Question

Let P(t) be a population at time t. A simple population model supposes the rate of growth of the population is proportional to the population (a) Express the last sentence in the language of calculus. (b) Solve for P(t) in the equation you found in part (a) (c) Suppose the net birthrate of the population is .1, and the initial population is 100. (i) Find the population at time t 10.(ii) Find when the population is 1000.

Answer 1

Answer:

$(a) \dfrac{dP}{dt} =k P(t)\\(b)P(t)=Ce^{kt}$

(c)$P(10)\approx 272$

(ii)$P(1000)\approx 26.88 \times 10^{44}$

Step-by-step explanation:

(a)The rate of growth of the population is proportional to the population, this is written as:

$\dfrac{dP}{dt} \propto P(t)\\ Introducing our proportionality constant, k\\ \dfrac{dP}{dt} =k P(t)$

(b)

$\dfrac{dP(t)}{P(t)} =k dt\\ Take the integral of both sides\\\int \dfrac{dP(t)}{P(t)} =\int k dt\\\ln P(t)=kt+C, C a constant of integration\\Take the exponential of both sides\\e^{\ln P(t)}=e^{kt+C}\\P(t)=e^{kt}\cdot e^C , (Since e^C is a constant, we then have:)\\P(t)=Ce^{kt}$

(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:

$100=Ce^{kX0}$

C=100

Therefore:

$P(t)=100e^{0.1t}$

(i)When t=10

$P(10)=100e^{0.1 \times 10}\\=271.8\\\approx 272$

(ii)When t=1000

$P(1000)=100e^{0.1 \times 1000}\\=26.88 \times 10^{44}$