AQA Paper 3 2023 June — Question 9 12 marks

Exam BoardAQA
ModulePaper 3 (Paper 3)
Year2023
SessionJune
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicParametric curves and Cartesian conversion
DifficultyStandard +0.3 This is a straightforward parametric equations question requiring routine techniques: substituting values to find coordinates (part a), applying the standard formula dy/dx = (dy/dt)/(dx/dt) (part b(i)), finding a minimum by setting dy/dx = 0 (part b(ii)), and using arctan to find an angle (part b(iii)). All steps are standard A-level procedures with no novel problem-solving required, making it slightly easier than average.
Spec1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation
A water slide is the shape of a curve \(PQ\) as shown in Figure 1 below. \includegraphics{figure_9} The curve can be modelled by the parametric equations $$x = t - \frac{1}{t} + 4.8$$ $$y = t + \frac{2}{t}$$ where \(0.2 \leq t \leq 3\) The horizontal distance from O is \(x\) metres. The vertical distance above the point O at ground level is \(y\) metres. P is the point where \(t = 0.2\) and Q is the point where \(t = 3\)
  1. To make sure speeds are safe at Q, the difference in height between P and Q must be less than 7 metres. Show that the slide meets this safety requirement. [3 marks]
    1. Find an expression for \(\frac{dy}{dx}\) in terms of \(t\) [3 marks]
    2. A vertical support, RS, is to be added between the ground and the lowest point on the slide as shown in Figure 2 below. \includegraphics{figure_9b} Find the length of RS [4 marks]
    3. Find the acute angle the slide makes with the horizontal at Q Give your answer to the nearest degree. [2 marks]
Question 9:
AnswerMarks
9(a)11
Obtains y =10.2 or y=
3
OE
11
AWFW [3.6, 3.7] for
AnswerMarks Guidance
33.4 B1
11
t =3, y=
3
11
10.2− =6.53< 7
3
The slide is safe.
Finds the difference between
AnswerMarks Guidance
their two values of y1.1b M1
Makes a comparison between
6.53 and 7 and states that the
safety requirement is met.
For 6.53, accept
AWFW [6.5,6.53]
AnswerMarks Guidance
OE3.2a R1
Subtotal3
QMarking instructions AO
AnswerMarks
9(b)(i)−2 −2
Obtains 1+t or 1−2t
OE
AnswerMarks Guidance
Ignore labels1.1b B1
=1+t −2
dt
dy
=1−2t −2
dt
dy dy dt
= ×
dx dt dx
dy 1−2t −2
=
dx 1+t −2
dy
Uses chain rule to obtain
dx
dx dy
using their and
dt dt
AnswerMarks Guidance
Condone missing brackets1.1a M1
Obtains a correct expression
AnswerMarks Guidance
ISW1.1b A1
Subtotal3
QMarking instructions AO
AnswerMarks
9(b)(ii)Forms equation for appropriate
derivative equal to zero.
dy dy
Their =0 or their =0
AnswerMarks Guidance
dx dt3.1a M1
t 2 =2
t = 2
2
y = 2+
2
=2 2
Length of RS = 2.83 metres
Obtains t = 2
Allow 1.4 or better for 2
t = 2must come from correct
dy dy
or
dx dt
AnswerMarks Guidance
PI by substituting 2 into y1.1b A1
Substitutes their value for t into
y and obtains a value for y
AnswerMarks Guidance
provided 0.2 < t < 33.4 M1
Obtains correct length with unit
e.g 2 2 metres or 2.8 metres or
or AWFW [2.82, 2.83] metres
Allow equivalent correct length
in different units
Do not ignore subsequent
AnswerMarks Guidance
incorrect rounding3.2a A1
Subtotal4
QMarking instructions AO
AnswerMarks
9(b)(iii)dy
States tanθ = value of their
dx
at t = 3
OE
PI by correct answer or 0.61 or
AnswerMarks Guidance
better or 553.1a M1
When t =3, =0.7
dx
tanθ=0.7
θ =35 
Obtains 35
AnswerMarks Guidance
CAO3.2a A1
Subtotal2
Question 9 Total12
QMarking instructions AO 
Question 9:
--- 9(a) ---
9(a) | 11
Obtains y =10.2 or y=
3
OE
11
AWFW [3.6, 3.7] for
3 | 3.4 | B1 | When t =0.2, y =10.2
11
t =3, y=
3
11
10.2− =6.53< 7
3
The slide is safe.
Finds the difference between
their two values of y | 1.1b | M1
Makes a comparison between
6.53 and 7 and states that the
safety requirement is met.
For 6.53, accept

AWFW [6.5,6.53]
OE | 3.2a | R1
Subtotal | 3
Q | Marking instructions | AO | Marks | Typical solution
--- 9(b)(i) ---
9(b)(i) | −2 −2
Obtains 1+t or 1−2t
OE
Ignore labels | 1.1b | B1 | dx
=1+t −2
dt
dy
=1−2t −2
dt
dy dy dt
= ×
dx dt dx
dy 1−2t −2
=
dx 1+t −2
dy
Uses chain rule to obtain
dx
dx dy
using their and
dt dt
Condone missing brackets | 1.1a | M1
Obtains a correct expression
ISW | 1.1b | A1
Subtotal | 3
Q | Marking instructions | AO | Marks | Typical solution
--- 9(b)(ii) ---
9(b)(ii) | Forms equation for appropriate
derivative equal to zero.
dy dy
Their =0 or their =0
dx dt | 3.1a | M1 | 1−2t −2 =0
t 2 =2
t = 2
2
y = 2+
2
=2 2
Length of RS = 2.83 metres
Obtains t = 2
Allow 1.4 or better for 2
t = 2must come from correct
dy dy
or
dx dt
PI by substituting 2 into y | 1.1b | A1
Substitutes their value for t into
y and obtains a value for y
provided 0.2 < t < 3 | 3.4 | M1
Obtains correct length with unit
e.g 2 2 metres or 2.8 metres or
or AWFW [2.82, 2.83] metres
Allow equivalent correct length
in different units
Do not ignore subsequent
incorrect rounding | 3.2a | A1
Subtotal | 4
Q | Marking instructions | AO | Marks | Typical solution
--- 9(b)(iii) ---
9(b)(iii) | dy
States tanθ = value of their
dx
at t = 3
OE
PI by correct answer or 0.61 or

better or 55 | 3.1a | M1 | dy
When t =3, =0.7
dx
tanθ=0.7
θ =35 

Obtains 35
CAO | 3.2a | A1
Subtotal | 2
Question 9 Total | 12
Q | Marking instructions | AO | Marks | Typical solution
A water slide is the shape of a curve $PQ$ as shown in Figure 1 below.

\includegraphics{figure_9}

The curve can be modelled by the parametric equations
$$x = t - \frac{1}{t} + 4.8$$
$$y = t + \frac{2}{t}$$
where $0.2 \leq t \leq 3$

The horizontal distance from O is $x$ metres.

The vertical distance above the point O at ground level is $y$ metres.

P is the point where $t = 0.2$ and Q is the point where $t = 3$

\begin{enumerate}[label=(\alph*)]
\item To make sure speeds are safe at Q, the difference in height between P and Q must be less than 7 metres.

Show that the slide meets this safety requirement.

[3 marks]

\item \begin{enumerate}[label=(\roman*)]
\item Find an expression for $\frac{dy}{dx}$ in terms of $t$

[3 marks]

\item A vertical support, RS, is to be added between the ground and the lowest point on the slide as shown in Figure 2 below.

\includegraphics{figure_9b}

Find the length of RS

[4 marks]

\item Find the acute angle the slide makes with the horizontal at Q

Give your answer to the nearest degree.

[2 marks]
\end{enumerate}
\end{enumerate}

\hfill \mbox{\textit{AQA Paper 3 2023 Q9 [12]}}
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