Question 2 - A-Level Maths

Edexcel M5 2015 June — Question 2 8 marks

Exam Board	Edexcel
Module	M5 (Mechanics 5)
Year	2015
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 standard integrating factor problem but in vector form with a non-trivial integrating factor (sec t) and requires integration of sec t tan t and sec t sin t. The mechanics context and vector notation add modest complexity beyond a typical C4 first-order DE, placing it somewhat above average difficulty.
Spec	1.10a Vectors in 2D: i,j notation and column vectors 4.10c Integrating factor: first order equations

2. A particle $P$ moves so that its position vector, $\mathbf { r }$ metres, at time $t$ seconds, where $0 \leqslant t < \frac { \pi } { 2 }$, satisfies the differential equation $$\frac { \mathrm { d } \mathbf { r } } { \mathrm {~d} t } - ( \tan t ) \mathbf { r } = ( \sin t ) \mathbf { i }$$ When $t = 0 , \mathbf { r } = - \frac { 1 } { 2 } \mathbf { i }$.
Find $\mathbf { r }$ in terms of $t$.

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Question 2:

Answer	Marks	Guidance
$\frac{d\mathbf{r}}{dt} - (\tan t)\mathbf{r} = (\sin t)\mathbf{i}$ — integrating factor: $e^{\int -\tan t \, dt}$	M1	Attempt integrating factor
$= e^{\ln \cos t} = \cos t$	A1	Correct integrating factor $\cos t$
$\frac{d}{dt}(\mathbf{r} \cos t) = \sin t \cos t \, \mathbf{i}$	M1	Multiply through and recognise exact derivative
$= \frac{1}{2}\sin 2t \, \mathbf{i}$	A1	Correct RHS form
$\mathbf{r}\cos t = \int \sin t \cos t \, dt \, \mathbf{i}$	M1	Integrate both sides
$= -\frac{1}{4}\cos 2t \, \mathbf{i} + \mathbf{C}$	A1	Correct integration $\left(-\frac{1}{4}\cos 2t\right)$
When $t = 0$, $\mathbf{r} = -\frac{1}{2}\mathbf{i}$: $-\frac{1}{2}\mathbf{i} = -\frac{1}{4}\mathbf{i} + \mathbf{C}$	M1	Apply initial condition
$\mathbf{C} = -\frac{1}{4}\mathbf{i}$
$\mathbf{r} = \frac{1}{\cos t}\left(-\frac{1}{4}\cos 2t - \frac{1}{4}\right)\mathbf{i}$
$\mathbf{r} = -\frac{\cos 2t + 1}{4\cos t}\mathbf{i} = -\frac{2\cos^2 t}{4\cos t}\mathbf{i} = -\frac{1}{2}\cos t \, \mathbf{i}$	A1	Correct final answer $\mathbf{r} = -\frac{1}{2}\cos t \, \mathbf{i}$

## Question 2:

| $\frac{d\mathbf{r}}{dt} - (\tan t)\mathbf{r} = (\sin t)\mathbf{i}$ — integrating factor: $e^{\int -\tan t \, dt}$ | M1 | Attempt integrating factor |
|---|---|---|
| $= e^{\ln \cos t} = \cos t$ | A1 | Correct integrating factor $\cos t$ |
| $\frac{d}{dt}(\mathbf{r} \cos t) = \sin t \cos t \, \mathbf{i}$ | M1 | Multiply through and recognise exact derivative |
| $= \frac{1}{2}\sin 2t \, \mathbf{i}$ | A1 | Correct RHS form |
| $\mathbf{r}\cos t = \int \sin t \cos t \, dt \, \mathbf{i}$ | M1 | Integrate both sides |
| $= -\frac{1}{4}\cos 2t \, \mathbf{i} + \mathbf{C}$ | A1 | Correct integration $\left(-\frac{1}{4}\cos 2t\right)$ |
| When $t = 0$, $\mathbf{r} = -\frac{1}{2}\mathbf{i}$: $-\frac{1}{2}\mathbf{i} = -\frac{1}{4}\mathbf{i} + \mathbf{C}$ | M1 | Apply initial condition |
| $\mathbf{C} = -\frac{1}{4}\mathbf{i}$ | | |
| $\mathbf{r} = \frac{1}{\cos t}\left(-\frac{1}{4}\cos 2t - \frac{1}{4}\right)\mathbf{i}$ | | |
| $\mathbf{r} = -\frac{\cos 2t + 1}{4\cos t}\mathbf{i} = -\frac{2\cos^2 t}{4\cos t}\mathbf{i} = -\frac{1}{2}\cos t \, \mathbf{i}$ | A1 | Correct final answer $\mathbf{r} = -\frac{1}{2}\cos t \, \mathbf{i}$ |

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2. A particle $P$ moves so that its position vector, $\mathbf { r }$ metres, at time $t$ seconds, where $0 \leqslant t < \frac { \pi } { 2 }$, satisfies the differential equation

$$\frac { \mathrm { d } \mathbf { r } } { \mathrm {~d} t } - ( \tan t ) \mathbf { r } = ( \sin t ) \mathbf { i }$$

When $t = 0 , \mathbf { r } = - \frac { 1 } { 2 } \mathbf { i }$.\\
Find $\mathbf { r }$ in terms of $t$.\\

\hfill \mbox{\textit{Edexcel M5 2015 Q2 [8]}}

This paper (7 questions)

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Q1 6 Q2 8 Q3 12 Q4 12 Q5 9 Q6 16 Q7 12

Answer	Marks	Guidance
\(\frac{d\mathbf{r}}{dt} - (\tan t)\mathbf{r} = (\sin t)\mathbf{i}\) — integrating factor: \(e^{\int -\tan t \, dt}\)	M1	Attempt integrating factor
\(= e^{\ln \cos t} = \cos t\)	A1	Correct integrating factor \(\cos t\)
\(\frac{d}{dt}(\mathbf{r} \cos t) = \sin t \cos t \, \mathbf{i}\)	M1	Multiply through and recognise exact derivative
\(= \frac{1}{2}\sin 2t \, \mathbf{i}\)	A1	Correct RHS form
\(\mathbf{r}\cos t = \int \sin t \cos t \, dt \, \mathbf{i}\)	M1	Integrate both sides
\(= -\frac{1}{4}\cos 2t \, \mathbf{i} + \mathbf{C}\)	A1	Correct integration \(\left(-\frac{1}{4}\cos 2t\right)\)
When \(t = 0\), \(\mathbf{r} = -\frac{1}{2}\mathbf{i}\): \(-\frac{1}{2}\mathbf{i} = -\frac{1}{4}\mathbf{i} + \mathbf{C}\)	M1	Apply initial condition
\(\mathbf{C} = -\frac{1}{4}\mathbf{i}\)
\(\mathbf{r} = \frac{1}{\cos t}\left(-\frac{1}{4}\cos 2t - \frac{1}{4}\right)\mathbf{i}\)
\(\mathbf{r} = -\frac{\cos 2t + 1}{4\cos t}\mathbf{i} = -\frac{2\cos^2 t}{4\cos t}\mathbf{i} = -\frac{1}{2}\cos t \, \mathbf{i}\)	A1	Correct final answer \(\mathbf{r} = -\frac{1}{2}\cos t \, \mathbf{i}\)