| Exam Board | AQA |
|---|---|
| Module | Paper 3 (Paper 3) |
| Year | 2022 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric curves and Cartesian conversion |
| Type | Find stationary/turning points |
| Difficulty | Standard +0.8 This question requires parametric arc length calculation (non-trivial integration), parametric differentiation, and finding maximum width by solving dy/dx=0. The arc length integral √(16t² + (9-1.4t)²) requires algebraic manipulation and careful evaluation. The multi-step reasoning connecting maximum width to length comparison elevates this above routine parametric questions, though the techniques themselves are standard A-level further maths content. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| 6(a) | Substitutes t = 9.5 into | |
| x=−2t2or 2t2 | 3.4 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| ISW | 1.1b | A1 |
| Subtotal | 2 | |
| Q | Marking instructions | AO |
| Answer | Marks |
|---|---|
| 6(b)(i) | Obtains 9 1.4t or 4t |
| Answer | Marks | Guidance |
|---|---|---|
| Ignore lab e l s | 1.1b | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Condone sign error | 3.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Do not ISW | 1.1b | A1 |
| Subtotal | 3 | |
| Q | Marking instructions | AO |
| Answer | Marks |
|---|---|
| 6(b)(ii) | dy dy |
| Answer | Marks | Guidance |
|---|---|---|
| dx | 3.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| dx | 1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| value for y provided 0 < t < 9.5 | 3.4 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Allow 180 for length | 3.2a | R1 |
| Subtotal | 4 | |
| Question 6 Total | 9 | |
| Q | Marking instructions | AO |
Question 6:
--- 6(a) ---
6(a) | Substitutes t = 9.5 into
x=−2t2or 2t2 | 3.4 | M1 | t =9.5⇒ x=−2×9.52= 180.5
Length = 180.5 cm −
Obtains 180.5
Condone incorrect or missing
units
ISW | 1.1b | A1
Subtotal | 2
Q | Marking instructions | AO | Marks | Typical solution
--- 6(b)(i) ---
6(b)(i) | Obtains 9 1.4t or 4t
OE
− −
Ignore lab e l s | 1.1b | B1 | dy
=9−1.4t
dt
dx
=−4t
dt
dy dy dt
= ×
dx dt dx
dy 9−1.4t
=
dx −4t
dy
Uses chain rule to obtain
dx
Condone sign error | 3.1a | M1
Obtains a correct expression
Do not ISW | 1.1b | A1
Subtotal | 3
Q | Marking instructions | AO | Marks | Typical solution
--- 6(b)(ii) ---
6(b)(ii) | dy dy
Equates their or their or
dx dt
dy
their numerator of their to 0
dx
dy
PI by correct t from correct
dx | 3.1a | M1 | 9−1.4t
=0
−4t
45
t = =6.43
7
y =9×6.43−0.7×( 6.43 )2
=28.9
Width of surfboard = 58 cm
180.5 ÷ 3 = 60.2 ≈ 58
Hence the width is approximately
one third of the length
Obtains correct value for t ACF
eg t = 6.4 or
9
dy
Must come fr1o.4m correct
dx | 1.1b | A1
Substitutes their value for t into
the model for y and obtains a
value for y provided 0 < t < 9.5 | 3.4 | M1
Compares correct width and
correct length and or 3 with a
1
correct concluding statement
3
OE
CSO
Allow 180 for length | 3.2a | R1
Subtotal | 4
Question 6 Total | 9
Q | Marking instructions | AO | Marks | Typical solutionA design for a surfboard is shown in Figure 1.
Figure 1
\includegraphics{figure_6_1}
The curve of the top half of the surfboard can be modelled by the parametric equations
$$x = -2t^2$$
$$y = 9t - 0.7t^2$$
for $0 \leq t \leq 9.5$ as shown in Figure 2, where $x$ and $y$ are measured in centimetres.
Figure 2
\includegraphics{figure_6_2}
\begin{enumerate}[label=(\alph*)]
\item Find the length of the surfboard.
[2 marks]
\item \begin{enumerate}[label=(\roman*)]
\item Find an expression for $\frac{dy}{dx}$ in terms of $t$.
[3 marks]
\item Hence, show that the width of the surfboard is approximately one third of its length.
[4 marks]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 3 2022 Q6 [9]}}