I've encountered this calculus problem:
1-A spherical balloon is being inflated in such a way that its radius is increasing at the constant rate of 3 cm/min. If the volume of the balloon is 0 at time 0, at what rate is the volume increasing after 5 minutes?
So I set the radius function as $r(t) = 3t$ and the volume function as $v(r) = 4\pi r^3/3$. I found the derivative to be $4\pi r^2$, used the chain rule and found the balloon to be increasing at a rate of $2700$ cm/min after $5$ minutes. Is that right?
Thanks!
Best Answer
Just to summarize what's been said in the comments:
You've obviously got the right volume as a function of the radius. Differentiating that function with respect to $t$ yields $$\frac{dv}{dt}=4\pi r^2\frac{dr}{dt}$$ Now we need to evaluate this expression at $t=5\mathrm{\:min}$. We can tell from the problem statement that $$r(5\mathrm{\:min})=15\mathrm{\:cm}$$ We can now plug this into our expression for $dv/dt$ along with the given value of $dr/dt$. Keeping the units attached to all the quantities is helpful here as a sanity check: $$\frac{dv}{dt}(5\mathrm{\:min})=4\pi(15\mathrm{\:cm})^2(3\mathrm{\:cm/min})=2700\pi\mathrm{\:cm^3/min}$$