| Exam Board | Edexcel |
|---|---|
| Module | M5 (Mechanics 5) |
| Year | 2010 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | First order differential equations (integrating factor) |
| Type | Standard linear first order - vector form |
| Difficulty | Standard +0.3 This is a standard first-order linear differential equation with integrating factor method applied to vectors. The setup is straightforward (recognizing v = dr/dt), finding the integrating factor e^(-2t) is routine, and solving with the given initial condition follows standard procedure. Slightly above average difficulty due to vector notation and exponential terms, but remains a textbook application of the technique. |
| Spec | 1.10b Vectors in 3D: i,j,k notation4.10d Second order homogeneous: auxiliary equation method |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Notes |
| \(\frac{d\mathbf{r}}{dt} - 2\mathbf{r} = 4e^t \mathbf{j}\), IF \(= e^{-2t}\) | ||
| \(e^{-2t}\left(\frac{d\mathbf{r}}{dt} - 2\mathbf{r}\right) = e^{-2t} \cdot 4e^t \mathbf{j}\) | M1 | |
| \(\frac{d(\mathbf{r}e^{-2t})}{dt} = 4e^{-t}\mathbf{j}\) | ||
| \(\mathbf{r}e^{-2t} = \int 4e^{-t}\mathbf{j}\,dt\) | DM1 | |
| \(= -4e^{-t}\mathbf{j} + \mathbf{C}\) | A1 | |
| \(t=0, \mathbf{r}=-3\mathbf{j} \Rightarrow \mathbf{C} = \mathbf{j}\) | DM1 | |
| \(e^{-2t}\mathbf{r} = (1-4e^{-t})\mathbf{j}\) or \(\mathbf{r} = (e^{2t}-4e^t)\mathbf{j}\) | A1 | |
| \((1-4e^{-t})=0\) or \((e^{2t}-4e^t)=0\) | DM1 | |
| \(t = \ln 4\), 1.4 or better | A1 | Total: 7 |
## Question 1:
| Working/Answer | Mark | Notes |
|---|---|---|
| $\frac{d\mathbf{r}}{dt} - 2\mathbf{r} = 4e^t \mathbf{j}$, IF $= e^{-2t}$ | | |
| $e^{-2t}\left(\frac{d\mathbf{r}}{dt} - 2\mathbf{r}\right) = e^{-2t} \cdot 4e^t \mathbf{j}$ | M1 | |
| $\frac{d(\mathbf{r}e^{-2t})}{dt} = 4e^{-t}\mathbf{j}$ | | |
| $\mathbf{r}e^{-2t} = \int 4e^{-t}\mathbf{j}\,dt$ | DM1 | |
| $= -4e^{-t}\mathbf{j} + \mathbf{C}$ | A1 | |
| $t=0, \mathbf{r}=-3\mathbf{j} \Rightarrow \mathbf{C} = \mathbf{j}$ | DM1 | |
| $e^{-2t}\mathbf{r} = (1-4e^{-t})\mathbf{j}$ or $\mathbf{r} = (e^{2t}-4e^t)\mathbf{j}$ | A1 | |
| $(1-4e^{-t})=0$ or $(e^{2t}-4e^t)=0$ | DM1 | |
| $t = \ln 4$, 1.4 or better | A1 | **Total: 7** |
---\begin{enumerate}
\item At time $t = 0$, the position vector of a particle $P$ is $- 3 \mathbf { j } \mathrm {~m}$. At time $t$ seconds, the position vector of $P$ is $\mathbf { r }$ metres and the velocity of $P$ is $\mathbf { v } \mathrm { m } \mathrm { s } ^ { - 1 }$. Given that
\end{enumerate}
$$\mathbf { v } - 2 \mathbf { r } = 4 \mathrm { e } ^ { t } \mathbf { j }$$
find the time when $P$ passes through the origin.\\
\hfill \mbox{\textit{Edexcel M5 2010 Q1 [7]}}