| Exam Board | AQA |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2010 |
| Session | January |
| Marks | 16 |
| 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 verification of intersection points by solving simultaneous equations, then applying the polar area formula with appropriate limits. While it involves multiple steps and careful integration, the techniques are routine for FP3 students: substituting equations, solving a quadratic in sin θ, and integrating standard polar forms. The algebraic manipulation is moderate but follows established procedures without requiring novel insight. |
| Spec | 4.09a Polar coordinates: convert to/from cartesian4.09c Area enclosed: by polar curve |
| Answer | Marks | Guidance |
|---|---|---|
| \((2\sin\theta - 1)^2 = 0 \Rightarrow \sin\theta = 0.5\) | M1, A1, m1 | Elimination of r or \(\theta\) [\(r = 4[(1-1/\pi)]]\); \(\{r^2 - 4r + 4 = 0\}\); Valid method to solve quadratic eqn. PI \(\{(r-2)^2 = 0 \Rightarrow r = 2\}\) |
| \(\theta = \frac{\pi}{6}, \theta = \frac{5\pi}{6}, r = 2\) | A2,1 | A1 for any two of the three |
| \(P\left(2, \frac{\pi}{6}\right), \quad Q\left(2, \frac{5\pi}{6}\right)\) | SC: Verification of \(P\left(2, \frac{\pi}{6}\right)\) scores max of B1 & a further B1 if \(Q\left(2, \frac{5\pi}{6}\right)\) stated; 5 marks total |
| Answer | Marks | Guidance |
|---|---|---|
| Area triangle \(OPQ = \frac{1}{2} \times 2 \times r_Q \times \sin POQ\) | M1 | Any valid method to correct (ft eg on \(r_0\)) expression with just one remaining unknown |
| Answer | Marks | Guidance |
|---|---|---|
| Area triangle \(OPQ = 2\sin\frac{2\pi}{3} = \sqrt{3}\) | m1, A1 | Valid method to find remaining unknown either relevant angle or relevant side |
| Unshaded area bounded by line \(OP\) and arc \(OP = \frac{1}{2}\int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} [4(1-\sin\theta)]^2 d\theta\) | M1 | Use of \(\frac{1}{2}\int r^2d\theta\) for relevant area(s) (condone missing/wrong limits) |
| \(= 8\int\left(1 - 2\sin\theta + \sin^2\theta\right)d\theta\) | B1 | Correct expn of \((1-\sin\theta)^2\) |
| \(= 8\int\left(1 - 2\sin\theta + \frac{1-\cos 2\theta}{2}\right)d\theta\) | M1 | Attempt to write \(\sin^2\theta\) in terms of \(\cos 2\theta\) |
| \(= 8\left[\theta + 2\cos\theta + \frac{\theta}{2} - \frac{\sin 2\theta}{4}\right] (+c)\) | A1F | Correct integration ft wrong coeffs |
| Answer | Marks | Guidance |
|---|---|---|
| \(= 8 \times \left(\frac{3\pi}{4} - \left(\frac{3\pi}{12} + 2\cos\frac{\pi}{6} - \frac{1}{4}\frac{2\pi}{6}\right)\right)\) | m1 | \(F\left(\frac{\pi}{2}\right) - F\left(\frac{\pi}{6}\right)\) OE for relevant area(s) |
| \(= 8 \times \left\{\frac{\pi}{2} - \sqrt{3} + \frac{\sqrt{3}}{8}\right\} = \{4\pi - 7\sqrt{3}\}\) | A1F | ft one slip; accept terms in \(\pi\) and \(\sqrt{3}\) left unsimplified |
| Shaded area = Area of triangle \(OPQ -\) \(2 \times \frac{1}{2}\int_{\frac{\pi}{6}}^{\frac{5\pi}{6}}[4(1-\sin\theta)]^2d\theta\) | M1 | OE |
| Shaded area \(= \sqrt{3} - 16\left(\frac{\pi}{2} - \sqrt{3} + \frac{\sqrt{3}}{8}\right) = 15\sqrt{3} - 8\pi\) | A1 | CSO Accept m = 15, n = -8; 11 marks total |
**(a)**
$4\sin\theta(1 - \sin\theta) = 1$
$4\sin^2\theta - 4\sin\theta + 1 = 0$
$(2\sin\theta - 1)^2 = 0 \Rightarrow \sin\theta = 0.5$ | M1, A1, m1 | Elimination of r or $\theta$ [$r = 4[(1-1/\pi)]]$; $\{r^2 - 4r + 4 = 0\}$; Valid method to solve quadratic eqn. PI $\{(r-2)^2 = 0 \Rightarrow r = 2\}$
$\theta = \frac{\pi}{6}, \theta = \frac{5\pi}{6}, r = 2$ | A2,1 | A1 for any two of the three
$P\left(2, \frac{\pi}{6}\right), \quad Q\left(2, \frac{5\pi}{6}\right)$ | | SC: Verification of $P\left(2, \frac{\pi}{6}\right)$ scores max of B1 & a further B1 if $Q\left(2, \frac{5\pi}{6}\right)$ stated; 5 marks total
**(b)**
Area triangle $OPQ = \frac{1}{2} \times 2 \times r_Q \times \sin POQ$ | M1 | Any valid method to correct (ft eg on $r_0$) expression with just one remaining unknown
Angle $POQ = \frac{5\pi}{6} - \frac{\pi}{6} = \frac{2\pi}{3}$
Area triangle $OPQ = 2\sin\frac{2\pi}{3} = \sqrt{3}$ | m1, A1 | Valid method to find remaining unknown either relevant angle or relevant side
Unshaded area bounded by line $OP$ and arc $OP = \frac{1}{2}\int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} [4(1-\sin\theta)]^2 d\theta$ | M1 | Use of $\frac{1}{2}\int r^2d\theta$ for relevant area(s) (condone missing/wrong limits)
$= 8\int\left(1 - 2\sin\theta + \sin^2\theta\right)d\theta$ | B1 | Correct expn of $(1-\sin\theta)^2$
$= 8\int\left(1 - 2\sin\theta + \frac{1-\cos 2\theta}{2}\right)d\theta$ | M1 | Attempt to write $\sin^2\theta$ in terms of $\cos 2\theta$
$= 8\left[\theta + 2\cos\theta + \frac{\theta}{2} - \frac{\sin 2\theta}{4}\right] (+c)$ | A1F | Correct integration ft wrong coeffs
$8\int_{\frac{\pi}{6}}^{\frac{5\pi}{6}}(1-\sin\theta)^2d\theta =$
$8 \times \left[\frac{3\theta}{2} + 2\cos\theta - \frac{\sin 2\theta}{4}\right]_{\frac{\pi}{6}}^{\frac{5\pi}{6}}$
$= 8 \times \left(\frac{3\pi}{4} - \left(\frac{3\pi}{12} + 2\cos\frac{\pi}{6} - \frac{1}{4}\frac{2\pi}{6}\right)\right)$ | m1 | $F\left(\frac{\pi}{2}\right) - F\left(\frac{\pi}{6}\right)$ OE for relevant area(s)
$= 8 \times \left\{\frac{\pi}{2} - \sqrt{3} + \frac{\sqrt{3}}{8}\right\} = \{4\pi - 7\sqrt{3}\}$ | A1F | ft one slip; accept terms in $\pi$ and $\sqrt{3}$ left unsimplified
Shaded area = Area of triangle $OPQ -$ $2 \times \frac{1}{2}\int_{\frac{\pi}{6}}^{\frac{5\pi}{6}}[4(1-\sin\theta)]^2d\theta$ | M1 | OE
Shaded area $= \sqrt{3} - 16\left(\frac{\pi}{2} - \sqrt{3} + \frac{\sqrt{3}}{8}\right) = 15\sqrt{3} - 8\pi$ | A1 | CSO Accept m = 15, n = -8; 11 marks total
**Total: 16 marks**
---
## TOTAL: 75 marks8 The diagram shows a sketch of a curve $C$ and a line $L$, which is parallel to the initial line and touches the curve at the points $P$ and $Q$.\\
\includegraphics[max width=\textwidth, alt={}, center]{32de7ef6-b7aa-4bfd-a73a-e12bfc0147e2-5_506_762_447_639}
The polar equation of the curve $C$ is
$$r = 4 ( 1 - \sin \theta ) , \quad 0 \leqslant \theta < 2 \pi$$
and the polar equation of the line $L$ is
$$r \sin \theta = 1$$
\begin{enumerate}[label=(\alph*)]
\item Show that the polar coordinates of $P$ are $\left( 2 , \frac { \pi } { 6 } \right)$ and find the polar coordinates of $Q$.
\item Find the area of the shaded region $R$ bounded by the line $L$ and the curve $C$. Give your answer in the form $m \sqrt { 3 } + n \pi$, where $m$ and $n$ are integers.
\end{enumerate}
\hfill \mbox{\textit{AQA FP3 2010 Q8 [16]}}