At time t minutes, the volume of rain water in the pit is $V$ meter cubic and the depth of rain water in the tank is $x$ meter.

Question

A water tank has horizontal base of 75 m by 20 m. Pumps are installed in the pit to avoid overflowing of water. At time $t$ minutes after the start of the test, rain water is released into the tank at a rate of $600e^{-\frac{t}{20}}$ cubic meters per minute and the pump will remove the rainwater at a rate of $30te^{-\frac{t}{20}}$ cubic meter per minute. Initially, the depth of the rain water in the tank is 0.1 meters. At time t minutes, the volume of rain water in the pit is $V$ meter cubic and the depth of rain water in the tank is $h$ meter. Formulate a ode relating the height $h$ and the time $t$.

Do i model the ode as $$\dfrac{dV}{dt} =600e^{-\frac{t}{20}} -30te^{-\frac{t}{20}} $$ and solve it? Then i use chain rule $$\dfrac{dV}{dt} = \dfrac{dV}{dh} \cdot \dfrac{dh}{dt} $$ where $$\dfrac{dV}{dh} = 75 \times 20?$$

Can anyone help me with this. The answer is $$\dfrac{dh}{dt} = \dfrac{1}{p}e^{-\frac{1}{20}}(20-t)$$ where $p$ is a constant to be determined.

Answer 1

Note that

• $V=Ah=1500h$

then

• $\dfrac{dV}{dt} =600e^{-\frac{t}{20}} -30te^{-\frac{t}{20}}\implies \dfrac{dh}{dt} =\frac25e^{-\frac{t}{20}} -\frac1{50}te^{-\frac{t}{20}}=\frac1{50}e^{-\frac{t}{20}}(20-t)$