| Exam Board | Edexcel |
|---|---|
| Module | FP1 AS (Further Pure 1 AS) |
| Year | 2019 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Separable variables - standard (applied/contextual) |
| Difficulty | Moderate -0.3 This is a straightforward numerical methods question requiring only two iterations of Euler's method with given formula and initial conditions. The calculation is mechanical with no conceptual challenges—students simply substitute values repeatedly. Slightly easier than average due to minimal steps (only 2 iterations) and explicit formula provided. |
| Spec | 1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams4.10a General/particular solutions: of differential equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Population after 4 months required over two iterations \(\Rightarrow h = \frac{1}{6}\) | B1 | Translates situation to state correct step length \(h\). AO3.3 |
| \(t_0 = 0,\ R_0 = 20 \Rightarrow \left(\frac{dR}{dt}\right)_0 = 2(20) + 4\sin 0\ \{= 40\}\) | M1 | Uses model to find initial value of \(\frac{dR}{dt}\) using \(t_0=0,\ R_0=20\). AO3.4 |
| \(\frac{R_1 - 20}{\frac{1}{6}} = 40 \Rightarrow R_1 = 20 + \frac{1}{6}(40)\) | M1 | Applies approximation formula with \(R_0=20\), stated \(h\), \(\left(\frac{dR}{dt}\right)_0\). AO1.1b |
| \(R_1 = \frac{80}{3}\) or awrt \(26.7\) or \(20 + (\text{their } h)(40)\) | A1ft | AO1.1b |
| \(\left(\frac{dR}{dt}\right)_1 = 2\left(\frac{80}{3}\right) + 4\sin\left(\frac{1}{6}\right)\ \{= 53.9969...\}\) | M1 | Attempts numerical expression for \(\left(\frac{dR}{dt}\right)_1\) with \(R_1 = \frac{80}{3}\) and \(t_1 = h\). AO1.1b |
| \(R_2 = R_1 + h\left(\frac{dR}{dt}\right)_1 = \frac{80}{3} + \frac{1}{6}(53.9969...) = 35.666... \approx 35\) or \(36\) rabbits | A1 | Applies approximation formula second time to give \(R_2\) as truncated 35 or value in \([35.5, 36]\). AO1.1b |
| \(R_2 = 35.666... \approx 35\) or \(36 < 40\). Julie will not be able to start to sell her rabbits after 4 months. | B1ft | Compares \(R_2\) with 40 and draws conclusion. AO3.2a. Give final B0 for more/fewer than two iterations before comparing. |
# Question 3:
| Answer | Mark | Guidance |
|--------|------|----------|
| Population after 4 months required over two iterations $\Rightarrow h = \frac{1}{6}$ | B1 | Translates situation to state correct step length $h$. AO3.3 |
| $t_0 = 0,\ R_0 = 20 \Rightarrow \left(\frac{dR}{dt}\right)_0 = 2(20) + 4\sin 0\ \{= 40\}$ | M1 | Uses model to find initial value of $\frac{dR}{dt}$ using $t_0=0,\ R_0=20$. AO3.4 |
| $\frac{R_1 - 20}{\frac{1}{6}} = 40 \Rightarrow R_1 = 20 + \frac{1}{6}(40)$ | M1 | Applies approximation formula with $R_0=20$, stated $h$, $\left(\frac{dR}{dt}\right)_0$. AO1.1b |
| $R_1 = \frac{80}{3}$ or awrt $26.7$ or $20 + (\text{their } h)(40)$ | A1ft | AO1.1b |
| $\left(\frac{dR}{dt}\right)_1 = 2\left(\frac{80}{3}\right) + 4\sin\left(\frac{1}{6}\right)\ \{= 53.9969...\}$ | M1 | Attempts numerical expression for $\left(\frac{dR}{dt}\right)_1$ with $R_1 = \frac{80}{3}$ and $t_1 = h$. AO1.1b |
| $R_2 = R_1 + h\left(\frac{dR}{dt}\right)_1 = \frac{80}{3} + \frac{1}{6}(53.9969...) = 35.666... \approx 35$ or $36$ rabbits | A1 | Applies approximation formula second time to give $R_2$ as truncated 35 or value in $[35.5, 36]$. AO1.1b |
| $R_2 = 35.666... \approx 35$ or $36 < 40$. Julie will **not** be able to start to sell her rabbits after 4 months. | B1ft | Compares $R_2$ with 40 and draws conclusion. AO3.2a. Give final B0 for more/fewer than two iterations before comparing. |\begin{enumerate}
\item Julie decides to start a business breeding rabbits to sell as pets.
\end{enumerate}
Initially she buys 20 rabbits. After $t$ years the number of rabbits, $R$, is modelled by the differential equation
$$\frac { \mathrm { d } R } { \mathrm {~d} t } = 2 R + 4 \sin t \quad t > 0$$
Julie needs to have at least 40 rabbits before she can start to sell them.\\
Use two iterations of the approximation formula
$$\left( \frac { \mathrm { d } y } { \mathrm {~d} x } \right) _ { n } \approx \frac { y _ { n + 1 } - y _ { n } } { h }$$
to find out if, according to the model, Julie will be able to start selling rabbits after 4 months.
\hfill \mbox{\textit{Edexcel FP1 AS 2019 Q3 [7]}}