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Math Help - Rate of change

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    Rate of change

    The edge of a cube is increasing at the rate of 2 cm/s. Find the rate of increase of the volume when each side is 8 cm
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    Quote Originally Posted by Punch View Post
    The edge of a cube is increasing at the rate of 2 cm/s. Find the rate of increase of the volume when each side is 8 cm
    I can't think of a pre-calc method for this one

    Let s be the side of the cube and let V be the volume.

    From the volume of a cube V = s^3

    it is known that \frac{ds}{dt} = 2

    We want to know \frac{dV}{dt}

    To get this use the chain rule:

    \frac{dV}{dt} = \frac{dV}{ds} \cdot \frac{ds}{dt}

    To find \frac{dV}{ds} differentiate the first equation

    \frac{dV}{ds} = 3s^2

    \frac{dV}{dt} = 3s^2 \times 2 = 6s^2

    To find the overall answer sub in s=8 in \frac{dV}{dt}

    I get an answer of 384 \text{cm}^3 \text{ s}^{-1}
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