M5 June 2009 Q2
2. At time \(t\) seconds, the position vector of a particle \(P\) is \(\mathbf{r}\) metres, where \(\mathbf{r}\) satisfies the vector differential equation
\[\frac{\mathrm{d}^2\mathbf{r}}{\mathrm{d}t^2} + 4\mathbf{r} = \mathrm{e}^{2t}\mathbf{j}.\]When \(t = 0\), \(P\) has position vector \((\mathbf{i} + \mathbf{j})\) m and velocity \(2\mathbf{i}\) m s\(^{-1}\).
Find an expression for \(\mathbf{r}\) in terms of \(t\). (11)
| Scheme | Marks |
|---|---|
| C.F. is \(\mathbf{r} = \mathbf{A}\cos 2t + \mathbf{B}\sin 2t\) | B1 |
| P.I. is \(\mathbf{r} = \mathbf{p}e^{2t}\) | B1 |
| \(\dot{\mathbf{r}} = 2\mathbf{p}e^{2t}\) \(\ddot{\mathbf{r}} = 4\mathbf{p}e^{2t}\) | B1 ft |
| \(4\mathbf{p}e^{2t} + 4\mathbf{p}e^{2t} = \mathbf{j}e^{2t}\) | M1 |
| so, (PI is) \(\mathbf{r} = \tfrac{1}{8}\mathbf{j}e^{2t}\) | A1 |
| GS is \(\mathbf{r} = \mathbf{A}\cos 2t + \mathbf{B}\sin 2t + \tfrac{1}{8}\mathbf{j}e^{2t}\) | A1 ft |
| \(t = 0,\ \mathbf{r} = \mathbf{i} + \mathbf{j} \Rightarrow \mathbf{i} + \mathbf{j} = \mathbf{A} + \tfrac{1}{8}\mathbf{j} \Rightarrow \mathbf{i} + \tfrac{7}{8}\mathbf{j} = \mathbf{A}\) | DM1 A1 |
| \(\dot{\mathbf{r}} = -2\mathbf{A}\sin 2t + 2\mathbf{B}\cos 2t + \tfrac{1}{4}\mathbf{j}e^{2t}\) | M1A1 |
| \(t = 0,\ \dot{\mathbf{r}} = 2\mathbf{i} \Rightarrow 2\mathbf{i} = 2\mathbf{B} + \tfrac{1}{4}\mathbf{j} \Rightarrow \mathbf{i} - \tfrac{1}{8}\mathbf{j} = \mathbf{B}\) \(\mathbf{r} = (\mathbf{i} + \tfrac{7}{8}\mathbf{j})\cos 2t + (\mathbf{i} - \tfrac{1}{8}\mathbf{j})\sin 2t + \tfrac{1}{8}\mathbf{j}e^{2t}\) | A1 |
| (11 marks) |