| Exam Board | WJEC |
|---|---|
| Module | Further Unit 4 (Further Unit 4) |
| Year | 2022 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polar coordinates |
| Type | Tangent parallel/perpendicular to initial line |
| Difficulty | Challenging +1.3 This is a Further Maths polar coordinates question requiring a sketch of a limaçon, derivation of a tangent condition using dr/dθ and the polar tangent formula, and solving a quadratic in cos θ. While it involves multiple steps and Further Maths content (polar curves), the techniques are standard: the sketch is routine for this curve type, the tangent condition follows a well-practiced formula (tan ψ = r/(dr/dθ)), and the final quadratic is straightforward. The 9 marks for part (b) reflect length rather than exceptional difficulty—this is a typical Further Maths examination question testing procedural fluency rather than requiring novel insight. |
| Spec | 4.09b Sketch polar curves: r = f(theta) |
| Answer | Marks | Guidance |
|---|---|---|
| [Graph showing circle with center at approximately (1, 0) and passing through origin, with reflection in initial line] | G1 | For shape, to include reflection in the initial line |
| G1 | Fully correct |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = r\sin\theta\) | M1 | |
| \(y = (2 - \cos\theta)\sin\theta\) | ||
| \(y = 2\sin\theta - \sin\theta\cos\theta\) | ||
| THEN | ||
| \(\left(y = 2\sin\theta - \frac{1}{2}\sin 2\theta\right)\) | ||
| \(\frac{dy}{d\theta} = 2\cos\theta - \cos 2\theta\) | A1 | |
| When parallel to initial line, | m1 | |
| \(2\cos\theta - \cos 2\theta = 0\) | ||
| \(2\cos\theta - (2\cos^2\theta - 1) = 0\) | A1 | convincing |
| \(2\cos^2\theta - 2\cos\theta - 1 = 0\) | ||
| OR | ||
| \(\frac{dy}{d\theta} = 2\cos\theta - (\cos^2\theta - \sin^2\theta)\) | (A1) | |
| When parallel to initial line, | ||
| \(2\cos\theta - (\cos^2\theta - \sin^2\theta) = 0\) | (m1) | |
| \(2\cos\theta - \cos^2\theta + (1 - \cos^2\theta) = 0\) | (A1) | convincing |
| \(2\cos^2\theta - 2\cos\theta - 1 = 0\) |
| Answer | Marks | Guidance |
|---|---|---|
| Solving | M1 | |
| \(\cos\theta = \frac{2 \pm \sqrt{4 + 8}}{4} = \frac{2 \pm \sqrt{4+8}}{4}\) | ||
| \(\cos\theta = 1.366\) therefore no solutions | A1 | |
| or | A1 | |
| \(\cos\theta = -0.366\) | A1 | Both values |
| B1 | FT their \(\theta\) | |
| \(\therefore \theta = 1.9455\) or \(4.3377\) | ||
| \(r = 2.366\) |
**Part a)**
[Graph showing circle with center at approximately (1, 0) and passing through origin, with reflection in initial line] | G1 | For shape, to include reflection in the initial line
| G1 | Fully correct
**Part b) i)**
$y = r\sin\theta$ | M1 |
$y = (2 - \cos\theta)\sin\theta$ | |
$y = 2\sin\theta - \sin\theta\cos\theta$ | |
THEN | |
$\left(y = 2\sin\theta - \frac{1}{2}\sin 2\theta\right)$ | |
$\frac{dy}{d\theta} = 2\cos\theta - \cos 2\theta$ | A1 |
When parallel to initial line, | m1 |
$2\cos\theta - \cos 2\theta = 0$ | |
$2\cos\theta - (2\cos^2\theta - 1) = 0$ | A1 | convincing
$2\cos^2\theta - 2\cos\theta - 1 = 0$ | |
OR | |
$\frac{dy}{d\theta} = 2\cos\theta - (\cos^2\theta - \sin^2\theta)$ | (A1) |
When parallel to initial line, | |
$2\cos\theta - (\cos^2\theta - \sin^2\theta) = 0$ | (m1) |
$2\cos\theta - \cos^2\theta + (1 - \cos^2\theta) = 0$ | (A1) | convincing
$2\cos^2\theta - 2\cos\theta - 1 = 0$ | |
**Part b) ii)**
Solving | M1 |
$\cos\theta = \frac{2 \pm \sqrt{4 + 8}}{4} = \frac{2 \pm \sqrt{4+8}}{4}$ | |
$\cos\theta = 1.366$ therefore no solutions | A1 |
or | A1 |
$\cos\theta = -0.366$ | A1 | Both values
| B1 | FT their $\theta$
$\therefore \theta = 1.9455$ or $4.3377$ | |
$r = 2.366$ | |
---The curve C has polar equation $r = 2 - \cos\theta$ for $0 \leq \theta \leq 2\pi$.
\begin{enumerate}[label=(\alph*)]
\item Sketch the curve C. [2]
\item \begin{enumerate}[label=(\roman*)]
\item Show that the values of $\theta$ at which the tangent to the curve $r = 2 - \cos\theta$ is parallel to the initial line satisfy the equation
$$2\cos^2\theta - 2\cos\theta - 1 = 0.$$
\item Find the polar coordinates of the points where the tangent to the curve $r = 2 - \cos\theta$ is parallel to the initial line. [9]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 4 2022 Q13 [11]}}