AS June 2023 Q4
4. 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
\[\frac{\mathrm{d}\theta}{\mathrm{d}t} + 0.05(\theta - 20) = 0\]Given that
- the initial temperature of the coffee was \(95^\circ\mathrm{C}\)
- the coffee can only be safely drunk when its temperature is below \(70^\circ\mathrm{C}\)
- the teacher made the cup of coffee at 1.15 pm
- the teacher needs to be able to start drinking the coffee by 1.20 pm
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 estimate whether the teacher will be able to start drinking the coffee at 1.20 pm.
(6)
| Scheme | Marks | AO |
|---|---|---|
| Identifies that \(\theta_0 = 95\) and \(t_0 = 0\) The temperature after 5 minutes is required so \(h = 2.5\) | B1 | 3.3 |
| \(t_0 = 0,\quad \theta_0 = 95,\quad \left(\dfrac{\mathrm{d}\theta}{\mathrm{d}t}\right)_0 = -0.05(95 - 20) = \ldots\left\{= -\dfrac{15}{4} = -3.75\right\}\) | M1 | 3.4 |
| \(\theta_1 \approx \theta_0 + h\left(\dfrac{\mathrm{d}\theta}{\mathrm{d}t}\right)_0 \approx 95 + 2.5 \times \text{“}-3.75\text{”} = \ldots\ \left\{\theta_1 = 85.625 = \dfrac{685}{8}\right\}\) | M1 | 1.1b |
| \(\left(\dfrac{\mathrm{d}\theta}{\mathrm{d}t}\right)_1 = -0.05(\text{“}85.625\text{”} - 20) = \ldots\left\{-3.28125 = -\dfrac{105}{32}\right\}\) \(\theta_2 \approx \theta_1 + h\left(\dfrac{\mathrm{d}\theta}{\mathrm{d}t}\right)_1 \approx 85.625 + 2.5 \times -3.28125\) | M1 | 1.1b |
| \(= \text{awrt } 77.4\) | A1 | 1.1b |
| \(\theta_2 = 77.4 \gt 70\) therefore, the teacher will not be able to start to drink the coffee at 1.20pm | B1ft | 3.2a |
| (6) | ||
| (6 marks) |
Notes
B1: Identifies the correct initial conditions and requirements for \(h\). These may be implicit in their work.
M1: Uses the model to evaluate \(\dfrac{\mathrm{d}\theta}{\mathrm{d}t}\) at \(t_0\), using their \(\theta_0\), allowing for slips if the intent is clear.
M1: Applies the approximation formula with their values for \(\theta_0\), \(h\), and \(\left(\dfrac{\mathrm{d}\theta}{\mathrm{d}t}\right)_0\) to find a value for \(\theta_1\)
M1: Attempts to find \(\left(\dfrac{\mathrm{d}\theta}{\mathrm{d}t}\right)_1\) with their ’85.625’ and applies the approximation formula with their values for \(\theta_1\), \(h\), and \(\left(\dfrac{\mathrm{d}\theta}{\mathrm{d}t}\right)_1\) to find a value for \(\theta_2\)
A1: Achieves awrt 77.4
B1ft: Obtains a value of \(\theta_2\) in the range \([40, 100]\), compares their value of \(\theta_2\) with 70 and draws an appropriate conclusion about whether the teacher will be able to start to drink the cup of coffee at 1.20pm.
The conclusion may be minimal, e.g. “ 77.4 > 70 so no” scores B1. Accept “it is too hot” or similar as either a comparison or a conclusion but not both.
E.g. Acceptable : “ 77.4 is too hot, so cannot drink the coffee”, 77.4 > 70, so too hot”
Unacceptable : “77.4 is too hot”.
(Brackets that are missing from the printed mark scheme because of a font fault have been restored.)