| Exam Board | AQA |
|---|---|
| Module | Paper 3 (Paper 3) |
| Session | Specimen |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Exponential growth/decay - direct proportionality (dN/dt = kN) |
| Difficulty | Moderate -0.3 This is a standard A-level differential equations modelling question with straightforward integration, substitution of initial conditions, and logarithmic manipulation. Part (a) requires basic separation of variables (dA/dt = kA), parts (b)(i-iii) involve routine algebraic manipulation and solving exponential equations, while parts (c-d) test understanding of model limitations—all typical textbook exercises with no novel problem-solving required. Slightly easier than average due to heavy scaffolding and standard techniques. |
| Spec | 1.06i Exponential growth/decay: in modelling context1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks |
|---|---|
| 3(a) | Translates proportionality into a |
| Answer | Marks | Guidance |
|---|---|---|
| proportionality. | AO3.3 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Separates variables | AO1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| correctly | AO1.1b | A1F |
| Answer | Marks | Guidance |
|---|---|---|
| contains no slips. | AO2.1 | R1 |
| (b)(i) | States correct value of B | AO1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| (b)(ii) | Uses t = 20 and A = 0.5 to find k | AO3.1b |
| Answer | Marks | Guidance |
|---|---|---|
| Finds correct value of k | AO1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| terms of t | AO1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| throughout. AG | AO2.1 | R1 |
| (b)(iii) | Uses the model to set up correct | |
| equation and attempt to find t | AO3.4 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Finds correct value of t | AO1.1b | A1 |
| (c) | States any sensible and relevant |
| Answer | Marks | Guidance |
|---|---|---|
| time. | AO3.5b | E1 |
| Answer | Marks |
|---|---|
| (d) | Any sensible and relevant |
| Answer | Marks | Guidance |
|---|---|---|
| time | AO3.5c | E1 |
| Answer | Marks | Guidance |
|---|---|---|
| Total | 13 | |
| Q | Marking Instructions | AO |
Question 3:
--- 3(a) ---
3(a) | Translates proportionality into a
differential equation involving
dA
, A and a constant of
dt
proportionality. | AO3.3 | M1 | dA
∝ A
dt
dA
⇒ =kA
dt
1
⇒∫ dA=∫k dt
A
⇒lnA=kt+c
⇒ A=ekt+c
⇒ A= Bekt AG
Separates variables | AO1.1a | M1
Integrates both of ‘their’ sides
correctly | AO1.1b | A1F
Constructs a rigorous
mathematical argument that
supports use of the given model.
AG
Only award if they have a
completely correct solution,
which is clear, easy to follow and
contains no slips. | AO2.1 | R1
(b)(i) | States correct value of B | AO1.1b | B1 | 1
B = 0.25 or B =
4
(b)(ii) | Uses t = 20 and A = 0.5 to find k | AO3.1b | M1 | When t = 20, A = 0.5
⇒0.5=0.25e20k
⇒20k =ln2
1
⇒k = ln2
20
1 t
⇒ A= (eln2)20
4
t
⇒ A=2−2×220
t
−2
⇒ A=220 AG
Finds correct value of k | AO1.1b | A1
Substitutes ‘their’ k to get A in
terms of t | AO1.1a | M1
Constructs rigorous and
convincing argument to show
t
−2
A=220
Using correct notation
throughout. AG | AO2.1 | R1
(b)(iii) | Uses the model to set up correct
equation and attempt to find t | AO3.4 | M1 | t
−2
2π=220
t = 93.03 days
Finds correct value of t | AO1.1b | A1
(c) | States any sensible and relevant
limitation of the model that is
specified in terms of the pond,
area, weed, rate of change or
time. | AO3.5b | E1 | Model predicts that the area of
weed will increase without limit
and this is not possible since the
area of the pond is 4π
(d) | Any sensible and relevant
refinement to the model that is
specified in terms of the pond,
area, weed, rate of change or
time | AO3.5c | E1 | Introduce a limiting factor such as
fish eating weed or rate of growth
decreases as surface area
covered
Total | 13
Q | Marking Instructions | AO | Marks | Typical SolutionA circular ornamental garden pond, of radius 2 metres, has weed starting to grow and cover its surface.
As the weed grows, it covers an area of $A$ square metres. A simple model assumes that the weed grows so that the rate of increase of its area is proportional to $A$.
\begin{enumerate}[label=(\alph*)]
\item Show that the area covered by the weed can be modelled by
$$A = Be^{kt}$$
where $B$ and $k$ are constants and $t$ is time in days since the weed was first noticed. [4 marks]
\item When it was first noticed, the weed covered an area of 0.25 m². Twenty days later the weed covered an area of 0.5 m²
\begin{enumerate}[label=(\roman*)]
\item State the value of $B$. [1 mark]
\item Show that the model for the area covered by the weed can be written as
$$A = 2^{\frac{t}{20} - 2}$$ [4 marks]
\item How many days does it take for the weed to cover half of the surface of the pond? [2 marks]
\end{enumerate}
\item State one limitation of the model. [1 mark]
\item Suggest one refinement that could be made to improve the model. [1 mark]
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 3 Q3 [13]}}