I just thought of this question and wondered if there was a nice way to solve it.
Question: The volume of a sphere is $100$ cubic metres and is losing $2\mathrm m^3/\mathrm s$. At what rate is the radius changing at $t = 2\mathrm{s}$?
I dont think I have a nice way of solving this,
but I thought about finding the volume after 2 seconds, which is 100-2x2 = 96 cubic metres. Then the radius, at this time.
Then find the volume at t = 2.00001 seconds, then the new radius at this time.
Then I use:
(new radius - old radius)/(2.00001 - 2)
to find the rate of change of the radius at t = 2s.
I believe there are better ways to solve this type of question. Thanks
Your methodology should produce an approximately correct result, but in practice most mathematicians use calculus to solve this type of problem.
Call the volume V and the time from the beginning of deflation $t=0$ then we have the rate of change of volume $$ {dV\over dt}=-2 $$ Now $V=\frac43\pi r^3$ where $r$ is the radius, so ${dV\over dr}=4\pi r^2$. The rate of change of the radius with time, ${dr\over dt}$, can be got from $${dr\over dt}={dV\over dt}/{dV\over dr}=-{1\over2\pi r^2}.$$ At $t=2$ seconds V=96 so calculate $r=\sqrt[3]{3 V/4\pi}$ (notice that is a cube root, use the $x^y$ button with $y=1/3$ on your calculator) and insert that into the above equation.
That should produce about the same result as your suggested methodology, so as a check why not try the above out, and try your method out, and see if you get the same result both ways.