Category Archives: Logistic Growth

Logistic Growth Worked Example

A brumby is a free-roaming wild horse found in large number in parts of Australia. The culling of brumbies was banned in the year 2000. At this time the estimated population of brumbies in
Kosciuszko National Park was 1600. Scientists have modelled the population, P(t), of brumbies in
Kosciuszko National Park t years since the ban, by

    \begin{equation*}P(t)=\frac{18000}{10.25e^{0.15t}+1}\end{equation}

(a) Use the model to determine how long it will take the brumbies to increase to a number that is triple the number when the ban came into effect.

(b) From this model, determine the estimated long run number of brumbies in Kosciuszko National Park.

It can be shown that the growth rate of the population of brumbies can be expressed as

    \begin{equation*}\frac{dP}{dt}=\frac{1}{r}P(k-P)\end{equation}

(c) Determine the values of the constants r and k.

(d) Determine the greatest growth rate for the population of brumbies.

ATAR 2024 Specialist Mathematics Question 13

(a) 1600\times 3=4800
4800=\frac{18000}{10.25e^{0.15t}+1}
t=8.8 years.

(b) \lim_{\limits_{t \to \infty}\frac{18000}{10.25e^{0.15t}+1}=18000

(c)k is the carrying capacity (long run number of Brumbies), therefore k=18000.
Remember,

    \begin{equation*}\frac{dP}{dt}=rP(k-P)\Longleftrightarrow P=\frac{KP_0}{(k-P_0)e^{-rkt}+P_0}\end{equation}

We have \frac{1}{r} instead of r.

Therefore,

    \begin{equation*}0.15=\frac{1}{r}\times 18000\end{equation}

r=120000

(d) The greatest growth rate occurs when P=\frac{k}{2}=9000

    \begin{equation*}\frac{dP}{dt}=\frac{1}{r}P(k-P)\end{equation}

    \begin{equation*}\frac{dP}{dt}=\frac{1}{120000}(9000)(9000)\end{equation}

The greatest growth rate is 675 Brumbies per year.

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Filed under Differential Equations, Logistic Growth, Year 12 Specialist Mathematics

Deriving the Logistic Growth Equation

The logistic differential equation

    \begin{equation*}\frac{dP}{dt}=rP(k-P)\end{equation}

where r is the growth parameter and k is the carrying capacity.

And the maximum rate of increase happens when P=\frac{k}{2}

    \begin{equation*}\frac{dP}{dt}=rP(k-P)\end{equation}

    \begin{equation*}\frac{dP}{P(k-P)}=r dt{\end{equation}

    \begin{equation*}\int \frac{dP}{P(k-P)}=\int r dt{\end{equation}

I am going to separate the denominator on the left hand side

\frac{1}{P(k-P)}=\frac{A}{P}+\frac{B}{k-P}
Hence,
\frac{1}{P(k-P)}=\frac{A(k-P)+BP}{P(k-P)}
1=A(k-P)+BP
When P=0,
1=Ak\Rightarrow A=\frac{1}{k}
When P=k,
1=BK\Rightarrow B=\frac{1}{k}

So our equation is,

    \begin{equation*}\int \frac{\frac{1}{k}}{P}+\frac{\frac{1}{k}}{k-P} dP=\int r dt\end{equation}

    \begin{equation*}\frac{1}{k}\int \frac{1}{P}+\frac{1}{k-P} dP=\int r dt\end{equation}

    \begin{equation*}\int \frac{1}{P}+\frac{1}{k-P} dP=\int kr dt\end{equation}

    \begin{equation*}ln\lvert{P}\rvert-ln\lvert{k-P}\rvert=krt+c\end{equation}

    \begin{equation*}ln\lvert{\frac{P}{k-P}\rvert=krt+c\end{equation}

    \begin{equation*}\frac{P}{k-P}=e^{krt+c}\end{equation}

    \begin{equation*}\frac{P}{k-P}=e^{krt}e^{c} \end{equation}

When t=0, P=P_0,

    \begin{equation*}\frac{P_0}{k-P_0}=e^{c} \end{equation}

The equation is now

    \begin{equation*}\frac{P}{k-P}=\frac{P_0}{k-P_0}e^{krt}\end{equation}

    \begin{equation*}P=\frac{P_0}{k-P_0}e^{krt}(k-P)\end{equation}

    \begin{equation*}P=k\frac{P_0}{k-P_0}e^{krt}-P\frac{P_0}{k-P_0}e^{krt}\end{equation}

    \begin{equation*}P+P\frac{P_0}{k-P_0}e^{krt}=k\frac{P_0}{k-P_0}e^{krt}\end{equation}

    \begin{equation*}P(1+\frac{P_0}{k-P_0}e^{krt})=k\frac{P_0}{k-P_0}e^{krt}\end{equation}

    \begin{equation*}P=\frac{k\frac{P_0}{k-P_0}e^{krt}}{1+\frac{P_0}{k-P_0}e^{krt}}\end{equation}

    \begin{equation*}P=\frac{k\frac{P_0}{k-P_0}e^{krt}}{\frac{k-P_0+P_0e^{krt}}{k-P_0}}\end{equation}

    \begin{equation*}P=\frac{kP_0e^{rkt}}{k-P_0+P_0e^{rkt}}\end{equation}

Divide by e^{rkt}

    \begin{equation*}P=\frac{kP_0}{(k-P_0)e^{-rkt}+P_0}\end{equation}

    \begin{equation*}}\frac{dP}{dt}=rP(k-P)\Longleftrightarrow P=\frac{kP_0}{(k-P_0)e^{-rkt}+P_0}\end{equation}

Proving the Maximum Rate of Increase Happens When P=\frac{k}{2}

    \begin{equation*}\frac{dP}{dt}=rP(k-P)\end{equation}

    \begin{equation*}\frac{d^2P}{dt^2}=r\frac{dP}{dt}(k-P)+rP(-\frac{dP}{dt})\end{equation}

    \begin{equation*}\frac{d^2P}{dt^2}=\frac{dP}{dt}(rk-rP-rP)\end{equation}

    \begin{equation*}\frac{d^2P}{dt^2}=0\end{equation}

    \begin{equation*}\frac{dP}{dt}(rk-rP-rP)=0\end{equation}

    \begin{equation*}r\frac{dP}{dt}(k-2P)=0\end{equation}

    \begin{equation*}\frac{dP}{dt}(k-2P)=0\end{equation}

    \begin{equation*}rP(k-P)(k-2P)=0\end{equation}

Hence P=k or P=\frac{k}{2}

(1)   \begin{equation*}\frac{d^3P}{dt^3}=\frac{dP^2}{dt^2}(rk-2rP)+\frac{dP}{dt}(-2\frac{dP}{dt})\end{equation*}

Substitute P=k into equation 1

    \begin{equation*}\frac{d^3P}{dt^3}=rk(k-k)(rk-2rk)(rk-2rk)-2(rk(k-k))^2=0\end{equation}

Hence, not a maximum.

Substitute P=\frac{k}{2} into equation 1

    \begin{equation*}\frac{d^3P}{dt^3}=rk(k-\frac{k}{2})(rk-2r\frac{k}{2})(rk-2r\frac{k}{2})-2(rk(k-\frac{k}{2}))^2=0\end{equation}

    \begin{equation*}\frac{d^3P}{dt^3}=-2(rk^2-\frac{rk^2}{2})^2\end{equation}

    \begin{equation*}\frac{d^3P}{dt^3}=-2\frac{r^2k^4}{4}\end{equation}

-2\frac{r^2k^4}{4}\le 0 For all values of P, r and k.

Hence maximum when P=\frac{k}{2}

We will look at a worked example in the next post.

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Filed under Differential Equations, Differentiation, Implicit, Logistic Growth, Optimisation, Product Rule, Uncategorized, Year 12 Specialist Mathematics

The Logistic Equation

My year 12 Specialist students are working on logistic growth at the moment. An example might be helpful.

A new viral disease was found to spread according to the equation \frac{dN}{dt}=kn(M-N), where M is the susceptible population, N is the number of people infected at time t months and k=1.5\times 10^{-9}. In March 2010, it was thought only 100 people out of a population of 18 million were infected. Use the logistic model to find the number infected in:

(a) March 2011

(b) June 2012

(c) January 2017

Specialist 12 – Nelson Senior Maths

\frac{dN}{dt}=kN(M-N)

\frac{dN}{dt}=(1.5\times10^{-9})N(18\times10^{6}-N)

\frac{dN}{N(18\times10^{6}-N)}=1.5\times10^{-9}dt

Use partial fractions to separate the denominator \frac{dN}{N(18\times10^{6}-N)}

\frac{1}{N(18\times10^{6}-N)}=\frac{A}{N}+\frac{B}{18\times10^{6}-N}

1=A(18\times10^{6}-N)+BN

When N=0

1=A(18\times10^{6})

A=\frac{1}{18\times10^{6}}

When N=18\times10^{6}

1=B(18\times10^{6})

B=\frac{1}{18\times10^{6}}

\frac{1}{18\times10^{6}}(\frac{1}{N}+\frac{1}{(18\times10^{6}-N)}dN)=1.5\times10^{-9}dt

\int\frac{1}{N}+\frac{1}{(18\times10^{6}-N)}dN=\int27\times10^{-3}dt

\ln|N|-\ln|18\times10^{6}-N|=(27\times10^{-3})t+c

\ln|\frac{N}{18\times10^{6}-N}|=(27\times10^{-3})t+c

\frac{N}{18\times10^{6}-N}=e^{(27\times10^{-3})t+c}

Let A=e^{c} and rearrange to make N the subject.

N=\frac{(18\times10^{6})Ae^{(27\times10^{-3})t}}{1+Ae^{(27\times10^{-3})t}}

Divide by Ae^{(27\times10^{-3})t}

N=\frac{(18\times10^{6})}{\frac{1}{A}e^{-(27\times10^{-3})t}+1}

Initially 100 people were infected.

100=\frac{(18\times10^{6})}{\frac{1}{A}+1}

A=\frac{1}{179999}

N=\frac{(18\times10^{6})}{179999e^{-(27\times10^{-3})t}+1}

(a) t=12, N=138.3, hence 138

(b) t=27, N=207.3, hence 207

(c) t=82, N=915.2, hence 915.

It is not necessary to solve the differential equation, you can use the formula

\frac{dP}{dt}=rP(k-P)\leftrightarrowP=\frac{kP_0}{P_0+(k-P_0)e^{-rkt}}

This formula is on the Year 12 Mathematics Specialist formula sheet for Western Australia.

For our question,

\frac{dN}{dt}=(1.5\times10^{-9})N(18\times10^6-N)

So, P=\frac{18\times10^{6}\times100}{100+(18\times10^6-100)e^{-(1.5\times10^{-9})(18\times10^6)t}}

And you can substitute values for t.

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Filed under Differential Equations, Integration, Logistic Growth