| Exam Board | SPS |
|---|---|
| Module | SPS SM Pure (SPS SM Pure) |
| Year | 2024 |
| Session | February |
| Marks | 10 |
| Topic | Parametric curves and Cartesian conversion |
| Type | Convert to Cartesian (polynomial/rational) |
| Difficulty | Standard +0.3 This is a straightforward parametric equations question requiring standard techniques: finding dy/dx using the chain rule, substituting a point to verify a gradient, solving a simple equation for when the gradient equals +1, and eliminating the parameter to get a Cartesian equation. All steps are routine A-level procedures 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 |
13.
A particle moves in the $x - y$ plane so that its position at time $t$ s is given by $x = t ^ { 3 } - 8 t , y = t ^ { 2 }$ for $- 3.5 < t < 3.5$. The units of distance are metres. The graph shows the path of the particle and the direction of travel at the point $\mathrm { P } ( 8,4 )$.\\
\includegraphics[max width=\textwidth, alt={}, center]{ede204ac-09c3-486b-8877-df935e6ed015-28_485_917_445_210}
\begin{enumerate}[label=(\alph*)]
\item Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $t$.
\item Hence show that the value of $\frac { \mathrm { d } y } { \mathrm {~d} x }$ at P is - 1 .
\item Find the time at which the particle is travelling in the direction opposite to that at P .
\item Find the cartesian equation of the path, giving $x ^ { 2 }$ as a function of $y$.
\end{enumerate}
\hfill \mbox{\textit{SPS SPS SM Pure 2024 Q13 [10]}}