The radius of a cone increases at the rate of 4 cm/sec and the altitude is decreasing at the rate of 3 cm/sec. Find the rate of change of the lateral surface area when the radius is 3 cm and the altitude is 4 cm.
Let $r= 3 \; cm =$ radius of the cone; $h = 4 \; cm =$ height (altitude) of the cone;
Given: $\dfrac{dr}{dt} = 4 \; cm/s$; $\dfrac{dh}{dt} = -3 \; cm/s$
$\ell = \sqrt{r^2 + h^2} =$ slant height of the cone
Lateral surface area of the cone $=A = \pi r \ell = \pi r \sqrt{r^2 + h^2}$
$\begin{aligned}
\dfrac{dA}{dt} & = \pi \sqrt{r^2+h^2} \times \dfrac{dr}{dt} + \pi r \times \dfrac{1}{2 \sqrt{r^2 + h^2}} \times \left(2r\dfrac{dr}{dt}+2h\dfrac{dh}{dt}\right) \\
& \\
& = \pi \dfrac{dr}{dt} \left[\sqrt{r^2 + h^2} + \dfrac{r^2}{\sqrt{r^2 + h^2}}\right] + \dfrac{\pi r h}{\sqrt{r^2 + h^2}} \dfrac{dh}{dt} \\
& \\
& = \pi \times 4 \left[\sqrt{3^2 + 4^2}+\dfrac{3^2}{\sqrt{3^2+4^2}}\right] + \dfrac{\pi \times 3 \times 4}{\sqrt{3^2+4^2}} \times \left(-3\right) \\
& \\
& = \dfrac{136 \pi}{5} - \dfrac{36\pi}{5} \\
& \\
& = 20 \pi \; cm^2 / s
\end{aligned}$