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A tank containing water is in the form of a cone with vertex \(C\). The axis is vertical and the semivertical angle is \(60 ^ { \circ }\), as shown in the diagram. At time \(t = 0\), the tank is full and the depth of water is \(H\). At this instant, a tap at \(C\) is opened and water begins to flow out. The volume of water in the tank decreases at a rate proportional to \(\sqrt { } h\), where \(h\) is the depth of water at time \(t\). The tank becomes empty when \(t = 60\).
- Show that \(h\) and \(t\) satisfy a differential equation of the form
$$\frac { \mathrm { d } h } { \mathrm {~d} t } = - A h ^ { - \frac { 3 } { 2 } } ,$$
where \(A\) is a positive constant.
- Solve the differential equation given in part (i) and obtain an expression for \(t\) in terms of \(h\) and \(H\).
- Find the time at which the depth reaches \(\frac { 1 } { 2 } H\).
[0pt]
[The volume \(V\) of a cone of vertical height \(h\) and base radius \(r\) is given by \(V = \frac { 1 } { 3 } \pi r ^ { 2 } h\).]