Differentiation

Water flows from a vertical cylinder of radius \(4\text{ cm}\) into an inverted cone of height \(12\text{ cm}\) and top radius \(6\text{ cm}\).
The water depth \(z\) in the cylinder decreases at \(0.5\text{ cm s}^{-1}\). No water is lost during transfer.
In the cone, the water has depth \(h\) and surface radius \(r\).

1. Use similar triangles to show that \[ r=\frac h2 \quad\text{and}\quad V=\frac{\pi h^3}{12}, \] where \(V\) is the volume of water in the cone.

2. Find the rate at which the volume of water in the cone increases.

3. Find \(\frac{\mathrm{d}h}{\mathrm{d}t}\) when \(h=4\text{ cm}\).

4. At that instant, find the rate of change of the area of the water’s horizontal surface.