| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2002 |
| Session | June |
| Marks | 15 |
| 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 (dr/dθ = r tan θ for tangent parallel to initial line), followed by straightforward applications of the cosine rule and polar area formula. While it involves multiple steps and FP2 content, the techniques are routine for this specification with no novel problem-solving required. |
| Spec | 4.09a Polar coordinates: convert to/from cartesian4.09b Sketch polar curves: r = f(theta)4.09c Area enclosed: by polar curve |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(y=r\sin\theta=a(3\sin\theta+\sqrt{5}\sin\theta\cos\theta)\) | ||
| \(\frac{dy}{d\theta}=a(3\cos\theta+\sqrt{5}\cos 2\theta)\) | M1, A1 | |
| \(2\sqrt{5}\cos^2\theta+3\cos\theta-\sqrt{5}=0\) | ||
| \(\cos\theta=\frac{-3\pm\sqrt{9+40}}{4\sqrt{5}}\), \(\cos\theta=\frac{1}{\sqrt{5}}\) | M1, A1 | |
| \(\theta=\pm 1.107...\) | A1 ft | |
| \(r=4a\) | A1 ft | (6) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(2r\sin\theta=20\) | M1 | |
| \(8a\sin\theta=20\), \(a=\frac{20}{8\sin\theta}=2.795...\) | M1, A1 | (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((3+\sqrt{5}\cos\theta)^2=9+6\sqrt{5}\cos\theta+5\cos^2\theta\) | B1 | |
| Integrate: \(9\theta+6\sqrt{5}\sin\theta+5\left(\frac{\sin 2\theta}{4}+\frac{\theta}{2}\right)\) | M1, A1 | |
| Limits used: \([\ldots]_0^{2\pi}=18\pi+5\pi\) (or upper limit: \(9\pi+\frac{5\pi}{2}\)) | A1 | |
| \(\frac{1}{2}\int_0^{2\pi}r^2\,d\theta=a^2(23\pi)\approx 282\text{ m}^2\) | M1, A1 | (6) |
| (15 marks) |
# Question 8:
## Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $y=r\sin\theta=a(3\sin\theta+\sqrt{5}\sin\theta\cos\theta)$ | | |
| $\frac{dy}{d\theta}=a(3\cos\theta+\sqrt{5}\cos 2\theta)$ | M1, A1 | |
| $2\sqrt{5}\cos^2\theta+3\cos\theta-\sqrt{5}=0$ | | |
| $\cos\theta=\frac{-3\pm\sqrt{9+40}}{4\sqrt{5}}$, $\cos\theta=\frac{1}{\sqrt{5}}$ | M1, A1 | |
| $\theta=\pm 1.107...$ | A1 ft | |
| $r=4a$ | A1 ft | **(6)** |
## Part (b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $2r\sin\theta=20$ | M1 | |
| $8a\sin\theta=20$, $a=\frac{20}{8\sin\theta}=2.795...$ | M1, A1 | **(3)** |
## Part (c)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(3+\sqrt{5}\cos\theta)^2=9+6\sqrt{5}\cos\theta+5\cos^2\theta$ | B1 | |
| Integrate: $9\theta+6\sqrt{5}\sin\theta+5\left(\frac{\sin 2\theta}{4}+\frac{\theta}{2}\right)$ | M1, A1 | |
| Limits used: $[\ldots]_0^{2\pi}=18\pi+5\pi$ (or upper limit: $9\pi+\frac{5\pi}{2}$) | A1 | |
| $\frac{1}{2}\int_0^{2\pi}r^2\,d\theta=a^2(23\pi)\approx 282\text{ m}^2$ | M1, A1 | **(6)** |
| | | **(15 marks)** |
---8.
\section*{Figure 1}
The curve $C$ shown in Fig. 1 has polar equation
$$r = a ( 3 + \sqrt { 5 } \cos \theta ) , \quad - \pi \leq \theta < \pi .$$
\includegraphics[max width=\textwidth, alt={}, center]{6d92bf8a-df0d-421c-8246-8160f5921ee6-2_460_792_1503_970}
\begin{enumerate}[label=(\alph*)]
\item Find the polar coordinates of the points $P$ and $Q$ where the tangents to $C$ are parallel to the initial line. (6) The curve $C$ represents the perimeter of the surface of a swimming pool. The direct distance from $P$ to $Q$ is 20 m.
\item Calculate the value of $a$.
\item Find the area of the surface of the pool. (6)
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP2 2002 Q8 [15]}}