| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2014 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polar coordinates |
| Type | Tangent parallel/perpendicular to initial line |
| Difficulty | Challenging +1.2 This is a Further Maths FP2 polar coordinates question requiring the tangent formula dy/dx = (r'sinθ + rcosθ)/(r'cosθ - rsinθ), setting the numerator to zero for horizontal tangent, then solving. It involves differentiation, trigonometric manipulation, and converting back to polar form. While systematic, it requires multiple techniques and careful algebra, making it moderately harder than average A-level questions but standard for FP2. |
| Spec | 1.07s Parametric and implicit differentiation4.09b Sketch polar curves: r = f(theta) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((y=)\, r\sin\theta = 2\cos 2\theta \sin\theta\) | M1 | Using \(y = r\sin\theta = 2\cos 2\theta \sin\theta\) |
| \(\frac{dy}{d\theta} = -4\sin 2\theta \sin\theta + 2\cos 2\theta \cos\theta\) | M1A1 | M1: differentiate \(r\sin\theta\) or \(r\cos\theta\) using product rule; A1: correct differentiation |
| \(2\sin 2\theta \sin\theta - \cos 2\theta \cos\theta = 0\) | ||
| \(4\sin^2\theta\cos\theta - (1-2\sin^2\theta)\cos\theta = 0\) | dM1 | Equate derivative to 0 and use \(\cos 2\theta = 1-2\sin^2\theta\) |
| \((6\sin^2\theta - 1)\cos\theta = 0\) | ||
| \((\cos\theta = 0\) no solutions in range\()\) | ||
| \(\therefore \sin\theta = \frac{1}{\sqrt{6}}\) | ddM1A1 | ddM1: solve resulting equation; A1: correct value |
| \(\sin\theta = \frac{1}{\sqrt{6}} \Rightarrow \cos 2\theta = 1 - 2\sin^2\theta = 1 - \frac{2}{6} = \frac{2}{3}\) | M1 | Use value to obtain exact value for \(\cos 2\theta\) |
| \(r\sin\theta = 2 \times \frac{2}{3} \times \frac{1}{\sqrt{6}} = \frac{4}{3\sqrt{6}}\) | M1 | Use values for \(\sin\theta\) and \(\cos 2\theta\) in \(r\sin\theta = 2\cos 2\theta\sin\theta\); requires \(0 \leq \sin\theta \leq \frac{1}{\sqrt{2}}\) |
| \(r = \frac{2\sqrt{6}}{9}\csc\theta\) \((0 < \theta < \pi)\) | A1 | Correct equation in exact form |
# Question 4:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(y=)\, r\sin\theta = 2\cos 2\theta \sin\theta$ | M1 | Using $y = r\sin\theta = 2\cos 2\theta \sin\theta$ |
| $\frac{dy}{d\theta} = -4\sin 2\theta \sin\theta + 2\cos 2\theta \cos\theta$ | M1A1 | M1: differentiate $r\sin\theta$ or $r\cos\theta$ using product rule; A1: correct differentiation |
| $2\sin 2\theta \sin\theta - \cos 2\theta \cos\theta = 0$ | | |
| $4\sin^2\theta\cos\theta - (1-2\sin^2\theta)\cos\theta = 0$ | dM1 | Equate derivative to 0 and use $\cos 2\theta = 1-2\sin^2\theta$ |
| $(6\sin^2\theta - 1)\cos\theta = 0$ | | |
| $(\cos\theta = 0$ no solutions in range$)$ | | |
| $\therefore \sin\theta = \frac{1}{\sqrt{6}}$ | ddM1A1 | ddM1: solve resulting equation; A1: correct value |
| $\sin\theta = \frac{1}{\sqrt{6}} \Rightarrow \cos 2\theta = 1 - 2\sin^2\theta = 1 - \frac{2}{6} = \frac{2}{3}$ | M1 | Use value to obtain exact value for $\cos 2\theta$ |
| $r\sin\theta = 2 \times \frac{2}{3} \times \frac{1}{\sqrt{6}} = \frac{4}{3\sqrt{6}}$ | M1 | Use values for $\sin\theta$ and $\cos 2\theta$ in $r\sin\theta = 2\cos 2\theta\sin\theta$; requires $0 \leq \sin\theta \leq \frac{1}{\sqrt{2}}$ |
| $r = \frac{2\sqrt{6}}{9}\csc\theta$ $(0 < \theta < \pi)$ | A1 | Correct equation in exact form |
---4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{c9fff982-d38b-42ff-ab4e-08008439a95b-06_456_1273_262_388}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows the curve $C$ with polar equation
$$r = 2 \cos 2 \theta , \quad 0 \leqslant \theta \leqslant \frac { \pi } { 4 }$$
The line $l$ is parallel to the initial line and is a tangent to $C$.
Find an equation of $l$, giving your answer in the form $r = \mathrm { f } ( \theta )$.\\
\hfill \mbox{\textit{Edexcel FP2 2014 Q4 [9]}}