| Exam Board | Edexcel |
|---|---|
| Module | F2 (Further Pure Mathematics 2) |
| Year | 2021 |
| Session | October |
| Marks | 11 |
| 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 question requiring: (a) finding dy/dx in polar form and solving for when it equals zero (perpendicular to initial line), and (b) computing area using polar integration then subtracting a triangle area. The calculus is non-trivial with multiple steps, and the geometric setup requires careful interpretation. However, it's a standard F2 polar tangent/area question type with clear structure. |
| 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 |
| \(x = r\cos\theta = (1+\sin\theta)\cos\theta \Rightarrow \frac{dx}{d\theta} = \cos^2\theta - (1+\sin\theta)\sin\theta\) or \(\frac{dx}{d\theta} = -\sin\theta + \cos 2\theta\) | M1 | Differentiates \(r\cos\theta\) using product rule or double angle formula |
| Correct derivative in any form | A1 | |
| \(\cos^2\theta - (1+\sin\theta)\sin\theta = 0 \Rightarrow 1-\sin^2\theta - \sin\theta - \sin^2\theta = 0 \Rightarrow 2\sin^2\theta + \sin\theta - 1 = 0\) or \(-\sin\theta + \cos 2\theta = 0 \Rightarrow -\sin\theta + 1 - 2\sin^2\theta = 0 \Rightarrow 2\sin^2\theta + \sin\theta - 1 = 0\) | dM1 | Sets \(\frac{dx}{d\theta} = 0\) and proceeds to a 3TQ in \(\sin\theta\). Depends on first M mark |
| \(\Rightarrow 2\sin^2\theta + \sin\theta - 1 = 0\) \(\Rightarrow \sin\theta = \frac{1}{2},\ (-1) \Rightarrow \theta = \ldots\) | ddM1 | Solves for \(\theta\). Depends on both M marks above |
| \(\left(\frac{3}{2},\ \frac{\pi}{6}\right)\) | A1 | Correct coordinates and no others. Need not be in coordinate brackets |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int(1+\sin\theta)^2\, d\theta = \int(1 + 2\sin\theta + \sin^2\theta)\, d\theta\) \(= \int\left(1 + 2\sin\theta + \frac{1}{2} - \frac{1}{2}\cos 2\theta\right)d\theta\) | M1 | Attempts \(\frac{1}{2}\int r^2\, d\theta\) and applies \(\sin^2\theta = \pm\frac{1}{2} \pm \frac{1}{2}\cos 2\theta\). Ignore any limits shown |
| \(\int(1+\sin\theta)^2\, d\theta = \frac{3}{2}\theta - 2\cos\theta - \frac{1}{4}\sin 2\theta\ (+c)\) | A1 | Correct integration (ignore limits) |
| \(\frac{1}{2}\left[\frac{3}{2}\theta - 2\cos\theta - \frac{1}{4}\sin 2\theta\right]_{\frac{\pi}{6}}^{\frac{\pi}{2}}\) \(= \frac{1}{2}\left[\frac{3\pi}{4} - \left(\frac{\pi}{4} - \sqrt{3} - \frac{\sqrt{3}}{8}\right)\right] = \left(\frac{\pi}{4} + \frac{9\sqrt{3}}{16}\right)\) | M1 | Applies limits of \(\frac{\pi}{2}\) and \(\frac{\pi}{6}\). Substitution must be shown but no simplification needed |
| Trapezium: \(\frac{1}{2}\left(2 + \left(2 - \frac{3}{2}\sin\frac{\pi}{6}\right)\right) \times \frac{3}{2}\cos\frac{\pi}{6}\) \(= \frac{39\sqrt{3}}{32}\) | M1 | Uses a correct strategy for the area of trapezium \(OQSP\) |
| Area of \(R = \frac{39\sqrt{3}}{32} - \frac{\pi}{4} - \frac{9\sqrt{3}}{16}\) | dM1 | Fully correct method for the required area. Depends on all previous method marks |
| \(\frac{1}{32}(21\sqrt{3} - 8\pi)\) | A1 | Cao |
## Question 8:
**Part (a):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x = r\cos\theta = (1+\sin\theta)\cos\theta \Rightarrow \frac{dx}{d\theta} = \cos^2\theta - (1+\sin\theta)\sin\theta$ or $\frac{dx}{d\theta} = -\sin\theta + \cos 2\theta$ | M1 | Differentiates $r\cos\theta$ using product rule or double angle formula |
| Correct derivative in any form | A1 | |
| $\cos^2\theta - (1+\sin\theta)\sin\theta = 0 \Rightarrow 1-\sin^2\theta - \sin\theta - \sin^2\theta = 0 \Rightarrow 2\sin^2\theta + \sin\theta - 1 = 0$ or $-\sin\theta + \cos 2\theta = 0 \Rightarrow -\sin\theta + 1 - 2\sin^2\theta = 0 \Rightarrow 2\sin^2\theta + \sin\theta - 1 = 0$ | dM1 | Sets $\frac{dx}{d\theta} = 0$ and proceeds to a 3TQ in $\sin\theta$. Depends on first M mark |
| $\Rightarrow 2\sin^2\theta + \sin\theta - 1 = 0$ $\Rightarrow \sin\theta = \frac{1}{2},\ (-1) \Rightarrow \theta = \ldots$ | ddM1 | Solves for $\theta$. Depends on both M marks above |
| $\left(\frac{3}{2},\ \frac{\pi}{6}\right)$ | A1 | Correct coordinates and no others. Need not be in coordinate brackets |
**Part (b):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int(1+\sin\theta)^2\, d\theta = \int(1 + 2\sin\theta + \sin^2\theta)\, d\theta$ $= \int\left(1 + 2\sin\theta + \frac{1}{2} - \frac{1}{2}\cos 2\theta\right)d\theta$ | M1 | Attempts $\frac{1}{2}\int r^2\, d\theta$ and applies $\sin^2\theta = \pm\frac{1}{2} \pm \frac{1}{2}\cos 2\theta$. Ignore any limits shown |
| $\int(1+\sin\theta)^2\, d\theta = \frac{3}{2}\theta - 2\cos\theta - \frac{1}{4}\sin 2\theta\ (+c)$ | A1 | Correct integration (ignore limits) |
| $\frac{1}{2}\left[\frac{3}{2}\theta - 2\cos\theta - \frac{1}{4}\sin 2\theta\right]_{\frac{\pi}{6}}^{\frac{\pi}{2}}$ $= \frac{1}{2}\left[\frac{3\pi}{4} - \left(\frac{\pi}{4} - \sqrt{3} - \frac{\sqrt{3}}{8}\right)\right] = \left(\frac{\pi}{4} + \frac{9\sqrt{3}}{16}\right)$ | M1 | Applies limits of $\frac{\pi}{2}$ and $\frac{\pi}{6}$. Substitution must be shown but no simplification needed |
| Trapezium: $\frac{1}{2}\left(2 + \left(2 - \frac{3}{2}\sin\frac{\pi}{6}\right)\right) \times \frac{3}{2}\cos\frac{\pi}{6}$ $= \frac{39\sqrt{3}}{32}$ | M1 | Uses a correct strategy for the area of trapezium $OQSP$ |
| Area of $R = \frac{39\sqrt{3}}{32} - \frac{\pi}{4} - \frac{9\sqrt{3}}{16}$ | dM1 | Fully correct method for the required area. Depends on all previous method marks |
| $\frac{1}{32}(21\sqrt{3} - 8\pi)$ | A1 | Cao |8.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{8fa1e7da-009f-4b7f-9fa8-21a1768bfd73-28_735_892_264_529}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
The curve $C$ shown in Figure 1 has polar equation
$$r = 1 + \sin \theta \quad - \frac { \pi } { 2 } < \theta \leqslant \frac { \pi } { 2 }$$
The point $P$ lies on $C$ such that the tangent to $C$ at $P$ is perpendicular to the initial line.
\begin{enumerate}[label=(\alph*)]
\item Use calculus to determine the polar coordinates of $P$.
The tangent to $C$ at the point $Q$ where $\theta = \frac { \pi } { 2 }$ is parallel to the initial line.\\
The tangent to $C$ at $Q$ meets the tangent to $C$ at $P$ at the point $S$, as shown in Figure 1.\\
The finite region $R$, shown shaded in Figure 1, is bounded by the line segments $Q S , S P$ and the curve $C$.
\item Use algebraic integration to show that the area of $R$ is
$$\frac { 1 } { 32 } ( a \sqrt { 3 } + b \pi )$$
where $a$ and $b$ are integers to be determined.\\
(6)
\end{enumerate}
\hfill \mbox{\textit{Edexcel F2 2021 Q8 [11]}}