A2 June 2022 Paper 2 Q14
14 On an isolated island some rabbits have been accidently introduced.
In order to eliminate them, conservationists have introduced some birds of prey.
At time \(t\) years \((t \geqslant 0)\) there are \(x\) rabbits and \(y\) birds of prey.
At time \(t = 0\) there are 1755 rabbits and 30 birds of prey.
When \(t \gt 0\) it is assumed that:
- the rabbits will reproduce at a rate of \(a\)% per year
- each bird of prey will kill, on average, \(b\) rabbits per year
- the death rate of the birds of prey is \(c\) birds per year
- the number of birds of prey will increase at a rate of \(d\)% of the rabbit population per year.
This system is represented by the coupled differential equations:
\[\frac{\mathrm{d}x}{\mathrm{d}t} = 0.4x - 13y \qquad (1)\]\[\frac{\mathrm{d}y}{\mathrm{d}t} = 0.01x - 1.95 \qquad (2)\](a) State the value of \(a\), the value of \(b\), the value of \(c\) and the value of \(d\) [2 marks]
(b) Solve the coupled differential equations to find both \(x\) and \(y\) in terms of \(t\) [9 marks]
(c) Given that \(x\) and \(y\) are both positive for \(0 \leqslant t \leqslant 5\), use your answer to part (b) to show that the conservationists’ plan will succeed. [3 marks]
| Scheme | Marks | AO |
|---|---|---|
| Obtains two correct values | B1 | 3.1b |
| Obtains four correct values Condone “40%” and “1%” Do not accept −1.95 | B1 | 3.1b |
| (2) |
Typical solution
\[a = 40\]\[b = 13\]\[c = 1.95\]\[d = 1\]| Scheme | Marks | AO |
|---|---|---|
| Differentiates one equation | M1 | 3.1a |
| Substitutes for \(\dot{y}\), or for \(x\) and \(\dot{x}\), in the other equation to eliminate one variable | M1 | 3.1a |
| Forms a correct simplified second order differential equation | A1 | 1.1b |
| Obtains roots of their auxiliary equation | M1 | 1.1a |
| Uses a valid method to find a particular integral for their DE | M1 | 2.2a |
| States general solution for either \(x\) or \(y\) with their non-zero particular integral | A1F | 1.1b |
| States general solutions for both \(x\) and \(y\) CAO | A1 | 1.1b |
| Uses initial conditions to find a value for each constant | M1 | 3.4 |
| Writes correct solutions for both \(x\) and \(y\) | A1 | 1.1b |
| (9) |
Typical solution
\[13y = 0.4x - \dot{x} \quad (1)\]\[y = \frac{2}{65}x - \frac{1}{13}\dot{x}\]\[\dot{y} = \frac{2}{65}\dot{x} - \frac{1}{13}\ddot{x}\]Sub into (2)
\[\frac{2}{65}\dot{x} - \frac{1}{13}\ddot{x} = 0.01x - 1.95\]\[5\ddot{x} - 2\dot{x} + 0.65x = 126.75\]CF: \(5m^2 - 2m + 0.65 = 0\)
\[m = 0.2 \pm 0.3\mathrm{i}\]PI: \(x = 195\)
\[\therefore x = A\mathrm{e}^{0.2t}\cos(0.3t) + B\mathrm{e}^{0.2t}\sin(0.3t) + 195\]\[y = \frac{2}{65}x - \frac{1}{13}\dot{x}\]\[\begin{aligned}\therefore y = {}&\frac{1}{65}A\mathrm{e}^{0.2t}\cos(0.3t) + \frac{3}{130}A\mathrm{e}^{0.2t}\sin(0.3t) \\ &+ \frac{1}{65}B\mathrm{e}^{0.2t}\sin(0.3t) - \frac{3}{130}B\mathrm{e}^{0.2t}\cos(0.3t) + 6\end{aligned}\]When \(t = 0\), \(x = 1755\) and \(y = 30\) so
\[1755 = A + 195 \text{ and } 30 = \frac{A}{65} - \frac{3B}{130} + 6\]\[\Rightarrow A = 1560,\ B = 0\]\[x = 1560\mathrm{e}^{0.2t}\cos(0.3t) + 195\]\[y = 24\mathrm{e}^{0.2t}\cos(0.3t) + 36\mathrm{e}^{0.2t}\sin(0.3t) + 6\]| Scheme | Marks | AO |
|---|---|---|
| Investigates values of \(x\) for \(t \gt 5\) PI by \(t = 5.38\), \(x = 0\) | M1 | 3.2a |
| Obtains a time when \(x = 0\) or \(x \lt 0\), and states that the rabbits die out | A1 | 3.2a |
| Obtains a positive value of \(y\) for a value of \(t\) for which \(x \leqslant 0\) and uses their correct answers to show that the rabbits die out first. Condone no investigation of values of \(y\) between \(t = 5\) and their 5.38 | E1 | 3.5a |
| (3) | ||
| (14 marks) |
Typical solution
Using calculator:
When \(t = 5.38\), \(x = 0\)
At this time \(y \approx 108\) so there are still birds of prey
The rabbits die out first.
So the conservationists’ plan succeeds.