| Exam Board | Edexcel |
|---|---|
| Module | FP1 AS (Further Pure 1 AS) |
| Year | 2022 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Logistic/bounded growth |
| Difficulty | Standard +0.3 This is a straightforward numerical methods question requiring two applications of Euler's method with a given differential equation. While the DE looks complex, students simply substitute values and perform arithmetic—no solving, no insight into logistic growth required. Easier than average FP1 as it's purely procedural calculation. |
| Spec | 1.09g Numerical methods in context |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(t_0 = \frac{1}{2}\) and steps are 2 months, so \(h = \frac{1}{6}\), \(\left(t_1 = \frac{2}{3}, t_2 = \frac{5}{6}\right)\) | B1 | Uses given information to set up correct parameters; \(t_0 = \frac{1}{2}\), \(h = \frac{1}{6}\) seen or implied |
| \(\left(\frac{dP}{dt}\right)_0 = \frac{540}{5000}\left(1000 - \frac{540 \times \left(\frac{1}{2}+1\right)}{6 \times \frac{1}{2}+5}\right) = 97.065 = \frac{19413}{200}\) | M1 | Uses \(P_0 = 540\) and their \(t_0\) in given equation to find \(\left(\frac{dP}{dt}\right)_0\) |
| So when \(t = \frac{2}{3}\), \(P_1 = 540 + \frac{1}{6} \times 97.065 = \ldots\) | M1 | Applies approximation formula with 540, their \(h\) and their \(\left(\frac{dP}{dt}\right)_0\) to find \(P_1\) |
| \(= \frac{222471}{400} = 556.1775\) | A1 | Correct approximation \(P\) at \(t = \frac{2}{3}\); accept awrt 556.2 |
| \(\left(\frac{dP}{dt}\right)_1 = \frac{556.1775}{5000}\left(1000 - \frac{556.1775 \times \left(\frac{2}{3}+1\right)}{6 \times \frac{2}{3}+5}\right) = 99.778\ldots\) | M1 | Uses \(t_1 = t_0 + h\) and their \(P_1\) in given equation to find \(\left(\frac{dP}{dt}\right)_1\) |
| So when \(t = \frac{5}{6}\), \(P_2 = 556.1775 + \frac{1}{6} \times 99.778\ldots = 572.807\ldots\) | M1 | Uses approximation a second time with their \(h\), \(P_1\), and \(\left(\frac{dP}{dt}\right)_1\) to find \(P_2\) |
| So there are estimated to be 572 or 573 deer after 10 months. | A1 | Correct answer; accept either 572 or 573 |
## Question 2:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $t_0 = \frac{1}{2}$ and steps are 2 months, so $h = \frac{1}{6}$, $\left(t_1 = \frac{2}{3}, t_2 = \frac{5}{6}\right)$ | **B1** | Uses given information to set up correct parameters; $t_0 = \frac{1}{2}$, $h = \frac{1}{6}$ seen or implied |
| $\left(\frac{dP}{dt}\right)_0 = \frac{540}{5000}\left(1000 - \frac{540 \times \left(\frac{1}{2}+1\right)}{6 \times \frac{1}{2}+5}\right) = 97.065 = \frac{19413}{200}$ | **M1** | Uses $P_0 = 540$ and their $t_0$ in given equation to find $\left(\frac{dP}{dt}\right)_0$ |
| So when $t = \frac{2}{3}$, $P_1 = 540 + \frac{1}{6} \times 97.065 = \ldots$ | **M1** | Applies approximation formula with 540, their $h$ and their $\left(\frac{dP}{dt}\right)_0$ to find $P_1$ |
| $= \frac{222471}{400} = 556.1775$ | **A1** | Correct approximation $P$ at $t = \frac{2}{3}$; accept awrt 556.2 |
| $\left(\frac{dP}{dt}\right)_1 = \frac{556.1775}{5000}\left(1000 - \frac{556.1775 \times \left(\frac{2}{3}+1\right)}{6 \times \frac{2}{3}+5}\right) = 99.778\ldots$ | **M1** | Uses $t_1 = t_0 + h$ and their $P_1$ in given equation to find $\left(\frac{dP}{dt}\right)_1$ |
| So when $t = \frac{5}{6}$, $P_2 = 556.1775 + \frac{1}{6} \times 99.778\ldots = 572.807\ldots$ | **M1** | Uses approximation a second time with their $h$, $P_1$, and $\left(\frac{dP}{dt}\right)_1$ to find $P_2$ |
| So there are estimated to be 572 or 573 deer after 10 months. | **A1** | Correct answer; accept either 572 or 573 |
**Note:** Use of $t_0 = 6$ and $h = 2$ leads to $\left(\frac{dP}{dt}\right)_0 = \frac{100494}{1025} = 98.04\ldots$, $P_1 = 736.08\ldots$, $\left(\frac{dP}{dt}\right)_1 = 128.8\ldots$, $P_2 = 993.7\ldots$ which scores maximum B0 M1 M1 A0 M1 M1 A0
---\begin{enumerate}
\item A population of deer was introduced onto an island.
\end{enumerate}
The number of deer, $P$, on the island at time $t$ years following their introduction is modelled by the differential equation
$$\frac { \mathrm { d } P } { \mathrm {~d} t } = \frac { P } { 5000 } \left( 1000 - \frac { P ( t + 1 ) } { 6 t + 5 } \right) \quad t > 0$$
It was estimated that there were 540 deer on the island six months after they were introduced.\\
Use two applications of the approximation formula $\left( \frac { \mathrm { d } y } { \mathrm {~d} x } \right) _ { n } \approx \frac { y _ { n + 1 } - y _ { n } } { h }$ to estimate the number of deer on the island 10 months after they were introduced.
\hfill \mbox{\textit{Edexcel FP1 AS 2022 Q2 [7]}}