In a cube shaped water tank of volume $8m^3$, water is poured into it at the rate of
$0.4 \,m^3/h$. Find the rate of change of increase water level with respect to time.
The following is the solution I obtained. I don't know if it's right or not; please let me know if it is correct or not and feel free to offer tips.
- water volume
$$V = x^2h \tag 1$$
- tank volume
$$V = x^3 , V = 8 \implies x = 2 \,m \tag 2$$
by equation $(2)$ in equation $(1)$
$$V = 4h$$
$$\frac {dv}{dt} = 4\frac {dh}{dt}$$
$$0.4 = 4\frac {dh}{dt}\implies \frac {dh}{dt} = 0.1 \,m/h$$