| Exam Board | AQA |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2009 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polar coordinates |
| Type | Polar curve with exponential function |
| Difficulty | Standard +0.8 This is a Further Maths FP3 polar coordinates question requiring integration of an exponential function for area, sketching a polar curve, and solving a transcendental equation. While the techniques are standard for FP3 (polar area formula, exponential manipulation), the combination of exponential polar curves and solving the intersection algebraically elevates it slightly above average difficulty for Further Maths content. |
| Spec | 4.09b Sketch polar curves: r = f(theta)4.09c Area enclosed: by polar curve |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\text{Area} = \frac{1}{2}\int\left(1 + 6e^{-\frac{\theta}{\pi}}\right)^2 d\theta\) | M1 | Use of \(\frac{1}{2}\int r^2\, d\theta\) |
| \(= \frac{1}{2}\int_0^{2\pi}\left(1 + 12e^{-\frac{\theta}{\pi}} + 36e^{-\frac{2\theta}{\pi}}\right)d\theta\) | B1 | Correct expansion |
| B1 | Correct limits | |
| \(= \frac{1}{2}\left[\theta - 12\pi e^{-\frac{\theta}{\pi}} - 18\pi e^{-\frac{2\theta}{\pi}}\right]_0^{2\pi}\) | m1 | Correct integration of at least two of the three terms: \(1\), \(p e^{-\frac{\theta}{\pi}}\), \(q e^{-\frac{2\theta}{\pi}}\) |
| \(= \pi(16 - 6e^{-2} - 9e^{-4})\) | A1 | ACF |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Spiral curve going the correct way round the pole | B1 | Going correct way round the pole |
| Curve increasing in distance from the pole | B1 | Increasing in distance from the pole |
| End-points \((1, 0)\) and \((e^2, 2\pi)\) | B2,1,0 | Correct end-points; B1 for each pair or for \(1\) and \(e^2\) shown on graph in correct positions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(e^{\frac{\theta}{\pi}} = 1 + 6e^{-\frac{\theta}{\pi}}\) | M1 | Elimination of \(r\) or \(\theta\): \([r = 1 + \frac{6}{r}]\) |
| \(\left(e^{\frac{\theta}{\pi}}\right)^2 - e^{\frac{\theta}{\pi}} - 6 = 0\) | m1 | Forming quadratic in \(e^{\frac{\theta}{\pi}}\) or in \(e^{-\frac{\theta}{\pi}}\) or in \(r\): \([r^2 - r - 6 = 0]\) |
| \(\left(e^{\frac{\theta}{\pi}} - 3\right)\left(e^{\frac{\theta}{\pi}} + 2\right) = 0\) | m1 | OE |
| \(e^{\frac{\theta}{\pi}} > 0\) so \(e^{\frac{\theta}{\pi}} = 3\) | E1 | Rejection of negative solution; PI \([r=3]\) |
| Polar coordinates of \(P\) are \((3,\, \pi \ln 3)\) | A1 |
# Question 7:
## Part 7(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\text{Area} = \frac{1}{2}\int\left(1 + 6e^{-\frac{\theta}{\pi}}\right)^2 d\theta$ | M1 | Use of $\frac{1}{2}\int r^2\, d\theta$ |
| $= \frac{1}{2}\int_0^{2\pi}\left(1 + 12e^{-\frac{\theta}{\pi}} + 36e^{-\frac{2\theta}{\pi}}\right)d\theta$ | B1 | Correct expansion |
| | B1 | Correct limits |
| $= \frac{1}{2}\left[\theta - 12\pi e^{-\frac{\theta}{\pi}} - 18\pi e^{-\frac{2\theta}{\pi}}\right]_0^{2\pi}$ | m1 | Correct integration of at least two of the three terms: $1$, $p e^{-\frac{\theta}{\pi}}$, $q e^{-\frac{2\theta}{\pi}}$ |
| $= \pi(16 - 6e^{-2} - 9e^{-4})$ | A1 | ACF | Total: 5 marks |
## Part 7(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Spiral curve going the correct way round the pole | B1 | Going correct way round the pole |
| Curve increasing in distance from the pole | B1 | Increasing in distance from the pole |
| End-points $(1, 0)$ and $(e^2, 2\pi)$ | B2,1,0 | Correct end-points; B1 for each pair or for $1$ and $e^2$ shown on graph in correct positions | Total: 4 marks |
## Part 7(c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $e^{\frac{\theta}{\pi}} = 1 + 6e^{-\frac{\theta}{\pi}}$ | M1 | Elimination of $r$ or $\theta$: $[r = 1 + \frac{6}{r}]$ |
| $\left(e^{\frac{\theta}{\pi}}\right)^2 - e^{\frac{\theta}{\pi}} - 6 = 0$ | m1 | Forming quadratic in $e^{\frac{\theta}{\pi}}$ or in $e^{-\frac{\theta}{\pi}}$ or in $r$: $[r^2 - r - 6 = 0]$ |
| $\left(e^{\frac{\theta}{\pi}} - 3\right)\left(e^{\frac{\theta}{\pi}} + 2\right) = 0$ | m1 | OE |
| $e^{\frac{\theta}{\pi}} > 0$ so $e^{\frac{\theta}{\pi}} = 3$ | E1 | Rejection of negative solution; PI $[r=3]$ |
| Polar coordinates of $P$ are $(3,\, \pi \ln 3)$ | A1 | | Total: 5 marks |
---7 The diagram shows the curve $C _ { 1 }$ with polar equation
$$r = 1 + 6 \mathrm { e } ^ { - \frac { \theta } { \pi } } , \quad 0 \leqslant \theta \leqslant 2 \pi$$
\includegraphics[max width=\textwidth, alt={}, center]{13cfb9ca-9495-4b69-80c5-9fb7e8e0f957-4_300_513_1414_760}
\begin{enumerate}[label=(\alph*)]
\item Find, in terms of $\pi$ and e , the area of the shaded region bounded by $C _ { 1 }$ and the initial line.
\item The polar equation of a curve $C _ { 2 }$ is
$$r = \mathrm { e } ^ { \frac { \theta } { \pi } } , \quad 0 \leqslant \theta \leqslant 2 \pi$$
Sketch the curve $C _ { 2 }$ and state the polar coordinates of the end-points of this curve.
\item The curves $C _ { 1 }$ and $C _ { 2 }$ intersect at the point $P$. Find the polar coordinates of $P$.
\end{enumerate}
\hfill \mbox{\textit{AQA FP3 2009 Q7 [14]}}