| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric differentiation |
| Type | Show gradient expression then find coordinates |
| Difficulty | Standard +0.3 This is a standard C4 parametric differentiation question with routine applications of the chain rule, solving trigonometric equations, and converting to Cartesian form using standard identities. All parts follow predictable patterns with no novel problem-solving 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 |
|---|---|---|
| \(\frac{dx}{dt} = -\sin t, \quad \frac{dy}{dt} = 2\cos 2t \quad \therefore \frac{dy}{dx} = \frac{2\cos 2t}{-\sin t}\) | M1 A1 A1 | (3) |
| Answer | Marks | Guidance |
|---|---|---|
| \(2\cos 2t = 0 \quad \therefore 2t = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \frac{7\pi}{2}\) | M1 | |
| \(\text{So } t = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}\) | A1 A1 | (3) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\left(\frac{1}{\sqrt{2}}, 1\right), \left(\frac{1}{\sqrt{2}}, -1\right), \left(-\frac{1}{\sqrt{2}}, 1\right), \left(-\frac{1}{\sqrt{2}}, -1\right)\) | M1 A1 | (2) |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = 2\sin t \cos t\) | M1 | |
| \(= 2\sqrt{1-\cos^2 t}\cos t = 2x\sqrt{1-x^2}\) | M1 A1 | (3) |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = -2x\sqrt{1-x^2}\) | B1 | (1) |
## (a)
$\frac{dx}{dt} = -\sin t, \quad \frac{dy}{dt} = 2\cos 2t \quad \therefore \frac{dy}{dx} = \frac{2\cos 2t}{-\sin t}$ | M1 A1 A1 | (3)
## (b)
$2\cos 2t = 0 \quad \therefore 2t = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \frac{7\pi}{2}$ | M1 |
$\text{So } t = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$ | A1 A1 | (3)
## (c)
$\left(\frac{1}{\sqrt{2}}, 1\right), \left(\frac{1}{\sqrt{2}}, -1\right), \left(-\frac{1}{\sqrt{2}}, 1\right), \left(-\frac{1}{\sqrt{2}}, -1\right)$ | M1 A1 | (2)
## (d)
$y = 2\sin t \cos t$ | M1 |
$= 2\sqrt{1-\cos^2 t}\cos t = 2x\sqrt{1-x^2}$ | M1 A1 | (3)
## (e)
$y = -2x\sqrt{1-x^2}$ | B1 | (1)
**Total: (12 marks)**
---\includegraphics{figure_1}
The curve shown in Fig. 1 has parametric equations
$$x = \cos t, \quad y = \sin 2t, \quad 0 \leq t < 2\pi.$$
\begin{enumerate}[label=(\alph*)]
\item Find an expression for $\frac{dy}{dx}$ in terms of the parameter $t$.
[3]
\item Find the values of the parameter $t$ at the points where $\frac{dy}{dx} = 0$.
[3]
\item Hence give the exact values of the coordinates of the points on the curve where the tangents are parallel to the $x$-axis.
[2]
\item Show that a cartesian equation for the part of the curve where $0 \leq t < \pi$ is
$$y = 2x\sqrt{(1 - x^2)}.$$
[3]
\item Write down a cartesian equation for the part of the curve where $\pi \leq t < 2\pi$.
[1]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 Q5 [12]}}