Question

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 1

`P(t) = (2t + 50)^2`

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:

`(dP)/(dt) = r√(P)`

In which r is the growth rate.

Solving by separation of variables, we have that:

`(dP)/(√(P)) = r dt`

Integrating both sides:

`2√(P) = rt + K`

In which K is a random constant

`√(P) = 0.5rt + K`

Finding the squares of both sides:

`P(t) = (0.5rt + K)^2`

Finding the value of K

The initial population is of 2500. That is, when
`t = 0, P = 2500`. So

`P(t) = (0.5rt + K)^2`

`2500 = (0.5r(0) + K)^2`

`K^2 = 2500`

`K = √(2500)`

`K = 50`

So

`P(t) = (0.5rt + 50)^2`

Finding the value of r

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

`P(t) = (0.5rt + 50)^2`

`3600 = (0.5r(5) + 50)^2`

`(2.5r + 50)^2 = 3600`

Taking the square root of both sides

`√((2.5r + 50)^2) = √(3600)`

`2.5r + 50 = 60`

`2.5r = 10`

`r = 4`

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

`P(t) = (0.5rt + 50)^2`

`P(t) = (0.5(4)t + 50)^2`

`P(t) = (2t + 50)^2`