M5 June 2014 (R) Q5
5. A particle moves in a plane so that its position vector \(\mathbf{r}\) metres at time \(t\) seconds satisfies the differential equation
\[\frac{\mathrm{d}\mathbf{r}}{\mathrm{d}t} + (\tan t)\,\mathbf{r} = (\cos^2 t)\,\mathbf{i} - (3\cos t)\mathbf{j}, \qquad 0 \leqslant t \lt \frac{\pi}{2}\]When \(t = 0\), the particle is at the point with position vector \(4\mathbf{j}\) m.
Find \(\mathbf{r}\) in terms of \(t\). (8)
| Scheme | Marks |
|---|---|
| \(R = \mathrm{e}^{\int \tan t\,\mathrm{d}t} = \mathrm{e}^{\ln\sec t} = \sec t\) | M1 A1 |
| \(\sec t\dfrac{\mathrm{d}\mathbf{r}}{\mathrm{d}t} + \sec t\tan t\,\mathbf{r} = \cos t\,\mathbf{i} - 3\mathbf{j}\) | |
| \(\dfrac{\mathrm{d}(\mathbf{r}\sec t)}{\mathrm{d}t} = \cos t\,\mathbf{i} - 3\mathbf{j}\) | M1 |
| \(\mathbf{r}\sec t = \displaystyle\int \cos t\,\mathbf{i} - 3\mathbf{j}\ \mathrm{d}t = \sin t\,\mathbf{i} - 3t\,\mathbf{j} + \mathbf{C}\) | M1 A1 |
| \(t = 0,\ \mathbf{r} = 4\mathbf{j} \Rightarrow \mathbf{C} = 4\mathbf{j}\) | M1 |
| \(\mathbf{r} = \cos t\sin t\,\mathbf{i} + (4\cos t - 3t\cos t)\,\mathbf{j}\) | M1 A1 |
| (8 marks) |