Question

# The population of a region is growing exponentially.

Exponential models

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?

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