A2 June 2023 Paper 2 Q8
8 A surge in the current, \(I\) units, through an electrical component at a time, \(t\) seconds, is to be modelled. The surge starts when \(t = 0\) and there is initially no current through the component. When the current has surged for 1 second it is measured as being 5 units. While the surge is occurring, \(I\) is modelled by the following differential equation.
\[\left(2t - t^2\right)\dfrac{\mathrm{d}I}{\mathrm{d}t} = \left(2t - t^2\right)^{\frac{3}{2}} - 2(t - 1)I\]
The surge lasts until there is again no current through the component.
| Scheme | Marks | AO |
|---|---|---|
| \(\left(2t - t^2\right)\dfrac{\mathrm{d}I}{\mathrm{d}t} = \left(2t - t^2\right)^{\frac{3}{2}} - 2(t - 1)I\) \(\therefore \dfrac{\mathrm{d}I}{\mathrm{d}t} = \left(2t - t^2\right)^{\frac{1}{2}} - \dfrac{2(t - 1)}{2t - t^2}I\) \(\therefore \dfrac{\mathrm{d}I}{\mathrm{d}t} + \dfrac{2(t - 1)}{2t - t^2}I = \left(2t - t^2\right)^{\frac{1}{2}}\) | *M1 | 3.3 |
| \(\text{IF} = \mathrm{e}^{\int\frac{2(t - 1)}{2t - t^2}\mathrm{d}t} = \mathrm{e}^{-\int\frac{2 - 2t}{2t - t^2}\mathrm{d}t} = \mathrm{e}^{-\ln\left(2t - t^2\right)} = \dfrac{1}{2t - t^2}\) | A1 | 2.2a |
| \(\therefore \left(2t - t^2\right)^{-1}\dfrac{\mathrm{d}I}{\mathrm{d}t} + 2(t - 1)\left(2t - t^2\right)^{-2}I = \left(2t - t^2\right)^{-\frac{1}{2}}\) \(\dfrac{\mathrm{d}}{\mathrm{d}t}\left(\left(2t - t^2\right)^{-1}I\right) = \left(2t - t^2\right)^{-\frac{1}{2}}\) | *dep*M1 | 1.1 |
| \(\left(2t - t^2\right)^{-1}I = \displaystyle\int\left(2t - t^2\right)^{-\frac{1}{2}}\mathrm{d}t = \int\frac{1}{\sqrt{1 - (t - 1)^2}}\,\mathrm{d}t\) | dep*M1 | 1.1 |
| \(= \sin^{-1}(t - 1) + c\) | *A1 | 1.1 |
| \(t = 1,\ I = 5 \Rightarrow (2 - 1)^{-1}5 = \sin^{-1}(1 - 1) + c\) \(\therefore c = 5\) \(\therefore \left(2t - t^2\right)^{-1}I = \sin^{-1}(t - 1) + 5\) \(\therefore I = \left(2t - t^2\right)\left(\sin^{-1}(t - 1) + 5\right)\) | dep*A1 | 3.3 |
| [6] |
Notes
*M1: Rearranging to the form \(\frac{\mathrm{d}I}{\mathrm{d}t} + P(t)I = Q(t)\)
A1: (1st) Finding correct integrating factor as an expression not involving exponentials and logs. Ignore unnecessary constant multiplier here
*dep*M1: Multiplying both sides by IF and recognising LHS as exact derivative of \(\left(2t - t^2\right)^{-1}I\). Can be implied by next M1. Can be awarded from slip in initial rearrangement if their IF works for their rearrangement
dep*M1: Taking integral of both sides and attempt to complete square in order to express RHS in a standard form (or using the substitution \(u = t - 1\) including \(\mathrm{d}u = \mathrm{d}t\)), \(= \int\frac{1}{\sqrt{1 - u^2}}\,\mathrm{d}u\). Condone \(\pm 1 \pm (t - 1)^2\) for M1
*A1: For correctly integrating RHS to a function of \(t\). “\(+\,c\)” not necessary here.
dep*A1: AG so use of relevant condition must be explicit. Verification of AG by substitution is not sufficient; value of \(c\) must be derived. Ignore workings using other condition (\(t = 0\), \(I = 0\))
| Scheme | Marks | AO |
|---|---|---|
| \(I = 0 \Rightarrow \left(2t - t^2\right)\left(\sin^{-1}(t - 1) + 5\right) = 0\) \(\therefore t(2 - t)\left(\sin^{-1}(t - 1) + 5\right) = 0\) \(\therefore t = 2\) So length of surge is \(2 - 0 = 2\) (seconds) | B1 | 3.1a |
| or \(\sin^{-1}(t - 1) + 5 = 0\) \(\therefore \sin^{-1}(t - 1) = -5\) which is not possible (since \(-\frac{1}{2}\pi \leqslant \sin^{-1}(t - 1) \leqslant \frac{1}{2}\pi\).) | B1 | 2.4 |
| [2] |
Notes
B1: (1st) If no other later comment or statement to the contrary then accept just \(t = 2\)
B1: (2nd) Do not accept incorrect explanation, e.g. \(-1 \leqslant \sin^{-1}(t - 1) \leqslant 1\). If B0B0 then Sc1 if length of surge determined to be awrt 1.96 s from \(\sin(-5\text{ rads}) + 1\) after \(t = 2\) found
| Scheme | Marks | AO |
|---|---|---|
| \(\left(I = \left(2t - t^2\right)\left(\sin^{-1}(t - 1) + 5\right)\right)\) & \(\left(2t - t^2\right)\dfrac{\mathrm{d}I}{\mathrm{d}t} = \left(2t - t^2\right)^{\frac{3}{2}} - 2(t - 1)I\) \(\therefore \dfrac{\mathrm{d}I}{\mathrm{d}t} = \dfrac{\left(2t - t^2\right)^{\frac{3}{2}} - 2(t - 1)\left(2t - t^2\right)\left(\sin^{-1}(t - 1) + 5\right)}{2t - t^2}\) | *M1 | 3.4 |
| \(\therefore \dfrac{\mathrm{d}I}{\mathrm{d}t} = \left(2t - t^2\right)^{\frac{1}{2}} - 2(t - 1)\left(\sin^{-1}(t - 1) + 5\right)\) | M1dep* | 2.2a |
| \(\therefore t = 0\) \(\Rightarrow \dfrac{\mathrm{d}I}{\mathrm{d}t} = 0 - 2(-1)\left(\sin^{-1}(-1) + 5\right)\) \(= 2\left(5 - \dfrac{1}{2}\pi\right) = 10 - \pi\) (units/s) | A1 | 3.4 |
| [3] |
Notes
*M1: Finding an expression for \(\frac{\mathrm{d}I}{\mathrm{d}t}\) in terms of \(t\) by either eliminating given \(I\) from given DE or differentiating given \(I(t)\) using product rule.
\(\left(I = \left(2t - t^2\right)\left(\sin^{-1}(t - 1) + 5\right) \Rightarrow\right) \dfrac{\mathrm{d}I}{\mathrm{d}t} = 2(1 - t)\left(\sin^{-1}(t - 1) + 5\right) + \dfrac{\left(2t - t^2\right)}{\sqrt{1 - (t - 1)^2}}\)
(The printed mark scheme also has a stray “\(\pm 1 \pm (t - 1)^2\)” after the expression for \(I\) in the first line of the answer; it has been left out.)
M1dep*: Simplifying their \(\frac{\mathrm{d}I}{\mathrm{d}t}\) so that there is no denominator which is zero at \(t = 0\)
\(\therefore \dfrac{\mathrm{d}I}{\mathrm{d}t} = 2(1 - t)\left(\sin^{-1}(t - 1) + 5\right) + \dfrac{\left(2t - t^2\right)}{\sqrt{2t - t^2}} = (2 - 2t)\left(\sin^{-1}(t - 1) + 5\right) + \left(2t - t^2\right)^{\frac{1}{2}}\)
A1: Must be in a simplified non-trigonometric form. Ignore units. If M1M0 or M0M0, then Sc B1 for correct answer