M5 June 2012 Q1
1. A particle \(P\) moves in a plane such that its position vector \(\mathbf{r}\) metres at time \(t\) seconds \((t > 0)\) satisfies the differential equation
\[\frac{\mathrm{d}\mathbf{r}}{\mathrm{d}t} - \frac{2}{t}\mathbf{r} = 4\mathbf{i}\]When \(t = 1\), the particle is at the point with position vector \((\mathbf{i} + \mathbf{j})\) m.
Find \(\mathbf{r}\) in terms of \(t\). (9)
| Scheme | Marks |
|---|---|
| \(\dfrac{\mathrm{d}\mathbf{r}}{\mathrm{d}t} - \dfrac{2}{t}\mathbf{r} = 4\mathbf{i}\) | |
| IF \(= \mathrm{e}^{\int -\frac{2}{t}\mathrm{d}t} = \dfrac{1}{t^2}\) | M1 A1 |
| \(\dfrac{\mathrm{d}}{\mathrm{d}t}\left(\dfrac{\mathbf{r}}{t^2}\right) = \dfrac{1}{t^2}4\mathbf{i}\) | M1 A1 |
| \(\dfrac{\mathbf{r}}{t^2} = \displaystyle\int \dfrac{1}{t^2}4\mathbf{i}\,\mathrm{d}t\) | M1 |
| \(= \dfrac{-1}{t}4\mathbf{i} + \mathbf{C}\) (\(\mathbf{C}\) not needed for A1) | A1 |
| \(\mathbf{r} = -4t\mathbf{i} + \mathbf{C}t^2\) \(t = 1,\ \mathbf{r} = \mathbf{i} + \mathbf{j} \Rightarrow \mathbf{i} + \mathbf{j} = -4\mathbf{i} + \mathbf{C} \Rightarrow 5\mathbf{i} + \mathbf{j} = \mathbf{C}\) | M1 A1 |
| \(\mathbf{r} = -4t\mathbf{i} + (5\mathbf{i} + \mathbf{j})t^2\) | A1 |
| (9 marks) |