Answer
(a) $\frac{dP}{dt} = 0.0033P(1-\frac{P}{20\times 10^9})$
(b) The model predicts that the population in the year 2010 is 6.24 billion people. This is a bit less than the actual population of 6.9 billion people.
(c) The model predicts that the population in the year 2100 will be 7.58 billion people.
The model predicts that the population in the year 2500 will be 13.9 billion people.
Work Step by Step
(a) The growth rate each year is the difference between the birth rate and the death rate. This difference is $20$ million.
Then: $\frac{dP}{dt} = 20,000,000 = 2.0\times 10^7$
We can find $k$:
$\frac{dP}{dt} = kP$
$k = \frac{\frac{dP}{dt}}{P}$
$k = \frac{2.0\times 10^7}{6.1\times 10^9}$
$k = 0.0033$
A logistic differential equation has this form:
$\frac{dP}{dt} = kP(1-\frac{P}{M})$
$\frac{dP}{dt} = 0.0033P(1-\frac{P}{20\times 10^9})$
(b) The solution has this form:
$P(t) = \frac{M}{1+Ae^{-kt}}~~$ where $~~A = \frac{M-P_0}{P_0}$
We can find $A$:
$A = \frac{M-P_0}{P_0} = \frac{20\times 10^9-6.1\times 10^9}{6.1\times 10^9} = 2.28$
We can write the solution:
$P(t) = \frac{20\times 10^9}{1+2.28e^{-0.0033t}}$
We can find the population after 10 years:
$P(10) = \frac{20\times 10^9}{1+2.28e^{-(0.0033)(10)}} = 6.24\times 10^9$
The model predicts that the population in the year 2010 is 6.24 billion people.
This is a bit less than the actual population of 6.9 billion people.
(c) We can find the population after 100 years:
$P(100) = \frac{20\times 10^9}{1+2.28e^{-(0.0033)(100)}} = 7.58\times 10^9$
The model predicts that the population in the year 2100 will be 7.58 billion people.
We can find the population after 500 years:
$P(500) = \frac{20\times 10^9}{1+2.28e^{-(0.0033)(500)}} = 13.9\times 10^9$
The model predicts that the population in the year 2500 will be 13.9 billion people.
