The population of a certain country is known to increase at

abreviatsjw 2022-01-03 Answered
The population of a certain country is known to increase at a rate proportional to the number of people presently living in the country. If after two years the population has doubled, and after three years the population is 20,000, find an expression for the approximate population of the country at any time and estimate the number of people initially living in the country.

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sirpsta3u
Answered 2022-01-04 Author has 4315 answers
Given that the rate of increase of population of country is proportional to thenumber of people livling presently in the country
\(\displaystyle{\frac{{{d}{P}}}{{{\left.{d}{t}\right.}}}}={k}{P}\), where P is the population
k is the proportionality constant
\(\displaystyle\int{\frac{{{1}}}{{{p}}}}{d}{P}=\int{k}{\left.{d}{t}\right.}\)
\(\displaystyle{\ln{{P}}}={k}{t}+{C}\)
Let \(\displaystyle{P}_{{{0}}}\) be the population initially
\(\displaystyle{t}={0}\Rightarrow{P}={P}_{{{0}}}\)
\(\displaystyle{P}={P}_{{{0}}}{e}^{{{k}{t}}}\)
Given that at 2 years the population is dobuled
\(\displaystyle{2}={e}^{{{2}{k}}}\)
\(\displaystyle{k}={\ln{{\left(\sqrt{{{2}}}\right)}}}\)
Given that at 3 years the population is 20000
\(\displaystyle{20000}={P}_{{{i}}}{e}^{{{3}{\ln{\sqrt{{{2}}}}}}}\)
Therefore the equation is \(\displaystyle{P}={P}_{{{0}}}{e}^{{{\frac{{{\ln{{\left({2}\right)}}}}}{{{2}}}}{t}}}\)
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Paul Mitchell
Answered 2022-01-05 Author has 201 answers
Rate of increase of population is proportional to Number of people living presently
After 2 years, Population has doubled
After 3 years, Population \(\displaystyle={20000}\)
Let P be population in the country.
Thus, according to the question:
\(\displaystyle{\frac{{{d}{P}}}{{{\left.{d}{t}\right.}}}}\infty{P}\)
\(\displaystyle{\frac{{{d}{P}}}{{{\left.{d}{t}\right.}}}}={k}{P}\), where k is any constant.
\(\displaystyle{\frac{{{d}{P}}}{{{P}}}}={k}.{\left.{d}{t}\right.}\)
\(\displaystyle\int{\frac{{{d}{P}}}{{{P}}}}=\int{k}.{\left.{d}{t}\right.}\)
\(\displaystyle{\ln{{P}}}={k}.{t}+{C}\)
When \(\displaystyle{t}={0}\) years, \(\displaystyle{P}={P}_{{{0}}}\), where \(\displaystyle{P}_{{{0}}}\) is the initial population.
Thus, \(\displaystyle{{\ln{{P}}}_{{{0}}}=}{0}+{C}\)
\(\displaystyle{C}={{\ln{{P}}}_{{{0}}}}\)
\(\displaystyle{\ln{{P}}}={k}{t}+{{\ln{{P}}}_{{{0}}}}\)
\(\displaystyle{\ln{{P}}}-{{\ln{{P}}}_{{{0}}}=}{k}{t}\)
\(\displaystyle{\ln{{\left({\frac{{{P}}}{{{P}_{{{0}}}}}}\right)}}}={k}{t}\)
When \(\displaystyle{t}={2}\) years, \(\displaystyle{P}={2}{P}_{{{0}}}\)
\(\displaystyle{\ln{{\left({\frac{{{2}{P}_{{{0}}}}}{{{P}_{{{0}}}}}}\right)}}}={k}{x}{2}\)
\(\displaystyle{k}={\frac{{{\ln{{2}}}}}{{{2}}}}\)
\(\displaystyle{\ln{{\left({\frac{{{P}}}{{{P}_{{{0}}}}}}\right)}}}={\frac{{{\ln{{2}}}}}{{{2}}}}.{t}\)
When \(\displaystyle{t}={3}\) years, \(\displaystyle{P}={20000}\)
\(\displaystyle{\ln{{\left({\frac{{{20000}}}{{{P}_{{{0}}}}}}\right)}}}={\frac{{{\ln{{2}}}}}{{{2}}}}{x}{3}\)
\(\displaystyle{\ln{{\left({\frac{{{20000}}}{{{P}_{{{0}}}}}}\right)}}}={1.0397}\)
\(\displaystyle{\frac{{{20000}}}{{{P}_{{{0}}}}}}={e}^{{{1.0397}}}\)
\(\displaystyle{P}_{{{0}}}={\frac{{{20000}}}{{{e}^{{{1.0397}}}}}}={7071.2}\)
\(\displaystyle{P}_{{{0}}}={7072}\)
So, initial population, \(\displaystyle{P}_{{{0}}}={7072}\)
0
karton
Answered 2022-01-09 Author has 8454 answers

Let make N represent the population of the country at any given time t and \(N_{o}\) as the initial population.
\(\frac{dN}{dt}=kN\)
where k is the constant of proportionality.
\(\int \frac{1}{N}dN=k \int dt\)
\(\ln N=kt+C\)
For \(t=0, N = N_{o}\)
Therefore, \(N=N_{o}e^{kt}\) (1)
For \(t=2\) yrs, \(N=2No\)
Substitute these values into (1)
\(2N_{o}=N_{o}e^{2t}\)
\(k=\frac{1}{2} \ln 2\)
\(k = 0.347\)
Therefore (1) becomes
\(N=N_{o} e^{0.347t}\) (2)
For \(t = 3\) yrs, \(N = 20,000\)
Substitute these values into (2)
\(20000=N_{o}e^{0.347(3)}\)
\(N_{o}=\frac{20000}{e^{0.347(3)}}\)
\(N_{o}=7062\)
Therefore the number of people initially living in the country is 7062

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