Innovative Forecasts: Population Models are Predicting the Future

To avoid a large-scale population count, we can use differential equations, more specifically, population growth models to estimate the population after a certain amount of time (years, days, …).

Utilizing the population models:

So assume we need the population of a country in the year \(2000\), (which we found by counting the numbers of people in each city in that year), be \(1,000,000\). From the statistical data related the our imaginary country, the population has a rate of \( 0.02% \) growth rate per year, and an additional \( 2% \) bonus after each ten years due to the population itself. if we want to find the population after \(20\) years, we can use:

Let \(P(t)\)  be the population at year \(t\),  with \(P(0)\)  being the population in \(2000\). The differential equation for population growth is:

\(\frac{dP}{dt} = 0.0002 \cdot P(t)\),

where \(0.0002\)  represents the \(0.02\%\)  growth rate.

Solve the differential equation using separation of variables and integration:

\(\int \frac{1}{P} dP = \int {0.0002} dt\),

which yields the solution:

\(P(t) = P(0) e^{0.0002t}\).

Every ten years, the population increases by an additional \(2\%\).  This can be modeled as:

\(P(10) = 1.02 \cdot P(0) e^{0.0002} \cdot 10\),

\(P(20) = 1.02 \cdot P(10) e^{0.0002} \cdot 10\).

Substituting back, the population after \(20 years is:

\(P(20) = 1.02^2 \cdot P(0) e^{0.0002} \cdot 20\).

This formula gives the population in the year \(2020\) considering both the continuous growth and the decade bonuses.

Assuming the initial population in \(2000 ( P(0) )\) is \(1,000,000\), the population in \(2020\) is calculated as:

\(P(20) = 1.02^2 \cdot P(0) e^{0.0002 \cdot 20\).

Substitute \(P(0) = 1,000,000\)  into the formula:

\(P(20) = 1.02^2 \cdot 1,000,000 \cdot e^{0.0002} \cdot 20\).

Given an initial population of \(1,000,000\)  in the year \(2000, the population in \(2020\) is calculated as:

\(P(20) = 1.02^2 \times 1,000,000 \times e^{0.0002 \times 20} \approx 1,044,570\).

This result accounts for the continuous growth rate of \(0.02\%\)  per year and an additional \(2\%\)  bonus every ten years.