# Rate of Change Calculus question for a sphere.

I am having a problem getting started on this problem.

Rate of inflation of a balloon. A spherical balloon is inflated at a rate of 10 cubic cm/min. At what rate is the diameter of the balloon increasing when the balloon has a diameter of 5 cm.

For the life of me I cannot figure out what formula to begin with so I can find dd/dt (this is an assumption that it is related to diameter) not sure it could be related to anything else. I am not looking for the problem to be solved just help getting started.

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## 2 Answers

HINT: volume of the balloon($V$) = $\frac{4}{3}\pi r^3=\frac{4}{3}\pi {(\frac{D}{2})}^3=\frac{1}{6}\pi D^3$

and you are given $\frac{dV}{dt}$ and need to find $\frac{dD}{dt}$

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thank you yes this helped. –  samack Sep 25 '12 at 4:01
@samack:i am glad :) –  Aang Sep 25 '12 at 4:02

We need a relationship between the diameter $x$ and the volume $V$. The standard formula for volume is $$V=\frac{4}{3}\pi r^3,$$ where $r$ is the radius. In terms of diameter, since $r=x/2$, we have $$V=\frac{\pi x^3}{6}.$$ It is I think best to differentiate immediately, using the Chain Rule. We get $$\frac{dV}{dt}=\frac{\pi x^2}{2}\frac{dx}{dt}.$$ Now can you finish?

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thank you yes this helped –  samack Sep 25 '12 at 4:04