| Exam Board | Edexcel |
|---|---|
| Module | Paper 2 (Paper 2) |
| Session | Specimen |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Functions |
| Type | Exponential growth/decay model setup |
| Difficulty | Moderate -0.8 This is a standard exponential growth modeling question requiring routine application of differential equations (dp/dt ∝ p leads to exponential form), reading values from a linearized graph, and basic interpretation. All steps are textbook procedures with no novel problem-solving required, making it easier than average but not trivial due to the multi-part structure. |
| Spec | 1.06i Exponential growth/decay: in modelling context1.07t Construct differential equations: in context |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\dfrac{dp}{dt} \propto p \Rightarrow \dfrac{dp}{dt} = kp\) | B1 | Translates proportionality statement into differential equation with constant of proportionality; AO 3.3 |
| \(\displaystyle\int \dfrac{1}{p}\,dp = \int k\,dt\) | M1 | Correct method of separating variables \(p\) and \(t\); AO 1.1b |
| \(\ln p = kt\ \{+ c\}\) | A1 | \(\ln p = kt\), with or without constant of integration; AO 1.1b |
| \(\ln p = kt + c \Rightarrow p = e^{kt+c} = e^{kt}e^c \Rightarrow p = ae^{kt}\) | A1* | Correct proof with no errors seen in working; AO 2.1 |
| (4 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(p = ae^{kt} \Rightarrow \ln p = \ln a + kt\); evidence of understanding that gradient \(= k\) or \("M" = k\), and vertical intercept \(= \ln a\) or \("C" = \ln a\) | M1 | AO 2.1 |
| gradient \(= k = 0.14\) | A1 | Correctly finds \(k = 0.14\); AO 1.1b |
| vertical intercept \(= \ln a = 3.95 \Rightarrow a = e^{3.95} = 51.935\ldots = 52\) (2 sf) | A1 | Correctly finds \(a = 52\); AO 1.1b |
| (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(p = ae^{kt} \Rightarrow p = a(e^k)^t = ab^t\); e.g. \(p = 52e^{0.14t} \Rightarrow p = 52(e^{0.14})^t\) | B1 | Uses algebra to correctly deduce \(p = ab^t\) from \(p = ae^{kt}\); AO 2.2a |
| \(b = 1.15\), which can be implied by \(p = 52(1.15)^t\) | B1 | AO 1.1b |
| (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Initial area (i.e. "52" mm²) of bacterial culture that was first placed onto the circular dish | B1 | AO 3.4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Rate of increase per hour of the area of bacterial culture; OR the area of bacterial culture increases by "15%" each hour | B1 | AO 3.4 |
| (2 marks total for d(i) and d(ii)) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| The model predicts that the area of the bacteria culture will increase indefinitely, but the size of the circular dish will be a constraint on this area | B1 | Gives a correct long-term limitation of the model for \(p\); AO 3.5b |
| (1 mark) |
# Question 7:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\dfrac{dp}{dt} \propto p \Rightarrow \dfrac{dp}{dt} = kp$ | B1 | Translates proportionality statement into differential equation with constant of proportionality; AO 3.3 |
| $\displaystyle\int \dfrac{1}{p}\,dp = \int k\,dt$ | M1 | Correct method of separating variables $p$ and $t$; AO 1.1b |
| $\ln p = kt\ \{+ c\}$ | A1 | $\ln p = kt$, with or without constant of integration; AO 1.1b |
| $\ln p = kt + c \Rightarrow p = e^{kt+c} = e^{kt}e^c \Rightarrow p = ae^{kt}$ | A1* | Correct proof with no errors seen in working; AO 2.1 |
| **(4 marks)** | | |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $p = ae^{kt} \Rightarrow \ln p = \ln a + kt$; evidence of understanding that gradient $= k$ or $"M" = k$, and vertical intercept $= \ln a$ or $"C" = \ln a$ | M1 | AO 2.1 |
| gradient $= k = 0.14$ | A1 | Correctly finds $k = 0.14$; AO 1.1b |
| vertical intercept $= \ln a = 3.95 \Rightarrow a = e^{3.95} = 51.935\ldots = 52$ (2 sf) | A1 | Correctly finds $a = 52$; AO 1.1b |
| **(3 marks)** | | |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $p = ae^{kt} \Rightarrow p = a(e^k)^t = ab^t$; e.g. $p = 52e^{0.14t} \Rightarrow p = 52(e^{0.14})^t$ | B1 | Uses algebra to correctly deduce $p = ab^t$ from $p = ae^{kt}$; AO 2.2a |
| $b = 1.15$, which can be implied by $p = 52(1.15)^t$ | B1 | AO 1.1b |
| **(2 marks)** | | |
## Part (d)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Initial area (i.e. "52" mm²) of bacterial culture that was first placed onto the circular dish | B1 | AO 3.4 |
## Part (d)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Rate of increase per hour of the area of bacterial culture; OR the area of bacterial culture increases by "15%" each hour | B1 | AO 3.4 |
| **(2 marks total for d(i) and d(ii))** | | |
## Part (e):
| Answer/Working | Mark | Guidance |
|---|---|---|
| The model predicts that the area of the bacteria culture will increase indefinitely, but the size of the circular dish will be a constraint on this area | B1 | Gives a correct long-term limitation of the model for $p$; AO 3.5b |
| **(1 mark)** | | |
**Total: 12 marks**\begin{enumerate}
\item A bacterial culture has area $p \mathrm {~mm} ^ { 2 }$ at time $t$ hours after the culture was placed onto a circular dish.
\end{enumerate}
A scientist states that at time $t$ hours, the rate of increase of the area of the culture can be modelled as being proportional to the area of the culture.\\
(a) Show that the scientist's model for $p$ leads to the equation
$$p = a \mathrm { e } ^ { k t }$$
where $a$ and $k$ are constants.
The scientist measures the values for $p$ at regular intervals during the first 24 hours after the culture was placed onto the dish.
She plots a graph of $\ln p$ against $t$ and finds that the points on the graph lie close to a straight line with gradient 0.14 and vertical intercept 3.95\\
(b) Estimate, to 2 significant figures, the value of $a$ and the value of $k$.\\
(c) Hence show that the model for $p$ can be rewritten as
$$p = a b ^ { t }$$
stating, to 3 significant figures, the value of the constant $b$.
With reference to this model,\\
(d) (i) interpret the value of the constant $a$,\\
(ii) interpret the value of the constant $b$.\\
(e) State a long term limitation of the model for $p$.
\hfill \mbox{\textit{Edexcel Paper 2 Q7 [12]}}