| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2007 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric curves and Cartesian conversion |
| Type | Convert to Cartesian (sin/cos identities) |
| Difficulty | Moderate -0.3 This is a standard parametric equations question requiring routine techniques: finding dy/dx using the chain rule, converting to Cartesian form using the double angle identity (cos 2t = 2cos²t - 1), and sketching. The gradient bound requires recognizing that -sin 2t/sin t = -2cos t has maximum value 2 when cos t = -1. All steps are textbook exercises with no novel insight required, making it slightly easier than average. |
| Spec | 1.02n Sketch curves: simple equations including polynomials1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation |
5 A curve $C$ has parametric equations
$$x = \cos t , \quad y = 3 + 2 \cos 2 t , \quad \text { where } 0 \leqslant t \leqslant \pi$$
(i) Express $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $t$ and hence show that the gradient at any point on $C$ cannot exceed 8 .\\
(ii) Show that all points on $C$ satisfy the cartesian equation $y = 4 x ^ { 2 } + 1$.\\
(iii) Sketch the curve $y = 4 x ^ { 2 } + 1$ and indicate on your sketch the part which represents $C$.
\hfill \mbox{\textit{OCR C4 2007 Q5 [9]}}