| Exam Board | Edexcel |
|---|---|
| Module | CP1 (Core Pure 1) |
| Year | 2024 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polar coordinates |
| Type | Find constant from given area |
| Difficulty | Challenging +1.2 This is a multi-part polar coordinates question requiring understanding of polar curves, finding a constant from a geometric constraint, and integrating to find area. Part (a) requires recognizing that the minimum r-value occurs when sin(3θ) = 1, giving a straightforward equation. Part (b) involves standard polar area integration with a trigonometric identity. While it requires several techniques and careful algebraic manipulation, the steps are relatively standard for Further Maths Core Pure content with no particularly novel insights needed. |
| Spec | 4.09b Sketch polar curves: r = f(theta)4.09c Area enclosed: by polar curve |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Max \(r = 4+a = 5.5 \Rightarrow a=\ldots\) | M1 | Uses maximum value of \(r\) is \((4+a)\) and uses radius of circle to find \(a\) |
| \(a=1.5\) | A1 | Correct value; note \(a=-1.5\) can potentially gain M1A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Pool area \(=\frac{1}{2}\int_0^{2\pi}(4-\text{"1.5"}\sin3\theta)^2\,d\theta\) | M1 | Correct strategy for area using polar area formula including the \(\frac{1}{2}\) |
| \((4-\text{"1.5"}\sin3\theta)^2=16-\text{"12"}\sin3\theta+\text{"2.25"}\sin^23\theta\) \(=16-\text{"12"}\sin3\theta+\text{"2.25"}\left(\frac{1-\cos6\theta}{2}\right)\) | M1 | Squares bracket achieving three terms and applies \(\sin^23\theta=\frac{\pm1\pm\cos6\theta}{2}\) |
| \(\int(4-\text{"1.5"}\sin3\theta)^2\,d\theta=16\theta+\text{"4"}\cos3\theta+\frac{\text{"9"}}{\text{"8"}}\left(\theta-\frac{\sin6\theta}{6}\right)\) | A1ft | Correct integration in any form (follow through on \(a\)) |
| \(\frac{1}{2}\left[\frac{\text{"137"}}{\text{"8"}}\theta+\text{"4"}\cos3\theta-\frac{\text{"3"}}{\text{"16"}}\sin6\theta\right]_0^{2\pi}=\ldots\left(\frac{137}{8}\pi\right)\) | dM1 | Applies appropriate limits to integrated function |
| Area of \(T=\pi\times36-\frac{137}{8}\pi\) | DM1 | Fully correct strategy for area of \(T\): correct circle area minus area inside curve |
| \(=\frac{151}{8}\pi\ (\text{m}^2)\) | A1 | Correct area from fully correct work (exact answer required) |
# Question 3:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Max $r = 4+a = 5.5 \Rightarrow a=\ldots$ | M1 | Uses maximum value of $r$ is $(4+a)$ and uses radius of circle to find $a$ |
| $a=1.5$ | A1 | Correct value; note $a=-1.5$ can potentially gain M1A0 |
**(2 marks)**
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Pool area $=\frac{1}{2}\int_0^{2\pi}(4-\text{"1.5"}\sin3\theta)^2\,d\theta$ | M1 | Correct strategy for area using polar area formula including the $\frac{1}{2}$ |
| $(4-\text{"1.5"}\sin3\theta)^2=16-\text{"12"}\sin3\theta+\text{"2.25"}\sin^23\theta$ $=16-\text{"12"}\sin3\theta+\text{"2.25"}\left(\frac{1-\cos6\theta}{2}\right)$ | M1 | Squares bracket achieving three terms and applies $\sin^23\theta=\frac{\pm1\pm\cos6\theta}{2}$ |
| $\int(4-\text{"1.5"}\sin3\theta)^2\,d\theta=16\theta+\text{"4"}\cos3\theta+\frac{\text{"9"}}{\text{"8"}}\left(\theta-\frac{\sin6\theta}{6}\right)$ | A1ft | Correct integration in any form (follow through on $a$) |
| $\frac{1}{2}\left[\frac{\text{"137"}}{\text{"8"}}\theta+\text{"4"}\cos3\theta-\frac{\text{"3"}}{\text{"16"}}\sin6\theta\right]_0^{2\pi}=\ldots\left(\frac{137}{8}\pi\right)$ | dM1 | Applies appropriate limits to integrated function |
| Area of $T=\pi\times36-\frac{137}{8}\pi$ | DM1 | Fully correct strategy for area of $T$: correct circle area minus area inside curve |
| $=\frac{151}{8}\pi\ (\text{m}^2)$ | A1 | Correct area from fully correct work (exact answer required) |
**(6 marks)**
**(8 marks)**\begin{enumerate}
\item In this question you must show all stages of your working.
\end{enumerate}
\section*{Solutions relying entirely on calculator technology are not acceptable.}
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{dc3e8e46-c60b-4263-9652-d7c2a322cfae-10_563_561_395_753}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows the design for a bathing pool.\\
The pool, $P$, shown unshaded in Figure 1, is surrounded by a tiled area, $T$, shown shaded in Figure 1.
The tiled area is bounded by the edge of the pool and by a circle, $C$, with radius 6 m .\\
The centre of the pool and the centre of the circle are the same point.\\
The edge of the pool is modelled by the curve with polar equation
$$r = 4 - a \sin 3 \theta \quad 0 \leqslant \theta \leqslant 2 \pi$$
where $a$ is a positive constant.\\
Given that the shortest distance between the edge of the pool and the circle $C$ is 0.5 m ,\\
(a) determine the value of $a$.\\
(b) Hence, using algebraic integration, determine, according to the model, the exact area of $T$.
\hfill \mbox{\textit{Edexcel CP1 2024 Q3 [8]}}