Spherical balloon: rate of change of the surface

Text

The volume of a spherical balloon grows at $6\ \mathrm{cm^{3}/s}$. How fast is its surface area increasing when the radius is $5$ cm?

Solution

With $V=\tfrac{4}{3}\pi r^{3}$ and $S=4\pi r^{2}$, differentiating with respect to time: $\frac{dV}{dt}=4\pi r^{2}\frac{dr}{dt},\qquad \frac{dS}{dt}=8\pi r\frac{dr}{dt}.$ Dividing gives $\dfrac{dS}{dt}=\dfrac{2}{r}\dfrac{dV}{dt}$. With $\dfrac{dV}{dt}=6$ and $r=5$: $\frac{dS}{dt}=\frac{2}{5}\cdot6=2.4\ \mathrm{cm^{2}/s}.$ $\boxed{\dfrac{dS}{dt}=2.4\ \mathrm{cm^{2}/s}}$