Question

The population of a mining city grows at a rate proportional to that population, in two years the population has doubled and a year later there were 10,000 inhabitants.

What was the initial population?

Answer 1

The initial population was approximatedly 3535 inhabitants.

Explanation:

The population of the city can be given by the following differential equation.

`(dP)/(dt) = Pr`,

In which r is the rate of growth of the population.

We can solve this diffential equation by the variable separation method.

`(dP)/(dt) = Pr`

`(dP)/(P) = r dt`

Integrating both sides:

`ln P = rt + c`

Since ln and the exponential are inverse operations, to write P in function of t, we apply ln to both sides.

`e^(ln P) = e^(rt + C)`

`P(t) = Ce^(rt)`

C is the initial population, so:

`P(t) = P(0)e^(rt)`

Now, we apply the problem's statements to first find the growth rate and then the initial population.

The problem states that:

In two years the population has doubled:

`P(2) = 2P(0)`

`P(t) = P(0)e^(rt)`

`2P(0) = P(0)e^(2r)`

`2 = e^(2r)`

To isolate r, we apply ln both sides

`e^(2r) = 2`

`ln e^(2r) = ln 2`

`2r = 0.69`

`r = (0.69)/(2)`

`r = 0.3466`

So

`P(t) = P(0)e^(0.3466t)`

In two years the population has doubled and a year later there were 10,000 inhabitants.

`P(3) = 10,000`

`P(t) = P(0)e^(0.3466t)`

`10,000= P(0)e^(0.3466*3)`

`P(0) = (10,000)/(e^(1.04))`

`P(0) = 3534.55`

The initial population was approximatedly 3535 inhabitants.