| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2006 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric integration |
| Type | Show integral then evaluate area |
| Difficulty | Standard +0.3 This is a standard parametric area question requiring routine techniques: finding where y=0 (simple trigonometry), setting up the parametric area integral (direct formula application), and evaluating a trigonometric integral using standard identities. All steps are textbook procedures with no novel insight required, making it slightly easier than average. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Solves \(y = 0 \Rightarrow \cos t = \frac{1}{2}\) to obtain \(t = \frac{\pi}{3}\) or \(\frac{5\pi}{3}\) | M1 A1 | Need both values for A1 |
| Or substitutes both values of \(t\) and shows that \(y = 0\) | (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dx}{dt} = 1 - 2\cos t\) | M1 A1 | |
| \(\text{Area} = \int y\,dx = \int_{\frac{\pi}{3}}^{\frac{5\pi}{3}}(1-2\cos t)(1-2\cos t)\,dt = \int_{\frac{\pi}{3}}^{\frac{5\pi}{3}}(1-2\cos t)^2\,dt\) | B1 | * AG (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\text{Area} = \int 1 - 4\cos t + 4\cos^2 t\,dt\) | M1 | 3 terms |
| \(= \int 1 - 4\cos t + 2(\cos 2t + 1)\,dt\) | M1 | Use of correct double angle formula |
| \(= \int 3 - 4\cos t + 2\cos 2t\,dt\) | ||
| \(= \left[3t - 4\sin t + \sin 2t\right]\) | M1 A1 | |
| Substitutes the two correct limits \(t = \frac{5\pi}{3}\) and \(\frac{\pi}{3}\) and subtracts | M1 | |
| \(= 4\pi + 3\sqrt{3}\) | A1A1 | (7) |
| [12] |
## Question 8:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Solves $y = 0 \Rightarrow \cos t = \frac{1}{2}$ to obtain $t = \frac{\pi}{3}$ or $\frac{5\pi}{3}$ | M1 A1 | Need both values for A1 |
| Or substitutes **both** values of $t$ and shows that $y = 0$ | | (2) |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dx}{dt} = 1 - 2\cos t$ | M1 A1 | |
| $\text{Area} = \int y\,dx = \int_{\frac{\pi}{3}}^{\frac{5\pi}{3}}(1-2\cos t)(1-2\cos t)\,dt = \int_{\frac{\pi}{3}}^{\frac{5\pi}{3}}(1-2\cos t)^2\,dt$ | B1 | * AG (3) |
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\text{Area} = \int 1 - 4\cos t + 4\cos^2 t\,dt$ | M1 | 3 terms |
| $= \int 1 - 4\cos t + 2(\cos 2t + 1)\,dt$ | M1 | Use of correct double angle formula |
| $= \int 3 - 4\cos t + 2\cos 2t\,dt$ | | |
| $= \left[3t - 4\sin t + \sin 2t\right]$ | M1 A1 | |
| Substitutes the two correct limits $t = \frac{5\pi}{3}$ and $\frac{\pi}{3}$ and subtracts | M1 | |
| $= 4\pi + 3\sqrt{3}$ | A1A1 | (7) |
| | | **[12]** |8.
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\includegraphics[alt={},max width=\textwidth]{9bf05d7e-7bb9-40f6-b626-69a8a6eda5a5-10_545_979_285_552}
\end{center}
\end{figure}
The curve shown in Figure 2 has parametric equations
$$x = t - 2 \sin t , \quad y = 1 - 2 \cos t , \quad 0 \leqslant t \leqslant 2 \pi$$
\begin{enumerate}[label=(\alph*)]
\item Show that the curve crosses the $x$-axis where $t = \frac { \pi } { 3 }$ and $t = \frac { 5 \pi } { 3 }$.
The finite region $R$ is enclosed by the curve and the $x$-axis, as shown shaded in Figure 2.
\item Show that the area of $R$ is given by the integral
$$\int _ { \frac { \pi } { 3 } } ^ { \frac { 5 \pi } { 3 } } ( 1 - 2 \cos t ) ^ { 2 } \mathrm {~d} t$$
\item Use this integral to find the exact value of the shaded area.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 2006 Q8 [12]}}