Question 3 - A-Level Maths

CAIE P2 2021 March — Question 3 5 marks

Exam Board	CAIE
Module	P2 (Pure Mathematics 2)
Year	2021
Session	March
Marks	5
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Parametric differentiation
Type	Find gradient at given parameter
Difficulty	Moderate -0.3 This is a straightforward application of the parametric differentiation formula dy/dx = (dy/dt)/(dx/dt) evaluated at a specific point. The derivatives require standard rules (product rule for x, chain rule for y) and the evaluation at t=0 simplifies nicely. Slightly easier than average due to being a single-step process with no algebraic complications.
Spec	1.03g Parametric equations: of curves and conversion to cartesian 1.07s Parametric and implicit differentiation

3 The parametric equations of a curve are $$x = \mathrm { e } ^ { 2 t } \cos 4 t , \quad y = 3 \sin 2 t$$ Find the gradient of the curve at the point for which $t = 0$.

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Question 3:

Answer	Marks	Guidance
Answer	Marks	Guidance
Attempt use of product rule to find $\frac{dx}{dt}$	M1
Obtain $2e^{2t}\cos 4t - 4e^{2t}\sin 4t$	A1
Obtain $\frac{dy}{dt} = 6\cos 2t$	B1
Divide to find $\frac{dy}{dx}$ with $t=0$ substituted	M1
Obtain 3	A1	CWO

## Question 3:

| Answer | Marks | Guidance |
|--------|-------|----------|
| Attempt use of product rule to find $\frac{dx}{dt}$ | M1 | |
| Obtain $2e^{2t}\cos 4t - 4e^{2t}\sin 4t$ | A1 | |
| Obtain $\frac{dy}{dt} = 6\cos 2t$ | B1 | |
| Divide to find $\frac{dy}{dx}$ with $t=0$ substituted | M1 | |
| Obtain 3 | A1 | CWO |

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3 The parametric equations of a curve are

$$x = \mathrm { e } ^ { 2 t } \cos 4 t , \quad y = 3 \sin 2 t$$

Find the gradient of the curve at the point for which $t = 0$.\\

\hfill \mbox{\textit{CAIE P2 2021 Q3 [5]}}

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Q1 5 Q2 5 Q3 5 Q4 8 Q5 8 Q6 10 Q7 9