Problem:
A spherical balloon leaks $0.2\mathrm m^3 / \mathrm{min}$.
How fast does the radius of the balloon decrease the moment the radius is $0.5\mathrm m$?
My progress:
Since we're dealing with the rate of change of the volume, I set up a function for volume wrt. the radius, which would be $$V(r) = \frac43\pi r^3$$
Then I differentiated it, and got $$V'(r) = 4\pi r^2$$
Now, I don't know quite how to use the first piece of information in the problem. Am I going to need to find a $V(t)$ here? As far as I can tell, we can say that $$V'(t) = -0.2$$ but how do I piece all this together? (Assuming I'm right so far.)
Any help appreciated!
Best Answer
At all times, $V=\frac43 \pi r^3$. So $\frac{dV}{dt}=4\pi r^2 \frac{dr}{dt}$.
At the moment of interest, you have a value for the radius and a value for $\frac{dV}{dt}$ (be careful here--the volume is decreasing). This should allow you to plug into the derivative equation above and find $\frac{dr}{dt}$ at the moment of interest.