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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 $\displaystyle l$ is:

    $\displaystyle V=l^3$.

    Surface area of the same cube is:

    $\displaystyle A=6 \, l^2$

    So:

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

    and:

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

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

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

    Also:

    $\displaystyle
    \frac{dA}{dt}=12 \sqrt{10}\, \frac{dl}{dt}=8\,\rm{cm^2/s}
    $,

    so:

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

    hence:

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

    RonL
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