The Natural Growth Model.
The Natural Growth Model is
\begin{equation*}
P(t)=P_0 e^{kt}
\end{equation*}
where \(P_0\) is the initial amount, \(k\) is the instantaneous or continuous growth rate expressed in decimal form, and \(t\) is time.
Section 2.5 Natural Growth and Logistic Growth
Subsection 2.5.1 Natural Growth
Consider a frog population that starts with 100 frogs and doubles each year, which is an annual increase of 100%. Is that the same as a population that increases by 50% every six months? Is that the same as a population that increases by 25% every three months? Let’s explore:
| Model | Number of Increases in 1 Year | Population after 1 Year |
|---|---|---|
| Increase 100% Each Year | \(1\) | \(100(1+1)^{1}=200\) |
| Increase 50% every 6 Months | \(2\) | \(100(1+0.50)^{2}=225\) |
| Increase by 25% every 3 Months | \(4\) | \(100(1+0.25)^{4}=244.14063\) |
The more frequently we apply the increases, the larger the population grows in one year.
Let’s generalize. If the frog population increases \(n\) times per year by \(\frac{100}{n}\%\text{,}\) then after one year the population will be \(100\left( 1+\frac{1}{n} \right)^{n}\text{.}\) If the value of \(n\) approaches infinity, meaning that the increases are applied continuously rather than at distinct points in time, the results approach about \(272\) frogs as seen in the table below.
| \(n\) | \(100\left( 1+\frac{1}{n} \right) ^{n}\) |
|---|---|
| \(1\) | \(200\) |
| \(12\) | \(261.3053\) |
| \(365\) | \(271.45675\) |
| \(1\,000\,000\) | \(271.82805\) |
The number \(e\) is a mathematical constant approximately equal to \(2.71828\text{.}\) Like the number \(\pi\text{,}\) it is irrational and also arises frequently in mathematics. The frog population is, in fact, approaching \(100e\text{.}\)
We can use the number \(e\) to develop a model for exponential growth in which the growth rate is applied continuously. This is called the Natural Growth Model because it arises frequently in the natural world when modeling population growth and exponential decay.
Example 2.5.3.
An elk herd begins with 27 animals and grows at a continuous rate of \(4\%\) per year. Write a function to model the population. Then find the population after three years.
Because this is a continuous growth rate, we should use the natural growth model with \(P_0=27\) and \(k=0.04\text{.}\) Therefore, the model is:
\begin{equation*}
P(t)=27e^{0.04t}
\end{equation*}
To find the population after 3 years, we evaluate \(P(3)\) on a calculator. The result is
\begin{equation*}
P(3)=27e^{0.04(3)}\approx 30.
\end{equation*}
Therefore, there will be about 30 elk after three years.
Note: your calculator should have an \(e\) button! You should never need to type in a decimal approximation for \(e\text{.}\)
Caution!
Be careful not to confuse yearly growth rates with continuous growth rates.
For example, a population with a continuous growth rate of \(20\%\) per year will end up increasing by about \(22\%\) from one year to the next. This is because the growth is, effectively, broken down into infinitely many shorter time spans and the growth compounds.
Example 2.5.4.
A population of bacteria grows at a continuous rate of \(50\%\) per hour from a starting population of 1000 bacteria. Write a function to model the population. Then find how long will it take for the population to double.
Solution.
Because this is a continuous growth rate, we should use the natural growth model with \(P_0\) as the initial population and \(k=0.5\text{.}\) Therefore, the model is:
\begin{equation*}
P(t)=1000 e^{0.5t}
\end{equation*}
To find how long it will take for the population to double, we set \(P(t)=2P_0=2000\) and solve for \(t\text{:}\)
\begin{align*}
1000 e^{0.5t} \amp= 2000 \\
e^{0.5t} \amp= 2 \\
\log\left( e^{0.5t} \right) \amp= \log(2) \\
0.5t \log(e) \amp= \log(2) \\
0.5t \amp= \frac{\log(2)}{\log(e) } \\
t \amp= \frac{\log(2)}{0.5\log(e) } \\
t \amp\approx 1.3863
\end{align*}
Therefore, it will take about 1.3863 hours for the population to double.
Subsection 2.5.2 Logistic Growth Model
The natural growth model is a good model for population growth when the population is small and resources are abundant. However, as the population grows, resources become more scarce and the growth rate slows down. The Logistic Growth Model is a more realistic model for population growth that takes into account the carrying capacity of the environment. The carrying capacity is the maximum population size that the environment can sustain indefinitely.
The Logistic Growth Model.
The logistic growth model is given by the formula:
\begin{equation*}
P(t)=\frac{c}{1+ae^{kt}}
\end{equation*}
where \(c\) is the carrying capacity, \(a\) and \(k\) are constants, and \(t\) is time.
Consider the function \(P(t)=\frac{100}{1+19e^{-0.5t}}\) that might represent a population of people in a paticular housing development where \(t\) is time in years. This function models a population with a carrying capacity of 100. The graph of this function is shown below.
Please note that the graph of the logistic growth curve appears exponential at first. However, as the population approaches the carrying capacity, the growth rate slows down and the graph levels off. The population will never reach or exceed the carrying capacity, but it will get closer and closer to it over time. (Students who have taken college algebra may recognize this as a horizontal asymptote.)
The initial population can be found by evaluating the function at \(t=0\text{.}\) Recall that \(b^0=1\) for any base \(b\ne 1\text{.}\)
\begin{gather*}
\end{gather*}
Thus, the initial population is 5 people.
Be careful: the value of \(a\) is related to the initial population, but it is not equal to the initial population. This means that if we are given the initial population, we cannot simply set \(a\) equal to that value. We must use the initial population and algebraic methods to find the value of \(a\) in order to write the logistic growth model as we will see in the next example.
What does the value of \(k\) represent in the logistic growth model? It is the opposite (negative) of the growth rate that would occur if the population were growing exponentially without any constraints. In other words, if the population were growing according to the natural growth model with a growth rate of \(r\text{,}\) then the value of \(k\) in the logistic growth model would be \(-r\text{.}\) In practice, the value of \(k\) is determine algebraically or through data fitting.
Example 2.5.5.
A population of rabbits has a carrying capacity of 500 rabbits. The initial population is 20 rabbits and after 10 years the population is 100 rabbits. Write a function to model the population of rabbits over time.
Solution.
Converting the given information into mathematical terms, we have: \(c = 500\text{,}\) \(P(0) = 20\text{,}\) and \(P(10) = 100\text{.}\) We can use these values to find the constants \(a\) and \(k\text{.}\)
First, we can find the value of \(a\) using the initial population:
\begin{align*}
P(0) \amp= \frac{500}{1+ae^{k \cdot 0}} \\
20 \amp= \frac{500}{1+a\cdot 1} \\
\frac{20}{1} \amp= \frac{500}{1+a} \\
20(1+a) \amp= 500(1) \\
1+a \amp= 25 \\
a \amp= 24
\end{align*}
Now that we have the value of \(a\text{,}\) we can find the value of \(k\) using the information that \(P(10) = 100\text{:}\)
\begin{align*}
P(10) \amp= \frac{500}{1+24e^{10k}} \\
100 \amp= \frac{500}{1+24e^{10k}} \\
100\left(1+24e^{10k}\right) \amp= 500 \\
1+24e^{10k} \amp= 5 \\
24e^{10k} \amp= 4 \\
e^{10k} \amp= \frac{1}{6} \\
10k \amp= \ln\left(\frac{1}{6}\right) \\
k \amp= \frac{1}{10}\ln\left(\frac{1}{6}\right) \approx -0.179176
\end{align*}
Therefore, the function to model the population of rabbits over time is:
\begin{equation*}
P(t)=\frac{500}{1+24e^{\frac{1}{10}\ln\left(\frac{1}{6}\right)t}}
\end{equation*}
or
\begin{equation*}
P(t)=\frac{500}{1+24e^{-0.179176t}}
\end{equation*}
Example 2.5.6.
A population of bacteria has a carrying capacity of 3000 bacteria. The initial population is 40 bacteria and after 5 hours the population is 200 bacteria. Write a function to model the population of bacteria over time and use the model to find how long it will take for the population to reach 500 bacteria.
Solution.
We set up the model as in the previous example. We have: \(c = 3000\text{,}\) \(P(0) = 40\text{,}\) and \(P(5) = 200\text{.}\)
Find the value of \(a\) using the initial population:
\begin{align*}
P(0) \amp= \frac{3000}{1+ae^{k \cdot 0}} \\
40 \amp= \frac{3000}{1+a\cdot 1} \\
\frac{40}{1} \amp= \frac{3000}{1+a} \\
40(1+a) \amp= 3000(1) \\
1+a \amp= 75 \\
a \amp= 74
\end{align*}
Find the value of \(k\) using the information that \(P(5) = 200\text{:}\)
\begin{align*}
P(5) \amp= \frac{3000}{1+74e^{5k}} \\
200 \amp= \frac{3000}{1+74e^{5k}} \\
200\left(1+74e^{5k}\right) \amp= 3000 \\
1+74e^{5k} \amp= 15 \\
74e^{5k} \amp= 14 \\
e^{5k} \amp= \frac{14}{74} \\
5k \amp= \ln\left(\frac{14}{74}\right) \\
k \amp= \frac{1}{5}\ln\left(\frac{14}{74}\right) \approx -0.336472
\end{align*}
Therefore, the function to model the population of bacteria over time is:
\begin{equation*}
P(t)=\frac{3000}{1+74e^{\frac{1}{5}\ln\left(\frac{14}{74}\right)t}}
\end{equation*}
or
\begin{equation*}
P(t)=\frac{3000}{1+74e^{-0.336472t}}
\end{equation*}
To find how long it will take for the population to reach 500 bacteria, we set \(P(t)=500\) and solve for \(t\text{:}\)
\begin{gather*}
\end{gather*}
