Water is leaking from a container. After \(t\) seconds, the depth of water in the container is \(x\) cm, and the volume of water is \(V\) cm\(^3\), where \(V = \frac{1}{3}x^3\). The rate at which water is lost is proportional to \(x\), so that \(\frac{dV}{dt} = -kx\), where \(k\) is a constant.
- Show that \(x \frac{dx}{dt} = -k\). [3]
Initially, the depth of water in the container is 10 cm.
- Show by integration that \(x = \sqrt{100 - 2kt}\). [4]
- Given that the container empties after 50 seconds, find \(k\). [2]
Once the container is empty, water is poured into it at a constant rate of 1 cm\(^3\) per second. The container continues to lose water as before.
- Show that, \(t\) seconds after starting to pour the water in, \(\frac{dx}{dt} = \frac{1-x}{x^2}\). [2]
- Show that \(\frac{1}{1-x} - x - 1 = \frac{x^2}{1-x}\).
Hence solve the differential equation in part (iv) to show that
$$t = \ln\left(\frac{1}{1-x}\right) - \frac{1}{2}x^2 - x.$$ [6]
- Show that the depth cannot reach 1 cm. [1]