Section 3.1 Population Growth
Subsection 3.1.1 A Motivating Problem
A very natural motivating problem is population growth. Here are some reasonable mathematical questions one might ask about population.
- How do I encode population growth as a function?
- What factors affect the growth of population?
- What is the long-term behaviour of a population?
- What is the growth rate of a population?
To encode population growth as a function, I chose a function \(p(t)\) representing population (\(p\)) in terms of the independant variable time (\(t\)). Let’s say I am given a data set, representing the values of a certain population over a period of time. Table 3.1.1 gives some example population data.
| Year | Population | Growth Rate |
| 0 | 1032 | Not Applicable |
| 1 | 1214 | 17.6 % |
| 2 | 1372 | 13.0 % |
| 3 | 1629 | 18.7 % |
| 4 | 2143 | 31.6 % |
| 5 | 2520 | 17.6 % |
| 6 | 2940 | 16.7 % |
| 7 | 3292 | 12.0 % |
| 8 | 3813 | 15.8 % |
| 9 | 4757 | 24.8 % |
| 10 | 5632 | 18.4 % |
| 11 | 6842 | 21.5 % |
| 12 | 8010 | 17.1 % |
I can try to match this data with a function, as I did with other functions in Subsection 2.5.2. The growth of this population is best matched by an exponential function, show in Figure 3.1.2.
So how did I come up with this exponential function? One way to do this regression is to look at doubling periods. By observation, the data looks like it doubles roughly every four years. (This isn’t perfect, since over some five year patterns is more than doubles and over other it fall short, but this isn’t a bad guess). For doubling, I can think of powers of \(2\text{,}\) so an exponential function with base \(2\) is appropriate. If I double every \(d\) units of time, the exponential function is
\begin{equation*}
p(t) = a 2^{\frac{t}{d}}
\end{equation*}
where \(a\) is the starting value of the poluation. In this case, I used the function
\begin{equation*}
p(t) = (1032) 2^{\frac{t}{4}}
\end{equation*}
to do the regression, and it fits reasonably well.
It is conventional to use the base \(e\) in calculus. I can change my previous exponential function using the fact that \(2
= e^{\ln 2}\text{.}\)
\begin{equation*}
p(t) = a \left( e^{\ln 2} \right)^{\frac{t}{d}}
\end{equation*}
Successive exponents are simplified by multiplying the exponents.
\begin{equation*}
p(t) = a e^{\ln 2\frac{1}{d} t} = ae^{\frac{t \ln 2}{d}}
\end{equation*}
This is a nice general way to write an exponential function with starting value \(a\) and doubling period \(d\text{.}\)
In the data, I included the percentage growth for each year. Though there is a lot of variability, there is also some consistency here. The percentage growth varies above and below 20%, but remains somewhat nearby that value. I could take the average of the yearly growth rates and get an average increase of $18.83\%$. This is an important observation: a consistent precentage growth matches with an exponential behaviour.
But why? Why does relatively consistent percentage growth require an exponential regression? What’s the connection between percentage growth and exponential functions? I’d like to take this as a motivating question for the course. Calculus can understand and describe growth-rates, like percentage growth, and relate them to functions. In the next section, I’ll start to sketch out the strategy to try to answer this motivating question.
