2015 H2 Maths Paper 2 Q1 — Worked Solution

Part (i)

1 mark

h_{\max} = 32

Explanation: the tree stops growing once $\frac{dh}{dt} = 0$ , i.e. once $16 - \frac{1}{2}h = 0$ , which gives $h = 32$ . Since $\frac{dh}{dt} > 0$ for $h < 32$ , the height increases towards this value and the model gives no greater height, so the maximum height according to this model is $32\ m$ .

Part (ii)

5 marks

Approach

the differential equation gives $\frac{dh}{dt}$ as a function of $h$ alone, so we separate variables and integrate directly, using $\int(a + bx)^{n}\, dx = \frac{(a + bx)^{n + 1}}{b(n + 1)} + C$ .

Since $\frac{dh}{dt} = \frac{1}{10}\sqrt{16 - \frac{1}{2}h}$ ,

\frac{dt}{dh} = \frac{10}{\sqrt{16 - \frac{1}{2}h}}

Integrating with respect to $h$ ,

t = \int 10\left( 16 - \frac{1}{2}h \right)^{- \frac{1}{2}}dh

= 10 \times \frac{\left( 16 - \frac{1}{2}h \right)^{\frac{1}{2}}}{- \frac{1}{2} \times \frac{1}{2}} + C

= - 40\sqrt{16 - \frac{1}{2}h} + C

Using $h = 0$ when $t = 0$ :

0 = - 40\sqrt{16} + C

C = 160

t = 160 - 40\sqrt{16 - \frac{1}{2}h}

Hence, at $h = 16$ (half the maximum height),

t = 160 - 40\sqrt{16 - 8}

= 160 - 40\sqrt{8}

= 160 - 80\sqrt{2}

By GC, $t \approx 46.9$ (3 s.f.).

To check your answer: at

h = 32

(the maximum height), the formula gives

t = 160 - 40\sqrt{0} = 160

, consistent with growth slowing to zero as

h \rightarrow 32

. ✓

$\therefore$ $t = 160 - 40\sqrt{16 - \frac{1}{2}h}$ , and the tree takes $160 - 80\sqrt{2} \approx 46.9$ years to reach half its maximum height.

2015 H2 Maths Paper 2, Question 1

Part (i)

Part (ii)

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