| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Session | Specimen |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Functions |
| Type | Rate of change in exponential model |
| Difficulty | Moderate -0.3 This is a straightforward application of exponential functions requiring substitution (part a), basic differentiation of e^(kt) (part b), solving a logarithmic equation (part c), and a simple model evaluation (part d). All techniques are standard AS-level procedures with no novel insight required, making it slightly easier than average due to its routine nature, though the multi-part structure and context provide some substance. |
| Spec | 1.06a Exponential function: a^x and e^x graphs and properties1.06i Exponential growth/decay: in modelling context1.07k Differentiate trig: sin(kx), cos(kx), tan(kx)1.08d Evaluate definite integrals: between limits |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(0.2 \text{ m}^2\) | B1 | Accept oe |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(A = 0.2e^{0.3t}\), rate of change \(= \frac{dA}{dt} = 0.06e^{0.3t}\) | M1 | Links rate of change to gradient; differentiates \(0.2e^{0.3t} \rightarrow ke^{0.3t}\) |
| At \(t=5 \Rightarrow\) Rate of growth is \(0.06e^{1.5} = 0.269 \text{ m}^2/\text{day}\) | A1 | Correct answer \(0.269 \text{ m}^2/\text{day}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(100 = 0.2e^{0.3t} \Rightarrow e^{0.3t} = 500\) | M1, A1 | Substitutes \(A=100\); proceeds to \(e^{0.3t}=k\) |
| \(t = \frac{\ln(500)}{0.3} = 20.7 \text{ days}\), i.e. 20 days 17 hours | M1, A1 | Correct method from \(e^{0.3t}=k \Rightarrow t=..\); correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Model suggests pond fully covered after 20 days 17 hours; observed data shows only 90% covered by end of one month (28/29/30/31 days), so model is not accurate | B1 | Valid conclusion following through on answer to (c) |
## Question 13:
### Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $0.2 \text{ m}^2$ | B1 | Accept oe |
### Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $A = 0.2e^{0.3t}$, rate of change $= \frac{dA}{dt} = 0.06e^{0.3t}$ | M1 | Links rate of change to gradient; differentiates $0.2e^{0.3t} \rightarrow ke^{0.3t}$ |
| At $t=5 \Rightarrow$ Rate of growth is $0.06e^{1.5} = 0.269 \text{ m}^2/\text{day}$ | A1 | Correct answer $0.269 \text{ m}^2/\text{day}$ |
### Part (c):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $100 = 0.2e^{0.3t} \Rightarrow e^{0.3t} = 500$ | M1, A1 | Substitutes $A=100$; proceeds to $e^{0.3t}=k$ |
| $t = \frac{\ln(500)}{0.3} = 20.7 \text{ days}$, i.e. 20 days 17 hours | M1, A1 | Correct method from $e^{0.3t}=k \Rightarrow t=..$; correct answer |
### Part (d):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Model suggests pond fully covered after 20 days 17 hours; observed data shows only 90% covered by end of one month (28/29/30/31 days), so model is not accurate | B1 | Valid conclusion following through on answer to (c) |
---\begin{enumerate}
\item The growth of pond weed on the surface of a pond is being investigated.
\end{enumerate}
The surface area of the pond covered by the weed, $A \mathrm {~m} ^ { 2 }$, can be modelled by the equation
$$A = 0.2 \mathrm { e } ^ { 0.3 t }$$
where $t$ is the number of days after the start of the investigation.\\
(a) State the surface area of the pond covered by the weed at the start of the investigation.\\
(b) Find the rate of increase of the surface area of the pond covered by the weed, in $\mathrm { m } ^ { 2 } /$ day, exactly 5 days after the start of the investigation.
Given that the pond has a surface area of $100 \mathrm {~m} ^ { 2 }$,\\
(c) find, to the nearest hour, the time taken, according to the model, for the surface of the pond to be fully covered by the weed.
The pond is observed for one month and by the end of the month $90 \%$ of the surface area of the pond was covered by the weed.\\
(d) Evaluate the model in light of this information, giving a reason for your answer.
\hfill \mbox{\textit{Edexcel AS Paper 1 Q13 [8]}}