| Exam Board | Edexcel |
|---|---|
| Module | F2 (Further Pure Mathematics 2) |
| Year | 2021 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polar coordinates |
| Type | Tangent parallel/perpendicular to initial line |
| Difficulty | Standard +0.8 This is a Further Maths polar coordinates question requiring (a) finding where dy/dx = 0 using the chain rule with polar derivatives, and (b) computing a polar area integral. While the techniques are standard for FM students, the calculus manipulation and exact evaluation require careful work, placing it moderately above average difficulty. |
| Spec | 1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates4.09a Polar coordinates: convert to/from cartesian4.09b Sketch polar curves: r = f(theta)4.09c Area enclosed: by polar curve |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(y = r\sin\theta = \sin\theta + \sin\theta\cos\theta\) OR \(r\sin\theta = \sin\theta + \frac{1}{2}\sin 2\theta\) | B1 | Use of \(r\sin\theta\); award if not seen explicitly but correct result follows from double angle formula |
| \(\frac{dy}{d\theta} = \cos\theta - \sin^2\theta + \cos^2\theta\) OR \(\frac{dy}{d\theta} = \cos\theta + \cos 2\theta\) | M1 | Differentiate \(r\sin\theta\) or \(r\cos\theta\) |
| \(0 = \cos\theta + 2\cos^2\theta - 1 = (2\cos\theta - 1)(\cos\theta + 1)\) | M1 | Set \(\frac{d(r\sin\theta)}{d\theta} = 0\) and solve; only solution used need be shown |
| \(\cos\theta = \frac{1}{2}\) (\(\cos\theta = -1\) outside range), \(\theta = \frac{\pi}{3}\) | ||
| \(A\) is \(\left(1\frac{1}{2}, \frac{\pi}{3}\right)\) | A1 (4) | Correct coordinates of \(A\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Area \(= \frac{1}{2}\int_0^{\frac{\pi}{3}}(1+\cos\theta)^2\,d\theta\) | B1 | Use of Area \(= \frac{1}{2}\int r^2\,d\theta\) with \(r = 1+\cos\theta\); limits not needed |
| \(= \frac{1}{2}\int\left(1 + 2\cos\theta + \frac{1}{2}(\cos 2\theta + 1)\right)d\theta\) | M1A1 | Attempt \((1+\cos\theta)^2\); minimum accepted is \((1 + k\cos\theta + \cos^2\theta)\) and change \(\cos^2\theta\) using \(\cos^2\theta = \frac{1}{2}(\pm\cos 2\theta \pm 1)\); A1 correct integrand, \(\frac{1}{2}\) may be missing |
| \(= \frac{1}{2}\left[\frac{3}{2}\theta + 2\sin\theta + \frac{1}{4}\sin 2\theta\right]_0^{\frac{\pi}{3}}\) | dM1A1 | Attempt to integrate all terms; \(\cos 2\theta \to \pm\frac{1}{k}\sin 2\theta\), \(k = \pm 1\) or \(\pm 2\); A1 correct integration and correct limits seen |
| \(= \frac{\pi}{4} + \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{16} = \frac{\pi}{4} + \frac{9\sqrt{3}}{16}\) | A1 (6) | Substitute correct limits and obtain correct answer in required form |
## Question 3(a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $y = r\sin\theta = \sin\theta + \sin\theta\cos\theta$ OR $r\sin\theta = \sin\theta + \frac{1}{2}\sin 2\theta$ | B1 | Use of $r\sin\theta$; award if not seen explicitly but correct result follows from double angle formula |
| $\frac{dy}{d\theta} = \cos\theta - \sin^2\theta + \cos^2\theta$ OR $\frac{dy}{d\theta} = \cos\theta + \cos 2\theta$ | M1 | Differentiate $r\sin\theta$ or $r\cos\theta$ |
| $0 = \cos\theta + 2\cos^2\theta - 1 = (2\cos\theta - 1)(\cos\theta + 1)$ | M1 | Set $\frac{d(r\sin\theta)}{d\theta} = 0$ and solve; only solution used need be shown |
| $\cos\theta = \frac{1}{2}$ ($\cos\theta = -1$ outside range), $\theta = \frac{\pi}{3}$ | | |
| $A$ is $\left(1\frac{1}{2}, \frac{\pi}{3}\right)$ | A1 (4) | Correct coordinates of $A$ |
## Question 3(b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Area $= \frac{1}{2}\int_0^{\frac{\pi}{3}}(1+\cos\theta)^2\,d\theta$ | B1 | Use of Area $= \frac{1}{2}\int r^2\,d\theta$ with $r = 1+\cos\theta$; limits not needed |
| $= \frac{1}{2}\int\left(1 + 2\cos\theta + \frac{1}{2}(\cos 2\theta + 1)\right)d\theta$ | M1A1 | Attempt $(1+\cos\theta)^2$; minimum accepted is $(1 + k\cos\theta + \cos^2\theta)$ and change $\cos^2\theta$ using $\cos^2\theta = \frac{1}{2}(\pm\cos 2\theta \pm 1)$; A1 correct integrand, $\frac{1}{2}$ may be missing |
| $= \frac{1}{2}\left[\frac{3}{2}\theta + 2\sin\theta + \frac{1}{4}\sin 2\theta\right]_0^{\frac{\pi}{3}}$ | dM1A1 | Attempt to integrate all terms; $\cos 2\theta \to \pm\frac{1}{k}\sin 2\theta$, $k = \pm 1$ or $\pm 2$; A1 correct integration and correct limits seen |
| $= \frac{\pi}{4} + \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{16} = \frac{\pi}{4} + \frac{9\sqrt{3}}{16}$ | A1 (6) | Substitute correct limits and obtain correct answer in required form |
---\begin{enumerate}
\item The curve $C$, with pole $O$, has polar equation
\end{enumerate}
$$r = 1 + \cos \theta , \quad 0 \leqslant \theta \leqslant \frac { \pi } { 2 }$$
At the point $A$ on $C$, the tangent to $C$ is parallel to the initial line.\\
(a) Find the polar coordinates of $A$.\\
(b) Find the finite area enclosed by the initial line, the line $O A$ and the curve $C$, giving your answer in the form $a \pi + b \sqrt { 3 }$, where $a$ and $b$ are rational constants to be found.
\hfill \mbox{\textit{Edexcel F2 2021 Q3 [10]}}