Rate of Change

• Oct 23rd 2012, 04:19 PM
smgrojean
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)
• Oct 23rd 2012, 06:07 PM
Prove It
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*}.