| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2006 |
| Session | January |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polar coordinates |
| Type | Tangent parallel/perpendicular to initial line |
| Difficulty | Challenging +1.8 This is a challenging Further Maths polar coordinates problem requiring multiple sophisticated techniques: expressing r²sin²θ in terms of sinθ only using the curve equation, implicit differentiation of polar equations to find tangent conditions (dy/dx = 0 for horizontal tangent), and polar area integration with trigonometric substitution. The multi-step nature, need for strategic algebraic manipulation, and integration complexity place this well above average difficulty, though the question provides significant scaffolding through its parts. |
| Spec | 4.09a Polar coordinates: convert to/from cartesian4.09c Area enclosed: by polar curve |
| Answer | Marks | Guidance |
|---|---|---|
| \(r^2\sin^2\theta = a^2\cos 2\theta\sin^2\theta = a^2(1 - 2\sin^2\theta)\sin^2\theta\) | B1 | 1 |
| \(= a^2(\sin^2\theta - 2\sin^4\theta)\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{d}{d\theta}(a^2(\sin^2\theta - 2\sin^4\theta)) = a^2(2\sin\theta\cos\theta - 8\sin^3\theta\cos\theta) = 0\) | M1, A1, M1 | |
| \(2 = 8\sin^2\theta\) | M1 | Proceed to \(\sin^2\theta = b\) |
| \(\sin\theta = \frac{1}{2} \Rightarrow \theta = \frac{\pi}{6}, r = \frac{a}{\sqrt{2}}\) | A1, A1 eso | 6 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{a^2}{2}\int\cos 2\theta d\theta = \frac{a^2}{4}\sin 2\theta\) | M1, A1 | M: Attempt \(\frac{1}{2}\int r^2 d\theta\), to get \(k\sin 2\theta\) M1 A1 |
| \(\left[.\right]_{\pi/6}^{a^2}{4}\left[1 - \frac{\sqrt{3}}{2}\right]\) | M1, A1 | M: Using correct limits |
| \(\Delta = \frac{1}{2}\left(\frac{a}{\sqrt{2}} \cdot \frac{1}{2}\right) \times \left(\frac{a}{\sqrt{2}} - \frac{\sqrt{3}}{2}\right) = \frac{\sqrt{3}a^2}{16}\) | M1, A1 | M: Full method for rectangle or triangle |
| \(R = \frac{\sqrt{3}a^2}{16} - \frac{a^2}{4}\left[1 - \frac{\sqrt{3}}{2}\right] = \frac{a^2}{16}(3\sqrt{3} - 4)\) | dM1, A1 eso | 8 |
**Part (a)(i)**
| $r^2\sin^2\theta = a^2\cos 2\theta\sin^2\theta = a^2(1 - 2\sin^2\theta)\sin^2\theta$ | B1 | 1 |
| $= a^2(\sin^2\theta - 2\sin^4\theta)$ | | |
**Part (a)(ii)**
| $\frac{d}{d\theta}(a^2(\sin^2\theta - 2\sin^4\theta)) = a^2(2\sin\theta\cos\theta - 8\sin^3\theta\cos\theta) = 0$ | M1, A1, M1 | |
| $2 = 8\sin^2\theta$ | M1 | Proceed to $\sin^2\theta = b$ |
| $\sin\theta = \frac{1}{2} \Rightarrow \theta = \frac{\pi}{6}, r = \frac{a}{\sqrt{2}}$ | A1, A1 eso | 6 |
**Part (b)**
| $\frac{a^2}{2}\int\cos 2\theta d\theta = \frac{a^2}{4}\sin 2\theta$ | M1, A1 | M: Attempt $\frac{1}{2}\int r^2 d\theta$, to get $k\sin 2\theta$ M1 A1 |
| $\left[.\right]_{\pi/6}^{a^2}{4}\left[1 - \frac{\sqrt{3}}{2}\right]$ | M1, A1 | M: Using correct limits |
| $\Delta = \frac{1}{2}\left(\frac{a}{\sqrt{2}} \cdot \frac{1}{2}\right) \times \left(\frac{a}{\sqrt{2}} - \frac{\sqrt{3}}{2}\right) = \frac{\sqrt{3}a^2}{16}$ | M1, A1 | M: Full method for rectangle or triangle |
| $R = \frac{\sqrt{3}a^2}{16} - \frac{a^2}{4}\left[1 - \frac{\sqrt{3}}{2}\right] = \frac{a^2}{16}(3\sqrt{3} - 4)$ | dM1, A1 eso | 8 |
[15]
**Notes:**
- Part (a)(ii): First A1 is for correct derivative of a correct expression for $r^2\sin^2\theta$ or $r\sin\theta$.
- Part (b): Final M mark is dependent on first and third M's. Attempts at triangle area by integration require full method for M1. Missing $a$ factors (or $a^2$): Maximum one mark penalty in the question.4. A curve $C$ has polar equation $r ^ { 2 } = a ^ { 2 } \cos 2 \theta , 0 \leq \theta \leq \frac { \pi } { 4 }$.
The line $l$ is parallel to the initial line, and $l$ is the tangent to $C$ at\\
above. above.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Show that, for any point on $C , r ^ { 2 } \sin ^ { 2 } \theta$ can be expressed in terms of $\sin \theta$ and $a$ only. (1)
\item Hence, using differentiation, show that the polar coordinates of $P$ are $\left( \frac { a } { \sqrt { 2 } } , \frac { \pi } { 6 } \right)$.(6)\\
\includegraphics[max width=\textwidth, alt={}, center]{2352f367-ddf9-4770-ace5-b561b0fbabbb-1_298_725_2163_1169}
The shaded region $R$, shown in the figure above, is bounded by $C$, the line $l$ and the half-line with equation\\
$\theta = \frac { \pi } { 2 }$.
\end{enumerate}\item Show that the area of $R$ is $\frac { a ^ { 2 } } { 16 } ( 3 \sqrt { 3 } - 4 )$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP2 2006 Q4 [15]}}