Rate of change of the surface radius of water poured into a hemispherical bowl

A bowl is a hemisphere of internal radius \(15\text{ cm}\), with its rim horizontal. It contains water to a depth of \(h\text{ cm}\), and the circular water surface has radius \(r\text{ cm}\). The diagram shows a vertical cross-section through the centre \(O\) of the rim.

1. Show that \(r=\sqrt{30h-h^2}\).[2]
2. Water is poured into the bowl so that the volume of water rises steadily by \(20\text{ cm}^3\) every minute.
   When the depth is \(h\text{ cm}\), the volume of water in the bowl is \(V=\frac{\pi}{3}(45h^2-h^3)\text{ cm}^3\).
   Find the rate at which \(r\) is changing at the instant when \(h=5\).[5]