M5 June 2015 Q2
2. A particle \(P\) moves so that its position vector, \(\mathbf{r}\) metres, at time \(t\) seconds, where \(0 \leqslant t \lt \dfrac{\pi}{2}\), satisfies the differential equation
\[\frac{\mathrm{d}\mathbf{r}}{\mathrm{d}t} - (\tan t)\mathbf{r} = (\sin t)\mathbf{i}\]When \(t = 0\), \(\mathbf{r} = -\dfrac{1}{2}\mathbf{i}\).
Find \(\mathbf{r}\) in terms of \(t\). (8)
| Scheme | Marks |
|---|---|
| IF \(= \mathrm{e}^{\int -\tan t\,\mathrm{d}t} = \cos t\) | M1 A1 |
| \(\dfrac{\mathrm{d}}{\mathrm{d}t}(\mathbf{r}\cos t) = \sin t\cos t\,\mathbf{i}\) | |
| \(\mathbf{r}\cos t = \displaystyle\int \sin t\cos t\,\mathbf{i}\,\mathrm{d}t\) | M1 |
| \(\mathbf{r}\cos t = \tfrac{1}{2}\sin^2 t\,\mathbf{i}\ (+\mathbf{C})\) | A1 |
| \(t = 0,\ \mathbf{r} = -\tfrac{1}{2}\mathbf{i} \Rightarrow \mathbf{C} = -\tfrac{1}{2}\mathbf{i}\) | M1 A1 |
| \(\mathbf{r}\cos t = \tfrac{1}{2}\sin^2 t\,\mathbf{i} - \tfrac{1}{2}\mathbf{i}\) | |
| \(\mathbf{r} = -\tfrac{1}{2}\cos t\,\mathbf{i}\) oe | DM1 A1 |
| (8 marks) |
Notes
First M1 for \(\mathrm{e}^{\int -\tan t\,\mathrm{d}t}\) (allow if – sign omitted)
First A1 for \(\cos t\)
Second M1 see scheme (multiply both sides by IF and integrate)
Second A1 for a correct equation (without \(\mathbf{C}\)) \(\left(-\tfrac{1}{2}\cos^2 t\right.\) or \(\left.-\tfrac{1}{4}\cos 2t\right)\)
Third M1 for use of initial conditions
Third A1 for a correct \(\mathbf{C}\)
Fourth M1 dependent on second M1 for producing an expression for \(\mathbf{r}\)
Fourth A1 for \(\mathbf{r} =\) any equivalent form (does not need to be simplified)