| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2013 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Connected Rates of Change |
| Type | Population or biological model |
| Difficulty | Standard +0.3 This is a standard logistic growth model question with straightforward substitutions and routine differentiation. Parts (a), (b), and (d) involve direct substitution into the given formula. Part (c) requires solving an exponential equation using logarithms—a standard C3 technique. Part (e) requires applying the chain rule to differentiate, which is routine for C3 students. The question is slightly above average difficulty due to the algebraic manipulation in part (c) and the multi-step nature, but all techniques are standard textbook exercises with no novel problem-solving required. |
| Spec | 1.02z Models in context: use functions in modelling1.06a Exponential function: a^x and e^x graphs and properties1.06i Exponential growth/decay: in modelling context1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(t = 0 \Rightarrow P = \frac{8000}{1+7} = 1000\) | M1A1 (cao) | Sets \(t=0\), giving \(e^{-k\times0}=1\). Award if candidate attempts \(\frac{8000}{1+7\times1}, \frac{8000}{8}\). Correct answer only 1000. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(t \to \infty \quad P \to \frac{8000}{1} = 8000\) | B1 | Accept \(P < 8000\). Condone \(P \leqslant 8000\) but not \(P > 8000\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(t=3, P=2500 \Rightarrow 2500 = \frac{8000}{1+7e^{-3k}}\) | B1 | Sets both \(t=3\) and \(P=2500\). May be implied by subsequent correct line. |
| \(e^{-3k} = \frac{2.2}{7} = (0.31..)\) | M1, A1 (oe) | Rearranges to make \(e^{\pm3k}\) subject, multiplying by \(1+7e^{-3k}\) term, proceeding to \(e^{\pm3k}=A\), \(A>0\). Correct intermediate answer \(e^{-3k}=\frac{2.2}{7}, \frac{11}{35}\) or equivalent. Accept awrt 0.31. Alternatively accept \(e^{3k}=\frac{35}{11}\), 3.18.. |
| \(k = -\frac{1}{3}\ln\left(\frac{2.2}{7}\right) =\) awrt \(0.386\) | M1A1 | Proceeds from \(e^{\pm3k}=A\), \(A>0\) by correctly taking \(\ln\)s and making \(k\) subject. Award for \(e^{-3k}=A \Rightarrow -3k=\ln(A) \Rightarrow k=\frac{\ln(A)}{-3}\). Awrt \(k=0.386\) 3dp |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Sub \(t=10\) into \(P = \frac{8000}{1+7e^{-0.386t}} \Rightarrow P = 6970\) | M1A1 (cao) | Substitutes \(t=10\) with their numerical value of \(k\) to find \(P\). \((P=)6970\) or other exact equivalents like \(6.97\times10^3\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dP}{dt} = -\frac{8000}{(1+7e^{-kt})^2} \times -7ke^{-kt}\) | M1, A1 | Differentiates using chain rule to form \(\frac{dP}{dt} = \frac{C}{(1+7e^{-kt})^2}\times e^{-kt}\). Accept quotient rule giving \(\frac{(1+7e^{-kt})\times0 - C\times-e^{-kt}}{(1+7e^{-kt})^2}\). A1 for correct unsimplified \(\frac{dP}{dt} = -\frac{8000}{(1+7e^{-kt})^2}\times-7ke^{-kt}\) |
| Sub \(t=10\): \(\left.\frac{dP}{dt}\right\ | _{t=10} = 346\) | A1 |
## Question 8:
### Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $t = 0 \Rightarrow P = \frac{8000}{1+7} = 1000$ | M1A1 (cao) | Sets $t=0$, giving $e^{-k\times0}=1$. Award if candidate attempts $\frac{8000}{1+7\times1}, \frac{8000}{8}$. Correct answer only 1000. |
### Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $t \to \infty \quad P \to \frac{8000}{1} = 8000$ | B1 | Accept $P < 8000$. Condone $P \leqslant 8000$ but not $P > 8000$ |
### Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $t=3, P=2500 \Rightarrow 2500 = \frac{8000}{1+7e^{-3k}}$ | B1 | Sets both $t=3$ and $P=2500$. May be implied by subsequent correct line. |
| $e^{-3k} = \frac{2.2}{7} = (0.31..)$ | M1, A1 (oe) | Rearranges to make $e^{\pm3k}$ subject, multiplying by $1+7e^{-3k}$ term, proceeding to $e^{\pm3k}=A$, $A>0$. Correct intermediate answer $e^{-3k}=\frac{2.2}{7}, \frac{11}{35}$ or equivalent. Accept awrt 0.31. Alternatively accept $e^{3k}=\frac{35}{11}$, 3.18.. |
| $k = -\frac{1}{3}\ln\left(\frac{2.2}{7}\right) =$ awrt $0.386$ | M1A1 | Proceeds from $e^{\pm3k}=A$, $A>0$ by correctly taking $\ln$s and making $k$ subject. Award for $e^{-3k}=A \Rightarrow -3k=\ln(A) \Rightarrow k=\frac{\ln(A)}{-3}$. Awrt $k=0.386$ 3dp |
### Part (d)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Sub $t=10$ into $P = \frac{8000}{1+7e^{-0.386t}} \Rightarrow P = 6970$ | M1A1 (cao) | Substitutes $t=10$ with their numerical value of $k$ to find $P$. $(P=)6970$ or other exact equivalents like $6.97\times10^3$ |
### Part (e)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dP}{dt} = -\frac{8000}{(1+7e^{-kt})^2} \times -7ke^{-kt}$ | M1, A1 | Differentiates using chain rule to form $\frac{dP}{dt} = \frac{C}{(1+7e^{-kt})^2}\times e^{-kt}$. Accept quotient rule giving $\frac{(1+7e^{-kt})\times0 - C\times-e^{-kt}}{(1+7e^{-kt})^2}$. A1 for correct unsimplified $\frac{dP}{dt} = -\frac{8000}{(1+7e^{-kt})^2}\times-7ke^{-kt}$ |
| Sub $t=10$: $\left.\frac{dP}{dt}\right\|_{t=10} = 346$ | A1 | Awrt 346. Note that M1 must have been achieved. Just the answer scores 0. |8.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{a80a71cb-42e0-4587-8f8e-bacd69b8d07a-13_721_1227_116_322}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
The population of a town is being studied. The population $P$, at time $t$ years from the start of the study, is assumed to be
$$P = \frac { 8000 } { 1 + 7 \mathrm { e } ^ { - k t } } , \quad t \geqslant 0$$
where $k$ is a positive constant.\\
The graph of $P$ against $t$ is shown in Figure 3.
Use the given equation to
\begin{enumerate}[label=(\alph*)]
\item find the population at the start of the study,
\item find a value for the expected upper limit of the population.
Given also that the population reaches 2500 at 3 years from the start of the study,
\item calculate the value of $k$ to 3 decimal places.
Using this value for $k$,
\item find the population at 10 years from the start of the study, giving your answer to 3 significant figures.
\item Find, using $\frac { \mathrm { d } P } { \mathrm {~d} t }$, the rate at which the population is growing at 10 years from the start of the study.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 2013 Q8 [13]}}