The surface area of a cube increases at the rate of $12 \; cm^2 /s$. Find the rate at which its volume increases, when its edge has a length of 5 cm.
Let length of edge of cube $=s=5 \; cm$ $\;\; \cdots$ (1)
Let surface area of the cube $=A = 6s^2$
Given: Rate of increase of surface area $=\dfrac{dA}{dt}=12 \; cm^2 /s$ $\;\; \cdots$ (2)
Now $\dfrac{dA}{dt} = 12 s \dfrac{ds}{dt}$ $\implies$ $\dfrac{ds}{dt} = \dfrac{1}{12s} \dfrac{dA}{dt}$ $\; \; \cdots$ (3)
Let volume of cube $=V=s^3 \;\; \cdots$ (4)
Differentiating both sides of equation (4) w.r.t time t gives
$\begin{aligned}
\dfrac{dV}{dt} & = 3 s^2 \dfrac{ds}{dt} \\
& = 3s^2 \times \dfrac{1}{12s} \dfrac{dA}{dt} \;\; \left[\text{From equation (3)}\right] \\
& = 3 \times 5^2 \times \dfrac{1}{12 \times 5} \times 12 \; \; \left[\text{From equations (1) and (2)}\right] \\
& = 15 \; cm^3 / s
\end{aligned}$
$\therefore$ Rate at which its volume increases $= 15 cm^3 /s$