| Exam Board | OCR MEI |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric differentiation |
| Type | Find dy/dx expression in terms of parameter |
| Difficulty | Challenging +1.2 This is a substantial multi-part parametric equations question requiring differentiation, finding stationary points, and applying tangent conditions. While it involves multiple steps and some algebraic manipulation (including verifying a trigonometric equation and working with the given θ = 2π/3), the techniques are standard C4 material. The question guides students through each stage systematically, making it moderately above average difficulty but not requiring exceptional insight. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.05o Trigonometric equations: solve in given intervals1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| At E, \(\theta = 2\pi \Rightarrow x = a(2\pi - \sin 2\pi) = 2a\pi\), so \(OE = 2a\pi\) | M1, A1 | |
| Max height when \(\theta = \pi\) | M1 | \(\theta = \pi\), \(180°\), \(\cos\theta = -1\) |
| \(\Rightarrow y = a(1 - \cos\pi) = 2a\) | A1 [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}\) | M1 | \(\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}\) for theirs |
| \(= \frac{a\sin\theta}{a(1-\cos\theta)}\) | M1 | \(\frac{d}{d\theta}(\sin\theta) = \cos\theta\), \(\frac{d}{d\theta}(\cos\theta) = -\sin\theta\) both, or equivalent; condone uncancelled \(a\) |
| \(= \frac{\sin\theta}{(1-\cos\theta)}\) | A1 [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\tan 30° = 1/\sqrt{3} \Rightarrow \frac{\sin\theta}{(1-\cos\theta)} = \frac{1}{\sqrt{3}}\) | M1 | Or gradient \(= 1/\sqrt{3}\) |
| \(\Rightarrow \sin\theta = \frac{1}{\sqrt{3}}(1-\cos\theta)\) | E1 | |
| When \(\theta = 2\pi/3\), \(\sin\theta = \sqrt{3}/2\) | M1 | \(\sin\theta = \sqrt{3}/2\), \(\cos\theta = -1/2\) |
| \((1-\cos\theta)/\sqrt{3} = (1+1/2)/\sqrt{3} = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2}\) | E1 | |
| \(BF = a(1 + 1/2) = 3a/2\) | E1 | |
| \(OF = a(2\pi/3 - \sqrt{3}/2)\) | B1 [6] | or equiv. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(BC = 2a\pi - 2a(2\pi/3 - \sqrt{3}/2) = a(2\pi/3 + \sqrt{3})\) | B1ft | their \(OE - 2\times\)their \(OF\) |
| \(AF = \sqrt{3} \times 3a/2 = 3\sqrt{3}a/2\) | M1 A1 | |
| \(AD = BC + 2AF\) | M1 | |
| \(= a(2\pi/3 + \sqrt{3} + 3\sqrt{3}) = a(2\pi/3 + 4\sqrt{3}) = 20\) | A1 [5] | |
| \(\Rightarrow a = 2.22\) m |
## Question 2:
### Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| At E, $\theta = 2\pi \Rightarrow x = a(2\pi - \sin 2\pi) = 2a\pi$, so $OE = 2a\pi$ | M1, A1 | |
| Max height when $\theta = \pi$ | M1 | $\theta = \pi$, $180°$, $\cos\theta = -1$ |
| $\Rightarrow y = a(1 - \cos\pi) = 2a$ | A1 [4] | |
### Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$ | M1 | $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$ for theirs |
| $= \frac{a\sin\theta}{a(1-\cos\theta)}$ | M1 | $\frac{d}{d\theta}(\sin\theta) = \cos\theta$, $\frac{d}{d\theta}(\cos\theta) = -\sin\theta$ both, or equivalent; condone uncancelled $a$ |
| $= \frac{\sin\theta}{(1-\cos\theta)}$ | A1 [3] | |
### Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\tan 30° = 1/\sqrt{3} \Rightarrow \frac{\sin\theta}{(1-\cos\theta)} = \frac{1}{\sqrt{3}}$ | M1 | Or gradient $= 1/\sqrt{3}$ |
| $\Rightarrow \sin\theta = \frac{1}{\sqrt{3}}(1-\cos\theta)$ | E1 | |
| When $\theta = 2\pi/3$, $\sin\theta = \sqrt{3}/2$ | M1 | $\sin\theta = \sqrt{3}/2$, $\cos\theta = -1/2$ |
| $(1-\cos\theta)/\sqrt{3} = (1+1/2)/\sqrt{3} = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2}$ | E1 | |
| $BF = a(1 + 1/2) = 3a/2$ | E1 | |
| $OF = a(2\pi/3 - \sqrt{3}/2)$ | B1 [6] | or equiv. |
### Part (iv)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $BC = 2a\pi - 2a(2\pi/3 - \sqrt{3}/2) = a(2\pi/3 + \sqrt{3})$ | B1ft | their $OE - 2\times$their $OF$ |
| $AF = \sqrt{3} \times 3a/2 = 3\sqrt{3}a/2$ | M1 A1 | |
| $AD = BC + 2AF$ | M1 | |
| $= a(2\pi/3 + \sqrt{3} + 3\sqrt{3}) = a(2\pi/3 + 4\sqrt{3}) = 20$ | A1 [5] | |
| $\Rightarrow a = 2.22$ m | | |
---2 Fig. 6 shows the arch ABCD of a bridge.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{0c0a2fe7-9e69-470a-af2e-fa5fd41e4a27-2_378_1630_397_132}
\captionsetup{labelformat=empty}
\caption{Fig. 6}
\end{center}
\end{figure}
The section from $B$ to $C$ is part of the curve $O B C E$ with parametric equations
$$x = a ( \theta - \sin \theta ) , y = a ( 1 - \cos \theta ) \text { for } 0 \leqslant \theta \leqslant 2 \pi \text {, }$$
where $a$ is a constant.
\begin{enumerate}[label=(\roman*)]
\item Find, in terms of $a$,\\
(A) the length of the straight line OE,\\
(B) the maximum height of the arch.
\item Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $\theta$.
The straight line sections AB and CD are inclined at $30 ^ { \circ }$ to the horizontal, and are tangents to the curve at B and C respectively. BC is parallel to the $x$-axis. BF is parallel to the $y$-axis.
\item Show that at the point B the parameter $\theta$ satisfies the equation
$$\sin \theta = \frac { 1 } { \sqrt { 3 } } ( 1 \quad \cos \theta ) .$$
Verify that $\theta = \frac { 2 } { 3 } \pi$ is a solution of this equation.\\
Hence show that $\mathrm { BF } = \frac { 3 } { 2 } a$, and find OF in terms of $a$, giving your answer exactly.
\item Find BC and AF in terms of $a$.
Given that the straight line distance AD is 20 metres, calculate the value of $a$.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI C4 Q2 [18]}}