| Exam Board | SPS |
|---|---|
| Module | SPS FM Pure (SPS FM Pure) |
| Year | 2021 |
| Session | June |
| Marks | 2 |
| Topic | Parametric differentiation |
| Type | Find gradient at given parameter |
| Difficulty | Moderate -0.8 This is a straightforward application of the parametric differentiation formula dy/dx = (dy/dt)/(dx/dt), requiring only basic differentiation of polynomials and evaluation at a given parameter value. It's a routine 2-mark question with no problem-solving element, making it easier than average but not trivial since it requires knowledge of the parametric chain rule. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation |
A curve is defined by the parametric equations
$$x = t^3 + 2, \quad y = t^2 - 1$$
Find the gradient of the curve at the point where $t = -2$
[2]
\hfill \mbox{\textit{SPS SPS FM Pure 2021 Q1 [2]}}