| Exam Board | SPS |
|---|---|
| Module | SPS SM (SPS SM) |
| Year | 2021 |
| Session | November |
| Marks | 7 |
| Topic | Parametric curves and Cartesian conversion |
| Type | Find dy/dx at a point |
| Difficulty | Standard +0.3 Part (a) is a standard parametric differentiation exercise requiring the chain rule (dy/dx = (dy/dθ)/(dx/dθ)) with straightforward derivatives and evaluation at a point. Part (b) requires eliminating the parameter using trigonometric identities (double angle and half angle formulas), which is more involved but still a textbook technique. Both parts are routine A-level further maths content with no novel problem-solving required, making this slightly easier than average overall. |
| Spec | 1.07s Parametric and implicit differentiation |
\begin{enumerate}[label=(\alph*)]
\item The parametric equations of a curve are $x = \theta \cos \theta$ and $y = \sin \theta$
Find the gradient of the curve at the point for which $\theta = \pi$ [3]
\item A curve is defined parametrically by the equations;
$$x = \cos \theta \qquad y = \left(\frac{\sin \theta}{2}\right)\left(\sin \frac{\theta}{2}\right)$$
Show that the cartesian equation of the curve can be written as $y^2 = \frac{1}{8}(1-x)^2(1+x)$ [4]
\end{enumerate}
\hfill \mbox{\textit{SPS SPS SM 2021 Q10 [7]}}