| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2010 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polar coordinates |
| Type | Area between two polar curves |
| Difficulty | Challenging +1.2 This is a standard Further Maths polar coordinates question requiring finding intersections by solving a trigonometric equation, then computing area between curves using the polar area formula. While it involves multiple steps and careful setup of integration limits, the techniques are routine for FP2 students: solving sin(3θ) = 0.5, applying ½∫(r₂² - r₁²)dθ, and integrating standard trigonometric functions. The algebraic manipulation to reach the specified form requires care but no novel insight. |
| Spec | 4.09a Polar coordinates: convert to/from cartesian4.09b Sketch polar curves: r = f(theta)4.09c Area enclosed: by polar curve |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(1.5 + \sin 3\theta = 2 \rightarrow \sin 3\theta = 0.5 \therefore 3\theta = \frac{\pi}{6}\) (or \(\frac{5\pi}{6}\)) | M1 A1 | |
| \(\therefore \theta = \frac{\pi}{18}\) or \(\frac{5\pi}{18}\) | A1 | (3 marks total) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\text{Area} = \frac{1}{2}\left[\int_{\frac{\pi}{18}}^{\frac{5\pi}{18}}(1.5 + \sin 3\theta)^2\, d\theta\right] - \frac{1}{9}\pi \times 2^2\) | M1, M1 | |
| \(= \frac{1}{2}\left[\int_{\frac{\pi}{18}}^{\frac{5\pi}{18}}(2.25 + 3\sin 3\theta + \frac{1}{2}(1-\cos 6\theta))\,d\theta\right] - \frac{1}{9}\pi \times 2^2\) | M1 | |
| \(= \frac{1}{2}\left[(2.25\theta - \cos 3\theta + \frac{1}{2}(\theta - \frac{1}{6}\sin 6\theta))\right]_{\frac{\pi}{18}}^{\frac{5\pi}{18}} - \frac{1}{9}\pi \times 2^2\) | M1 A1 | |
| \(= \frac{13\sqrt{3}}{24} - \frac{5\pi}{36}\) | M1 A1 | (7 marks total) |
## Question 5:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $1.5 + \sin 3\theta = 2 \rightarrow \sin 3\theta = 0.5 \therefore 3\theta = \frac{\pi}{6}$ (or $\frac{5\pi}{6}$) | M1 A1 | |
| $\therefore \theta = \frac{\pi}{18}$ or $\frac{5\pi}{18}$ | A1 | (3 marks total) |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\text{Area} = \frac{1}{2}\left[\int_{\frac{\pi}{18}}^{\frac{5\pi}{18}}(1.5 + \sin 3\theta)^2\, d\theta\right] - \frac{1}{9}\pi \times 2^2$ | M1, M1 | |
| $= \frac{1}{2}\left[\int_{\frac{\pi}{18}}^{\frac{5\pi}{18}}(2.25 + 3\sin 3\theta + \frac{1}{2}(1-\cos 6\theta))\,d\theta\right] - \frac{1}{9}\pi \times 2^2$ | M1 | |
| $= \frac{1}{2}\left[(2.25\theta - \cos 3\theta + \frac{1}{2}(\theta - \frac{1}{6}\sin 6\theta))\right]_{\frac{\pi}{18}}^{\frac{5\pi}{18}} - \frac{1}{9}\pi \times 2^2$ | M1 A1 | |
| $= \frac{13\sqrt{3}}{24} - \frac{5\pi}{36}$ | M1 A1 | (7 marks total) |
---5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{3ff7c42d-40b0-4d59-8716-14de4890ac1b-06_524_750_219_610}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows the curves given by the polar equations
$$r = 2 , \quad 0 \leqslant \theta \leqslant \frac { \pi } { 2 } ,$$
and
$$r = 1.5 + \sin 3 \theta , \quad 0 \leqslant \theta \leqslant \frac { \pi } { 2 }$$
\begin{enumerate}[label=(\alph*)]
\item Find the coordinates of the points where the curves intersect.
The region $S$, between the curves, for which $r > 2$ and for which $r < ( 1.5 + \sin 3 \theta )$, is shown shaded in Figure 1.
\item Find, by integration, the area of the shaded region $S$, giving your answer in the form $a \pi + b \sqrt { 3 }$, where $a$ and $b$ are simplified fractions.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP2 2010 Q5 [10]}}