CAIE P3 2019 November — Question 3 5 marks

Exam BoardCAIE
ModuleP3 (Pure Mathematics 3)
Year2019
SessionNovember
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 chain rule application (dy/dx = dy/dt รท dx/dt) with routine trigonometric differentiation and simplification using the identity sin 2t = 2sin t cos t. The 'show that' format removes problem-solving burden, making it slightly easier than average despite requiring algebraic manipulation.
Spec1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation
3 The parametric equations of a curve are $$x = 2 t + \sin 2 t , \quad y = \ln ( 1 - \cos 2 t )$$ Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \operatorname { cosec } 2 t\).
Question 3:
AnswerMarks Guidance
AnswerMark Guidance
State \(\frac{dx}{dt} = 2 + 2\cos 2t\)B1
Use the chain rule to find the derivative of \(y\)M1
Obtain \(\frac{dy}{dt} = \frac{2\sin 2t}{1 - \cos 2t}\)A1 OE
Use \(\frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt}\)M1
Obtain \(\frac{dy}{dx} = \cosec 2t\) correctlyA1 AG 
## Question 3:

| Answer | Mark | Guidance |
|--------|------|----------|
| State $\frac{dx}{dt} = 2 + 2\cos 2t$ | B1 | |
| Use the chain rule to find the derivative of $y$ | M1 | |
| Obtain $\frac{dy}{dt} = \frac{2\sin 2t}{1 - \cos 2t}$ | A1 | OE |
| Use $\frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt}$ | M1 | |
| Obtain $\frac{dy}{dx} = \cosec 2t$ correctly | A1 | AG |

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3 The parametric equations of a curve are

$$x = 2 t + \sin 2 t , \quad y = \ln ( 1 - \cos 2 t )$$

Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \operatorname { cosec } 2 t$.\\

\hfill \mbox{\textit{CAIE P3 2019 Q3 [5]}}
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