| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2009 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Functions |
| Type | Exponential growth/decay model setup |
| Difficulty | Moderate -0.8 This is a straightforward exponential model question requiring only routine techniques: reading initial value from the equation, solving a simple exponential equation using logarithms, differentiating e^(kt), and substituting into the derivative. All parts are standard textbook exercises with no problem-solving or novel insight required. |
| Spec | 1.06a Exponential function: a^x and e^x graphs and properties1.06i Exponential growth/decay: in modelling context1.07j Differentiate exponentials: e^(kx) and a^(kx) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(t = 0 \Rightarrow P = 80e^0 = 80(1) = 80\) | B1 | \(80\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P = 1000 \Rightarrow \frac{1000}{80} = e^{\frac{t}{5}}\) | M1 | Substitutes \(P = 1000\) and rearranges to make \(e^{\frac{t}{5}}\) the subject |
| \(t = 5\ln\left(\frac{1000}{80}\right) = 12.6286...\) | A1 | awrt \(12.6\) or \(13\) years. Note \(t = 12\) or \(t\) = awrt \(12.6 \Rightarrow t = 12\) will score A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dP}{dt} = 16e^{\frac{t}{5}}\) | M1 | \(ke^{\frac{t}{5}}\) and \(k \neq 80\) |
| A1 | \(16e^{\frac{t}{5}}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(50 = 16e^{\frac{t}{5}} \Rightarrow t = 5\ln\left(\frac{50}{16}\right)\) \(\{= 5.69717...\}\) | M1 | Using \(50 = \frac{dP}{dt}\) and attempt to solve to find value of \(t\) or \(\frac{t}{5}\) |
| \(P = 80e^{\frac{1}{5}\left(5\ln\left(\frac{50}{16}\right)\right)}\) or \(P = 80e^{\frac{1}{5}(5.69717...)}\) | dM1 | Substitutes their value of \(t\) back into the equation for \(P\) |
| \(P = \frac{80(50)}{16} = 250\) | A1 | \(250\) or awrt \(250\) |
# Question 3:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $t = 0 \Rightarrow P = 80e^0 = 80(1) = 80$ | B1 | $80$ |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P = 1000 \Rightarrow \frac{1000}{80} = e^{\frac{t}{5}}$ | M1 | Substitutes $P = 1000$ and rearranges to make $e^{\frac{t}{5}}$ the subject |
| $t = 5\ln\left(\frac{1000}{80}\right) = 12.6286...$ | A1 | awrt $12.6$ or $13$ years. Note $t = 12$ or $t$ = awrt $12.6 \Rightarrow t = 12$ will score A0 |
## Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dP}{dt} = 16e^{\frac{t}{5}}$ | M1 | $ke^{\frac{t}{5}}$ and $k \neq 80$ |
| | A1 | $16e^{\frac{t}{5}}$ |
## Part (d)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $50 = 16e^{\frac{t}{5}} \Rightarrow t = 5\ln\left(\frac{50}{16}\right)$ $\{= 5.69717...\}$ | M1 | Using $50 = \frac{dP}{dt}$ and attempt to solve to find value of $t$ or $\frac{t}{5}$ |
| $P = 80e^{\frac{1}{5}\left(5\ln\left(\frac{50}{16}\right)\right)}$ or $P = 80e^{\frac{1}{5}(5.69717...)}$ | dM1 | Substitutes their value of $t$ back into the equation for $P$ |
| $P = \frac{80(50)}{16} = 250$ | A1 | $250$ or awrt $250$ |
---\begin{enumerate}
\item Rabbits were introduced onto an island. The number of rabbits, $P , t$ years after they were introduced is modelled by the equation
\end{enumerate}
$$P = 80 \mathrm { e } ^ { \frac { 1 } { 5 } t } , \quad t \in \mathbb { R } , t \geqslant 0$$
(a) Write down the number of rabbits that were introduced to the island.\\
(b) Find the number of years it would take for the number of rabbits to first exceed 1000.\\
(c) Find $\frac { \mathrm { d } P } { \mathrm {~d} t }$.\\
(d) Find $P$ when $\frac { \mathrm { d } P } { \mathrm {~d} t } = 50$.\\
\hfill \mbox{\textit{Edexcel C3 2009 Q3 [8]}}