Challenging +1.2 This is a standard integrating factor question with vector notation, requiring recognition of the linear form, finding the integrating factor involving sec and tan functions, and integrating both components separately. While the trigonometric functions are more complex than typical examples, the method is entirely routine for Further Maths students who have practiced this technique.
3. A particle \(P\) moves in the \(x y\)-plane in such a way that its position vector \(\mathbf { r }\) metres at time \(t\) seconds, where \(0 \leqslant t < \pi\), satisfies the differential equation
$$\sec ^ { 2 } \left( \frac { 1 } { 2 } t \right) \frac { \mathrm { d } \mathbf { r } } { \mathrm {~d} t } + \sec ^ { 3 } \left( \frac { 1 } { 2 } t \right) \sin \left( \frac { 1 } { 2 } t \right) \mathbf { r } = \sin \left( \frac { 1 } { 2 } t \right) \mathbf { i } + \sec ^ { 2 } \left( \frac { 1 } { 2 } t \right) \mathbf { j }$$
When \(t = 0\), the particle is at the point with position vector \(( - \mathbf { i } + \mathbf { j } ) \mathrm { m }\).
Find \(\mathbf { r }\) in terms of \(t\).
N.B. If components used, marks can only be scored once components have been put together.
## Question 3:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{\mathrm{d}\mathbf{r}}{\mathrm{d}t} + \tan\!\left(\tfrac{1}{2}t\right)\mathbf{r} = \sin\!\left(\tfrac{1}{2}t\right)\cos^2\!\left(\tfrac{1}{2}t\right)\mathbf{i} + \mathbf{j}$ | M1 | M1 for dividing by $\sec^2\!\left(\tfrac{1}{2}t\right)$ |
| $R = e^{\int\tan(\frac{1}{2}t)\,\mathrm{d}t} = \sec^2\!\left(\tfrac{1}{2}t\right)$ | M1 A1 | M1 for correct IF formula; A1 for correct integrating factor in terms of $t$ |
| $\mathbf{r}\sec^2\!\left(\tfrac{1}{2}t\right) = \int\sin\!\left(\tfrac{1}{2}t\right)\mathbf{i} + \sec^2\!\left(\tfrac{1}{2}t\right)\mathbf{j}\;\mathrm{d}t$ | M1 A1 | M1 for multiplying through by IF and attempting to integrate both sides; A1 for correct equation with LHS integrated correctly |
| $= -2\cos\!\left(\tfrac{1}{2}t\right)\mathbf{i} + 2\tan\!\left(\tfrac{1}{2}t\right)\mathbf{j}\ (+\mathbf{C})$ | A1 | A1 fully correct equation, **C not needed** |
| $t=0,\ \mathbf{r} = -\mathbf{i}+\mathbf{j} \Rightarrow \mathbf{C} = \mathbf{i}+\mathbf{j}$ | M1 | M1 for use of limits to find **C** |
| $\mathbf{r} = \left(\cos^2\!\left(\tfrac{1}{2}t\right)-2\cos^3\!\left(\tfrac{1}{2}t\right)\right)\mathbf{i}+\left(2\sin\!\left(\tfrac{1}{2}t\right)\cos\!\left(\tfrac{1}{2}t\right)+\cos^2\!\left(\tfrac{1}{2}t\right)\right)\mathbf{j}$ | A1 | A1 for answer in any form |
**Total: [8]** | N.B. If components used, marks can only be scored once components have been put together.
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3. A particle $P$ moves in the $x y$-plane in such a way that its position vector $\mathbf { r }$ metres at time $t$ seconds, where $0 \leqslant t < \pi$, satisfies the differential equation
$$\sec ^ { 2 } \left( \frac { 1 } { 2 } t \right) \frac { \mathrm { d } \mathbf { r } } { \mathrm {~d} t } + \sec ^ { 3 } \left( \frac { 1 } { 2 } t \right) \sin \left( \frac { 1 } { 2 } t \right) \mathbf { r } = \sin \left( \frac { 1 } { 2 } t \right) \mathbf { i } + \sec ^ { 2 } \left( \frac { 1 } { 2 } t \right) \mathbf { j }$$
When $t = 0$, the particle is at the point with position vector $( - \mathbf { i } + \mathbf { j } ) \mathrm { m }$.\\
Find $\mathbf { r }$ in terms of $t$.\\
\hfill \mbox{\textit{Edexcel M5 2018 Q3 [8]}}