| Exam Board | OCR MEI |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2014 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric curves and Cartesian conversion |
| Difficulty | Standard +0.3 This is a standard C4 parametric differentiation question requiring the chain rule (dy/dx = (dy/dt)/(dx/dt)) and elimination of the parameter. Both parts follow routine procedures taught in the module 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 |
A curve has parametric equations $x = e^{2t}, y = te^{2t}$.
\begin{enumerate}[label=(\roman*)]
\item Find $\frac{dy}{dx}$ in terms of $t$. Hence find the exact gradient of the curve at the point with parameter $t = 1$. [4]
\item Find the cartesian equation of the curve in the form $y = ax^b \ln x$, where $a$ and $b$ are constants to be determined. [3]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI C4 2014 Q5 [7]}}