| Exam Board | AQA |
|---|---|
| Module | Paper 3 (Paper 3) |
| Year | 2021 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Radians, Arc Length and Sector Area |
| Type | Optimization with sectors |
| Difficulty | Moderate -0.8 This is a straightforward applied calculus question using sector formulas (area = ½r²θ, arc length = rθ) and basic optimization. Part (a) requires simple substitution into standard formulas. Part (b) involves algebraic manipulation to derive a given expression and then routine differentiation to find a minimum. All techniques are standard A-level procedures with no novel problem-solving required, making it easier than average but not trivial due to the multi-step nature and algebraic manipulation involved. |
| Spec | 1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta1.07n Stationary points: find maxima, minima using derivatives1.07o Increasing/decreasing: functions using sign of dy/dx |
| Answer | Marks | Guidance |
|---|---|---|
| 5(a)(i) | Uses formula correctly for area | |
| of sector | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| units | 1.1b | A1 |
| Subtotal | 2 | |
| Q | Marking instructions | AO |
| Answer | Marks | Guidance |
|---|---|---|
| 5(a)(ii) | Uses formula for arc length | 1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| perimeter by 1.80 | 3.1b | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| CAO | 3.2a | A1 |
| Subtotal | 3 | |
| Q | Marking instructions | AO |
| Answer | Marks |
|---|---|
| 5(b)(i) | Forms at least one correct |
| Answer | Marks | Guidance |
|---|---|---|
| formulae for C | 3.3 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| expression for P in terms of r | 3.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | 2.1 | R1 |
| Subtotal | 3 | |
| Q | Marking instructions | AO |
| Answer | Marks |
|---|---|
| 5(b)(ii) | Recognises the use |
| Answer | Marks | Guidance |
|---|---|---|
| 𝑑𝑑𝑑𝑑 | 3.4 | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| OE | 1.1b | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| 𝑑𝑑𝐶𝐶 | 2.4 | E1 |
| Answer | Marks | Guidance |
|---|---|---|
| r to at le𝑑𝑑a𝑑𝑑st= two decimal places | 1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| 𝑟𝑟 ≈ 4.5 | 2.1 | R1 |
| Subtotal | 5 | |
| Question Total | 13 | |
| Q | Marking instructions | AO |
Question 5:
--- 5(a)(i) ---
5(a)(i) | Uses formula correctly for area
of sector | 1.1a | M1 | 1
A= ×52×0.7
2
=8.75m2
Obtains 8.75
Condone incorrect or missing
units | 1.1b | A1
Subtotal | 2
Q | Marking instructions | AO | Marks | Typical solution
--- 5(a)(ii) ---
5(a)(ii) | Uses formula for arc length | 1.1b | B1 | P =5×0.7+2×5
=13.5
Cost =13.5×1.80
=£24.30
Obtains the perimeter by adding
twice the radius to their arc
length and multiplies their
perimeter by 1.80 | 3.1b | M1
Obtains correct cost £24.30
CAO | 3.2a | A1
Subtotal | 3
Q | Marking instructions | AO | Marks | Typical solution
--- 5(b)(i) ---
5(b)(i) | Forms at least one correct
equation for area or perimeter
May be embedded in the
formulae for C | 3.3 | M1 | P=rθ+2r
1
r2θ=20
2
40
⇒θ=
r2
40
P = +2r
r
40×1.8
C = +2×1.8r
r
72 18
= + r
r 5
1820
= +r
5 r
Eliminates θ from two fully
correct equations for area and
perimeter to obtain an
expression for P in terms of r | 3.1b | A1
Completes argument to show
the required result
Accept 3.6 for
18
5 | 2.1 | R1
Subtotal | 3
Q | Marking instructions | AO | Marks | Typical solution
--- 5(b)(ii) ---
5(b)(ii) | Recognises the use
differentiation in the model
PI if seen
𝑑𝑑𝑑𝑑 | 3.4 | B1 | 72 18
𝐶𝐶 = + 𝑟𝑟
𝑟𝑟 5
𝑑𝑑𝐶𝐶 72 18
= − 2 +
Minimu𝑑𝑑m𝑟𝑟 occur𝑟𝑟s whe5n 0
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑 =
72 18
− 2 + = 0
𝑟𝑟 = 520
2
𝑟𝑟
4.472..
𝑟𝑟 =H e√n2c0e r≈ 4.5
≈
2
𝑑𝑑 𝐶𝐶 144
2 = 3
𝑑𝑑𝑟𝑟 𝑟𝑟
When r = ,
2
𝑑𝑑 𝑑𝑑
2
√20 𝑑𝑑𝑑𝑑 > 0
Therefore minimum at
𝑟𝑟 ≈ 4.5
Diffe𝑑𝑑re𝑑𝑑ntiates given model with
at least one term correct
Condone sign error
OE | 1.1b | M1
Explains that a
minimum/stationary/turning point
occurs when 0
𝑑𝑑𝐶𝐶 | 2.4 | E1
𝑑𝑑𝑟𝑟 =
Solves 0 to find correct
𝑑𝑑𝑑𝑑
exact value or decimal value for
r to at le𝑑𝑑a𝑑𝑑st= two decimal places | 1.1b | A1
Uses a gradient test or second
derivative or sketches graph to
determine nature of stationary
point
Completes argument to show
minimum occurs when
Must have shown in
previous step 𝑟𝑟 ≈ 4.5
𝑟𝑟 ≈ 4.5 | 2.1 | R1
Subtotal | 5
Question Total | 13
Q | Marking instructions | AO | Marks | Typical solutionA gardener is creating flowerbeds in the shape of sectors of circles.
The gardener uses an edging strip around the perimeter of each of the flowerbeds.
The cost of the edging strip is £1.80 per metre and can be purchased for any length.
One of the flowerbeds has a radius of 5 metres and an angle at the centre of 0.7 radians as shown in the diagram below.
\includegraphics{figure_5}
\begin{enumerate}[label=(\alph*)]
\item
\begin{enumerate}[label=(\roman*)]
\item Find the area of this flowerbed.
[2 marks]
\item Find the cost of the edging strip required for this flowerbed.
[3 marks]
\end{enumerate}
\item A flowerbed is to be made with an area of 20 m²
\begin{enumerate}[label=(\roman*)]
\item Show that the cost, £$C$, of the edging strip required for this flowerbed is given by
$$C = \frac{18}{5}\left(\frac{20}{r} + r\right)$$
where $r$ is the radius measured in metres.
[3 marks]
\item Hence, show that the minimum cost of the edging strip for this flowerbed occurs when $r \approx 4.5$
Fully justify your answer.
[5 marks]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 3 2021 Q5 [13]}}