2016 H2 Maths Paper 2 Q1 — Worked Solution

Solution

7 marks

Approach

This is a related-rates problem. We are given $\frac{dV}{dt}$ and asked for $\frac{dh}{dt}$ , so we need $V$ as a function of $h$ alone. The diagram shows the semi-vertical angle $\alpha$ is measured from the axis of the cone to the slant, so $\tan\alpha = \frac{r}{h}$ ; this lets us eliminate $r$ from the volume formula before differentiating.

Since $\tan\alpha = \frac{r}{h} = 0.5$ , we have $r = \frac{1}{2}h$ .

Substituting into the given volume formula:

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

= \frac{1}{3}\pi\left( \frac{1}{2}h \right)^{2}h

= \frac{1}{12}\pi h^{3}

Differentiating with respect to $h$ :

\frac{dV}{dh} = \frac{1}{4}\pi h^{2}

When $V = 3$ :

3 = \frac{1}{12}\pi h^{3}

h^{3} = \frac{36}{\pi}

h = \left( \frac{36}{\pi} \right)^{\frac{1}{3}}

By GC, $h = 2.2545\ldots$

Using the chain rule, with $\frac{dV}{dt} = 0.1$ :

\frac{dh}{dt} = \frac{dh}{dV} \times \frac{dV}{dt}

= \frac{1}{\frac{1}{4}\pi h^{2}} \times 0.1

= \frac{0.4}{\pi h^{2}}

By GC, substituting $h = 2.2545\ldots$ :

\frac{dh}{dt} = 0.0251\ \left( 3\ \text{s.f.} \right)

To check your answer: near

V = 3

, a small increase

\Delta V = 0.1

should raise

h

by roughly

0.025

; since

h \approx 2.25

, this is under

2\%

h

over a full minute, consistent with the cone widening quickly as it fills. ✓

$\therefore$ the depth of water is increasing at $0.0251\ \text{m min}^{- 1}$ (3 s.f.) when $V = 3\ m^{3}$ .

2016 H2 Maths Paper 2, Question 1

Solution

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