| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2015 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polar coordinates |
| Type | Area of region with line boundary |
| Difficulty | Challenging +1.2 This is a standard Further Maths polar coordinates question requiring (a) finding where dy/dx = 0 using the polar tangent condition, and (b) computing a polar area integral. While it involves FP2 content and multi-step working, the techniques are routine applications of standard formulas with straightforward integration of trigonometric functions. |
| Spec | 4.09b Sketch polar curves: r = f(theta)4.09c Area enclosed: by polar curve |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(r\sin\theta = 3a\sin\theta + 3a\sin\theta\cos\theta\) OR \(3a\sin\theta + \frac{3}{2}a\sin 2\theta\) | M1 | Using \(r\sin\theta\); \(r\cos\theta\) scores M0 |
| \(\frac{d(r\sin\theta)}{d\theta} = 3a\cos\theta + 3a\cos^2\theta - 3a\sin^2\theta\) OR \(3a\cos\theta + 3a\cos 2\theta\) | dM1 | Attempt differentiation of \(r\sin\theta\), inc product rule or \(\sin 2\theta = 2\sin\theta\cos\theta\) |
| \(2\cos^2\theta + \cos\theta - 1 = 0\), i.e. \((2\cos\theta-1)(\cos\theta+1)=0\) | A1 | Correct 3-term quadratic in \(\cos\theta\) |
| \(\cos\theta = \frac{1}{2}\), \(\theta = \frac{\pi}{3}\) (\(\theta=\pi\) need not be seen) | ddM1A1 | Dep on both M marks; solve quadratic; A1 for \(\theta=\frac{\pi}{3}\) |
| \(r = 3a \times \frac{3}{2} = \frac{9}{2}a\) | A1 (6) | Correct \(r\) obtained |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| Area \(= \frac{1}{2}\int r^2\, d\theta = \frac{1}{2}\int_0^{\pi/3} 9a^2(1+\cos\theta)^2\, d\theta\) | M1 | Use of correct area formula; \(\frac{1}{2}\) may appear later; squaring bracket to get 3 terms |
| \(= \frac{9a^2}{2}\int_0^{\pi/3}\left(1+2\cos\theta+\frac{1}{2}(\cos 2\theta+1)\right)d\theta\) | M1 | Use double angle formula \(\cos^2\theta = \pm\frac{1}{2}(\cos 2\theta \pm 1)\) to get integrable function |
| \(= \frac{9a^2}{2}\left[\theta + 2\sin\theta + \frac{1}{2}\left(\frac{1}{2}\sin 2\theta+\theta\right)\right]_0^{\pi/3}\) | dM1A1 | Attempt integration; dep on 2nd M; A1 correct integration |
| \(= \frac{9a^2}{2}\left[\frac{\pi}{3}+\sqrt{3}+\frac{1}{4}\times\frac{\sqrt{3}}{2}+\frac{\pi}{6}\right]\) | ||
| \(= \frac{9a^2}{2}\left[\frac{\pi}{2}+\frac{9\sqrt{3}}{8}\right] = \left(\frac{9\pi}{4}+\frac{81\sqrt{3}}{16}\right)a^2\) | A1 (5)[11] | Correct final answer; any equivalent form |
## Question 6:
### Part (a):
| Working | Marks | Guidance |
|---------|-------|----------|
| $r\sin\theta = 3a\sin\theta + 3a\sin\theta\cos\theta$ OR $3a\sin\theta + \frac{3}{2}a\sin 2\theta$ | M1 | Using $r\sin\theta$; $r\cos\theta$ scores M0 |
| $\frac{d(r\sin\theta)}{d\theta} = 3a\cos\theta + 3a\cos^2\theta - 3a\sin^2\theta$ OR $3a\cos\theta + 3a\cos 2\theta$ | dM1 | Attempt differentiation of $r\sin\theta$, inc product rule or $\sin 2\theta = 2\sin\theta\cos\theta$ |
| $2\cos^2\theta + \cos\theta - 1 = 0$, i.e. $(2\cos\theta-1)(\cos\theta+1)=0$ | A1 | Correct 3-term quadratic in $\cos\theta$ |
| $\cos\theta = \frac{1}{2}$, $\theta = \frac{\pi}{3}$ ($\theta=\pi$ need not be seen) | ddM1A1 | Dep on both M marks; solve quadratic; A1 for $\theta=\frac{\pi}{3}$ |
| $r = 3a \times \frac{3}{2} = \frac{9}{2}a$ | A1 (6) | Correct $r$ obtained |
### Part (b):
| Working | Marks | Guidance |
|---------|-------|----------|
| Area $= \frac{1}{2}\int r^2\, d\theta = \frac{1}{2}\int_0^{\pi/3} 9a^2(1+\cos\theta)^2\, d\theta$ | M1 | Use of correct area formula; $\frac{1}{2}$ may appear later; squaring bracket to get 3 terms |
| $= \frac{9a^2}{2}\int_0^{\pi/3}\left(1+2\cos\theta+\frac{1}{2}(\cos 2\theta+1)\right)d\theta$ | M1 | Use double angle formula $\cos^2\theta = \pm\frac{1}{2}(\cos 2\theta \pm 1)$ to get integrable function |
| $= \frac{9a^2}{2}\left[\theta + 2\sin\theta + \frac{1}{2}\left(\frac{1}{2}\sin 2\theta+\theta\right)\right]_0^{\pi/3}$ | dM1A1 | Attempt integration; dep on 2nd M; A1 correct integration |
| $= \frac{9a^2}{2}\left[\frac{\pi}{3}+\sqrt{3}+\frac{1}{4}\times\frac{\sqrt{3}}{2}+\frac{\pi}{6}\right]$ | | |
| $= \frac{9a^2}{2}\left[\frac{\pi}{2}+\frac{9\sqrt{3}}{8}\right] = \left(\frac{9\pi}{4}+\frac{81\sqrt{3}}{16}\right)a^2$ | A1 (5)[11] | Correct final answer; any equivalent form |
---6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{49da3c56-ccd1-4599-95d8-d1395461bcca-11_451_1063_237_438}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
The curve $C$, shown in Figure 1, has polar equation
$$r = 3 a ( 1 + \cos \theta ) , \quad 0 \leqslant \theta < \pi$$
The tangent to $C$ at the point $A$ is parallel to the initial line.
\begin{enumerate}[label=(\alph*)]
\item Find the polar coordinates of $A$.
The finite region $R$, shown shaded in Figure 1, is bounded by the curve $C$, the initial line and the line $O A$.
\item Use calculus to find the area of the shaded region $R$, giving your answer in the form $a ^ { 2 } ( p \pi + q \sqrt { 3 } )$, where $p$ and $q$ are rational numbers.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP2 2015 Q6 [11]}}