| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2009 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric curves and Cartesian conversion |
| Type | Find dy/dx at a point |
| Difficulty | Standard +0.3 This is a standard C4 parametric equations question requiring routine differentiation using dy/dx = (dy/dt)/(dx/dt) for part (a), application of the double angle formula cos(2t) = 1 - 2sin²(t) to eliminate the parameter in part (b), and reading off the range from the given domain in part (c). All techniques are textbook exercises with no novel insight required, making it slightly easier than average. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dx}{dt}=-4\sin 2t\), \(\frac{dy}{dt}=6\cos t\) | B1, B1 | |
| \(\frac{dy}{dx}=\frac{6\cos t}{-4\sin 2t}\left(=-\frac{3}{4\sin t}\right)\) | M1 | |
| At \(t=\frac{\pi}{3}\): \(m=-\frac{3}{4\times\frac{\sqrt{3}}{2}}=-\frac{\sqrt{3}}{2}\) | A1 (4) | accept equivalents, awrt \(-0.87\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Use of \(\cos 2t = 1-2\sin^2 t\) | M1 | |
| \(\cos 2t=\frac{x}{2}\), \(\sin t=\frac{y}{6}\); \(\frac{x}{2}=1-2\left(\frac{y}{6}\right)^2\) | M1 | |
| Leading to \(y=\sqrt{18-9x}\ \left(=3\sqrt{2-x}\right)\) | A1 | cao |
| \(-2\leq x\leq 2\), \(k=2\) | B1 (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(0\leq f(x)\leq 6\) | B1 | either \(0\leq f(x)\) or \(f(x)\leq 6\) |
| Fully correct. Accept \(0\leq y\leq 6\), \([0,6]\) | B1 (2) |
# Question 5:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dx}{dt}=-4\sin 2t$, $\frac{dy}{dt}=6\cos t$ | B1, B1 | |
| $\frac{dy}{dx}=\frac{6\cos t}{-4\sin 2t}\left(=-\frac{3}{4\sin t}\right)$ | M1 | |
| At $t=\frac{\pi}{3}$: $m=-\frac{3}{4\times\frac{\sqrt{3}}{2}}=-\frac{\sqrt{3}}{2}$ | A1 (4) | accept equivalents, awrt $-0.87$ |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Use of $\cos 2t = 1-2\sin^2 t$ | M1 | |
| $\cos 2t=\frac{x}{2}$, $\sin t=\frac{y}{6}$; $\frac{x}{2}=1-2\left(\frac{y}{6}\right)^2$ | M1 | |
| Leading to $y=\sqrt{18-9x}\ \left(=3\sqrt{2-x}\right)$ | A1 | cao |
| $-2\leq x\leq 2$, $k=2$ | B1 (4) | |
## Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $0\leq f(x)\leq 6$ | B1 | either $0\leq f(x)$ or $f(x)\leq 6$ |
| Fully correct. Accept $0\leq y\leq 6$, $[0,6]$ | B1 (2) | |5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{c2622c33-9436-4254-a728-10ba4703a28c-09_735_1222_205_358}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 2 shows a sketch of the curve with parametric equations
$$x = 2 \cos 2 t , \quad y = 6 \sin t , \quad 0 \leqslant t \leqslant \frac { \pi } { 2 }$$
\begin{enumerate}[label=(\alph*)]
\item Find the gradient of the curve at the point where $t = \frac { \pi } { 3 }$.
\item Find a cartesian equation of the curve in the form
$$y = \mathrm { f } ( x ) , \quad - k \leqslant x \leqslant k$$
stating the value of the constant $k$.
\item Write down the range of $\mathrm { f } ( x )$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 2009 Q5 [10]}}