Question 2 - A-Level Maths

CAIE P3 2010 November — Question 2 5 marks

Exam Board	CAIE
Module	P3 (Pure Mathematics 3)
Year	2010
Session	November
Marks	5
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Parametric differentiation
Type	Find gradient at given parameter
Difficulty	Moderate -0.8 This is a straightforward application of the parametric differentiation formula dy/dx = (dy/dt)/(dx/dt), requiring only basic differentiation of an exponential and a quotient, followed by substitution of t=0. It's a routine single-step calculation with no conceptual challenges, making it easier than average.
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 $$x = \frac { t } { 2 t + 3 } , \quad y = \mathrm { e } ^ { - 2 t }$$ Find the gradient of the curve at the point for which $t = 0$.

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Answer	Marks	Guidance
Use of correct quotient or product rule to differentiate $x$ or $t$	M1
Obtain correct $\frac{3}{(2t+3)^2}$ or unsimplified equivalent	A1
Obtain $-2e^{-y}$ for derivative of $y$	B1
Use $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ or equivalent	M1
Obtain $-6$	A1	cwo

Alternative:

Answer	Marks	Guidance
Eliminate parameter and attempt differentiation $\left(y = e^{-6x/(1-2x)}\right)$	B1
Use correct quotient or product rule	M1
Use chain rule	M1
Obtain $\frac{dy}{dx} = \frac{-6}{(1-2x)^2}e^{-6x/(1-2x)}$	A1
Obtain $-6$	A1	cwo

Use of correct quotient or product rule to differentiate $x$ or $t$ | M1 |
Obtain correct $\frac{3}{(2t+3)^2}$ or unsimplified equivalent | A1 |
Obtain $-2e^{-y}$ for derivative of $y$ | B1 |
Use $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ or equivalent | M1 |
Obtain $-6$ | A1 | cwo | [5]

**Alternative:**

Eliminate parameter and attempt differentiation $\left(y = e^{-6x/(1-2x)}\right)$ | B1 |
Use correct quotient or product rule | M1 |
Use chain rule | M1 |
Obtain $\frac{dy}{dx} = \frac{-6}{(1-2x)^2}e^{-6x/(1-2x)}$ | A1 |
Obtain $-6$ | A1 | cwo |

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2 The parametric equations of a curve are

$$x = \frac { t } { 2 t + 3 } , \quad y = \mathrm { e } ^ { - 2 t }$$

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

\hfill \mbox{\textit{CAIE P3 2010 Q2 [5]}}

This paper (10 questions)

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Q1 3 Q2 5 Q3 6 Q4 6 Q5 7 Q6 8 Q7 8 Q8 9 Q9 10 Q10 13

Answer	Marks	Guidance
Use of correct quotient or product rule to differentiate \(x\) or \(t\)	M1
Obtain correct \(\frac{3}{(2t+3)^2}\) or unsimplified equivalent	A1
Obtain \(-2e^{-y}\) for derivative of \(y\)	B1
Use \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt}\) or equivalent	M1
Obtain \(-6\)	A1	cwo