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Math Help - Rate of Change

  1. #1
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    Rate of Change

    In my class we recently went over rates of change and while I understood a few of the problems in the homework I just took an exam and really bombed on this one. Where do you even start with this one?

    I put down that f'(x) = 3*pi*R^2 (marked wrong for the 3) and that dr/dt = 3*pi*R^2 / 72*pi

    but then I didn't know what to do next (assuming that was the right place to start)
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  2. #2
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    Re: Rate of Change

    The volume of the snowball is increasing at \displaystyle \begin{align*} 72\pi \textrm{ cm}^3 / \textrm{min} \end{align*}, so \displaystyle \begin{align*} \frac{dV}{dt} = 72\pi \end{align*}, as long as t is measured in minutes.

    You also know that \displaystyle \begin{align*} V = \frac{4}{3}\pi \, r^3 \end{align*}, which means \displaystyle \begin{align*} \frac{dV}{dr} = 4\pi \, r^2 \end{align*}.

    By the chain rule, \displaystyle \begin{align*} \frac{dV}{dt} = \frac{dV}{dr} \cdot \frac{dr}{dt} \end{align*}, so

    \displaystyle \begin{align*} 72\pi &= 4\pi \, r^2 \, \frac{dr}{dt} \\ 72 &= 4r^2 \, \frac{dr}{dt} \\ 18 &= r^2 \,\frac{dr}{dt} \\ \frac{18}{r^2} &= \frac{dr}{dt} \end{align*}

    So now it's just a case of evaluating what the radius is when the volume is \displaystyle \begin{align*} 36\pi \textrm{ cm}^3 \end{align*} and substituting in to evaluate \displaystyle \begin{align*} \frac{dr}{dt} \end{align*}.
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