| Exam Board | AQA |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2007 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polar coordinates |
| Type | Area enclosed by polar curve |
| Difficulty | Challenging +1.2 This is a structured Further Maths question on polar coordinates with clear scaffolding. Part (a) is trivial algebra, part (b) is a standard conversion using r²=x²+y², and part (c) involves routine polar area integration. While the topic is Further Maths, the question guides students through each step methodically with no novel insights required, making it only moderately above average difficulty. |
| Spec | 4.09a Polar coordinates: convert to/from cartesian4.09c Area enclosed: by polar curve |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((\cos\theta + \sin\theta)^2 = \cos^2\theta + \sin^2\theta + 2\cos\theta\sin\theta = 1 + \sin 2\theta\) | B1 | 1 mark |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((x^2 + y^2)^3 = (x+y)^4\) | ||
| \((r^2)^3 = (r\cos\theta + r\sin\theta)^4\) | M2,1,0 | M1 for one of \(x^2+y^2=r^2\) OE, \(x = r\cos\theta\), \(y = r\sin\theta\) used |
| \(r^6 = r^4(\cos\theta + \sin\theta)^4\) | ||
| \(r^6 = r^4(1 + \sin 2\theta)^2\) | M1 | Uses (a) OE at any stage |
| \(r^2 = (1 + \sin 2\theta)^2\) | ||
| \(\Rightarrow r = (1 + \sin 2\theta)\ \{r \geq 0\}\) | A1 | 4 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(r = 0 \Rightarrow \sin 2\theta = -1\) | ||
| \(2\theta = \sin^{-1}(-1);\ = -\frac{\pi}{2},\ \frac{3\pi}{2}\) | M1 | |
| \(\theta = -\frac{\pi}{4};\ \frac{3\pi}{4}\) | A1A1ft | 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Area \(= \frac{1}{2}\int(1 + \sin 2\theta)^2\, d\theta\) | M1 | Use of \(\frac{1}{2}\int r^2\, d\theta\) |
| \(= \frac{1}{2}\int(1 + 2\sin 2\theta + \sin^2 2\theta)\, d\theta\) | B1 | Correct expansion of \((1+\sin 2\theta)^2\) |
| \(= \frac{1}{2}\int\left(1 + 2\sin 2\theta + \frac{1}{2}(1 - \cos 4\theta)\right)d\theta\) | M1 | Attempt to write \(\sin^2 2\theta\) in terms of \(\cos 4\theta\) |
| \(= \left[\frac{3}{4}\theta - \frac{1}{2}\cos 2\theta - \frac{1}{16}\sin 4\theta\right]\) | A1ft | Correct integration ft wrong coefficients only |
| \(= \left[\frac{3}{4}\theta - \frac{1}{2}\cos 2\theta - \frac{1}{16}\sin 4\theta\right]_{-\pi/4}^{3\pi/4}\) | ||
| \(= \left(\frac{9\pi}{16}\right) - \left(-\frac{3\pi}{16}\right)\) | m1 | Using c's values from (c)(i) as limits or the correct limits |
| \(= \frac{3\pi}{4}\) | A1 | 6 marks |
## Question 4:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(\cos\theta + \sin\theta)^2 = \cos^2\theta + \sin^2\theta + 2\cos\theta\sin\theta = 1 + \sin 2\theta$ | B1 | **1 mark** | AG (be convinced) |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(x^2 + y^2)^3 = (x+y)^4$ | | |
| $(r^2)^3 = (r\cos\theta + r\sin\theta)^4$ | M2,1,0 | M1 for one of $x^2+y^2=r^2$ OE, $x = r\cos\theta$, $y = r\sin\theta$ used |
| $r^6 = r^4(\cos\theta + \sin\theta)^4$ | | |
| $r^6 = r^4(1 + \sin 2\theta)^2$ | M1 | Uses (a) OE at any stage |
| $r^2 = (1 + \sin 2\theta)^2$ | | |
| $\Rightarrow r = (1 + \sin 2\theta)\ \{r \geq 0\}$ | A1 | **4 marks** | CSO; AG |
### Part (c)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $r = 0 \Rightarrow \sin 2\theta = -1$ | | |
| $2\theta = \sin^{-1}(-1);\ = -\frac{\pi}{2},\ \frac{3\pi}{2}$ | M1 | |
| $\theta = -\frac{\pi}{4};\ \frac{3\pi}{4}$ | A1A1ft | **3 marks** | A1 for either |
### Part (c)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Area $= \frac{1}{2}\int(1 + \sin 2\theta)^2\, d\theta$ | M1 | Use of $\frac{1}{2}\int r^2\, d\theta$ |
| $= \frac{1}{2}\int(1 + 2\sin 2\theta + \sin^2 2\theta)\, d\theta$ | B1 | Correct expansion of $(1+\sin 2\theta)^2$ |
| $= \frac{1}{2}\int\left(1 + 2\sin 2\theta + \frac{1}{2}(1 - \cos 4\theta)\right)d\theta$ | M1 | Attempt to write $\sin^2 2\theta$ in terms of $\cos 4\theta$ |
| $= \left[\frac{3}{4}\theta - \frac{1}{2}\cos 2\theta - \frac{1}{16}\sin 4\theta\right]$ | A1ft | Correct integration ft wrong coefficients only |
| $= \left[\frac{3}{4}\theta - \frac{1}{2}\cos 2\theta - \frac{1}{16}\sin 4\theta\right]_{-\pi/4}^{3\pi/4}$ | | |
| $= \left(\frac{9\pi}{16}\right) - \left(-\frac{3\pi}{16}\right)$ | m1 | Using c's values from (c)(i) as limits or the correct limits |
| $= \frac{3\pi}{4}$ | A1 | **6 marks** | CSO |
---4
\begin{enumerate}[label=(\alph*)]
\item Show that $( \cos \theta + \sin \theta ) ^ { 2 } = 1 + \sin 2 \theta$.
\item A curve has cartesian equation
$$\left( x ^ { 2 } + y ^ { 2 } \right) ^ { 3 } = ( x + y ) ^ { 4 }$$
Given that $r \geqslant 0$, show that the polar equation of the curve is
$$r = 1 + \sin 2 \theta$$
\item The curve with polar equation
$$r = 1 + \sin 2 \theta , \quad - \pi \leqslant \theta \leqslant \pi$$
is shown in the diagram.\\
\includegraphics[max width=\textwidth, alt={}, center]{f90167c3-2ffd-464a-b2d2-9f86a8d64887-3_389_611_1062_708}
\begin{enumerate}[label=(\roman*)]
\item Find the two values of $\theta$ for which $r = 0$.
\item Find the area of one of the loops.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA FP3 2007 Q4 [14]}}