AS June 2022 Q2
2. A population of deer was introduced onto an island.
The number of deer, \(P\), on the island at time \(t\) years following their introduction is modelled by the differential equation
\[\frac{\mathrm{d}P}{\mathrm{d}t} = \frac{P}{5000}\left(1000 - \frac{P(t + 1)}{6t + 5}\right) \qquad t \gt 0\]It was estimated that there were 540 deer on the island six months after they were introduced.
Use two applications of the approximation formula \(\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)_n \approx \dfrac{y_{n+1} - y_n}{h}\) to estimate the number of deer on the island 10 months after they were introduced.
(7)
| Scheme | Marks | AO |
|---|---|---|
| \(t_0 = \dfrac{1}{2}\) and steps are 2 months, so \(h = \dfrac{1}{6}\quad \left(t_1 = \dfrac{2}{3}, t_2 = \dfrac{5}{6}\right)\) | B1 | 3.3 |
| \(\left(\dfrac{\mathrm{d}P}{\mathrm{d}t}\right)_0 = \dfrac{540}{5000}\left(1000 - \dfrac{540 \times \left(\frac{1}{2} + 1\right)}{6 \times \frac{1}{2} + 5}\right) = \ldots\left(97.065 = \dfrac{19413}{200}\right)\) | M1 | 3.4 |
| So when \(t = \dfrac{2}{3}\), \(P_1 = 540 + \dfrac{\text{'}1\text{'}}{6} \times \text{'}97.065\text{'} = \ldots\) Or starts with \(\text{'}97.065\text{'} = \dfrac{y_1 - 540}{\frac{1}{6}}\) and rearranges to find \(P_1 = \ldots\) | M1 | 1.1b |
| \(= \dfrac{222471}{400} = 556.1775\) | A1 | 1.1b |
| \(\left(\dfrac{\mathrm{d}P}{\mathrm{d}t}\right)_1 = \dfrac{\text{'}556.1775\text{'}}{5000}\left(1000 - \dfrac{\text{'}556.1775\text{'} \times \left(\frac{2}{3} + 1\right)}{6 \times \frac{2}{3} + 5}\right) = \ldots(99.778\ldots)\) | M1 | 3.4 |
| So when \(t = \dfrac{5}{6}\), \(P_2 = \text{'}556.1775\text{'} + \dfrac{1}{6} \times \text{'}99.778\ldots\text{'} = \ldots(572.807\ldots)\) | M1 | 1.1b |
| So there are estimated to be 572 or 573 deer after 10 months. | A1 | 3.2a |
| (7) | ||
| (7 marks) |
Notes
B1: Uses the given information to set up correct parameters for the model, \(t_0 = \dfrac{1}{2}\), \(\left(t_1 = \dfrac{2}{3}, t_2 = \dfrac{5}{6}\right)\) and \(h = \dfrac{1}{6}\) seen or implied.
M1: Uses \(P_0 = 540\) and their value for \(t_0\) in the given equation to find a value for \(\left(\dfrac{\mathrm{d}P}{\mathrm{d}t}\right)_0\)
M1: Applies the approximation formula with 540, their \(h\) and their \(\left(\dfrac{\mathrm{d}P}{\mathrm{d}t}\right)_0\) to find a value for \(P_1\)
A1: Correct approximation \(P\) at \(t = \dfrac{2}{3}\). Accept awrt 556.2
M1: Uses \(t_1 = t_0 + h\) and their \(P_1\) in the given equation to find a value for \(\left(\dfrac{\mathrm{d}P}{\mathrm{d}t}\right)_1\).
M1: Uses the approximation a second time with their \(h\), their \(P_1\) and their \(\left(\dfrac{\mathrm{d}P}{\mathrm{d}t}\right)_1\). to find a value for \(P_2\)
A1: Correct answer. Accept either 572 or 573.
Useful table of values for reference
| \(n\) | \(P_n\) | \(t\) | \(\dfrac{\mathrm{d}P}{\mathrm{d}t}\) | \(h\dfrac{\mathrm{d}P}{\mathrm{d}t}\) |
|---|---|---|---|---|
| 0 | 540 | \(\dfrac{1}{2}\) | 97.065 | 16.1775 |
| 1 | 556.1775 | \(\dfrac{2}{3}\) | 99.77870698 | 16.6297845 |
| 2 | 572.807 | \(\dfrac{5}{6}\) |
Note use of \(t_0 = 6\) and \(h = 2\) leads to \(\left(\dfrac{\mathrm{d}P}{\mathrm{d}t}\right)_0 = \dfrac{100494}{1025} = 98.04\ldots\quad P_1 = 736.08\ldots\quad \left(\dfrac{\mathrm{d}P}{\mathrm{d}t}\right)_1 = 128.8\ldots\)
\(P_2 = 993.7\ldots\) which scores a maximum of B0 M1 M1 A0 M1 M1 A0