Question

The population of California was 29.76 million in 1990 and 33.87 million in 2000. Assume that the population grows exponentially.
(a) Find a function that models the population t years after 1990.
(b) Find the time required for the population to double.
(c) Use the function from part (a) to predict the population of California in the year 2010. Look up California’s actual population in 2010, and compare.

Answer 1

The population of California was 29.76 million in 1990 and 33.87 million in 2000. Assume that the population grows exponentially.
Let \(\displaystyle{A}_{{0}}\) and r be the initial population & growth rate of population respectively. Then population in California t years after 1990 is,
\(\displaystyle{P}{\left({t}\right)}={A}_{{0}}{e}^{{{r}{t}}}\) (1)
a. Since the population in 1990 is 29.76 million, therefore \(\displaystyle{A}_{{0}}={29.76}\)
In 2000, the population is 33.87 million. That is,
\(\displaystyle{P}{\left({t}\right)}={33.87}\) when \(\displaystyle{t}={10}\)
Therefore from equation (1), we get
\(\displaystyle{33.87}={29.76}{e}^{{{10}{r}}}\)
\(\displaystyle\Rightarrow{e}^{{{10}{r}}}={1.13810483}\)
\(\displaystyle\Rightarrow{10}{r}={\ln{{\left({1.13810483}\right)}}}\)
\(\displaystyle\Rightarrow{r}={\frac{{{\ln{{\left({1.13810483}\right)}}}}}{{{10}}}}\)
\(\displaystyle\Rightarrow{r}={0.012936}\)
Therefore the population t years after 1990 is,
\(\displaystyle{P}{\left({t}\right)}={29.76}{e}^{{{0.012936}{t}}}\)
b) The double of 29.76 million is 59.52 million.
So put \(P(t)=59.52\) in \(\displaystyle{P}{\left({t}\right)}={29.76}{e}^{{{0.012936}{t}}}\), we get
\(\displaystyle{59.52}={29.76}{e}^{{{0.012936}{t}}}\)
\(\displaystyle\Rightarrow{e}^{{{0.012936}{t}}}={2}\)
\(\displaystyle\Rightarrow{0.012936}{t}={\ln{{\left({2}\right)}}}\)
\(\displaystyle\Rightarrow{t}={\frac{{{\ln{{\left({2}\right)}}}}}{{{0.012936}}}}\)
\(\displaystyle\Rightarrow{t}={53.5828}\)
Hence the population will double in 53.5828 years.
c)Since 2010−1990=20, therefore to predict the population in 2010, put \(t=20\) in \(\displaystyle{P}{\left({t}\right)}={29.76}{e}^{{{0.012936}{t}}}\) , we get
\(\displaystyle{P}{\left({20}\right)}={29.76}{e}^{{{0.012936}{\left({20}\right)}}}\)
\(\displaystyle={38.5472}\) million
The population in California in 2010 is 37.3 million, therefore the actual population is approximately 1 million less than the obtained population.