Question

The population of a region is growing exponentially.

Exponential models
ANSWERED
asked 2021-02-20

The population of a region is growing exponentially. There were 10 million people in 1980 (when \(t=0\)) and 75 million people in 1990. Find an exponential model for the population (in millions of people) at any time tt, in years after 1980.
P(t)=?
What population do you predict for the year 2000?
Predicted population in the year 2000 =million people. What is the doubling time?

Answers (1)

2021-02-21

Given:
\(p(0)=10\) million ... at \( t=0\), 1980
\(p(10)=75\) million ... at \(t=10\) 1990
So, \(p(0)=p_{0}=10\) million
\(\displaystyle\Rightarrow{p}{\left({t}\right)}={p}_{{0}}{e}^{{{k}{t}}}\)
\(\displaystyle\Rightarrow{p}{\left({t}\right)}={10}{e}^{{{k}{t}}}\)
when \(t=10\), (1990) \(\displaystyle\Rightarrow{p}{\left({10}\right)}={75}\)
\(\displaystyle\Rightarrow{p}{\left({10}\right)}={10}{e}^{{{10}{k}}}\Rightarrow{75}={10}{e}^{{{10}{k}}}\)
\(\displaystyle\Rightarrow{e}^{{{10}{k}}}={\frac{{{75}}}{{{10}}}}\)
\(\displaystyle\Rightarrow{e}^{{{10}{k}}}={7.5}\)
Take log on both sides
\(\displaystyle\Rightarrow{10}{k}={{\log}_{{e}}{\left({7.5}\right)}}\)
\(\displaystyle{10}{k}={2.02}\)
\(\displaystyle{k}={\frac{{{2.02}}}{{{10}}}}\)
\(\displaystyle{k}={0.202}\)
\(\displaystyle\Rightarrow{k}={0.2}\)
\(\displaystyle{p}{\left({t}\right)}={10}{e}^{{{0}\cdot{2}{t}}}\)
predicted population in year 2000
\(\displaystyle\Rightarrow{t}={20}\)
\(\displaystyle\Rightarrow{p}{\left({20}\right)}={10}{e}^{{{0}\cdot{2}{\left({20}\right)}}}\)
\(\displaystyle\Rightarrow{p}{\left({20}\right)}={10}{e}^{{4}}\)
\(\displaystyle\Rightarrow{p}{\left({20}\right)}={545.98}\approx{546}\)
Population in 2000 \(p(20)= 546\) million
we, \(\displaystyle{p}{\left({t}\right)}={10}{e}^{{{0}\cdot{2}{t}}}\)
Now,
The population during doubling time \(\displaystyle{p}{\left({t}\right)}={2}\times{10}\) at \(t=0\)
\(\displaystyle{p}{\left({t}\right)}={20}\)
\(\displaystyle\Rightarrow{20}={100}{e}^{{{0}\cdot{2}{t}}}\)
\(\displaystyle\Rightarrow{e}^{{{0}\cdot{2}{t}}}={2}\)
\(\displaystyle{0}\cdot{2}{t}={\ln{{\left({2}\right)}}}\)
\(\displaystyle{0}\cdot{2}{t}={0.69}\)
\(\displaystyle{t}={\frac{{{0.69}}}{{{0.2}}}}\)
\(\displaystyle{t}={3.47}\) doubling time

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