M5 June 2008 Q2
2. The velocity \(\mathbf{v}\) m s\(^{-1}\) of a particle \(P\) at time \(t\) seconds satisfies the vector differential equation
\[\frac{\mathrm{d}\mathbf{v}}{\mathrm{d}t} + 4\mathbf{v} = \mathbf{0}.\]The position vector of \(P\) at time \(t\) seconds is \(\mathbf{r}\) metres.
Given that at \(t = 0\), \(\mathbf{r} = (\mathbf{i} - \mathbf{j})\) and \(\mathbf{v} = (-8\mathbf{i} + 4\mathbf{j})\), find \(\mathbf{r}\) at time \(t\) seconds.
| Scheme | Marks |
|---|---|
| Aux Equn: \(m^2 + 4m = 0 \Rightarrow m = 0\) or \(-4\) | M1 |
| \(\mathbf{r} = \mathbf{A} + \mathbf{B}\mathrm{e}^{-4t}\) | A1 |
| \(t = 0,\ \mathbf{r} = \mathbf{i} - \mathbf{j}\): \(\mathbf{A} + \mathbf{B} = \mathbf{i} - \mathbf{j}\) | M1 |
| \(\mathbf{v} = -4\mathbf{B}\mathrm{e}^{-4t}\) | M1 |
| \(t = 0,\ \mathbf{v} = -8\mathbf{i} + 4\mathbf{j}\): \(-4\mathbf{B} = -8\mathbf{i} + 4\mathbf{j}\) | |
| \(\mathbf{B} = 2\mathbf{i} - \mathbf{j} \Rightarrow \mathbf{A} = -\mathbf{i}\) | A1 A1 |
| so, \(\mathbf{r} = -\mathbf{i} + (2\mathbf{i} - \mathbf{j})\mathrm{e}^{-4t}\) \(= \left(2\mathrm{e}^{-4t} - 1\right)\mathbf{i} - \mathrm{e}^{-4t}\mathbf{j}\) | A1 |
| (7 marks) |