Question 9 - A-Level Maths

Edexcel C34 2016 June — Question 9 11 marks

Exam Board	Edexcel
Module	C34 (Core Mathematics 3 & 4)
Year	2016
Session	June
Marks	11
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Differential equations
Type	Long-term behaviour analysis
Difficulty	Standard +0.3 This is a structured multi-part question on a logistic growth model requiring routine techniques: substitution (parts a,b), solving an exponential equation (part c), and differentiation using quotient rule followed by substitution (part d). While it involves several steps and the quotient rule differentiation requires care, each part follows standard procedures without requiring novel insight or particularly challenging algebraic manipulation. It's slightly easier than average due to its guided structure and straightforward application of techniques.
Spec	1.06a Exponential function: a^x and e^x graphs and properties 1.06i Exponential growth/decay: in modelling context 1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates

9. \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{8d67f716-c8af-4460-8a6b-62073ba9b825-17_574_1333_260_303} \captionsetup{labelformat=empty} \caption{Figure 2}

\end{figure} The population of a species of animal is being studied. The population $P$, at time $t$ years from the start of the study, is assumed to be $$P = \frac { 9000 \mathrm { e } ^ { k t } } { 3 \mathrm { e } ^ { k t } + 7 } , \quad t \geqslant 0$$ where $k$ is a positive constant.
A sketch of the graph of $P$ against $t$ is shown in Figure 2 .
Use the given equation to

find the population at the start of the study,
find the value for the upper limit of the population. Given that $P = 2500$ when $t = 4$
calculate the value of $k$, giving your answer to 3 decimal places. Using this value for $k$,
find, using $\frac { \mathrm { d } P } { \mathrm {~d} t }$, the rate at which the population is increasing when $t = 10$ Give your answer to the nearest integer.

Show mark scheme Show mark scheme source

Question 9(a):

Answer	Marks	Guidance
$t = 0 \Rightarrow P = \frac{9000}{3+7} = 900$	M1A1	M1: Sets $t=0$, may be implied by $e^0 = 1$ or $\frac{9000}{3+7}$ or correct answer of 900. A1: 900

Question 9(b):

Answer	Marks	Guidance
$t \to \infty \quad P \to \frac{9000}{3} = 3000$	B1	Sight of 3000

Question 9(c):

Answer	Marks	Guidance
$t=4, P=2500 \Rightarrow 2500 = \frac{9000e^{4k}}{3e^{4k}+7}$	B1	Correct equation with $t=4$ and $P=2500$
$e^{4k} = \frac{17500}{1500}$ (awrt 11.7 or 11.6) or $e^{-4k} = \frac{1500}{17500}$ (awrt 0.857)	M1A1	M1: Rearranges to make $e^{\pm4k}$ the subject, multiply by $3e^{4k}+7$ term and collect terms. A1: Achieves intermediate answer shown.
$k = \frac{1}{4}\ln\left(\frac{35}{3}\right)$ or awrt 0.614	dM1A1	dM1: Proceeds from $e^{\pm4k} = C$, $C>0$ by correctly taking ln's and making $k$ the subject. A1: cao awrt 0.614 or correct exact answer.

Alternative:

Answer	Marks	Guidance
$t=4, P=2500 \Rightarrow 2500 = \frac{9000e^{4k}}{3e^{4k}+7}$	B1	Correct equation
$\ln 1500 + \ln e^{4k} = \ln 17500$	M1A1	M1: Takes ln's correctly. A1: Correct equation.
$k = \frac{\ln 17500 - \ln 1500}{4}$	M1A1	Makes $k$ the subject
$k = \frac{1}{4}\ln\left(\frac{35}{3}\right)$ or awrt 0.614		cao

Question 9(d):

Answer	Marks	Guidance
$\frac{dP}{dt} = \frac{(3e^{kt}+7)\times 9000ke^{kt} - 9000e^{kt}\times 3ke^{kt}}{(3e^{kt}+7)^2} \left(= \frac{63000ke^{kt}}{(3e^{kt}+7)^2}\right)$	M1	Differentiates using quotient rule achieving correct form, or product rule, or chain rule on $P = 9000(3+7e^{-kt})^{-1}$. Watch for $e^{kt} \to kte^{kt}$ which is M0
Sub $t=10$ and $k=0.614 \Rightarrow \frac{dP}{dt} = \ldots$	dM1 (A1 on Epen)	Substitutes $t=10$ and their $k$. If value incorrect then substitution of $t=10$ must be seen explicitly.
$\frac{dP}{dt} = 9$	A1	Awrt 9 (NB $\frac{dP}{dt} = 9.1694\ldots$)

## Question 9(a):

$t = 0 \Rightarrow P = \frac{9000}{3+7} = 900$ | M1A1 | M1: Sets $t=0$, may be implied by $e^0 = 1$ or $\frac{9000}{3+7}$ or correct answer of 900. A1: 900

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## Question 9(b):

$t \to \infty \quad P \to \frac{9000}{3} = 3000$ | B1 | Sight of 3000

---

## Question 9(c):

$t=4, P=2500 \Rightarrow 2500 = \frac{9000e^{4k}}{3e^{4k}+7}$ | B1 | Correct equation with $t=4$ and $P=2500$

$e^{4k} = \frac{17500}{1500}$ (awrt 11.7 or 11.6) or $e^{-4k} = \frac{1500}{17500}$ (awrt 0.857) | M1A1 | M1: Rearranges to make $e^{\pm4k}$ the subject, multiply by $3e^{4k}+7$ term and collect terms. A1: Achieves intermediate answer shown.

$k = \frac{1}{4}\ln\left(\frac{35}{3}\right)$ **or** awrt 0.614 | dM1A1 | dM1: Proceeds from $e^{\pm4k} = C$, $C>0$ by correctly taking ln's and making $k$ the subject. A1: cao awrt 0.614 or correct exact answer.

**Alternative:**

$t=4, P=2500 \Rightarrow 2500 = \frac{9000e^{4k}}{3e^{4k}+7}$ | B1 | Correct equation

$\ln 1500 + \ln e^{4k} = \ln 17500$ | M1A1 | M1: Takes ln's correctly. A1: Correct equation.

$k = \frac{\ln 17500 - \ln 1500}{4}$ | M1A1 | Makes $k$ the subject

$k = \frac{1}{4}\ln\left(\frac{35}{3}\right)$ **or** awrt 0.614 | | cao

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## Question 9(d):

$\frac{dP}{dt} = \frac{(3e^{kt}+7)\times 9000ke^{kt} - 9000e^{kt}\times 3ke^{kt}}{(3e^{kt}+7)^2} \left(= \frac{63000ke^{kt}}{(3e^{kt}+7)^2}\right)$ | M1 | Differentiates using quotient rule achieving correct form, or product rule, or chain rule on $P = 9000(3+7e^{-kt})^{-1}$. **Watch for $e^{kt} \to kte^{kt}$ which is M0**

Sub $t=10$ and $k=0.614 \Rightarrow \frac{dP}{dt} = \ldots$ | dM1 (A1 on Epen) | Substitutes $t=10$ and their $k$. If value incorrect then substitution of $t=10$ must be seen explicitly.

$\frac{dP}{dt} = 9$ | A1 | Awrt 9 (NB $\frac{dP}{dt} = 9.1694\ldots$)

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Show LaTeX source

9.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{8d67f716-c8af-4460-8a6b-62073ba9b825-17_574_1333_260_303}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}

The population of a species of animal is being studied. The population $P$, at time $t$ years from the start of the study, is assumed to be

$$P = \frac { 9000 \mathrm { e } ^ { k t } } { 3 \mathrm { e } ^ { k t } + 7 } , \quad t \geqslant 0$$

where $k$ is a positive constant.\\
A sketch of the graph of $P$ against $t$ is shown in Figure 2 .\\
Use the given equation to
\begin{enumerate}[label=(\alph*)]
\item find the population at the start of the study,
\item find the value for the upper limit of the population.

Given that $P = 2500$ when $t = 4$
\item calculate the value of $k$, giving your answer to 3 decimal places.

Using this value for $k$,
\item find, using $\frac { \mathrm { d } P } { \mathrm {~d} t }$, the rate at which the population is increasing when $t = 10$

Give your answer to the nearest integer.

\begin{center}

\end{center}
\end{enumerate}

\hfill \mbox{\textit{Edexcel C34 2016 Q9 [11]}}

This paper (13 questions)

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Q1 9 Q2 6 Q3 9 Q4 9 Q5 6 Q6 9 Q7 9 Q8 11 Q9 11 Q10 9 Q11 12 Q12 11 Q13 14

Answer	Marks	Guidance
\(t=4, P=2500 \Rightarrow 2500 = \frac{9000e^{4k}}{3e^{4k}+7}\)	B1	Correct equation with \(t=4\) and \(P=2500\)
\(e^{4k} = \frac{17500}{1500}\) (awrt 11.7 or 11.6) or \(e^{-4k} = \frac{1500}{17500}\) (awrt 0.857)	M1A1	M1: Rearranges to make \(e^{\pm4k}\) the subject, multiply by \(3e^{4k}+7\) term and collect terms. A1: Achieves intermediate answer shown.
\(k = \frac{1}{4}\ln\left(\frac{35}{3}\right)\) or awrt 0.614	dM1A1	dM1: Proceeds from \(e^{\pm4k} = C\), \(C>0\) by correctly taking ln's and making \(k\) the subject. A1: cao awrt 0.614 or correct exact answer.

Answer	Marks	Guidance
\(t=4, P=2500 \Rightarrow 2500 = \frac{9000e^{4k}}{3e^{4k}+7}\)	B1	Correct equation
\(\ln 1500 + \ln e^{4k} = \ln 17500\)	M1A1	M1: Takes ln's correctly. A1: Correct equation.
\(k = \frac{\ln 17500 - \ln 1500}{4}\)	M1A1	Makes \(k\) the subject
\(k = \frac{1}{4}\ln\left(\frac{35}{3}\right)\) or awrt 0.614		cao

Answer	Marks	Guidance
\(\frac{dP}{dt} = \frac{(3e^{kt}+7)\times 9000ke^{kt} - 9000e^{kt}\times 3ke^{kt}}{(3e^{kt}+7)^2} \left(= \frac{63000ke^{kt}}{(3e^{kt}+7)^2}\right)\)	M1	Differentiates using quotient rule achieving correct form, or product rule, or chain rule on \(P = 9000(3+7e^{-kt})^{-1}\). Watch for \(e^{kt} \to kte^{kt}\) which is M0
Sub \(t=10\) and \(k=0.614 \Rightarrow \frac{dP}{dt} = \ldots\)	dM1 (A1 on Epen)	Substitutes \(t=10\) and their \(k\). If value incorrect then substitution of \(t=10\) must be seen explicitly.
\(\frac{dP}{dt} = 9\)	A1	Awrt 9 (NB \(\frac{dP}{dt} = 9.1694\ldots\))