| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2003 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric differentiation |
| Type | Show dy/dx simplifies to given form |
| Difficulty | Standard +0.3 This is a straightforward parametric differentiation question requiring standard chain rule application (dx/dθ and dy/dθ), then finding a tangent equation and horizontal tangents. The trigonometric simplification to get cot θ is routine, and all parts follow predictable patterns for this topic, making it slightly easier than average. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^21.07m Tangents and normals: gradient and equations1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| State \(\dfrac{dx}{d\theta} = 2 - 2\cos 2\theta\) or \(\dfrac{dy}{d\theta} = 2\sin 2\theta\) | B1 | |
| Use \(\dfrac{dy}{dx} = \dfrac{dy}{d\theta} \div \dfrac{dx}{d\theta}\) | M1 | |
| Obtain answer \(\dfrac{dy}{dx} = \dfrac{2\sin 2\theta}{2 - 2\cos 2\theta}\) or equivalent | A1 | |
| Make relevant use of \(\sin 2A\) and \(\cos 2A\) formulae | M1 | indep. |
| Obtain given answer correctly | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Substitute \(\theta = \frac{1}{4}\pi\) in \(\dfrac{dy}{dx}\) and both parametric equations | M1 | |
| Obtain \(\dfrac{dy}{dx} = 1\), \(x = \frac{1}{2}\pi - 1\), \(y = 2\) | A1 | |
| Obtain equation \(y = x + 1.43\), or any exact equivalent | A1\(\sqrt{}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| State or imply that tangent is horizontal when \(\theta = \frac{1}{2}\pi\) or \(\frac{3}{2}\pi\) | B1 | |
| Obtain a correct pair of \(x\), \(y\) or \(x\)- or \(y\)-coordinates | B1 | |
| State correct answers \((\pi, 3)\) and \((3\pi, 3)\) | B1 |
# Question 7(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| State $\dfrac{dx}{d\theta} = 2 - 2\cos 2\theta$ or $\dfrac{dy}{d\theta} = 2\sin 2\theta$ | B1 | |
| Use $\dfrac{dy}{dx} = \dfrac{dy}{d\theta} \div \dfrac{dx}{d\theta}$ | M1 | |
| Obtain answer $\dfrac{dy}{dx} = \dfrac{2\sin 2\theta}{2 - 2\cos 2\theta}$ or equivalent | A1 | |
| Make relevant use of $\sin 2A$ and $\cos 2A$ formulae | M1 | indep. |
| Obtain given answer correctly | A1 | |
**Total: [5]**
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# Question 7(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Substitute $\theta = \frac{1}{4}\pi$ in $\dfrac{dy}{dx}$ and both parametric equations | M1 | |
| Obtain $\dfrac{dy}{dx} = 1$, $x = \frac{1}{2}\pi - 1$, $y = 2$ | A1 | |
| Obtain equation $y = x + 1.43$, or any exact equivalent | A1$\sqrt{}$ | |
**Total: [3]**
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# Question 7(iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| State or imply that tangent is horizontal when $\theta = \frac{1}{2}\pi$ or $\frac{3}{2}\pi$ | B1 | |
| Obtain a correct pair of $x$, $y$ or $x$- or $y$-coordinates | B1 | |
| State correct answers $(\pi, 3)$ and $(3\pi, 3)$ | B1 | |
**Total: [3]**7 The parametric equations of a curve are
$$x = 2 \theta - \sin 2 \theta , \quad y = 2 - \cos 2 \theta$$
(i) Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \cot \theta$.\\
(ii) Find the equation of the tangent to the curve at the point where $\theta = \frac { 1 } { 4 } \pi$.\\
(iii) For the part of the curve where $0 < \theta < 2 \pi$, find the coordinates of the points where the tangent is parallel to the $x$-axis.
\hfill \mbox{\textit{CAIE P2 2003 Q7 [11]}}