Inflating spherical balloon

Text

A spherical balloon is inflated with air at the constant rate of $100\ \text{cm}^3/\text{s}$. When the radius is $r=5\text{ cm}$, how fast is the radius increasing? And the surface area?

Solution

Volume: $V=\tfrac43\pi r^{3}$, so $\dfrac{dV}{dt}=4\pi r^{2}\dfrac{dr}{dt}$. $\dfrac{dr}{dt}=\dfrac{100}{4\pi\cdot25}=\dfrac{1}{\pi}\approx0.318\text{ cm/s}$. Surface: $S=4\pi r^{2}$, $\dfrac{dS}{dt}=8\pi r\dfrac{dr}{dt}=8\pi\cdot5\cdot\dfrac1\pi=40\text{ cm}^2/\text{s}$. $\boxed{\dfrac{dr}{dt}=\tfrac1\pi\approx0.318\ \text{cm/s};\quad \dfrac{dS}{dt}=40\ \text{cm}^2/\text{s}}$