2 Water is leaking from a container. After \(t\) seconds, the depth of water in the container is \(x \mathrm {~cm}\), and the volume of water is \(V \mathrm {~cm} ^ { 3 }\), where \(V = \frac { 1 } { 3 } x ^ { 3 }\). The rate at which water is lost is proportional to \(x\), so that \(\frac { \mathrm { d } V } { \mathrm {~d} t } = - k x\), where \(k\) is a constant.

1. Show that \(x \frac { \mathrm {~d} x } { \mathrm {~d} t } = - k\). Initially, the depth of water in the container is 10 cm .
2. Show by integration that \(x = \sqrt { 100 - 2 k t }\).
3. Given that the container empties after 50 seconds, find \(k\). Once the container is empty, water is poured into it at a constant rate of \(1 \mathrm {~cm} ^ { 3 }\) per second. The container continues to lose water as before.
4. Show that, \(t\) seconds after starting to pour the water in, \(\frac { \mathrm { d } x } { \mathrm {~d} t } = \frac { 1 - x } { x ^ { 2 } }\).
5. 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 .