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 results are not the same! The more times we apply the increases, the larger the population grows in one year. This is happening even when the growth rate is scaled down proportionally.
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.
