| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2011 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric differentiation |
| Type | Find tangent equation at parameter |
| Difficulty | Standard +0.3 This is a straightforward parametric differentiation question requiring standard techniques: differentiate both equations with respect to θ, apply the chain rule (dy/dx = dy/dθ ÷ dx/dθ), and simplify using trigonometric identities. Part (ii) is routine substitution into y = mx + c. Slightly above average due to the algebraic manipulation needed, but follows a well-practiced procedure with no novel insight required. |
| Spec | 1.07m Tangents and normals: gradient and equations1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| (i) State \(\frac{dx}{dt} = 4\sin\theta\cos\theta\) or equivalent (nothing for \(\frac{dy}{dx} = 4\sec^2\theta\)) | B1 | |
| Use \(\frac{dy}{dx} = \frac{dy}{d\theta} \div \frac{dx}{d\theta}\) | M1 | |
| Obtain given answer correctly | A1 | [3] |
| (ii) Substitute \(\theta = \frac{\pi}{4}\) in \(\frac{dy}{dx}\) and both parametric equations | M1 | |
| Obtain \(\frac{dy}{dx} = 4\) and coordinates \((2, 4)\) | A1 | |
| Form equation of tangent at their point | M1 | |
| State equation of tangent in correct form \(y = 4x - 4\) | A1 | [4] |
**(i)** State $\frac{dx}{dt} = 4\sin\theta\cos\theta$ or equivalent (nothing for $\frac{dy}{dx} = 4\sec^2\theta$) | B1 |
Use $\frac{dy}{dx} = \frac{dy}{d\theta} \div \frac{dx}{d\theta}$ | M1 |
Obtain given answer correctly | A1 | [3] |
**(ii)** Substitute $\theta = \frac{\pi}{4}$ in $\frac{dy}{dx}$ and both parametric equations | M1 |
Obtain $\frac{dy}{dx} = 4$ and coordinates $(2, 4)$ | A1 |
Form equation of tangent at their point | M1 |
State equation of tangent in correct form $y = 4x - 4$ | A1 | [4] |6 The parametric equations of a curve are
$$x = 1 + 2 \sin ^ { 2 } \theta , \quad y = 4 \tan \theta$$
(i) Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { \sin \theta \cos ^ { 3 } \theta }$.\\
(ii) Find the equation of the tangent to the curve at the point where $\theta = \frac { 1 } { 4 } \pi$, giving your answer in the form $y = m x + c$.
\hfill \mbox{\textit{CAIE P2 2011 Q6 [7]}}