A water cup is in the shape of a cone with its axis vertical. The vertical angle of the cone is \(\frac{\pi}{6}\) radians. The depth of water in the cup is \(h\), and the surface of the water is a circle of radius \(r\).

The volume of a cone of height \(h\) and base radius \(r\) is \(V=\frac13\pi r^2h\).

It is known that

\(\sin\frac{\pi}{12}=\frac{\sqrt6-\sqrt2}{4}, \qquad \cos\frac{\pi}{12}=\frac{\sqrt6+\sqrt2}{4}, \qquad \tan\frac{\pi}{12}=2-\sqrt3.\)

(i) Find an expression for \(r\) in terms of \(h\) and show that the volume of water in the cup is

\(V=\frac{\pi(7-4\sqrt3)h^3}{3}.\)

(ii) Water is poured into the cup at a rate of \(30\text{ cm}^3\text{ s}^{-1}\). Find, correct to 2 decimal places, the rate at which the depth of water is increasing when \(h=5\).