| Exam Board | AQA |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2007 |
| Session | January |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polar coordinates |
| Type | Polar curve intersection points |
| Difficulty | Challenging +1.8 This is a substantial multi-part polar coordinates question requiring: (a) standard polar area integration, (b) coordinate geometry proof using polar-to-Cartesian conversion, and (c) finding intersection points of lines with polar curves, then computing distances and angles. Part (c) requires solving a non-trivial system to find point R, then using geometric properties. While the techniques are within FP3 scope, the extended multi-step reasoning across several parts and the need to track multiple intersection points elevates this above routine exercises. |
| Spec | 4.09b Sketch polar curves: r = f(theta)4.09c Area enclosed: by polar curve |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance Notes |
| Area \(= \frac{1}{2}\int(6+4\cos\theta)^2\,d\theta\) | M1 | use of \(\frac{1}{2}\int r^2\,d\theta\) |
| \(= \frac{1}{2}\left(\int_{-\pi}^{\pi} 36 + 48\cos\theta + 16\cos^2\theta\,d\theta\right)\) | B1 | for correct expansion of \([6+4\cos\theta]^2\) |
| B1 | for limits | |
| \(= \left(\int_{-\pi}^{\pi} 18 + 24\cos\theta + 4(\cos 2\theta + 1)\,d\theta\right)\) | M1 | Attempt to write \(\cos^2\theta\) in terms of \(\cos 2\theta\) |
| \(= \left[22\theta + 24\sin\theta + 2\sin 2\theta\right]_{-\pi}^{\pi}\) | A1F | correct integration ft wrong coefficients |
| \(= 44\pi\) | A1 | CSO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance Notes |
| At \(P\), \(r=4\); At \(Q\), \(r=2\) | B1 | PI |
| \(P\{x=\}\) \(r\cos\theta = 4\cos\frac{2\pi}{3} = -2\) | M1 | Attempt to use \(r\cos\theta\) |
| \(Q\{x=\}\) \(r\cos\theta = 2\cos\pi = -2\) | A1 | Both |
| Since \(P\) and \(Q\) have same \(x\), \(PQ\) is vertical so \(QP\) is parallel to the vertical line \(\theta = \frac{\pi}{2}\) | E1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance Notes |
| \(OP = 4\); \(OS = 8\) | B1 | |
| Angle \(POS = \frac{\pi}{3}\) | B1 | or \(S\,(4,\,4\sqrt{3})\) and \(P\,(-2,\,2\sqrt{3})\) |
| \(PS^2 = 4^2 + 8^2 - 2\times4\times8\times\cos\frac{\pi}{3}\) oe | M1 | Cosine rule used in triangle \(POS\); OE \(PS^2 = (4+2)^2 + (4\sqrt{3}-2\sqrt{3})^2\) |
| \(PS = \sqrt{48}\ \left\{=4\sqrt{3}\right\}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance Notes |
| Since \(8^2 = 4^2 + (\sqrt{48})^2\), \(OS^2 = OP^2 + PS^2 \Rightarrow OPS\) is a right angle. (Converse of Pythagoras Theorem) | E1 | Accept valid equivalents e.g. \(PR = 2PQ = 2(2\sqrt{3}) = PS\); \(\angle SRP = \angle RSP = \angle RPO = \frac{\pi}{6} \Rightarrow OPS\) is a right angle |
# Question 7:
## Part (a):
| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| Area $= \frac{1}{2}\int(6+4\cos\theta)^2\,d\theta$ | M1 | use of $\frac{1}{2}\int r^2\,d\theta$ |
| $= \frac{1}{2}\left(\int_{-\pi}^{\pi} 36 + 48\cos\theta + 16\cos^2\theta\,d\theta\right)$ | B1 | for correct expansion of $[6+4\cos\theta]^2$ |
| | B1 | for limits |
| $= \left(\int_{-\pi}^{\pi} 18 + 24\cos\theta + 4(\cos 2\theta + 1)\,d\theta\right)$ | M1 | Attempt to write $\cos^2\theta$ in terms of $\cos 2\theta$ |
| $= \left[22\theta + 24\sin\theta + 2\sin 2\theta\right]_{-\pi}^{\pi}$ | A1F | correct integration ft wrong coefficients |
| $= 44\pi$ | A1 | CSO | **Total: 6 marks** |
## Part (b):
| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| At $P$, $r=4$; At $Q$, $r=2$ | B1 | PI |
| $P\{x=\}$ $r\cos\theta = 4\cos\frac{2\pi}{3} = -2$ | M1 | Attempt to use $r\cos\theta$ |
| $Q\{x=\}$ $r\cos\theta = 2\cos\pi = -2$ | A1 | Both |
| Since $P$ and $Q$ have same $x$, $PQ$ is vertical so $QP$ is parallel to the vertical line $\theta = \frac{\pi}{2}$ | E1 | | **Total: 4 marks** |
## Part (c)(i):
| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| $OP = 4$; $OS = 8$ | B1 | |
| Angle $POS = \frac{\pi}{3}$ | B1 | or $S\,(4,\,4\sqrt{3})$ and $P\,(-2,\,2\sqrt{3})$ |
| $PS^2 = 4^2 + 8^2 - 2\times4\times8\times\cos\frac{\pi}{3}$ oe | M1 | Cosine rule used in triangle $POS$; OE $PS^2 = (4+2)^2 + (4\sqrt{3}-2\sqrt{3})^2$ |
| $PS = \sqrt{48}\ \left\{=4\sqrt{3}\right\}$ | A1 | | **Total: 4 marks** |
## Part (c)(ii):
| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| Since $8^2 = 4^2 + (\sqrt{48})^2$, $OS^2 = OP^2 + PS^2 \Rightarrow OPS$ is a right angle. (Converse of Pythagoras Theorem) | E1 | Accept valid equivalents e.g. $PR = 2PQ = 2(2\sqrt{3}) = PS$; $\angle SRP = \angle RSP = \angle RPO = \frac{\pi}{6} \Rightarrow OPS$ is a right angle | **Total: 1 mark** |
**Question 7 Total: 15 marks**
**TOTAL: 75 marks**7 A curve $C$ has polar equation
$$r = 6 + 4 \cos \theta , \quad - \pi \leqslant \theta \leqslant \pi$$
The diagram shows a sketch of the curve $C$, the pole $O$ and the initial line.\\
\includegraphics[max width=\textwidth, alt={}, center]{0d894ac0-8d96-4182-8454-c306e1fdad8f-4_599_866_612_587}
\begin{enumerate}[label=(\alph*)]
\item Calculate the area of the region bounded by the curve $C$.
\item The point $P$ is the point on the curve $C$ for which $\theta = \frac { 2 \pi } { 3 }$.
The point $Q$ is the point on $C$ for which $\theta = \pi$.\\
Show that $Q P$ is parallel to the line $\theta = \frac { \pi } { 2 }$.
\item The line $P Q$ intersects the curve $C$ again at a point $R$.
The line $R O$ intersects $C$ again at a point $S$.
\begin{enumerate}[label=(\roman*)]
\item Find, in surd form, the length of $P S$.
\item Show that the angle $O P S$ is a right angle.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA FP3 2007 Q7 [15]}}