Question

The population of mosquitoes in a certain area increases at a rate proportional to the current population, and in the absence of other factors, the population doubles each week. There are 400,000 mosquitoes in the area initially, and predators (birds, bats, and so forth) eat 50,000 mosquitoes/day. Determine the population of mosquitoes in the area at any time. (Note that the variable t represents days.) Enter an exact answer. Enclose arguments of functions in parentheses. For example, sin(2x).

Answer 1

Answer:

Population of mosquitoes in the area at any time t is:

$P(t) =504,943.26 -104,943.26e^{0.693t}$

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

 $\implies \frac{dP}{dt} =kP$----(1)

where k is constant rate at which population is doubled

solving (1)

$ln|P(t)|=kt +C\\P(t)= e^{kt+C}\\P(t)=Ce^{kt}$

$t=0\\P(0) = P_{o}\\\implies C= P_{o}\\P(t) =P_{o}e^{kt}$ ---- (2)

initial population = 400,000

population is doubled every week

                                                 ⇒P(1)=2P(0)

Using (2)

                                 $P_{o}e^{k(1)} = 2P_{o} e^{k(0)}$

                                            $e^{k} =2\\k=ln|2|$

In presence of predators amount is decreased by 50,000 per day

Then amount decreased per week = 350,000

In this case (1) becomes

$\frac{dP}{dt}=kP-350,000\\\frac{dP}{dt} - kP=-350,000$ ---(3)

solving (3) by calculating integrating factor

                                          $I.F=e^{\int-k dt}$

Multiplying I.F with all terms of (3)

$e^{-kt}\frac{dP}{dt} - ke^{-kt}P =-350,000 e^{-kt}\\\frac{d}{dt}(e^{-kt}P) = -350,000 e^{-kt}$

Integrating w.r.to t

                         $e^{-kt}P(t)= \frac{350,000e^{-kt}}{k} +C$

                         $P(t) =\frac{350,000}{k} +Ce^{kt}$

                                          $k=ln|2| =0.693$

                          $P(t) =504,943.26 + Ce^{0.693t}$

at t=0

                        $P(0) =504,943.26 + Ce^{0.693(0)}$

                        $400,000 =504,943.26 + C$

                           $C = -104,943.26$

So, population of mosquitoes in the area at any time t is

                  $P(t) =504,943.26 -104,943.26e^{0.693t}$