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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?

User ZAfLu
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1 Answer

1 vote

Answer:

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.

User Nelstaar
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