| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric curves and Cartesian conversion |
| Type | Properties of specific curves |
| Difficulty | Standard +0.3 This is a straightforward parametric differentiation question requiring standard chain rule application (dy/dx = (dy/dt)/(dx/dt)), followed by finding where y=0, computing the tangent equation, and calculating a triangle area. All techniques are routine C4 material with clear signposting and no novel insights required. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation |
\includegraphics{figure_5}
The diagram shows the curve with parametric equations
$$x = a\sqrt{t}, \quad y = at(1-t), \quad t \geq 0,$$
where $a$ is a positive constant.
\begin{enumerate}[label=(\roman*)]
\item Find $\frac{dy}{dx}$ in terms of $t$. [3]
\end{enumerate}
The curve meets the $x$-axis at the origin, $O$, and at the point $A$. The tangent to the curve at $A$ meets the $y$-axis at the point $B$ as shown.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Show that the area of triangle $OAB$ is $a^2$. [5]
\end{enumerate}
\hfill \mbox{\textit{OCR C4 Q5 [8]}}