A2 June 2023 Q2
2. The vertical height, \(h\) m, above horizontal ground, of a passenger on a fairground ride, \(t\) seconds after the ride starts, where \(t \leqslant 5\), is modelled by the differential equation
\[t^2\frac{\mathrm{d}^2 h}{\mathrm{d}t^2} - 2t\frac{\mathrm{d}h}{\mathrm{d}t} + 2h = t^3 \qquad \text{(I)}\]Given that when \(t = 1\), \(h = 2.5\) and when \(t = 2\), \(\dfrac{\mathrm{d}h}{\mathrm{d}t} = -1\)
| Scheme | Marks | AO |
|---|---|---|
| (i) \(t = \mathrm{e}^x \Rightarrow \dfrac{\mathrm{d}t}{\mathrm{d}h} = \mathrm{e}^x\dfrac{\mathrm{d}x}{\mathrm{d}h} \Rightarrow \dfrac{\mathrm{d}h}{\mathrm{d}t} = \mathrm{e}^{-x}\dfrac{\mathrm{d}h}{\mathrm{d}x}\) | M1 | 1.1b |
| \(\Rightarrow t\dfrac{\mathrm{d}h}{\mathrm{d}t} = \dfrac{\mathrm{d}h}{\mathrm{d}x}\ *\) | A1* | 2.1 |
| (ii) \(t\dfrac{\mathrm{d}h}{\mathrm{d}t} = \dfrac{\mathrm{d}h}{\mathrm{d}x} \Rightarrow t\dfrac{\mathrm{d}^{2}h}{\mathrm{d}t^{2}} + \dfrac{\mathrm{d}h}{\mathrm{d}t} = \dfrac{\mathrm{d}^{2}h}{\mathrm{d}x^{2}}\dfrac{\mathrm{d}x}{\mathrm{d}t}\) | M1 | 1.1b |
| \(t\dfrac{\mathrm{d}^{2}h}{\mathrm{d}t^{2}} + \dfrac{\mathrm{d}h}{\mathrm{d}t} = \dfrac{\mathrm{d}^{2}h}{\mathrm{d}x^{2}}\dfrac{\mathrm{d}x}{\mathrm{d}t} \Rightarrow t\dfrac{\mathrm{d}^{2}h}{\mathrm{d}t^{2}} = \dfrac{1}{t}\dfrac{\mathrm{d}^{2}h}{\mathrm{d}x^{2}} - \dfrac{\mathrm{d}h}{\mathrm{d}t}\) \(\Rightarrow t^2\dfrac{\mathrm{d}^{2}h}{\mathrm{d}t^{2}} = \dfrac{\mathrm{d}^{2}h}{\mathrm{d}x^{2}} - t\dfrac{\mathrm{d}h}{\mathrm{d}t} = \dfrac{\mathrm{d}^{2}h}{\mathrm{d}x^{2}} - \dfrac{\mathrm{d}h}{\mathrm{d}x}\ *\) | A1* | 2.1 |
| (4) |
Notes
(a)(i)
M1: Uses the chain rule (must be a clear statement of this before substitution or clear implication of it use) with an attempt to differentiate \(t = \mathrm{e}^x\) to form an equation linking \(\dfrac{\mathrm{d}h}{\mathrm{d}t}\) and \(\dfrac{\mathrm{d}h}{\mathrm{d}x}\) (or their reciprocals) in terms of either \(x\) or \(t\). Note they may take \(\ln t\) first, which is fine. Note that \(\dfrac{\mathrm{d}h}{\mathrm{d}x} = \mathrm{e}^x\dfrac{\mathrm{d}h}{\mathrm{d}t}\) with no supporting working is M0.
A1*: Correct proof with no errors.
(ii)
M1: Differentiates again with the product and chain rule, fully and correctly on at least one product, in order to establish any second derivative equation linking \(\dfrac{\mathrm{d}^{2}h}{\mathrm{d}t^{2}}\) and \(\dfrac{\mathrm{d}^{2}h}{\mathrm{d}x^{2}}\)
A1*: Correct proof with no errors.
| Scheme | Marks | AO |
|---|---|---|
| \(t^2\dfrac{\mathrm{d}^{2}h}{\mathrm{d}t^{2}} - 2t\dfrac{\mathrm{d}h}{\mathrm{d}t} + 2h = t^3 \Rightarrow \dfrac{\mathrm{d}^{2}h}{\mathrm{d}x^{2}} - \dfrac{\mathrm{d}h}{\mathrm{d}x} - 2\dfrac{\mathrm{d}h}{\mathrm{d}x} + 2h = \mathrm{e}^{3x}\) \(\Rightarrow \dfrac{\mathrm{d}^{2}h}{\mathrm{d}x^{2}} - 3\dfrac{\mathrm{d}h}{\mathrm{d}x} + 2h = \mathrm{e}^{3x}\ *\) | B1* | 1.1b |
| (1) |
Notes
B1*: Shows clearly the substitution of the given results into the differential equation and obtains the given transformed equation with no errors.
| Scheme | Marks | AO |
|---|---|---|
| \(m^2 - 3m + 2 = 0 \Rightarrow m = 1, 2\) | M1 | 1.1b |
| \((h =)\,A\mathrm{e}^x + B\mathrm{e}^{2x}\) | A1ft | 1.1b |
| PI is \(C = k\mathrm{e}^{3x}\) | B1 | 2.2a |
| \(\dfrac{\mathrm{d}h}{\mathrm{d}x} = 3k\mathrm{e}^{3x},\ \dfrac{\mathrm{d}^{2}h}{\mathrm{d}x^{2}} = 9k\mathrm{e}^{3x} \Rightarrow 9k - 9k + 2k = 1 \Rightarrow k = \tfrac{1}{2} \Rightarrow \text{PI: } (h =)\,\dfrac{1}{2}\mathrm{e}^{3x}\) | M1 A1 | 1.1b 1.1b |
| \(h = A\mathrm{e}^x + B\mathrm{e}^{2x} + \tfrac{1}{2}\mathrm{e}^{3x} \Rightarrow h = At + Bt^2 + \tfrac{1}{2}t^3\ *\) | A1* | 2.2a |
| (6) |
Notes
M1: Forms and solves a quadratic auxiliary equation.
A1ft: Correct form for the CF for their AE solutions which must be \(\pm 1, \pm 2\) (ie accept sign errors solving the AE). The \(h =\) is not needed and condone e.g. \(A\mathrm{e}^x + B\mathrm{e}^{2x} = 0\)
B1: Deduces the correct form for the PI. Must be in terms of \(x\) for the equation II.
M1: Differentiates their PI twice and substitutes their derivatives into the DE to find “\(k\)”.
A1: Correct PI (for equation II) seen or implied by working. May or may not be combined with the CF for this mark.
A1*: Forms the correct GS for \(h\) in terms of \(x\) and deduces the correct GS for the height in terms of \(t\) with no errors. This is a given answer, so must have seen the GS in terms of \(h\) before proceeding to this answer. Accept with the constants \(A\) and \(B\) either way round (or with other constants).
| Scheme | Marks | AO |
|---|---|---|
| \(t = 1, h = 2.5 \Rightarrow 2.5 = A + B + \tfrac{1}{2}\) | M1 | 3.4 |
| \(\dfrac{\mathrm{d}h}{\mathrm{d}t} = A + 2Bt + \tfrac{3}{2}t^2 \Rightarrow -1 = A + 4B + 6\) | M1 | 3.4 |
| \(A = 5, B = -3 \Rightarrow h = 5t - 3t^2 + \tfrac{1}{2}t^3\) | A1 | 1.1b |
| \(t = 5 \Rightarrow h = 5 \times 5 - 3 \times 5^2 + \dfrac{5^3}{2}\) | M1 | 1.1b |
| height is 12.5 m (including units) | A1 | 3.2a |
| (5) | ||
| (16 marks) |
Notes
M1: Uses the conditions of the model (\(t = 1\), \(h = 2.5\)) to form an equation in \(A\) and \(B\),
M1: Differentiates and shows of evidence of using the conditions of the model \(\left(t = 2,\ \dfrac{\mathrm{d}h}{\mathrm{d}t} = -1\right)\) to form another equation in \(A\) and \(B\).
A1: Solves simultaneously to obtain correct constants and hence a correct equation connecting \(h\) with \(t\).
M1: Substitutes \(t = 5\). If substitution is not seen you will need to check their answer matches their expression.
A1: Obtains 12.5 m using the model. Must include the units.