Water rises through a conical container · source reference

Printed question

The diagram shows a vertical cross-section of a container in the form of an inverted cone of height 24 cm and base radius 9 cm. The container is initially empty. Water is poured into it at a constant rate of $30\,\mathrm{cm^3\,s^{-1}}$ and the height of water level in the container at time $t$ is $h$ cm.

(i) Show that the volume of water in the container is given by $V=\frac3{64}\pi h^3$. [2]

(ii) Find the value of $h$ at the instant when the rate of increase in the depth of water is $5.65\,\mathrm{cm\,s^{-1}}$, leaving your answer in 2 decimal places. [3]

[2009/SRJC/Prelim/II/1]

Printed working

(i) Let $r$ cm be the radius of the circular water surface at time $t$ sec. Similar triangles give $r/9=h/24\Rightarrow r=9h/24=3h/8$.

$$V=\frac13\pi r^2h=\frac13\pi(3h/8)^2h=\frac3{64}\pi h^3$$

(shown).

(ii) $dV/dh=\pi(9/64)h^2$, and $dV/dt=(dV/dh)(dh/dt)$. Thus $30=\pi(9/64)h^2(dh/dt)$.

When $dh/dt=5.65$, $30=\pi(9/64)h^2(5.65)\Rightarrow h=3.47$ cm (2 d.p.).

Teacher notes

Given in the question, $dV/dt=30$.

Answer key

(ii) $h=3.47$ (2 d.p.).

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