| Exam Board | Edexcel |
|---|---|
| Module | FP1 AS (Further Pure 1 AS) |
| Year | 2023 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Newton's law of cooling |
| Difficulty | Standard +0.3 This is a straightforward application of Euler's method to a first-order differential equation with clear initial conditions. Students need to rearrange the approximation formula, substitute values, and perform two iterations of basic arithmetic. The context is standard Newton's law of cooling, and the question requires only mechanical application of a given numerical method rather than analytical solution or conceptual insight. Slightly easier than average due to the step-by-step nature and limited iterations required. |
| Spec | 1.09g Numerical methods in context |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Identifies \(\theta_0 = 95\) and \(t_0 = 0\); temperature after 5 minutes required so \(h = 2.5\) | B1 | Identifies correct initial conditions and value of \(h\); may be implicit |
| \(t_0 = 0,\ \theta_0 = 95,\ \left(\frac{d\theta}{dt}\right)_0 = -0.05(95-20) = ... \left\{= -\frac{15}{4} = -3.75\right\}\) | M1 | Uses model to evaluate \(\frac{d\theta}{dt}\) at \(t_0\) using their \(\theta_0\) |
| \(\theta_1 \approx \theta_0 + h\left(\frac{d\theta}{dt}\right)_0 \approx 95 + 2.5 \times (-3.75) = ... \left\{\theta_1 = 85.625 = \frac{685}{8}\right\}\) | M1 | Applies approximation formula with \(\theta_0\), \(h\), and \(\left(\frac{d\theta}{dt}\right)_0\) to find \(\theta_1\) |
| \(\left(\frac{d\theta}{dt}\right)_1 = -0.05(85.625 - 20) = ... \left\{-3.28125 = -\frac{105}{32}\right\}\), then \(\theta_2 \approx 85.625 + 2.5 \times (-3.28125)\) | M1 | Finds \(\left(\frac{d\theta}{dt}\right)_1\) with '85.625' and applies approximation formula to find \(\theta_2\) |
| \(= \text{awrt } 77.4\) | A1 | Achieves awrt 77.4 |
| \(\theta_2 = 77.4 > 70\) therefore the teacher will not be able to start to drink the coffee at 1.20pm | B1ft | Obtains \(\theta_2 \in [40,100]\), compares with 70 and draws appropriate conclusion; e.g. "77.4 > 70 so too hot" scores B1; "77.4 is too hot" alone is unacceptable |
## Question 4:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Identifies $\theta_0 = 95$ and $t_0 = 0$; temperature after 5 minutes required so $h = 2.5$ | B1 | Identifies correct initial conditions and value of $h$; may be implicit |
| $t_0 = 0,\ \theta_0 = 95,\ \left(\frac{d\theta}{dt}\right)_0 = -0.05(95-20) = ... \left\{= -\frac{15}{4} = -3.75\right\}$ | M1 | Uses model to evaluate $\frac{d\theta}{dt}$ at $t_0$ using their $\theta_0$ |
| $\theta_1 \approx \theta_0 + h\left(\frac{d\theta}{dt}\right)_0 \approx 95 + 2.5 \times (-3.75) = ... \left\{\theta_1 = 85.625 = \frac{685}{8}\right\}$ | M1 | Applies approximation formula with $\theta_0$, $h$, and $\left(\frac{d\theta}{dt}\right)_0$ to find $\theta_1$ |
| $\left(\frac{d\theta}{dt}\right)_1 = -0.05(85.625 - 20) = ... \left\{-3.28125 = -\frac{105}{32}\right\}$, then $\theta_2 \approx 85.625 + 2.5 \times (-3.28125)$ | M1 | Finds $\left(\frac{d\theta}{dt}\right)_1$ with '85.625' and applies approximation formula to find $\theta_2$ |
| $= \text{awrt } 77.4$ | A1 | Achieves awrt 77.4 |
| $\theta_2 = 77.4 > 70$ therefore the teacher will not be able to start to drink the coffee at 1.20pm | B1ft | Obtains $\theta_2 \in [40,100]$, compares with 70 and draws appropriate conclusion; e.g. "77.4 > 70 so too hot" scores B1; "77.4 is too hot" alone is unacceptable |\begin{enumerate}
\item A teacher made a cup of coffee. The temperature $\theta ^ { \circ } \mathrm { C }$ of the coffee, $t$ minutes after it was made, is modelled by the differential equation
\end{enumerate}
$$\frac { \mathrm { d } \theta } { \mathrm {~d} t } + 0.05 ( \theta - 20 ) = 0$$
Given that
\begin{itemize}
\item the initial temperature of the coffee was $95 ^ { \circ } \mathrm { C }$
\item the coffee can only be safely drunk when its temperature is below $70 ^ { \circ } \mathrm { C }$
\item the teacher made the cup of coffee at 1.15 pm
\item the teacher needs to be able to start drinking the coffee by 1.20 pm\\
use two iterations of the approximation formula
\end{itemize}
$$\left( \frac { \mathrm { d } y } { \mathrm {~d} x } \right) _ { n } \approx \frac { y _ { n + 1 } - y _ { n } } { h }$$
to estimate whether the teacher will be able to start drinking the coffee at 1.20 pm .
\hfill \mbox{\textit{Edexcel FP1 AS 2023 Q4 [6]}}