the rate of change of the length??
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Tascja View Post
    The surface area of a cube is changing at a rate of 8 cm^2/s. How fast is the volume changing when the surface area is 60cm^2?

    How would i find the length and the rate of change of the length??
    Volume of cube, of side length l is:

    V=l^3.

    Surface area of the same cube is:

    A=6 \, l^2

    So:

    \frac{dV}{dt}=3\,l^2\, \frac{dl}{dt}

    and:

    \frac{dA}{dt}=12\,l\, \frac{dl}{dt}.

    Now when the area is 60\ \rm{cm^2} l^2=10\ \rm{cm^2}, so:

    \frac{dV}{dt}=30\, \frac{dl}{dt}.

    Also:

    <br />
\frac{dA}{dt}=12 \sqrt{10}\, \frac{dl}{dt}=8\,\rm{cm^2/s}<br />
,

    so:

    \frac{dl}{dt}=\frac{2}{3\, \sqrt{10}},

    hence:

    \frac{dV}{dt}=\frac{30\times 2}{3\,\sqrt(10}=2\,\sqrt{10}

    RonL
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