AQA Paper 2 Specimen — Question 3 6 marks

Exam BoardAQA
ModulePaper 2 (Paper 2)
SessionSpecimen
Marks6
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Mark schemeDownload PDF ↗
TopicParametric curves and Cartesian conversion
TypeFind dy/dx at a point
DifficultyModerate -0.3 Part (a) requires standard application of the parametric differentiation formula dy/dx = (dy/dt)/(dx/dt) with straightforward polynomial derivatives. Part (b) involves eliminating the parameter by expressing t from one equation and substituting—a routine technique. Both parts are textbook exercises requiring recall of standard methods with minimal problem-solving, making this slightly easier than average.
Spec1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation
A curve is defined by the parametric equations $$x = t^3 + 2, \quad y = t^2 - 1$$
  1. Find the gradient of the curve at the point where \(t = -2\) [4 marks]
  2. Find a Cartesian equation of the curve. [2 marks]
Question 3:
AnswerMarks
3(a)Uses a correct method for finding
dy
dx
evidence for this includes sight of
dy dx
or and chain rule
dx dt
OR an attempt at implicit or explicit
differentiation of a correct Cartesian
equation or ‘their’ equation from
AnswerMarks Guidance
part (b)AO1.1a M1
3t2 2t
dt dt
dy 2t
dx 3t2
dy 1
When t 2 
dx 3
ALT
2
y(x2)3 1
1
dy 2(x2) 3
dx 3
When t = –2 , x = –6
1
dy 2(62) 3 1
 
dx 3 3
dy
Obtains correct
AnswerMarks Guidance
dxAO1.1b A1
Substitutes t = –2 (or x = –6) into
dy
‘their’ equation for
AnswerMarks Guidance
dxAO1.1a M1
Obtains correct simplified gradient of
the curve
dy
FT ‘their’ equation for
AnswerMarks Guidance
dxAO1.1b A1F
(b)Eliminates t or makes t the subject in
one expression
(evidence for this includes one
equation with t as the subject or two
AnswerMarks Guidance
equations with equal powers of t.)AO1.1a M1
t6 (x2)2, t6 (y1)3
(x2)2  (y1)3
ALT
t3 (x2)
1
t (x2)3
2
y(x2)3 1
Finds a correct Cartesian equation in
AnswerMarks Guidance
any formAO1.1b A1
Total6
QMarking Instructions AO 
Question 3:
--- 3(a) ---
3(a) | Uses a correct method for finding
dy
dx
evidence for this includes sight of
dy dx
or and chain rule
dx dt
OR an attempt at implicit or explicit
differentiation of a correct Cartesian
equation or ‘their’ equation from
part (b) | AO1.1a | M1 | dx dy
3t2 2t
dt dt
dy 2t

dx 3t2
dy 1
When t 2 
dx 3
ALT
2
y(x2)3 1
1

dy 2(x2) 3

dx 3
When t = –2 , x = –6
1

dy 2(62) 3 1
 
dx 3 3
dy
Obtains correct
dx | AO1.1b | A1
Substitutes t = –2 (or x = –6) into
dy
‘their’ equation for
dx | AO1.1a | M1
Obtains correct simplified gradient of
the curve
dy
FT ‘their’ equation for
dx | AO1.1b | A1F
(b) | Eliminates t or makes t the subject in
one expression
(evidence for this includes one
equation with t as the subject or two
equations with equal powers of t.) | AO1.1a | M1 | t3 (x2), t2 (y1)
t6 (x2)2, t6 (y1)3
(x2)2  (y1)3
ALT
t3 (x2)
1
t (x2)3
2
y(x2)3 1
Finds a correct Cartesian equation in
any form | AO1.1b | A1
Total | 6
Q | Marking Instructions | AO | Marks | Typical Solution
A curve is defined by the parametric equations
$$x = t^3 + 2, \quad y = t^2 - 1$$

\begin{enumerate}[label=(\alph*)]
\item Find the gradient of the curve at the point where $t = -2$
[4 marks]

\item Find a Cartesian equation of the curve.
[2 marks]
\end{enumerate}

\hfill \mbox{\textit{AQA Paper 2  Q3 [6]}}
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