| Exam Board | OCR |
|---|---|
| Module | Further Pure Core 1 (Further Pure Core 1) |
| Year | 2018 |
| Session | December |
| Marks | 9 |
| Topic | Polar coordinates |
| Type | Area enclosed by polar curve |
| Difficulty | Standard +0.8 This is a Further Maths polar coordinates question requiring students to find maxima using calculus (dr/dθ = 0), then apply the polar area formula ∫½r²dθ with a rose curve. While the integration itself is standard (using double angle formulas), the context of Further Maths polar curves and the need for exact answers with trigonometric manipulation places it moderately above average difficulty. |
| Spec | 4.09b Sketch polar curves: r = f(theta)4.09c Area enclosed: by polar curve |
| Answer | Marks | Guidance |
|---|---|---|
| (a)(i) Maximum value of \(\sin 2\theta = 1\) when \(2\theta = \frac{\pi}{2} \Rightarrow \theta = \frac{\pi}{4}\) | M1, A1 | Alternatively: Differentiate \(r\) and set equal to 0; \(r = 3\sin 2\theta \Rightarrow \frac{dr}{d\theta} = 6\cos 2\theta = 0\) when \(\cos 2\theta = 0 \Rightarrow 2\theta = \frac{\pi}{2} \Rightarrow \theta = \frac{\pi}{4}\) |
| (a)(ii) \(r = 3\) | B1 | \(\Rightarrow r = 3\sin\frac{\pi}{2} = 3\) |
| (a)(iii) \(P\) marked in correct place | B1 | |
| (b) \(A = \frac{1}{2}\int_0^{\pi} r^2 d\theta = \frac{1}{2}\int_0^{\pi}(3\sin 2\theta)^2 d\theta = \frac{9}{2}\int_0^{\pi}\sin^2 2\theta d\theta\) | M1 | Use of correct formula, ignore limits |
| \(= \frac{9}{4}\int_0^{\pi}(1-\cos 4\theta)d\theta = \frac{9}{4}\left[\theta - \frac{1}{4}\sin 4\theta\right]_0^{\pi}\) | M1, A1 | Use of double angle formula; Integration |
| \(= \frac{9}{4}\left(\frac{\pi}{2} - \frac{1}{4}\sin 2\pi - 0 + \frac{1}{4}\sin 0\right) = \frac{9\pi}{8}\) | M1, A1 | Substitution of their limits into their integrated function |
**(a)(i)** Maximum value of $\sin 2\theta = 1$ when $2\theta = \frac{\pi}{2} \Rightarrow \theta = \frac{\pi}{4}$ | M1, A1 | Alternatively: Differentiate $r$ and set equal to 0; $r = 3\sin 2\theta \Rightarrow \frac{dr}{d\theta} = 6\cos 2\theta = 0$ when $\cos 2\theta = 0 \Rightarrow 2\theta = \frac{\pi}{2} \Rightarrow \theta = \frac{\pi}{4}$
**(a)(ii)** $r = 3$ | B1 | $\Rightarrow r = 3\sin\frac{\pi}{2} = 3$
**(a)(iii)** $P$ marked in correct place | B1 |
**(b)** $A = \frac{1}{2}\int_0^{\pi} r^2 d\theta = \frac{1}{2}\int_0^{\pi}(3\sin 2\theta)^2 d\theta = \frac{9}{2}\int_0^{\pi}\sin^2 2\theta d\theta$ | M1 | Use of correct formula, ignore limits
$= \frac{9}{4}\int_0^{\pi}(1-\cos 4\theta)d\theta = \frac{9}{4}\left[\theta - \frac{1}{4}\sin 4\theta\right]_0^{\pi}$ | M1, A1 | Use of double angle formula; Integration
$= \frac{9}{4}\left(\frac{\pi}{2} - \frac{1}{4}\sin 2\pi - 0 + \frac{1}{4}\sin 0\right) = \frac{9\pi}{8}$ | M1, A1 | Substitution of their limits into their integrated function
---2 The equation of the curve shown on the graph is, in polar coordinates, $r = 3 \sin 2 \theta$ for $0 \leqslant \theta \leqslant \frac { 1 } { 2 } \pi$.\\
\includegraphics[max width=\textwidth, alt={}, center]{8315a796-0e7d-464f-8604-9fe3ab7af359-2_470_657_913_319}
\begin{enumerate}[label=(\alph*)]
\item The greatest value of $r$ on the curve occurs at the point $P$.
\begin{enumerate}[label=(\roman*)]
\item Show that $\theta = \frac { 1 } { 4 } \pi$ at the point $P$.
\item Find the value of $r$ at the point $P$.
\item Mark the point $P$ on the copy of the graph in the Printed Answer Booklet.
\end{enumerate}\item In this question you must show detailed reasoning.
Find the exact area of the region enclosed by the curve.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Pure Core 1 2018 Q2 [9]}}