Respuesta :
Answer:
[tex](a) \dfrac{dP}{dt} =k P(t)\\(b)P(t)=Ce^{kt}[/tex]
(c)[tex]P(10)\approx 272[/tex]
(ii)[tex]P(1000)\approx 26.88 \times 10^{44}\\[/tex]
Step-by-step explanation:
(a)The rate of growth of the population is proportional to the population, this is written as:
[tex]\dfrac{dP}{dt} \propto P(t)\\$Introducing our proportionality constant, k\\ \dfrac{dP}{dt} =k P(t)[/tex]
(b)
[tex]\dfrac{dP(t)}{P(t)} =k dt\\$Take the integral of both sides\\\int \dfrac{dP(t)}{P(t)} =\int k dt\\\ln P(t)=kt+C, $C a constant of integration\\Take the exponential of both sides\\e^{\ln P(t)}=e^{kt+C}\\P(t)=e^{kt}\cdot e^C $, (Since e^C$ is a constant, we then have:)\\P(t)=Ce^{kt}[/tex]
(c)
Suppose the net birthrate of the population is .1, and the initial population is 100.
k=0.1
P(0)=100
Substitution into P(t) gives:
[tex]100=Ce^{kX0}[/tex]
C=100
Therefore:
[tex]P(t)=100e^{0.1t}[/tex]
(i)When t=10
[tex]P(10)=100e^{0.1 \times 10}\\=271.8\\\approx 272[/tex]
(ii)When t=1000
[tex]P(1000)=100e^{0.1 \times 1000}\\=26.88 \times 10^{44}\\[/tex]
