| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2013 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric differentiation |
| Type | Find parameter value given gradient condition |
| Difficulty | Standard +0.3 This is a straightforward parametric differentiation question requiring standard techniques: differentiate both x and y with respect to θ, apply the chain rule dy/dx = (dy/dθ)/(dx/dθ), then simplify using the double angle formula cos 2θ = 1 - 2sin²θ. Part (ii) involves routine equation solving. Slightly easier than average due to the 'show that' structure guiding students and the standard nature of the techniques. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.05a Sine, cosine, tangent: definitions for all arguments1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| (i) State \(\frac{dx}{d\theta} = -2\sin 2\theta + \sin \theta\) or \(\frac{dy}{d\theta} = 8\sin \theta \cos \theta\) | B1 | |
| Use \(\frac{dy}{dx} = \frac{dy}{d\theta} \div \frac{dx}{d\theta}\) | M1 | |
| Use \(\sin 2\theta = 2\sin\theta \cos\theta\) | M1 | |
| Obtain given answer correctly | A1 | [4] |
| (ii) Equate derivative to \(-4\) and solve for \(\cos \theta\) | M1 | |
| Obtain \(\cos \theta = \frac{1}{2}\) | A1 | |
| Obtain \(x = -1\) | A1 | |
| Obtain \(y = 3\) | A1 | [4] |
**(i)** State $\frac{dx}{d\theta} = -2\sin 2\theta + \sin \theta$ or $\frac{dy}{d\theta} = 8\sin \theta \cos \theta$ | B1 |
Use $\frac{dy}{dx} = \frac{dy}{d\theta} \div \frac{dx}{d\theta}$ | M1 |
Use $\sin 2\theta = 2\sin\theta \cos\theta$ | M1 |
Obtain given answer correctly | A1 | [4]
**(ii)** Equate derivative to $-4$ and solve for $\cos \theta$ | M1 |
Obtain $\cos \theta = \frac{1}{2}$ | A1 |
Obtain $x = -1$ | A1 |
Obtain $y = 3$ | A1 | [4]
---5 The parametric equations of a curve are
$$x = \cos 2 \theta - \cos \theta , \quad y = 4 \sin ^ { 2 } \theta$$
for $0 \leqslant \theta \leqslant \pi$.\\
(i) Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 8 \cos \theta } { 1 - 4 \cos \theta }$.\\
(ii) Find the coordinates of the point on the curve at which the gradient is - 4 .
\hfill \mbox{\textit{CAIE P2 2013 Q5 [8]}}