2.39: Volume and related rates in a conical cup

0606/22/M/J/18 — Question 12 · 8 marks

In this question all lengths are in centimetres.

The volume of a cone of height

h

and base radius

r

is given by

V = \tfrac{1}{3}\pi r^{2}h

It is known that

\sin\tfrac{\pi}{12} = \tfrac{\sqrt{6} - \sqrt{2}}{4}

\cos\tfrac{\pi}{12} = \tfrac{\sqrt{6} + \sqrt{2}}{4}

\tan\tfrac{\pi}{12} = 2 - \sqrt{3}

A water cup is in the shape of a cone with its axis vertical. The vertical angle of the cone is

\tfrac{\pi}{6}

radians. The depth of water in the cup is

h

. The surface of the water is a circle of radius

r

(i)

Find an expression for

r

in terms of

h

and show that the volume of water in the cup is given by

V = \dfrac{\pi(7 - 4\sqrt{3})h^{3}}{3}

[4]

(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

[4]