Activity 7.6.2.
Using some recent population data for planet Earth, we want to model the growth of the population over time.
| Year | 1998 | 1999 | 2000 | 2001 | 2002 | 2005 | 2006 | 2007 | 2008 | 2009 | 2010 |
| Pop | \(5.932\) | \(6.008\) | \(6.084\) | \(6.159\) | \(6.234\) | \(6.456\) | \(6.531\) | \(6.606\) | \(6.681\) | \(6.756\) | \(6.831\) |
Our first model will be based on the following assumption:
The rate of change of the population is proportional to the population.
On the face of it, this seems pretty reasonable. When there is a relatively small number of people, there will be fewer births and deaths so the rate of change will be small. When there is a larger number of people, there will be more births and deaths so we expect a larger rate of change. If \(P(t)\) is the population \(t\) years after the year 2000, we may express this assumption as
\begin{equation*}
\frac{dP}{dt} = kP
\end{equation*}
where \(k\) is a constant of proportionality.
(a)
Use the data in the table to estimate the derivative \(P'(0)\) using a central difference. Assume that \(t=0\) corresponds to the year 2000.
(b)
What is the population \(P(0)\text{?}\)
(c)
Use your results from (a) and (b) to estimate the constant of proportionality \(k\) in the differential equation.
(d)
Now that we know the value of \(k\text{,}\) we have the initial value problem
\begin{equation*}
\frac{dP}{dt} = kP, \ P(0) = 6.084\text{.}
\end{equation*}
Find the solution to this initial value problem.
(e)
What does your solution predict for the population in the year 2010? Is this close to the actual population given in the table?
(f)
When does your solution predict that the population will reach 12 billion?
(g)
What does your solution predict for the population in the year 2500?
