| Exam Board | Edexcel |
|---|---|
| Module | FP1 AS (Further Pure 1 AS) |
| Session | Specimen |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Separable variables - standard (applied/contextual) |
| Difficulty | Standard +0.3 This is a straightforward numerical methods question requiring two iterations of Euler's method with given initial conditions. While it involves a non-standard differential equation, students only need to apply a formula mechanically without solving analytically or requiring deep insight. The calculation is routine for FP1 standard, slightly easier than average due to the explicit step-by-step nature. |
| Spec | 1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| vity SIHI NI JIUM I ON OC | VIUV SHIL NI JIHM I ON OC | VIIV SIHI NI JIIIM ION OC |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| £300 purchased one hour after opening \(\Rightarrow V_0 = 3\) and \(t_0 = 1\); half an hour after purchase \(\Rightarrow t_2 = 1.5\), so step \(h\) required is \(0.25\) | B1 | Identifies correct initial conditions and requirement for \(h\) |
| \(t_0 = 1,\ V_0 = 3,\ \left(\frac{dV}{dt}\right)_0 \approx \frac{3^2-1}{1^2+3} = 2\) | M1 | Uses the model to evaluate \(\frac{dV}{dt}\) at \(t_0\), using their \(t_0\) and \(V_0\) |
| \(V_1 \approx V_0 + h\left(\frac{dV}{dt}\right)_0 = 3 + 0.25 \times 2 = \ldots\) | M1 | Applies the approximation formula with their values |
| \(= 3.5\) | A1ft | 3.5 or exact equivalent. Follow through their step value |
| \(\left(\frac{dV}{dt}\right)_1 \approx \frac{3.5^2-1.25}{1.25^2+1.25\times 3.5}\left(=\frac{176}{95}\right)\) | M1 | Attempt to find \(\left(\frac{dV}{dt}\right)_1\) with their 3.5 |
| \(V_2 \approx V_1 + h\left(\frac{dV}{dt}\right)_1 = 3.5 + 0.25 \times \frac{176}{95} = 3.963\ldots\), so £396 (nearest £) | A1 | Applies the approximation and interprets the result to give £396 |
## Question 2:
| Answer/Working | Mark | Guidance |
|---|---|---|
| £300 purchased one hour after opening $\Rightarrow V_0 = 3$ and $t_0 = 1$; half an hour after purchase $\Rightarrow t_2 = 1.5$, so step $h$ required is $0.25$ | B1 | Identifies correct initial conditions and requirement for $h$ |
| $t_0 = 1,\ V_0 = 3,\ \left(\frac{dV}{dt}\right)_0 \approx \frac{3^2-1}{1^2+3} = 2$ | M1 | Uses the model to evaluate $\frac{dV}{dt}$ at $t_0$, using their $t_0$ and $V_0$ |
| $V_1 \approx V_0 + h\left(\frac{dV}{dt}\right)_0 = 3 + 0.25 \times 2 = \ldots$ | M1 | Applies the approximation formula with their values |
| $= 3.5$ | A1ft | 3.5 or exact equivalent. Follow through their step value |
| $\left(\frac{dV}{dt}\right)_1 \approx \frac{3.5^2-1.25}{1.25^2+1.25\times 3.5}\left(=\frac{176}{95}\right)$ | M1 | Attempt to find $\left(\frac{dV}{dt}\right)_1$ with their 3.5 |
| $V_2 \approx V_1 + h\left(\frac{dV}{dt}\right)_1 = 3.5 + 0.25 \times \frac{176}{95} = 3.963\ldots$, so £396 (nearest £) | A1 | Applies the approximation and interprets the result to give £396 |
---\begin{enumerate}
\item The value, V hundred pounds, of a particular stock thours after the opening of trading on a given day is modelled by the differential equation
\end{enumerate}
$$\frac { d V } { d t } = \frac { V ^ { 2 } - t } { t ^ { 2 } + t V } \quad 0 < t < 8.5$$
A trader purchases $\pounds 300$ of the stock one hour after the opening of trading.\\
Use two iterations of the approximation formula $\left( \frac { \mathrm { dy } } { \mathrm { dx } } \right) _ { 0 } \approx \frac { \mathrm { y } _ { 1 } - \mathrm { y } _ { 0 } } { \mathrm {~h} }$ to estimate, to the nearest $\pounds$, the value of the trader's stock half an hour after it was purchased.
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vity SIHI NI JIUM I ON OC & VIUV SHIL NI JIHM I ON OC & VIIV SIHI NI JIIIM ION OC \\
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\hfill \mbox{\textit{Edexcel FP1 AS Q2 [6]}}