A cylinder is heated so that its radius remains twice of its height at any moment. Find the rate of increase of its total surface area and volume, when the radius is 3 cm and the radius increases at the rate 2 cm/s.
Let height of the cylinder $= h \; cm$
Radius of cylinder $= r = 2h \; cm$
$\therefore$ $h = \dfrac{r}{2}$
Given: $r = 3 \; cm$ $\;\; \cdots$ (1)
%$\therefore$ $h = \dfrac{r}{2} = \dfrac{3}{2} \; cm$ $\;\; \cdots$ (2)
Rate of change of radius $= \dfrac{dr}{dt} = 2 \; cm/s$ $\;\; \cdots$ (2)
Total surface area of cylinder $=A=2\pi r \left(r+h\right) = 2 \pi r \left(r+\dfrac{r}{2}\right) = 3 \pi r^2$ $\;\; \cdots$ (3)
Volume of cylinder $=V = \pi r^2 h = \dfrac{1}{2}\pi r^3$ $\;\; \cdots$ (4)
Differentiating equation (3) w.r.t time t gives
$\begin{aligned}
\dfrac{dA}{dt} & = 6 \pi r \dfrac{dr}{dt} \\
& = 6 \pi \times 3 \times 2 \;\; \left[\text{from equations (1) and (2)}\right] \\
& = 36 \pi \; cm^2 /s
\end{aligned}$
Differentiating equation (4) w.r.t time t gives
$\begin{aligned}
\dfrac{dV}{dt} & = \dfrac{3}{2} \pi r^2 \dfrac{dr}{dt} \\
& = \dfrac{3}{2} \pi \times 3^2 \times 2 \;\; \left[\text{from equations (1) and (2)}\right] \\
& = 27 \pi \; cm^3 /s
\end{aligned}$