CAIE P3 2006 June — Question 3 5 marks

Exam BoardCAIE
ModuleP3 (Pure Mathematics 3)
Year2006
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicParametric differentiation
TypeShow dy/dx simplifies to given form
DifficultyModerate -0.3 This is a straightforward parametric differentiation question requiring standard application of the chain rule (dy/dx = dy/dθ ÷ dx/dθ) with basic trigonometric derivatives. The algebra simplifies nicely using the double angle formula sin 2θ = 2sin θ cos θ, making it slightly easier than average but still requiring correct technique and trigonometric manipulation.
Spec1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation
The parametric equations of a curve are $$x = 2\theta + \sin 2\theta, \quad y = 1 - \cos 2\theta.$$ Show that \(\frac{dy}{dx} = \tan \theta\). [5]
AnswerMarks Guidance
State that \(\frac{dx}{d\theta} = 2 - 2\cos 2\theta\) or \(\frac{dy}{d\theta} = 2\sin 2\theta\)B1
Use \(\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}\)M1
Obtain answer in any correct form, e.g. \(\frac{2\sin 2\theta}{2 + 2\cos 2\theta}\)A1
Make relevant use of \(\sin 2\theta\) and \(\cos 2\theta\) formulaeM1
Obtain given answer correctlyA1 5 
| State that $\frac{dx}{d\theta} = 2 - 2\cos 2\theta$ or $\frac{dy}{d\theta} = 2\sin 2\theta$ | B1 |
| Use $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$ | M1 |
| Obtain answer in any correct form, e.g. $\frac{2\sin 2\theta}{2 + 2\cos 2\theta}$ | A1 |
| Make relevant use of $\sin 2\theta$ and $\cos 2\theta$ formulae | M1 |
| Obtain given answer correctly | A1 | 5 |
The parametric equations of a curve are
$$x = 2\theta + \sin 2\theta, \quad y = 1 - \cos 2\theta.$$
Show that $\frac{dy}{dx} = \tan \theta$. [5]

\hfill \mbox{\textit{CAIE P3 2006 Q3 [5]}}
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