# 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.

Question
Exponential models
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.

2021-02-26
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.

### Relevant Questions

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.
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?
Several models have been proposed to explain the diversification of life during geological periods. According to Benton (1997), The diversification of marine families in the past 600 million years (Myr) appears to have followed two or three logistic curves, with equilibrium levels that lasted for up to 200 Myr. In contrast, continental organisms clearly show an exponential pattern of diversification, and although it is not clear whether the empirical diversification patterns are real or are artifacts of a poor fossil record, the latter explanation seems unlikely. In this problem, we will investigate three models fordiversification. They are analogous to models for populationgrowth, however, the quantities involved have a differentinterpretation. We denote by N(t) the diversification function,which counts the number of taxa as a function of time, and by rthe intrinsic rate of diversification.
(a) (Exponential Model) This model is described by $$\displaystyle{\frac{{{d}{N}}}{{{\left.{d}{t}\right.}}}}={r}_{{{e}}}{N}\ {\left({8.86}\right)}.$$ Solve (8.86) with the initial condition N(0) at time 0, and show that $$\displaystyle{r}_{{{e}}}$$ can be estimated from $$\displaystyle{r}_{{{e}}}={\frac{{{1}}}{{{t}}}}\ {\ln{\ }}{\left[{\frac{{{N}{\left({t}\right)}}}{{{N}{\left({0}\right)}}}}\right]}\ {\left({8.87}\right)}$$
(b) (Logistic Growth) This model is described by $$\displaystyle{\frac{{{d}{N}}}{{{\left.{d}{t}\right.}}}}={r}_{{{l}}}{N}\ {\left({1}\ -\ {\frac{{{N}}}{{{K}}}}\right)}\ {\left({8.88}\right)}$$ where K is the equilibrium value. Solve (8.88) with the initial condition N(0) at time 0, and show that $$\displaystyle{r}_{{{l}}}$$ can be estimated from $$\displaystyle{r}_{{{l}}}={\frac{{{1}}}{{{t}}}}\ {\ln{\ }}{\left[{\frac{{{K}\ -\ {N}{\left({0}\right)}}}{{{N}{\left({0}\right)}}}}\right]}\ +\ {\frac{{{1}}}{{{t}}}}\ {\ln{\ }}{\left[{\frac{{{N}{\left({t}\right)}}}{{{K}\ -\ {N}{\left({t}\right)}}}}\right]}\ {\left({8.89}\right)}$$ for $$\displaystyle{N}{\left({t}\right)}\ {<}\ {K}.$$
(c) Assume that $$\displaystyle{N}{\left({0}\right)}={1}$$ and $$\displaystyle{N}{\left({10}\right)}={1000}.$$ Estimate $$\displaystyle{r}_{{{e}}}$$ and $$\displaystyle{r}_{{{l}}}$$ for both $$\displaystyle{K}={1001}$$ and $$\displaystyle{K}={10000}.$$
(d) Use your answer in (c) to explain the following quote from Stanley (1979): There must be a general tendency for calculated values of $$\displaystyle{\left[{r}\right]}$$ to represent underestimates of exponential rates,because some radiation will have followed distinctly sigmoid paths during the interval evaluated.
(e) Explain why the exponential model is a good approximation to the logistic model when $$\displaystyle\frac{{N}}{{K}}$$ is small compared with 1.
The fox population in a certain region has a continuous growth rate of 6 percent per year. It is estimated that the population in the year 2000 was 18900.
(a) Find a function that models the population t years after 2000 (t=0 for 2000). Hint: Use an exponential function with base e.
(b) Use the function from part (a) to estimate the fox population in the year 2008.
The fox population in a certain region has a continuous growth rate of 6 percent per year. It is estimated that the population in the year 2000 was 18900.
(a) Find a function that models the population t years after 2000 (t=0 for 2000). Hint: Use an exponential function with base e.
(b) Use the function from part (a) to estimate the fox population in the year 2008.
Often new technology spreads exponentially. Between 1995 and 2005, each year the number of Internet domain hosts was 1.43 times the number of hosts in the preceding year. In 1995, the number of hosts was 8.2 million.
(a) Explain why the number of hosts is an exponential function of time. The number of hosts grows by a factor of -----? each year, this is an exponential function because the number is growing by ------? decreasing constant increasing multiples.
(b) Find a formula for the exponential function that gives the number N of hosts, in millions, as a function of the time t in years since 1995.
(c) According to this model, in what year did the number of hosts reach 49 million?
Solve the given problem related to population growth.
During the first decade of this century the population of a certain city grew exponentially the population of the city was 146210 in 2000 and 217245 in 2010. Find the exponential growth function that models the population growth of the city use t= 0 to represent 2000 and t= 10 to represent 2010 and so on.
N(t)=?
Use your exponential growth function to predict the population of the city in 2016. Round to the nearest thousand.
$$\displaystyle{b}{e}{g}\in{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}{\left\lbrace{\left|{c}\right|}{c}{\mid}\right\rbrace}{h}{l}\in{e}\text{Year}&\text{Population}\backslash{h}{l}\in{e}{1960}&{94.092}\backslash{h}{l}\in{e}{1965}&{98.883}\backslash{h}{l}\in{e}{1970}&{104.345}\backslash{h}{l}\in{e}{1975}&{111.573}\backslash{h}{l}\in{e}{1980}&{116.807}\backslash{h}{l}\in{e}{1985}&{120.754}\backslash{h}{l}\in{e}{1990}&{123.537}\backslash{h}{l}\in{e}{1995}&{125.327}\backslash{h}{l}\in{e}{2000}&{126.776}\backslash{h}{l}\in{e}{2005}&{127.715}\backslash{h}{l}\in{e}{2010}&{127.579}\backslash{h}{l}\in{e}{e}{n}{d}{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}$$
Use a calculator to fit both an exponential function and a logistic function to these data. Graph the data points and both functions, and comment on the accuracy of the models. [Hint: Subtract 94,000 from each of the population figures. Then, after obtaining a model from your calculator, add 94,000 to get your final model. It might be helpful to choose $$\displaystyle{t}={0}$$ to correspond to 1960 or 1980.]
$$\displaystyle{b}{e}{g}\in{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}{\left\lbrace{\left|{c}\right|}{c}{\mid}\right\rbrace}{h}{l}\in{e}\text{Year}&\text{Population}\backslash{h}{l}\in{e}{1960}&{3581}\backslash{h}{l}\in{e}{1965}&{3723}\backslash{h}{l}\in{e}{1970}&{3877}\backslash{h}{l}\in{e}{1975}&{4007}\backslash{h}{l}\in{e}{1980}&{4086}\backslash{h}{l}\in{e}{1985}&{4152}\backslash{h}{l}\in{e}{1990}&{4242}\backslash{h}{l}\in{e}{1995}&{4359}\backslash{h}{l}\in{e}{2000}&{4492}\backslash{h}{l}\in{e}{2005}&{4625}\backslash{h}{l}\in{e}{2010}&{4891}\backslash{h}{l}\in{e}{e}{n}{d}{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}$$
Use a calculator to fit both an exponential function and a logistic function to these data. Graph the data points and both functions, and comment on the accuracy of the models. [Hint: Subtract 3500 from each of the population figures. Then, after obtaining a model from your calculator, add 3500 to get your final model. It might be helpful to choose $$\displaystyle{t}={0}$$ to correspond to 1960.]