A certain population increases at a rate proportional to the square root of the population. If the population goes from 2500 to 3600 in five years, what is the population at the end of t years

Answer:

[tex]P(t) = (2t + 50)^2[/tex]

Step-by-step explanation:

A certain population increases at a rate proportional to the square root of the population.

This means that the population can be described by the following differential equation:

[tex]\frac{dP}{dt} = r\sqrt{P}[/tex]

In which r is the growth rate.

Solving by separation of variables, we have that:

[tex]\frac{dP}{\sqrt{P}} = r dt[/tex]

Integrating both sides:

[tex]2\sqrt{P} = rt + K[/tex]

In which K is a random constant

[tex]\sqrt{P} = 0.5rt + K[/tex]

Finding the squares of both sides:

[tex]P(t) = (0.5rt + K)^2[/tex]

Finding the value of K

The initial population is of 2500. That is, when [tex]t = 0, P = 2500[/tex]. So

[tex]P(t) = (0.5rt + K)^2[/tex]

[tex]2500 = (0.5r(0) + K)^2[/tex]

[tex]K^2 = 2500[/tex]

[tex]K = \sqrt{2500}[/tex]

[tex]K = 50[/tex]

So

[tex]P(t) = (0.5rt + 50)^2[/tex]

Finding the value of r

Population of 3600 in 5 years(when t = 5). So

[tex]P(t) = (0.5rt + 50)^2[/tex]

[tex]3600 = (0.5r(5) + 50)^2[/tex]

[tex](2.5r + 50)^2 = 3600[/tex]

Taking the square root of both sides

[tex]\sqrt{(2.5r + 50)^2} = \sqrt{3600}[/tex]

[tex]2.5r + 50 = 60[/tex]

[tex]2.5r = 10[/tex]

[tex]r = 4[/tex]

So, the equation for the population at the end of t years is given by:

[tex]P(t) = (0.5rt + 50)^2[/tex]

[tex]P(t) = (0.5(4)t + 50)^2[/tex]

[tex]P(t) = (2t + 50)^2[/tex]

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