| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2018 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric curves and Cartesian conversion |
| Type | Find dy/dx at a point |
| Difficulty | Moderate -0.3 This is a straightforward parametric equations question requiring standard techniques: differentiation using the chain rule (dy/dx = (dy/dt)/(dx/dt)) and algebraic manipulation to eliminate the parameter. Part (a) is routine calculus, and part (b) involves rearranging to make t the subject from one equation and substituting into the other—a standard textbook exercise with no novel insight required. Slightly easier than average due to the simple algebraic forms involved. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation |
2. A curve $C$ has parametric equations
$$x = \frac { 3 } { 2 } t - 5 , \quad y = 4 - \frac { 6 } { t } \quad t \neq 0$$
\begin{enumerate}[label=(\alph*)]
\item Find the value of $\frac { \mathrm { d } y } { \mathrm {~d} x }$ at $t = 3$, giving your answer as a fraction in its simplest form.
\item Show that a cartesian equation of $C$ can be expressed in the form
$$y = \frac { a x + b } { x + 5 } \quad x \neq k$$
where $a , b$ and $k$ are integers to be found.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C34 2018 Q2 [7]}}