A2 June 2025 Paper 2 Q17
17 A sample of biological material is placed in a freezer, and the freezer is then turned on.
The initial temperature of the sample is 38 °C
The temperature \(u\) °C of the freezer, at time \(t\) minutes after it is turned on, is modelled by
\[u = 18 - 2t\]The temperature \(y\) °C of the sample is modelled as decreasing at a rate which is proportional to the difference between the temperature of the sample and the temperature of the freezer.
Initially, the temperature of the sample is decreasing at a rate of 0.8 °C per minute.
(a) Show that \(y\) satisfies the differential equation\[\frac{\mathrm{d}y}{\mathrm{d}t} + 0.04y = 0.72 - 0.08t\] [3 marks]
(b) Find an expression for \(y\) in terms of \(t\) [7 marks]
(c) Find the time at which the freezer is 28 °C colder than the sample.
Give your answer in minutes and seconds. [4 marks]
(d) State one limitation of the model used. [1 mark]
| Scheme | Marks | AO |
|---|---|---|
| Obtains \(\dfrac{\mathrm{d}y}{\mathrm{d}t} = \pm k(y - u)\) OE | M1 | 3.3 |
| Substitutes initial conditions into their differential equation to obtain \(k = \pm 0.04\) | M1 | 3.4 |
| Completes a reasoned argument to obtain \(\dfrac{\mathrm{d}y}{\mathrm{d}t} + 0.04y = 0.72 - 0.08t\) | R1 | 2.1 |
| (3) |
Typical solution
\[\frac{\mathrm{d}y}{\mathrm{d}t} = k(y - u)\]\[\frac{\mathrm{d}y}{\mathrm{d}t} = k(y - 18 + 2t)\]Initially
\[-0.8 = k(38 - 18) = 20k\]\[k = -0.04\]\[\begin{aligned} \frac{\mathrm{d}y}{\mathrm{d}t} &= -0.04(y - 18 + 2t) \\ &= -0.04y + 0.72 - 0.08t \end{aligned}\]\[\frac{\mathrm{d}y}{\mathrm{d}t} + 0.04y = 0.72 - 0.08t\]| Scheme | Marks | AO |
|---|---|---|
| Selects a suitable method eg obtains an integrating factor or Obtains a complementary function. | M1 | 3.1a |
| Obtains correct integrating factor or Obtains correct complementary function. | A1 | 1.1b |
| Multiplies equation by their integrating factor or Uses correct form of particular integral. | M1 | 1.1a |
| Uses integration by parts on RHS or Equates coefficients of \(t\) and equates constant coefficients. | M1 | 3.1a |
| Obtains \(18\mathrm{e}^{0.04t} - 2t\mathrm{e}^{0.04t} + 50\mathrm{e}^{0.04t}\) or Obtains correct particular integral \(y = 68 - 2t\) | A1 | 1.1b |
| Uses initial conditions to obtain constant of integration or coefficient of exponential term. | M1 | 3.1b |
| Obtains \(y = 68 - 2t - 30\mathrm{e}^{-0.04t}\) | A1 | 1.1b |
| (7) |
Typical solution
\[\frac{\mathrm{d}y}{\mathrm{d}t} + 0.04y = 0.72 - 0.08t\]\[\mathrm{IF} = \mathrm{e}^{\int 0.04\,\mathrm{d}t} = \mathrm{e}^{0.04t}\]\[\mathrm{e}^{0.04t}\left(\frac{\mathrm{d}y}{\mathrm{d}t} + 0.04y\right) = \mathrm{e}^{0.04t}(0.72 - 0.08t)\]\[\frac{\mathrm{d}}{\mathrm{d}t}\left(y\mathrm{e}^{0.04t}\right) = 0.72\mathrm{e}^{0.04t} - 0.08t\mathrm{e}^{0.04t}\]\[y\mathrm{e}^{0.04t} = 0.72\int \mathrm{e}^{0.04t}\,\mathrm{d}t - 0.08\int t\mathrm{e}^{0.04t}\,\mathrm{d}t\]\[\begin{aligned} u &= t & v^{\prime} &= e^{0.04t} \\ u^{\prime} &= 1 & v &= 25e^{0.04t} \end{aligned}\]\[\begin{aligned} y\mathrm{e}^{0.04t} &= 0.72\left(25\mathrm{e}^{0.04t}\right) - 0.08\left(25t\mathrm{e}^{0.04t}\right) + 0.08(25)\int \mathrm{e}^{0.04t}\,\mathrm{d}t \\ &= 18\mathrm{e}^{0.04t} - 2t\mathrm{e}^{0.04t} + 50\mathrm{e}^{0.04t} + c \end{aligned}\]\[y = 68 - 2t + c\mathrm{e}^{-0.04t}\]When \(t = 0\), \(y = 38 \Rightarrow c = -30\)
\[y = 68 - 2t - 30\mathrm{e}^{-0.04t}\]| Scheme | Marks | AO |
|---|---|---|
| Obtains \(y - u = 28\) OE | M1 | 3.3 |
| Forms a correct equation from their \(y\), which is in the form \(y = A\mathrm{e}^{\alpha t} + B + Ct\) | A1F | 1.1b |
| Solves their exponential equation. | M1 | 3.1a |
| Obtains 7 minutes 45 seconds. Condone 7 minutes 46 seconds. | A1 | 3.2a |
| (4) |
Typical solution
\[y - u = 28\]\[68 - 2t - 30\mathrm{e}^{-0.04t} - (18 - 2t) = 28\]\[22 = 30\mathrm{e}^{-0.04t}\]\[\mathrm{e}^{-0.04t} = \frac{22}{30}\]\[t = 7.75387\]7 minutes 45 seconds
| Scheme | Marks | AO |
|---|---|---|
| States a limitation of the model, with reference to the fact that the model implies that the temperature of the freezer or the sample could decrease indefinitely. | E1 | 3.5b |
| (1) | ||
| (15 marks) |
Typical solution
The model would not work for large values of \(t\) since the temperature of the freezer would not decrease indefinitely.