| Exam Board | Edexcel |
|---|---|
| Module | F2 (Further Pure Mathematics 2) |
| Year | 2020 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polar coordinates |
| Type | Region bounded by curve and tangent lines |
| Difficulty | Challenging +1.8 This is a challenging Further Maths polar coordinates question requiring: (a) finding where dy/dx = 0 using the polar tangent condition, and (b) computing area using polar integration then subtracting a triangular region. The calculus is non-trivial with trigonometric manipulation, but follows standard FM2 polar techniques without requiring exceptional insight. |
| Spec | 1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates4.09b Sketch polar curves: r = f(theta)4.09c Area enclosed: by polar curve |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(r\sin\theta = 2a\sin\theta + 2a\sin\theta\cos\theta\) OR \(r\sin\theta = 2a\sin\theta + a\sin 2\theta\) | B1 | Multiply \(r\) by \(\sin\theta\); award if not seen explicitly but correct result follows from double angle formula |
| \(\frac{d(r\sin\theta)}{d\theta} = 2a\cos\theta + 2a\cos^2\theta - 2a\sin^2\theta\) | M1 A1 | Differentiate \(r\sin\theta\) or \(r\cos\theta\) using product rule or double angle formula first; correct derivative |
| \(2\cos^2\theta+\cos\theta-1=0\), \((2\cos\theta-1)(\cos\theta+1)=0\) | dM1 | Use \(\sin^2\theta+\cos^2\theta=1\) to form a 3TQ in \(\cos\theta\) and solve by valid method |
| \(\cos\theta=\frac{1}{2}\), \(\theta=\frac{\pi}{3}\) (\(\theta=\pi\) need not be seen) | A1 | Correct value of \(\theta\) |
| \(r = 2a\times\frac{3}{2} = 3a\) | A1 (6) | Correct \(r\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\text{Area}=\frac{1}{2}\int r^2\,d\theta = \frac{1}{2}\int_{\pi/6}^{\pi/3}4a^2(1+\cos\theta)^2\,d\theta\) | M1 | Use area \(=\frac{1}{2}\int r^2\,d\theta\) with \(r=2a+2a\cos\theta\); no limits needed |
| \(=2a^2\int_{\pi/6}^{\pi/3}(1+2\cos\theta+\cos^2\theta)\,d\theta\) | ||
| \(=2a^2\int_{\pi/6}^{\pi/3}\left(1+2\cos\theta+\frac{1}{2}(\cos 2\theta+1)\right)d\theta\) | M1 | Use double angle formula to obtain function ready for integrating |
| \(=2a^2\left[\theta+2\sin\theta+\frac{1}{2}\left(\frac{1}{2}\sin 2\theta+\theta\right)\right]_{\pi/6}^{\pi/3}\) | dM1 A1 | Attempt integration with \(\cos 2\theta\to\frac{1}{k}\sin 2\theta\), \(k=\pm2\) or \(\pm1\); correct integration |
| \(=2a^2\left[\frac{\pi}{3}+\sqrt{3}+\frac{1}{4}\times\frac{\sqrt{3}}{2}+\frac{\pi}{6}-\left(\frac{\pi}{6}+1+\frac{1}{4}\times\frac{\sqrt{3}}{2}+\frac{\pi}{12}\right)\right]\) | M1 | Substitute limits (\(\frac{\pi}{6}\) and \(\theta\) from (a)); this must be \(>\frac{\pi}{6}\) |
| \(=2a^2\left(\frac{\pi}{4}+\sqrt{3}-1\right)\) | ||
| Area of \(\triangle OAB = \frac{1}{2}\times 3a\times(2+\sqrt{3})a\times\sin\frac{\pi}{6} = \frac{3}{4}a^2(2+\sqrt{3})\) | M1 | Obtain area of \(\triangle OAB\) and subtract from previous area |
| Shaded area \(= 2a^2\left(\frac{\pi}{4}+\sqrt{3}-1\right)-\frac{3}{4}a^2(2+\sqrt{3}) = \frac{a^2}{4}(2\pi-14+5\sqrt{3})\) | M1 A1cao (7) | Correct answer |
## Question 7:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $r\sin\theta = 2a\sin\theta + 2a\sin\theta\cos\theta$ OR $r\sin\theta = 2a\sin\theta + a\sin 2\theta$ | B1 | Multiply $r$ by $\sin\theta$; award if not seen explicitly but correct result follows from double angle formula |
| $\frac{d(r\sin\theta)}{d\theta} = 2a\cos\theta + 2a\cos^2\theta - 2a\sin^2\theta$ | M1 A1 | Differentiate $r\sin\theta$ or $r\cos\theta$ using product rule or double angle formula first; correct derivative |
| $2\cos^2\theta+\cos\theta-1=0$, $(2\cos\theta-1)(\cos\theta+1)=0$ | dM1 | Use $\sin^2\theta+\cos^2\theta=1$ to form a 3TQ in $\cos\theta$ and solve by valid method |
| $\cos\theta=\frac{1}{2}$, $\theta=\frac{\pi}{3}$ ($\theta=\pi$ need not be seen) | A1 | Correct value of $\theta$ |
| $r = 2a\times\frac{3}{2} = 3a$ | A1 (6) | Correct $r$ |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\text{Area}=\frac{1}{2}\int r^2\,d\theta = \frac{1}{2}\int_{\pi/6}^{\pi/3}4a^2(1+\cos\theta)^2\,d\theta$ | M1 | Use area $=\frac{1}{2}\int r^2\,d\theta$ with $r=2a+2a\cos\theta$; no limits needed |
| $=2a^2\int_{\pi/6}^{\pi/3}(1+2\cos\theta+\cos^2\theta)\,d\theta$ | | |
| $=2a^2\int_{\pi/6}^{\pi/3}\left(1+2\cos\theta+\frac{1}{2}(\cos 2\theta+1)\right)d\theta$ | M1 | Use double angle formula to obtain function ready for integrating |
| $=2a^2\left[\theta+2\sin\theta+\frac{1}{2}\left(\frac{1}{2}\sin 2\theta+\theta\right)\right]_{\pi/6}^{\pi/3}$ | dM1 A1 | Attempt integration with $\cos 2\theta\to\frac{1}{k}\sin 2\theta$, $k=\pm2$ or $\pm1$; correct integration |
| $=2a^2\left[\frac{\pi}{3}+\sqrt{3}+\frac{1}{4}\times\frac{\sqrt{3}}{2}+\frac{\pi}{6}-\left(\frac{\pi}{6}+1+\frac{1}{4}\times\frac{\sqrt{3}}{2}+\frac{\pi}{12}\right)\right]$ | M1 | Substitute limits ($\frac{\pi}{6}$ and $\theta$ from (a)); this must be $>\frac{\pi}{6}$ |
| $=2a^2\left(\frac{\pi}{4}+\sqrt{3}-1\right)$ | | |
| Area of $\triangle OAB = \frac{1}{2}\times 3a\times(2+\sqrt{3})a\times\sin\frac{\pi}{6} = \frac{3}{4}a^2(2+\sqrt{3})$ | M1 | Obtain area of $\triangle OAB$ and subtract from previous area |
| Shaded area $= 2a^2\left(\frac{\pi}{4}+\sqrt{3}-1\right)-\frac{3}{4}a^2(2+\sqrt{3}) = \frac{a^2}{4}(2\pi-14+5\sqrt{3})$ | M1 A1cao (7) | Correct answer |
---7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{17b48fd7-5e88-4a62-beb9-8590a363e70f-20_476_972_251_488}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
The curve $C$, shown in Figure 1, has polar equation
$$r = 2 a ( 1 + \cos \theta ) \quad 0 \leqslant \theta \leqslant \pi$$
where $a$ is a positive constant.
The tangent to $C$ at the point $A$ is parallel to the initial line.
\begin{enumerate}[label=(\alph*)]
\item Determine the polar coordinates of $A$.
The point $B$ on the curve has polar coordinates $\quad a ( 2 + \sqrt { 3 } ) , \frac { \pi } { 6 }$
The finite region $R$, shown shaded in Figure 1, is bounded by the curve $C$ and the line $A B$.
\item Use calculus to determine the exact area of the shaded region $R$.
Give your answer in the form
$$\frac { a ^ { 2 } } { 4 } ( d \pi - e + f \sqrt { 3 } )$$
where $d , e$ and $f$ are integers.
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel F2 2020 Q7 [13]}}