Question

This problem should be solved using a differential equation:
A person is trying to fill a bathtub with water. Water is flowing into the bathtub from the tap at a constant rate of k litres/sec. However, there is a hole in the bottom of the bathtub and water is flowing out of the bathtub at a rate proportional to the square of the volume of water present in the bathtub. If V(t) is the volume of water (in litres) present in the bathtub at time t (in seconds) and the bathtub initially contains V(0) litres of water.

How can a differential equation of this problem b e written? (not solve just writing the equation)

Answer 1

Answer on Question #46851 – Math – Differential Calculus | Equations

This problem should be solved using a differential equation:

A person is trying to fill a bathtub with water. Water is flowing into the bathtub from the tap at a constant rate of $k \frac{L}{s}$ . However, there is a hole in the bottom of the bathtub and water is flowing out of the bathtub at a rate proportional to the square of the volume of water present in the bathtub. If $V(t)$ is the volume of water (in liters) present in the bathtub at time $t$ (in seconds) and the bathtub initially contains $V(0)$ litters of water.

How can a differential equation of this problem be written? (not solve just writing the equation)

Solution

The conservation equations dictate that the rate of change in volume is given by

$$\frac {d V (t)}{d t} = \dot {V _ {t n}} - \dot {V _ {o u t}},$$

where $\dot{V}_{in}$ and $\dot{V}_{out}$ , are the corresponding inflow and outflow. Note that the units of both magnitudes are $\frac{[V]}{[t]}$ , which in your case are $\frac{L}{s}$ . So we have $\dot{V}_{in} = k$ and $\dot{V}_{out} = \alpha V^2$ , where $\alpha$ is a constant of proportionality (measured in $\frac{s}{L}$ ). Thus, we come up with the following nonlinear first order ODE:

$$\frac {d V}{d t} + \alpha V ^ {2} = k, V (0) = V _ {0}.$$

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