Respuesta :
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]
