A2 October 2020 Paper 2 Q3
3. A scientist is investigating the concentration of antibodies in the bloodstream of a patient following a vaccination.
The concentration of antibodies, \(x\), measured in micrograms (μg) per millilitre (ml) of blood, is modelled by the differential equation
where \(t\) is the number of weeks since the vaccination was given.
Initially,
- there are no antibodies in the bloodstream of the patient
- the concentration of antibodies is estimated to be increasing at 10 μg/ml per week
A second dose of the vaccine has to be given to try to ensure that it is fully effective. It is only safe to give the second dose if the concentration of antibodies in the bloodstream of the patient is less than 5 μg/ml.
| Scheme | Marks | AO |
|---|---|---|
| \(100m^2 + 60m + 13 = 0 \Rightarrow m = -0.3 \pm 0.2\mathrm{i}\) | M1 | 1.1b |
| \(x = \mathrm{e}^{-0.3t}(A\cos 0.2t + B\sin 0.2t)\) | A1 | 1.1b |
| PI: \(x = 2\) | B1 | 1.1b |
| \(x = \mathrm{e}^{-0.3t}(A\cos 0.2t + B\sin 0.2t) + 2\) | A1ft | 2.2a |
| (4) |
Notes
(a)
M1: Uses the model to form and solve the auxiliary equation
A1: Correct CF, does not need \(x =\)
B1: Correct PI
A1ft: Deduces the correct GS (follow through their CF + PI). Must have \(x = \mathrm{f}(t)\) and PI not 0
| Scheme | Marks | AO |
|---|---|---|
| \(t = 0,\ x = 0 \Rightarrow A = -2\) | M1 | 3.4 |
| \(\dfrac{\mathrm{d}x}{\mathrm{d}t} = -0.3\mathrm{e}^{-0.3t}(-2\cos 0.2t + B\sin 0.2t) + \mathrm{e}^{-0.3t}(0.4\sin 0.2t + 0.2B\cos 0.2t)\) \(t = 0,\ \dfrac{\mathrm{d}x}{\mathrm{d}t} = 10 \Rightarrow B = \ldots\) (NB \(B = 47\)) | M1 | 3.4 |
| \(x = \mathrm{e}^{-0.3t}(47\sin 0.2t - 2\cos 0.2t) + 2\) | A1 | 1.1b |
| \(-0.3\mathrm{e}^{-0.3t}(47\sin 0.2t - 2\cos 0.2t) + \mathrm{e}^{-0.3t}(9.4\cos 0.2t + 0.4\sin 0.2t) = 0\) \(\Rightarrow t = \ldots\) or \(x = \sqrt{2213}\mathrm{e}^{-0.3t}\sin(0.2t - 0.0425) + 2\) \(\Rightarrow \dfrac{\mathrm{d}x}{\mathrm{d}t} = -0.3\sqrt{2213}\mathrm{e}^{-0.3t}\sin(0.2t - 0.0425) + 0.2\sqrt{2213}\mathrm{e}^{-0.3t}\cos(0.2t - 0.0425)\) \(\Rightarrow t = \ldots\) | M1 | 3.1b |
| \(\tan 0.2t = \dfrac{100}{137} \Rightarrow 0.2t = 0.630\ldots\) or \(\tan(0.2t - 0.0425) = \dfrac{2}{3} \Rightarrow 0.2t = 0.630\) | M1 | 2.1 |
| \(t = 3.15\ldots\) weeks | A1 | 1.1b |
| \(x = \mathrm{e}^{-0.3 \times \text{"}3.15\ldots\text{"}}\left(47\sin(0.2 \times \text{"}3.15\ldots\text{"}) - 2\cos(0.2 \times \text{"}3.15\ldots\text{"})\right) + 2\) | M1 | 3.4 |
| = awrt 12.1 {μg/ml} | A1 | 3.2a |
| (8) |
Notes
(b)
M1: Uses the model and the initial conditions to establish the value of “\(A\)”
M1: Differentiates their model using the product rule and uses the initial conditions to establish the value of “\(B\)”. Must be using \(x = 0\) and \(\dfrac{\mathrm{d}x}{\mathrm{d}t} = 10\)
A1: Correct particular solution. This can be implied by the correct constants found following a correct answer to part (a).
M1: Uses their solution to the model with a correct strategy to obtain the required value of \(t\) e.g. differentiates, sets equal to zero and solves for \(t\)
M1: Uses a correct trigonometric approach that leads to a value for \(t\)
A1: Correct value for \(t\)
M1: Uses the model and their value for \(t\) to find the maximum concentration.
A1: Correct value
| Scheme | Marks | AO |
|---|---|---|
| \(t = 10 \Rightarrow x = \mathrm{e}^{-3}(47\sin(2) - 2\cos(2)) + 2 = 4.16\ldots\) | M1 | 3.4 |
| The model suggests that it would be safe to give the second dose | A1ft | 2.2a |
| (2) | ||
| (14 marks) |
Notes
(c)
M1: Uses the model to find the concentration when \(t = 10\)
A1ft: Makes a suitable comment that is consistent with their calculated value
Special case: If the candidate’s maximum value is less than 5 then
M1: never reaches 5 as maximum is…. or max is less than 5
A1: yes, it is safe