Determining the rate of change of a radius as a sphere loses volume

Question

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!

Answer 1

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.

Answer 2

You started correctly with $$ V = \frac{4}{3} \pi r^3 $$ But your time derivative misses $\dot{r}$, that is the quantity you want $$ \dot{V} = 4 \pi r^2 \dot{r} \iff \dot{r} = \frac{\dot{V}}{4\pi r^2} $$ You had $\dot{V} = -0.2 \,\mbox{m}^3/\mbox{min} = \mbox{const}$ and $r = 0.5 \,\mbox{m}$ which gives $$ \dot{r} = -0.064 \,\mbox{m}/\mbox{min} $$