Respuesta :
Answer:
[tex]P(t) = 3000e^{0.1155t}[/tex]
Step-by-step explanation:
The growth of the population can be modeled by the following differential equation:
[tex]\frac{dP}{dt} = rP[/tex]
Where r is the growth rate, P is the population, and t is the time measures in months.
I am going to solve the above differential equation with the separation of variables method.
[tex]\frac{dP}{P} = rdt[/tex]
Integrating both sides:
[tex]ln P = rt + P(0)[/tex]
Where P(0) is the initial condition
We need to isolate P in this equation, so we do this
[tex]e^{ln P} = e^{rt + P(0)}[/tex]
So
[tex]P(t) = P(0)e^{rt}[/tex]
The problem states that P(0)=3000, so:
[tex]P(t) = 3000e^{rt}[/tex]
The problem wants us to find the value of r:
It states that the population doubles in size from 3000 to 6000 in a 6- month period, meaning that P(6) = 6000. So
[tex]6000 = 3000e^{6r}[/tex]
[tex]e^{6r} = 2[/tex]
To isolate 6r, we apply ln to both sides.
[tex]ln (e^{6r}) = ln 2[/tex]
[tex]6r = 0.693[/tex]
[tex]r = \frac{0.693}{6}[/tex]
r = 0.1155
The particular solution to the differential equation with the initial condition P(0)=3000 is:
[tex]P(t) = 3000e^{0.1155t}[/tex]
