| Exam Board | Edexcel |
|---|---|
| Module | F2 (Further Pure Mathematics 2) |
| Year | 2023 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polar coordinates |
| Type | Tangent parallel/perpendicular to initial line |
| Difficulty | Challenging +1.2 This is a standard Further Maths polar coordinates question requiring the tangent condition dy/dx = 0, then area integration. While it involves multiple steps and algebraic manipulation to reach the given form, the techniques are routine for FM students: differentiating polar curves and integrating r² for area. The calculus is mechanical rather than requiring novel insight. |
| Spec | 1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.08h Integration by substitution4.09a Polar coordinates: convert to/from cartesian4.09b Sketch polar curves: r = f(theta)4.09c Area enclosed: by polar curve |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(y = r\sin\theta = (1-\sin\theta)\sin\theta = \sin\theta - \sin^2\theta\), \(\Rightarrow \frac{dy}{d\theta} = \cos\theta - 2\sin\theta\cos\theta\) or \(\frac{dy}{d\theta} = \cos\theta - \sin 2\theta\) | M1 | Differentiates \((1-\sin\theta)\sin\theta\) to achieve \(\pm\cos\theta \pm k\sin\theta\cos\theta\). Use of \(y=r\cos\theta\) or \(x=r\cos\theta\) scores M0 |
| Correct derivative in any form | A1 | |
| \(\cos\theta - 2\sin\theta\cos\theta = 0 \Rightarrow \cos\theta(1-2\sin\theta)=0 \Rightarrow \sin\theta = \frac{1}{2} \Rightarrow \theta = ...\) | dM1 | Solves to find a value for \(\theta\). Depends on first M |
| \(\left(\frac{1}{2}, \frac{\pi}{6}\right)\) correct coordinates and no others. Allow e.g. \(\theta=\frac{\pi}{6},\ r=\frac{1}{2}\) | A1 | isw if necessary. Value of \(r\) must be seen in (a) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\int(1-\sin\theta)^2\,d\theta = \int\left(1-2\sin\theta+\frac{1}{2}-\frac{1}{2}\cos 2\theta\right)d\theta\) | M1 | Attempts \(\frac{1}{2}\int r^2\,d\theta\) and applies \(\sin^2\theta = \pm\frac{1}{2}\pm\frac{1}{2}\cos 2\theta\) |
| \(\int(1-\sin\theta)^2\,d\theta = \frac{3}{2}\theta + 2\cos\theta - \frac{1}{4}\sin 2\theta\ (+c)\) | A1 | Correct integration. Condone mixed variables |
| \(\left(\frac{1}{2}\right)\left[\frac{3}{2}\theta + 2\cos\theta - \frac{1}{4}\sin 2\theta\right]_0^{\frac{\pi}{6}} = \left(\frac{1}{2}\right)\left[\left(\frac{\pi}{4}+\sqrt{3}-\frac{\sqrt{3}}{8}\right)-(2)\right]\) | M1 | Applies limits of \(0\) and \(\frac{\pi}{6}\). The \(\frac{1}{2}\) is not required. For integration look for at least \(\pm\int\sin\theta\,d\theta \to \pm\cos\theta\) |
| Triangle: \(\frac{1}{2}\times\frac{1}{2}\sin\frac{\pi}{6}\times\frac{1}{2}\cos\frac{\pi}{6}\left(=\frac{\sqrt{3}}{32}\right)\) | M1 | Uses a correct strategy for the area of the triangle |
| Area of \(R = \frac{\pi}{8}+\frac{7\sqrt{3}}{16}-1+\frac{\sqrt{3}}{32}\) | dM1 | Fully correct method for required area. Depends on all previous method marks |
| \(\frac{1}{32}(4\pi+15\sqrt{3}-32)\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\int(1-\sin\theta)^2\,d\theta = \int\left(1-2\sin\theta+\frac{1}{2}-\frac{1}{2}\cos 2\theta\right)d\theta\) | M1 | Attempts \(\frac{1}{2}\int r^2\,d\theta\) and applies \(\sin^2\theta=\pm\frac{1}{2}\pm\frac{1}{2}\cos 2\theta\) |
| \(\int(1-\sin\theta)^2\,d\theta = \frac{3}{2}\theta+2\cos\theta-\frac{1}{4}\sin 2\theta\ (+c)\) | A1 | Correct integration. Condone mixed variables |
| \(\left(\frac{1}{2}\right)\left[\frac{3}{2}\theta+2\cos\theta-\frac{1}{4}\sin 2\theta\right]_0^{\frac{\pi}{2}} = \left(\frac{1}{2}\right)\left[\left(\frac{3\pi}{4}+0-0\right)-(2)\right] = \frac{3\pi}{8}-1\) | M1 | Evidence of use of both limits \(0\) and \(\frac{\pi}{2}\). The \(\frac{1}{2}\) is not required |
| Triangle "Segment": \(\frac{1}{2}\times\frac{1}{2}\sin\frac{\pi}{6}\times\frac{1}{2}\cos\frac{\pi}{6} - \frac{1}{2}\int_{\frac{\pi}{6}}^{\frac{\pi}{2}}(1-\sin\theta)^2\,d\theta\) | M1 | Uses fully correct strategy for area above curve between \(O\) and \(P\). Requires correct method for triangle and "segment" using both \(\frac{\pi}{6}\) and \(\frac{\pi}{2}\) |
| Area of \(R = \frac{3\pi}{8}-1+\frac{15\sqrt{3}}{32}-\frac{\pi}{4}\) | dM1 | Fully correct method. Depends on all previous method marks |
| \(\frac{1}{32}(4\pi+15\sqrt{3}-32)\) | A1 | cao |
# Question 8:
## Part (a):
| Working | Mark | Guidance |
|---------|------|----------|
| $y = r\sin\theta = (1-\sin\theta)\sin\theta = \sin\theta - \sin^2\theta$, $\Rightarrow \frac{dy}{d\theta} = \cos\theta - 2\sin\theta\cos\theta$ or $\frac{dy}{d\theta} = \cos\theta - \sin 2\theta$ | M1 | Differentiates $(1-\sin\theta)\sin\theta$ to achieve $\pm\cos\theta \pm k\sin\theta\cos\theta$. Use of $y=r\cos\theta$ or $x=r\cos\theta$ scores M0 |
| Correct derivative in any form | A1 | |
| $\cos\theta - 2\sin\theta\cos\theta = 0 \Rightarrow \cos\theta(1-2\sin\theta)=0 \Rightarrow \sin\theta = \frac{1}{2} \Rightarrow \theta = ...$ | dM1 | Solves to find a value for $\theta$. Depends on first M |
| $\left(\frac{1}{2}, \frac{\pi}{6}\right)$ correct coordinates and no others. Allow e.g. $\theta=\frac{\pi}{6},\ r=\frac{1}{2}$ | A1 | isw if necessary. **Value of $r$ must be seen in (a)** |
## Part (b) Way 1:
| Working | Mark | Guidance |
|---------|------|----------|
| $\int(1-\sin\theta)^2\,d\theta = \int\left(1-2\sin\theta+\frac{1}{2}-\frac{1}{2}\cos 2\theta\right)d\theta$ | M1 | Attempts $\frac{1}{2}\int r^2\,d\theta$ and applies $\sin^2\theta = \pm\frac{1}{2}\pm\frac{1}{2}\cos 2\theta$ |
| $\int(1-\sin\theta)^2\,d\theta = \frac{3}{2}\theta + 2\cos\theta - \frac{1}{4}\sin 2\theta\ (+c)$ | A1 | Correct integration. Condone mixed variables |
| $\left(\frac{1}{2}\right)\left[\frac{3}{2}\theta + 2\cos\theta - \frac{1}{4}\sin 2\theta\right]_0^{\frac{\pi}{6}} = \left(\frac{1}{2}\right)\left[\left(\frac{\pi}{4}+\sqrt{3}-\frac{\sqrt{3}}{8}\right)-(2)\right]$ | M1 | Applies limits of $0$ and $\frac{\pi}{6}$. The $\frac{1}{2}$ is not required. For integration look for at least $\pm\int\sin\theta\,d\theta \to \pm\cos\theta$ |
| Triangle: $\frac{1}{2}\times\frac{1}{2}\sin\frac{\pi}{6}\times\frac{1}{2}\cos\frac{\pi}{6}\left(=\frac{\sqrt{3}}{32}\right)$ | M1 | Uses a correct strategy for the area of the triangle |
| Area of $R = \frac{\pi}{8}+\frac{7\sqrt{3}}{16}-1+\frac{\sqrt{3}}{32}$ | dM1 | Fully correct method for required area. **Depends on all previous method marks** |
| $\frac{1}{32}(4\pi+15\sqrt{3}-32)$ | A1 | cao |
## Part (b) Way 2:
| Working | Mark | Guidance |
|---------|------|----------|
| $\int(1-\sin\theta)^2\,d\theta = \int\left(1-2\sin\theta+\frac{1}{2}-\frac{1}{2}\cos 2\theta\right)d\theta$ | M1 | Attempts $\frac{1}{2}\int r^2\,d\theta$ and applies $\sin^2\theta=\pm\frac{1}{2}\pm\frac{1}{2}\cos 2\theta$ |
| $\int(1-\sin\theta)^2\,d\theta = \frac{3}{2}\theta+2\cos\theta-\frac{1}{4}\sin 2\theta\ (+c)$ | A1 | Correct integration. Condone mixed variables |
| $\left(\frac{1}{2}\right)\left[\frac{3}{2}\theta+2\cos\theta-\frac{1}{4}\sin 2\theta\right]_0^{\frac{\pi}{2}} = \left(\frac{1}{2}\right)\left[\left(\frac{3\pi}{4}+0-0\right)-(2)\right] = \frac{3\pi}{8}-1$ | M1 | Evidence of use of **both** limits $0$ and $\frac{\pi}{2}$. The $\frac{1}{2}$ is not required |
| Triangle "Segment": $\frac{1}{2}\times\frac{1}{2}\sin\frac{\pi}{6}\times\frac{1}{2}\cos\frac{\pi}{6} - \frac{1}{2}\int_{\frac{\pi}{6}}^{\frac{\pi}{2}}(1-\sin\theta)^2\,d\theta$ | M1 | Uses fully correct strategy for area above curve between $O$ and $P$. Requires correct method for triangle and "segment" using **both** $\frac{\pi}{6}$ and $\frac{\pi}{2}$ |
| Area of $R = \frac{3\pi}{8}-1+\frac{15\sqrt{3}}{32}-\frac{\pi}{4}$ | dM1 | Fully correct method. **Depends on all previous method marks** |
| $\frac{1}{32}(4\pi+15\sqrt{3}-32)$ | A1 | cao |
---8.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{ed3689f7-b3f0-447b-baa5-e44b8d8342d0-28_522_1084_260_495}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
The curve $C$ shown in Figure 1 has polar equation
$$r = 1 - \sin \theta \quad 0 \leqslant \theta < \frac { \pi } { 2 }$$
The point $P$ lies on $C$, such that the tangent to $C$ at $P$ is parallel to the initial line.
\begin{enumerate}[label=(\alph*)]
\item Use calculus to determine the polar coordinates of $P$
The finite region $R$, shown shaded in Figure 1, is bounded by
\begin{itemize}
\item the line with equation $\theta = \frac { \pi } { 2 }$
\item the tangent to $C$ at $P$
\item part of the curve $C$
\item the initial line
\item Use algebraic integration to show that the area of $R$ is
\end{itemize}
$$\frac { 1 } { 32 } ( a \pi + b \sqrt { 3 } + c )$$
where $a$, $b$ and $c$ are integers to be determined.
\end{enumerate}
\hfill \mbox{\textit{Edexcel F2 2023 Q8 [10]}}