Respuesta :
Answer:
Population of mosquitoes in the area at any time t is:
[tex]P(t) =504,943.26 -104,943.26e^{0.693t}[/tex]
Step-by-step explanation:
assume population at any time t = P(t)
population increases at a rate proportional to the current population:
⇒dP/dt ∝ P
[tex]\implies \frac{dP}{dt} =kP[/tex]----(1)
where k is constant rate at which population is doubled
solving (1)
[tex]ln|P(t)|=kt +C\\P(t)= e^{kt+C}\\P(t)=Ce^{kt}[/tex]
[tex] t=0\\P(0) = P_{o}\\\implies C= P_{o}\\P(t) =P_{o}e^{kt}\\[/tex] ---- (2)
initial population = 400,000
population is doubled every week
⇒P(1)=2P(0)
Using (2)
[tex]P_{o}e^{k(1)} = 2P_{o} e^{k(0)}\\[/tex]
[tex]e^{k} =2\\k=ln|2|\\[/tex]
In presence of predators amount is decreased by 50,000 per day
Then amount decreased per week = 350,000
In this case (1) becomes
[tex]\frac{dP}{dt}=kP-350,000\\\frac{dP}{dt} - kP=-350,000\\[/tex] ---(3)
solving (3) by calculating integrating factor
[tex]I.F=e^{\int-k dt}[/tex]
Multiplying I.F with all terms of (3)
[tex]e^{-kt}\frac{dP}{dt} - ke^{-kt}P =-350,000 e^{-kt}\\\frac{d}{dt}(e^{-kt}P) = -350,000 e^{-kt}[/tex]
Integrating w.r.to t
[tex]e^{-kt}P(t)= \frac{350,000e^{-kt}}{k} +C[/tex]
[tex]P(t) =\frac{350,000}{k} +Ce^{kt}\\[/tex]
[tex]k=ln|2| =0.693[/tex]
[tex]P(t) =504,943.26 + Ce^{0.693t}\\[/tex]
at t=0
[tex]P(0) =504,943.26 + Ce^{0.693(0)}[/tex]
[tex]400,000 =504,943.26 + C[/tex]
[tex]C = -104,943.26 [/tex]
So, population of mosquitoes in the area at any time t is
[tex]P(t) =504,943.26 -104,943.26e^{0.693t}[/tex]
