A2 June 2024 Paper 2 Q6
6. The motion of a particle \(P\) along the \(x\)-axis is modelled by the differential equation
\[2\frac{\mathrm{d}^2x}{\mathrm{d}t^2} + 5\frac{\mathrm{d}x}{\mathrm{d}t} + 2x = 4t + 12\]where \(P\) is \(x\) metres from the origin \(O\) at time \(t\) seconds, \(t \geqslant 0\)
For large values of \(t\) the particle is expected to move with constant speed.
| Scheme | Marks | AO |
|---|---|---|
| \(2m^2 + 5m + 2 = 0 \Rightarrow m = -\dfrac{1}{2},\ -2\) | M1 | 3.4 |
| \(x = A\mathrm{e}^{-0.5t} + B\mathrm{e}^{-2t}\) | A1 | 1.1b |
| PI is of the form \(x = pt + q\) | B1 | 1.1b |
| \(\dfrac{\mathrm{d}x}{\mathrm{d}t} = p,\ \dfrac{\mathrm{d}^2x}{\mathrm{d}t^2} = 0 \Rightarrow 5p + 2pt + 2q = 4t + 12 \Rightarrow p = \ldots, q = \ldots\) | M1 | 3.4 |
| \(p = 2, q = 1\) | A1 | 1.1b |
| \(x = A\mathrm{e}^{-0.5t} + B\mathrm{e}^{-2t} + 2t + 1\) | A1ft | 1.1b |
| (6) |
Notes
M1: Attempts to solve \(2m^2 + 5m + 2 = 0\), usual rules apply for solving a quadratic equation.
A1: Correct CF. Do not need “\(x =\)” here, but must be in terms of \(t\)
B1: Correct form for the PI i.e. \(x = pt + q\) or \(x = at^2 + bt + c\)
M1: Differentiates their PI (of the forms \(x = pt + q\) or \(x = at^2 + bt + c\)) twice and substitutes into the given differential equation finding values for their constants to obtain a PI of the form \(pt + q\), \(p, q \neq 0\)
A1: Correct PI
A1ft: Correct GS or correct ft GS, which is the sum of their CF and PI. This is dependent on achieving both previous M marks. Must have \(x =\) and their GS must be in terms of t.
| Scheme | Marks | AO |
|---|---|---|
| \(t = 0,\ x = 3 \Rightarrow 3 = A + B + 1\) | M1 | 3.4 |
| \(\dfrac{\mathrm{d}x}{\mathrm{d}t} = -0.5A\mathrm{e}^{-0.5t} - 2B\mathrm{e}^{-2t} + 2\) \(t = 0,\ \dfrac{\mathrm{d}x}{\mathrm{d}t} = -2 \Rightarrow -\dfrac{1}{2}A - 2B + 2 = -2\) \(\Rightarrow A = \ldots,\ B = \ldots\) | M1 | 3.4 |
| \(x = 2\mathrm{e}^{-2t} + 2t + 1\) | A1 | 1.1b |
| (3) |
Notes
M1: Substitutes \(x = 3\) when \(t = 0\) into their GS to establish an equation in \(A\) and \(B\), allow minor slips if the intention is clear.
M1: Differentiates their answer to part (a) which must be in terms of \(t\) only, and sets \(= -2\) with \(t = 0\) to establish another equation in \(A\) and \(B\) and solves simultaneously to find \(A\) and \(B\).
Do not be concerned about how their simultaneous equations are solved; award this mark if they then go on to write values for \(A\) and \(B\).
Functions which require the use of product rule or trigonometric functions must be differentiated appropriately.
A1: Correct PS. Need “\(x =\)” here, their answer must be in terms of \(t\) and no other variable.
| Scheme | Marks | AO |
|---|---|---|
| (i) \(\dfrac{\mathrm{d}x}{\mathrm{d}t} = -4\mathrm{e}^{-2t} + 2\) | M1 | 3.1b |
| \(\mathrm{e}^{-2t} = \dfrac{1}{2} \Rightarrow -2t = \ln\dfrac{1}{2} \Rightarrow t = \dfrac{1}{2}\ln 2\) | dM1 | 2.1 |
| \(x = 2\mathrm{e}^{-2t} + 2t + 1 = 1 + \ln 2 + 1 = 2 + \ln 2\,*\) | A1* | 1.1b |
| (ii) \(\dfrac{\mathrm{d}^2x}{\mathrm{d}t^2} = 8\mathrm{e}^{-2t}\) is \(\gt 0\) for all values of \(t\) so distance is a minimum | B1ft | 2.4 |
| (4) |
Notes
(c) Mark (i) and (ii) together but only award for work done in (c)
(i)
M1: Differentiates their Particular Solution of the form \(x = a\mathrm{e}^{-kt} + bt + c\) where c may be 0 to obtain an expression of the form \(C\mathrm{e}^{-kt} + D\)
dM1: Solves an equation of the form \(C\mathrm{e}^{-kt} + D = 0,\ \ C \times D \lt 0\), to obtain \(t = -\dfrac{1}{k}\ln\left(\dfrac{-D}{C}\right)\)
A1*: Substitutes \(t = \dfrac{1}{2}\ln 2\) oe to obtain the printed answer with no errors.
Condone going from \(2\mathrm{e}^{-2\left(\frac{1}{2}\ln 2\right)} + 2\left(\dfrac{1}{2}\ln 2\right) + 1 = 2 + \ln 2\)
(ii)
B1ft: Obtains a second derivative of the form \(\dfrac{\mathrm{d}^2x}{\mathrm{d}t^2} = \lambda\mathrm{e}^{-\mu t},\ \ \lambda, \mu \gt 0\) and makes a conclusion e.g.
- \(\dfrac{\mathrm{d}^2x}{\mathrm{d}t^2} \gt 0\) (for all values of \(t\)) hence minimum.
- or substitutes their value of \(t\) (even if incorrect) and states \(\dfrac{\mathrm{d}^2x}{\mathrm{d}t^2} \gt 0\) hence minimum
| Scheme | Marks | AO |
|---|---|---|
| Examples: For large values of \(t\), \(\left[\mathrm{e}^{-2t} \to 0 \Rightarrow\right] x \to 2t + 1\) so constant speed For large values of \(t\), \(\left[\mathrm{e}^{-2t} \to 0 \Rightarrow\right] \dfrac{\mathrm{d}x}{\mathrm{d}t} \to 2\) so constant speed For large values of \(t\), \(\left[\mathrm{e}^{-2t} \to 0 \Rightarrow\right] \dfrac{\mathrm{d}^2x}{\mathrm{d}t^2} \to 0\) so constant speed Conclusion: so the model is suitable | B1ft | 3.2b |
| (1) | ||
| (14 marks) |
Notes
B1ft: Dependent on having obtained a Particular Solution of the form \(\mathrm{f}(t) + bt + c\) where \(\mathrm{f}(t)\) only has terms in \(a\mathrm{e}^{-kt}\) where \(k \gt 0\) and \(a, b \neq 0\)
This mark is awarded for the candidate demonstrating that in the model for “large values” or “as \(t \to \infty\)”, the value of their \(\mathrm{e}^{-kt} \to 0\), so there is constant speed and states that the model is suitable, or equivalent statement. (See scheme for examples)