Water is poured at a rate of \(0.1\text{ m}^3\) per minute into a container in the form of an open cone. The semi-vertical angle of the cone is \(\alpha\), where \(\tan\alpha=0.5\). At time \(t\) minutes after the start, the radius of the water surface is \(r\text{ m}\) (see diagram). Find the rate of increase of the depth of the water when the volume of water in the container is \(3\text{ m}^3\).[7]
[The volume of a cone of base radius \(r\) and height \(h\) is given by \(V=\frac13\pi r^2h\).]
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