Respuesta :
Answer:
The population would be 34007.
Step-by-step explanation:
The given equation that shows the population growth rate of a certain city,
[tex]r(t)=1800(1+\frac{6t}{3+t^2}); 0\leq t\leq 5[/tex]
Let P represents the population,
Since, change rate in population = r(t)
[tex]\frac{dP(t)}{dt}=r(t)[/tex]
[tex]\frac{dP(t)}{dt}=1800(1+\frac{6t}{3+t^2})[/tex]
[tex]dP(t)=1800(1+\frac{6t}{3+t^2})dt[/tex]
Integrating both sides,
[tex]\int dP(t)=\int 1800(1+\frac{6t}{3+t^2})dt[/tex]
[tex]P(t)=\int 1800 dt+\int \frac{6t}{3+t^2}dt[/tex]
[tex]P(t)=1800t + \int \frac{6t}{3+t^2}dt[/tex]
Put [tex]3+t^2 = u\implies 2t dt = du[/tex]
[tex]P(t)=1800t + \int \frac{3}{u}du[/tex]
[tex]P(t)=1800t + 3\ln u + C[/tex]
[tex]P(t)=1800t + 3\ln (3+t^2) + C[/tex]
We have P(0) = 25000
[tex]25000 = 1800(0) + 3\ln (3) + C[/tex]
[tex]25000 = 3\ln (3) + C[/tex]
[tex]\implies C = 25000 - 3\ln (3) =24996.70[/tex]
Hence, the function that shows the population after t years,
[tex]P(t)=1800t + 3\ln (3+t^2) + 24996.70[/tex]
At t = 5,
Population after 5 years would be,
[tex]P(5) = 1800(5) + 3\ln (3+5^2) + 24996.70 = 34006.70\approx 34007[/tex]
