| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2007 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polar coordinates |
| Type | Area of region with line boundary |
| Difficulty | Challenging +1.2 This is a multi-part Further Maths polar coordinates question requiring: (a) finding where tangent is perpendicular to initial line using dy/dx=0, (b) setting up area integral and algebraic manipulation to match given form, (c) integration using standard techniques. While it involves several steps and FP2 content, each part follows standard procedures without requiring novel insight—the algebraic form in part (b) is given to guide students, and the integration uses routine trigonometric identities. |
| Spec | 1.05l Double angle formulae: and compound angle formulae1.08d Evaluate definite integrals: between limits4.09a Polar coordinates: convert to/from cartesian4.09b Sketch polar curves: r = f(theta)4.09c Area enclosed: by polar curve |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(x = r\sin\theta \Rightarrow \frac{dx}{d\theta} = 4\sin\theta\cos^3\theta\) | M1 | |
| \(\frac{dr}{d\theta} = 4\cos^3\theta - 12\cos\theta\sin^2\theta\) | ||
| \(\frac{dx}{d\theta} = 0 \left[\frac{dx}{d\theta} = 0 \Rightarrow 4\cos^2\theta(\cos^2\theta - 3\sin^2\theta) = 0\right]\) | M1A1 | |
| Solving | M1 | |
| \(\sin\theta = \frac{1}{2}\) or \(\cos\theta = \frac{\sqrt{3}}{2}\) or \(\tan\theta = \frac{1}{\sqrt{3}} \Rightarrow \theta = \frac{\pi}{6}\) | AG | A1cso |
| \(r = \frac{4\sin\frac{\pi}{6}\cos^2\frac{\pi}{6}}{} = \frac{3}{2}\) | AG | A1cso6 |
| So many ways x may be expressing e.g. \(2\sin 2\theta\cos^2\theta\), \(\sin 2A1 + \cos 2\theta\), \(\sin 2\theta + (1/2)\sin 4\theta\) leading to many results for \(\frac{dx}{d\theta}\). Some relevant equations in solving: \([(1 - 4\sin^2\theta) = 0, (4\cos^2\theta - 3) = 0, (1 - 3\tan^2\theta) = 0, \cos 3\theta = 0]\). Showing that \(\theta = \frac{\pi}{6}\) satisfies \(\frac{dx}{d\theta} = 0\), allow M1A1, providing \(\frac{dx}{d\theta}\) correct. Starting with \(x = r\sin\theta\) can gain M0M1M1 | ||
| (b) \(A = \frac{1}{2}\int_{\pi/6}^{\pi/4} r^2 d\theta = \frac{1}{2} \cdot 16 \int_{\pi/6}^{\pi/4} \sin^2\theta\cos^4 t d\theta\) | ||
| \(8\sin^2\theta\cos^4\theta = 2\cos^2\theta A(4\sin^2\theta\cos^2\theta) = 2\cos^2\theta\sin^2 2\theta\) | M1 | |
| \(= (\cos 2\theta + 1)\sin^2 2\theta\) | M1 | |
| \(= \cos 2\theta\sin^2 2\theta + \frac{1 - \cos 4\theta}{2}\) | AG | A1cso3 |
| First M1 for use of double angle formula for \(\sin 2A\). Second M1 for use of \(\cos 2A = 2\cos^2 A - 1\). Answer given: must be intermediate step, as shown, and no incorrect work | ||
| (c) Area \(= \left[\frac{1}{6}\sin^3 2\theta + \frac{\theta}{2} - \frac{\sin 4\theta}{8}\right]_{\pi/4}^{\pi/6}\) | M1A1 | ignore limits |
| M1 | (sub. limits) | |
| \(= \left(\frac{1}{6}\sin^3\frac{\pi}{2} + \frac{\pi}{8} - \frac{\sin\pi}{8}\right) - \left(\frac{1}{6}\sin^3\frac{\pi}{3} + \frac{\pi}{12} - \frac{\sin\frac{2\pi}{3}}{8}\right)\) | ||
| \(= \left(\frac{1}{6} + \frac{\pi}{8} - \left(\frac{\sqrt{3}}{16} + \frac{\pi}{12} - \frac{\sqrt{3}}{16}\right)\right) = \frac{1}{6} + \frac{\pi}{8} - \frac{\pi}{12}\) | both cao | A1, A15 |
| \(= \frac{1 + \pi}{6} + \frac{\pi}{24}\) | ||
| For first M, of the form \(a\sin^3 2\theta + \frac{\theta}{2} \pm b\sin 4\theta\) (Allow if two of correct form). On ePen the order of the As in answer is as written |
**(a)** $x = r\sin\theta \Rightarrow \frac{dx}{d\theta} = 4\sin\theta\cos^3\theta$ | M1 |
$\frac{dr}{d\theta} = 4\cos^3\theta - 12\cos\theta\sin^2\theta$ | |
$\frac{dx}{d\theta} = 0 \left[\frac{dx}{d\theta} = 0 \Rightarrow 4\cos^2\theta(\cos^2\theta - 3\sin^2\theta) = 0\right]$ | M1A1 |
Solving | M1 |
$\sin\theta = \frac{1}{2}$ or $\cos\theta = \frac{\sqrt{3}}{2}$ or $\tan\theta = \frac{1}{\sqrt{3}} \Rightarrow \theta = \frac{\pi}{6}$ | AG | A1cso |
$r = \frac{4\sin\frac{\pi}{6}\cos^2\frac{\pi}{6}}{} = \frac{3}{2}$ | AG | A1cso6 |
So many ways x may be expressing e.g. $2\sin 2\theta\cos^2\theta$, $\sin 2A1 + \cos 2\theta$, $\sin 2\theta + (1/2)\sin 4\theta$ leading to many results for $\frac{dx}{d\theta}$. Some relevant equations in solving: $[(1 - 4\sin^2\theta) = 0, (4\cos^2\theta - 3) = 0, (1 - 3\tan^2\theta) = 0, \cos 3\theta = 0]$. Showing that $\theta = \frac{\pi}{6}$ satisfies $\frac{dx}{d\theta} = 0$, allow M1A1, providing $\frac{dx}{d\theta}$ correct. Starting with $x = r\sin\theta$ can gain M0M1M1 | |
**(b)** $A = \frac{1}{2}\int_{\pi/6}^{\pi/4} r^2 d\theta = \frac{1}{2} \cdot 16 \int_{\pi/6}^{\pi/4} \sin^2\theta\cos^4 t d\theta$ | |
$8\sin^2\theta\cos^4\theta = 2\cos^2\theta A(4\sin^2\theta\cos^2\theta) = 2\cos^2\theta\sin^2 2\theta$ | M1 |
$= (\cos 2\theta + 1)\sin^2 2\theta$ | M1 |
$= \cos 2\theta\sin^2 2\theta + \frac{1 - \cos 4\theta}{2}$ | AG | A1cso3 |
First M1 for use of double angle formula for $\sin 2A$. Second M1 for use of $\cos 2A = 2\cos^2 A - 1$. Answer given: must be intermediate step, as shown, and no incorrect work | |
**(c)** Area $= \left[\frac{1}{6}\sin^3 2\theta + \frac{\theta}{2} - \frac{\sin 4\theta}{8}\right]_{\pi/4}^{\pi/6}$ | M1A1 | ignore limits
| | M1 | (sub. limits) |
$= \left(\frac{1}{6}\sin^3\frac{\pi}{2} + \frac{\pi}{8} - \frac{\sin\pi}{8}\right) - \left(\frac{1}{6}\sin^3\frac{\pi}{3} + \frac{\pi}{12} - \frac{\sin\frac{2\pi}{3}}{8}\right)$ | |
| | | |
$= \left(\frac{1}{6} + \frac{\pi}{8} - \left(\frac{\sqrt{3}}{16} + \frac{\pi}{12} - \frac{\sqrt{3}}{16}\right)\right) = \frac{1}{6} + \frac{\pi}{8} - \frac{\pi}{12}$ | both cao | A1, A15 |
$= \frac{1 + \pi}{6} + \frac{\pi}{24}$ | |
For first M, of the form $a\sin^3 2\theta + \frac{\theta}{2} \pm b\sin 4\theta$ (Allow if two of correct form). On ePen the order of the As in answer is as written | |
**Total: [14]**
---4.\\
\includegraphics[max width=\textwidth, alt={}, center]{d6befd60-de40-41b6-8ae5-48656dbca40c-3_535_1027_276_577}
The diagram above shows a sketch of the curve $C$ with polar equation
$$r = 4 \sin \theta \cos ^ { 2 } \theta , \quad 0 \leq \theta < \frac { \pi } { 2 }$$
The tangent to $C$ at the point $P$ is perpendicular to the initial line.
\begin{enumerate}[label=(\alph*)]
\item Show that $P$ has polar coordinates $\left( \frac { 3 } { 2 } , \frac { \pi } { 6 } \right)$.
The point $Q$ on $C$ has polar coordinates $\left( \sqrt { 2 } , \frac { \pi } { 4 } \right)$.\\
The shaded region $R$ is bounded by $O P , O Q$ and $C$, as shown in the diagram above.
\item Show that the area of $R$ is given by
$$\int _ { \frac { \pi } { 6 } } ^ { \frac { \pi } { 4 } } \left( \sin ^ { 2 } 2 \theta \cos 2 \theta + \frac { 1 } { 2 } - \frac { 1 } { 2 } \cos 4 \theta \right) \mathrm { d } \theta$$
\item Hence, or otherwise, find the area of $R$, giving your answer in the form $a + b \pi$, where $a$ and $b$ are rational numbers.\\
(Total 14 marks)
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP2 2007 Q4 [14]}}