| Exam Board | Edexcel |
|---|---|
| Module | M5 (Mechanics 5) |
| Year | 2014 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | First order differential equations (integrating factor) |
| Type | Standard linear first order - vector form |
| Difficulty | Standard +0.8 This is a vector differential equation requiring integrating factor method with non-trivial integration (integrating factor is sec t, leading to integrals involving sec t tan t and sec³t). While the technique is standard for Further Maths, the vector context and the specific trigonometric integrals required elevate this above a routine first-order DE question, making it moderately challenging but still within expected FM scope. |
| Spec | 4.10c Integrating factor: first order equations |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Integrating factor: \(e^{\int \tan t \, dt} = e^{-\ln \cos t} = \sec t\) (or \(\frac{1}{\cos t}\)) | M1 | Attempt to find integrating factor |
| \(\frac{d}{dt}(\sec t \cdot \mathbf{r}) = \sec t(\cos^2 t \, \mathbf{i} - 3\cos t \, \mathbf{j})\) | A1 | Correct form after multiplying through |
| \(= \cos t \, \mathbf{i} - 3\mathbf{j}\) | A1 | Simplification |
| \(\sec t \cdot \mathbf{r} = \sin t \, \mathbf{i} - 3t \, \mathbf{j} + \mathbf{c}\) | M1 A1 | Integration of both components |
| At \(t=0\): \(\mathbf{r} = 4\mathbf{j}\), so \(1 \cdot 4\mathbf{j} = \mathbf{0}\,\mathbf{i} + \mathbf{0}\,\mathbf{j} + \mathbf{c}\), giving \(\mathbf{c} = 4\mathbf{j}\) | M1 | Apply initial condition |
| \(\mathbf{r} = \cos t(\sin t \, \mathbf{i} + (4 - 3t)\mathbf{j})\) | A1 A1 | Final answer, one mark each component |
## Question 5:
**Integrating Factor Method**
| Working/Answer | Mark | Guidance |
|---|---|---|
| Integrating factor: $e^{\int \tan t \, dt} = e^{-\ln \cos t} = \sec t$ (or $\frac{1}{\cos t}$) | M1 | Attempt to find integrating factor |
| $\frac{d}{dt}(\sec t \cdot \mathbf{r}) = \sec t(\cos^2 t \, \mathbf{i} - 3\cos t \, \mathbf{j})$ | A1 | Correct form after multiplying through |
| $= \cos t \, \mathbf{i} - 3\mathbf{j}$ | A1 | Simplification |
| $\sec t \cdot \mathbf{r} = \sin t \, \mathbf{i} - 3t \, \mathbf{j} + \mathbf{c}$ | M1 A1 | Integration of both components |
| At $t=0$: $\mathbf{r} = 4\mathbf{j}$, so $1 \cdot 4\mathbf{j} = \mathbf{0}\,\mathbf{i} + \mathbf{0}\,\mathbf{j} + \mathbf{c}$, giving $\mathbf{c} = 4\mathbf{j}$ | M1 | Apply initial condition |
| $\mathbf{r} = \cos t(\sin t \, \mathbf{i} + (4 - 3t)\mathbf{j})$ | A1 A1 | Final answer, one mark each component |
---\begin{enumerate}
\item A particle moves in a plane so that its position vector $\mathbf { r }$ metres at time $t$ seconds satisfies the differential equation
\end{enumerate}
$$\frac { \mathrm { d } \mathbf { r } } { \mathrm {~d} t } + ( \tan t ) \mathbf { r } = \left( \cos ^ { 2 } t \right) \mathbf { i } - ( 3 \cos t ) \mathbf { j } , \quad 0 \leqslant t < \frac { \pi } { 2 }$$
When $t = 0$, the particle is at the point with position vector $4 \mathbf { j } \mathrm {~m}$.
Find $\mathbf { r }$ in terms of $t$.\\
\hfill \mbox{\textit{Edexcel M5 2014 Q5 [8]}}