Question 2 - A-Level Maths

CAIE P3 2011 November — Question 2 5 marks

Exam Board	CAIE
Module	P3 (Pure Mathematics 3)
Year	2011
Session	November
Marks	5
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Parametric differentiation
Type	Find dy/dx expression in terms of parameter
Difficulty	Moderate -0.3 This is a straightforward parametric differentiation question requiring the chain rule (dy/dx = dy/dt ÷ dx/dt) with standard trigonometric derivatives. The simplification involves basic trigonometric identities (sin²t + cos²t = 1) but follows a routine pattern. Slightly easier than average as it's a direct application of a standard technique with no conceptual challenges.
Spec	1.03g Parametric equations: of curves and conversion to cartesian 1.07s Parametric and implicit differentiation

2 The parametric equations of a curve are $$x = 3 \left( 1 + \sin ^ { 2 } t \right) , \quad y = 2 \cos ^ { 3 } t$$ Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $t$, simplifying your answer as far as possible.

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EITHER: Use chain rule

Answer	Marks
obtain $\frac{dv}{dt} = 6\sin t \cos t$, or equivalent	A1
obtain $\frac{dv}{dt} = -6\cos^2 t \sin t$, or equivalent	A1
Use $\frac{dv}{dx} = \frac{dv}{dt} \div \frac{dx}{dt}$	M1
Obtain final answer $\frac{dy}{dx} = -\cos t$	A1

OR: Express $y$ in terms of $x$ and use chain rule

Answer	Marks	Guidance
Obtain $\frac{dv}{dx} = k(2 - \frac{x}{3})^{-\frac{1}{2}}$, or equivalent	M1 A1
Obtain $\frac{dv}{dx} = -(2 - \frac{x}{3})^{-\frac{1}{2}}$, or equivalent	A1
Express derivative in terms of $t$	M1
Obtain final answer $\frac{dy}{dx} = -\cos t$	A1	[5]

**EITHER: Use chain rule**

obtain $\frac{dv}{dt} = 6\sin t \cos t$, or equivalent | A1 |
obtain $\frac{dv}{dt} = -6\cos^2 t \sin t$, or equivalent | A1 |
Use $\frac{dv}{dx} = \frac{dv}{dt} \div \frac{dx}{dt}$ | M1 |
Obtain final answer $\frac{dy}{dx} = -\cos t$ | A1 |

**OR: Express $y$ in terms of $x$ and use chain rule**

Obtain $\frac{dv}{dx} = k(2 - \frac{x}{3})^{-\frac{1}{2}}$, or equivalent | M1 A1 |
Obtain $\frac{dv}{dx} = -(2 - \frac{x}{3})^{-\frac{1}{2}}$, or equivalent | A1 |
Express derivative in terms of $t$ | M1 |
Obtain final answer $\frac{dy}{dx} = -\cos t$ | A1 | [5]

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2 The parametric equations of a curve are

$$x = 3 \left( 1 + \sin ^ { 2 } t \right) , \quad y = 2 \cos ^ { 3 } t$$

Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $t$, simplifying your answer as far as possible.

\hfill \mbox{\textit{CAIE P3 2011 Q2 [5]}}

This paper (10 questions)

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Q1 4 Q2 5 Q3 6 Q4 7 Q5 8 Q6 8 Q7 8 Q8 9 Q9 10 Q10 10

Answer	Marks
obtain \(\frac{dv}{dt} = 6\sin t \cos t\), or equivalent	A1
obtain \(\frac{dv}{dt} = -6\cos^2 t \sin t\), or equivalent	A1
Use \(\frac{dv}{dx} = \frac{dv}{dt} \div \frac{dx}{dt}\)	M1
Obtain final answer \(\frac{dy}{dx} = -\cos t\)	A1

Answer	Marks	Guidance
Obtain \(\frac{dv}{dx} = k(2 - \frac{x}{3})^{-\frac{1}{2}}\), or equivalent	M1 A1
Obtain \(\frac{dv}{dx} = -(2 - \frac{x}{3})^{-\frac{1}{2}}\), or equivalent	A1
Express derivative in terms of \(t\)	M1
Obtain final answer \(\frac{dy}{dx} = -\cos t\)	A1	[5]

CAIE P3 2011 November — Question 2 5 marks

EITHER: Use chain rule

OR: Express \(y\) in terms of \(x\) and use chain rule

This paper (10 questions)