| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric differentiation |
| Type | Tangent parallel to axis condition |
| Difficulty | Standard +0.3 This is a straightforward parametric differentiation question requiring the chain rule (dy/dx = (dy/dt)/(dx/dt)) and finding where dy/dx = 0. The trigonometric derivatives are standard, and solving cos t = 0 in the given range is routine. Slightly easier than average due to the simple functions involved and clear method. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation |
4.
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The diagram shows the curve with parametric equations
$$x = t + \sin t , \quad y = \sin t , \quad 0 \leq t \leq \pi$$
(i) Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $t$.\\
(ii) Find, in exact form, the coordinates of the point where the tangent to the curve is parallel to the $x$-axis.\\
\hfill \mbox{\textit{OCR C4 Q4 [6]}}