An upright vessel is in the shape of a right circular cone with vertical height $12\text{ cm}$ and

Question

An upright vessel is in the shape of a right circular cone with vertical height $12\text{ cm}$ and base radius $5\text{ cm}$. The vessel is initially completely filled with water. Water then leaks from a small hole at the base at a constant rate of $20\pi\text{ cm}^3\text{/s}$. At time $t$ seconds, the water has depth $h\text{ cm}$ and its surface has radius $r\text{ cm}$.

Using similar triangles, show that the volume $V\text{ cm}^3$ of water remaining in the vessel is given by $V=25\pi \left(h- \frac{h^2}{12}+\frac{h^3}{432} \right)$.
[4]
Find the rate at which the depth of the water is decreasing at the instant when $h=6$.[3]

Accepted Answer

(i) $V=25\pi\left(h-\frac{h^2}{12}+\frac{h^3}{432}\right)$ $(\text{shown})$ (ii) $\frac{\mathrm{d}h}{\mathrm{d}t}=-\frac{16}{5}\text{ cm s}^{-1}$, so the depth is decreasing at $\frac{16}{5}\text{ cm s}^{-1}$

Water Leaking from an Upright Conical Vessel

Water Leaking from an Upright Conical Vessel

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