Question

the bottom of a rectangle swimming tank is 25 m by 40 m. water is pumped into the tank at the rate of 500 cubic meters per hour.find the rate at which the level of water in the tank is rising.

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{V}\mathrm{be}\mathrm{the}\mathrm{volume}\mathrm{and}\mathrm{h}\mathrm{be}\mathrm{the}\mathrm{height}\mathrm{of}\mathrm{the}\mathrm{water}\mathrm{inside}\mathrm{the}\mathrm{tank}\mathrm{at}\mathrm{any}\mathrm{instant}\mathrm{t}. \\ \mathrm{length},l=25\mathrm{m},\mathrm{breadth},b=40\mathrm{m} \\ \mathrm{Volume}=\mathrm{lbh}=1000\mathrm{h}{\mathrm{m}}^3 \\ \mathrm{Now}, \\ \frac{\mathrm{dV}}{\mathrm{dt}}=500{\mathrm{m}}^3/\mathrm{hr} \\ \mathrm{V}=1000\mathrm{h} \\ \mathrm{Differentiating}\mathrm{both}\mathrm{sides}\mathrm{with}\mathrm{respect}\mathrm{to}\mathrm{t}, \\ \Rightarrow \frac{\mathrm{dV}}{\mathrm{dt}}=1000\frac{\mathrm{dh}}{\mathrm{dt}} \\ \Rightarrow 500=1000\frac{\mathrm{dh}}{\mathrm{dt}} \\ \Rightarrow \frac{\mathrm{dh}}{\mathrm{dt}}=0.5\mathrm{m}/\mathrm{hr} \end{gathered}$$