2020 H2 Maths Paper 1 Q10 — Worked Solution

Part (i)

2 marks

Since the death rate exceeds the birth rate by $3\%$ of the population every year, $P$ decreases at a rate proportional to itself, with constant of proportionality $0.03$ .

\frac{dP}{dt} = - 0.03P

Part (ii)

4 marks

Separating variables and integrating,

\int\frac{1}{P}\, dP = \int - 0.03\, dt

\Rightarrow \ln P = - 0.03t + c

\Rightarrow P = Ae^{- 0.03t},\ \text{where}\ A = e^{c}

Let $P_{0}$ be the initial number of sheep, at $t = 0$ . Then $A = P_{0}$ :

P = P_{0}e^{- 0.03t}

To check your answer: differentiating gives

\frac{dP}{dt} = - 0.03P_{0}e^{- 0.03t} = - 0.03P

, matching the differential equation of part (i). ✓

As $t \rightarrow \infty$ , $e^{- 0.03t} \rightarrow 0$ , so $P \rightarrow 0$ .

$\therefore$ $P = P_{0}e^{- 0.03t}$ ; if this situation continues over many years, the sheep population decreases indefinitely towards zero, i.e. it eventually dies out.

Part (iii)

2 marks

Sheep are now imported at a constant rate of $n$ per year, added to the existing $3\%$ decline.

\frac{dP}{dt} = - 0.03P + n

Part (iv)

4 marks

Approach

with a constant term added, we still separate variables, but the left side integrates to a logarithm of $n - 0.03P$ rather than of $P$ alone.

\int\frac{1}{n - 0.03P}\, dP = \int dt

\Rightarrow - \frac{1}{0.03}\ln|n - 0.03P| = t + c

\Rightarrow \ln|n - 0.03P| = - 0.03t - 0.03c

\Rightarrow n - 0.03P = Be^{- 0.03t},\ \text{where}\ B = \pm e^{- 0.03c}

P = \frac{n - Be^{- 0.03t}}{0.03}

= \frac{n}{0.03} - \frac{B}{0.03}e^{- 0.03t}

Using the initial condition $P = P_{0}$ at $t = 0$ ,

P_{0} = \frac{n}{0.03} - \frac{B}{0.03}

\Rightarrow \frac{B}{0.03} = \frac{n}{0.03} - P_{0}

Substituting back,

P = \frac{n}{0.03} - \left( \frac{n}{0.03} - P_{0} \right)e^{- 0.03t}

= \frac{n}{0.03} + \left( P_{0} - \frac{n}{0.03} \right)e^{- 0.03t}

To check your answer: at

t = 0

,

P = \frac{n}{0.03} + \left( P_{0} - \frac{n}{0.03} \right) = P_{0}

✓; differentiating gives

\frac{dP}{dt} = - 0.03\left( P_{0} - \frac{n}{0.03} \right)e^{- 0.03t} = - 0.03\left( P - \frac{n}{0.03} \right) = - 0.03P + n

, matching the differential equation of part (iii). ✓

$\therefore$ $P = \frac{n}{0.03} + \left( P_{0} - \frac{n}{0.03} \right)e^{- 0.03t}$ .

Part (v)

2 marks

As $t \rightarrow \infty$ , $e^{- 0.03t} \rightarrow 0$ , so $P$ settles to the steady-state value $\frac{n}{0.03}$ . Setting this equal to $500$ ,

\frac{n}{0.03} = 500

\Rightarrow n = 500 \times 0.03

= 15

To check your answer: with

n = 15

, the steady-state population is

\frac{15}{0.03} = 500

, matching the given value. ✓

$\therefore$ $n = 15$ .

2020 H2 Maths Paper 1, Question 10

Part (i)

Part (ii)

Part (iii)

Part (iv)

Part (v)

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