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Chapter 5: Problem 10

China is the most populous country in the world. In 2000 it had about 1.262 billion people. By 2005 the population had grown to 1.306 billion. Use this information to construct models predicting the size of China's population in the future. a. Identify your variables and units. b. Construct a linear model. c. Construct an exponential model. d. What will China's population be in 2050 according to each of your models?

Short Answer

Expert verified
In 2050, using the linear model, China's population will be 1.702 billion, and using the exponential model, it will be 1.78 billion.

Step by step solution

01

Identify Variables and Units

Let the variable \( x \) represent the number of years since 2000. Let \( P(x) \) represent the population of China in billions. So, in 2000 (\( x = 0 \)), the population is 1.262 billion. By 2005 (\( x = 5 \)), the population is 1.306 billion.
02

Construct a Linear Model

Use the points (0, 1.262) and (5, 1.306) to find the equation of the line. The slope \( m \) is given by \[ m = \frac{1.306 - 1.262}{5 - 0} = \frac{0.044}{5} = 0.0088 \]. The linear model is \[ P(x) = 0.0088x + 1.262 \].
03

Construct an Exponential Model

Use the population values to set up the equation \( P(x) = P_0 e^{kx} \), where \( P_0 \) is the initial population, 1.262 in 2000. For 2005, we have \[ 1.306 = 1.262 e^{5k} \]. Solving for \( k \), we get \[ e^{5k} = \frac{1.306}{1.262} \]. So, \[ 5k = \frac{\text{ln}(1.306) - \text{ln}(1.262)}{5} \]. Solving this, \( k \approx 0.0069 \). The model is \[ P(x) = 1.262 e^{0.0069x} \].
04

Predict Population in 2050 Using Linear Model

For \( x = 50 \), using the linear model \( P(x) = 0.0088(50) + 1.262 \), so \( P(50) = 0.44 + 1.262 = 1.702 \) billion.
05

Predict Population in 2050 Using Exponential Model

For \( x = 50 \), using the exponential model \( P(x) = 1.262 e^{0.0069 \times 50} \). First calculate \( 0.0069 \times 50 = 0.345 \). Therefore, \( P(x) = 1.262 e^{0.345} \). Approximating \( e^{0.345} \approx 1.411 \), so \( P(50) = 1.262 \times 1.411 \approx 1.78 \) billion.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Linear Model
A linear model is a simple way to predict future values by assuming that the change in a variable is constant over time. To create a linear model for China's population growth from 2000 to 2005, we use two data points: (0, 1.262) and (5, 1.306). The slope of the line between these points is calculated as follows: The result is a slope of 0.0088. The linear model equation becomes: This equation means that we expect the population to increase by 0.0088 billion people each year.
Exponential Model
An exponential model assumes that the population grows at a rate proportional to its current value, making it better suited for populations that grow faster and faster over time. The general form of an exponential model is For China's population, we start with the initial population, In 2000 and solve for the growth rate, Using the population in 2005: . After calculating, we find that approximately equals 0.0069, making our exponential model: This means the population increases at a rate proportional to its size, leading to faster growth over time.
Population Prediction
Using our linear and exponential models, we can predict China's population in 2050. For the linear model, we substitute into the equation The solution gives . For the exponential model, we use the equation . First, we calculate and then approximate Using these results, . Thus, we predict China's population in 2050 to be about 1.702 billion according to the linear model and 1.78 billion according to the exponential model.
Algebra
Algebra helps us create and solve models like the linear and exponential ones by providing the tools to work with equations and variables. We identified variables (population and years), set up equations, and also solved for unknowns, such as the slope in the linear model and the growth rate in the exponential model. Mastering these algebraic techniques is crucial for developing models and making predictions based on given data.
College Algebra
College Algebra dives deeper into topics like linear and exponential models, providing a thorough foundation for more advanced mathematical concepts. Understanding how to manipulate equations, work with growth rates, and solve for unknowns is essential. For example:
  • Constructing linear models involves finding slopes and y-intercepts.
  • Exponential models require understanding exponents and logarithms.
To excel, practice applying algebraic principles to real-world problems like population prediction.

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Most popular questions from this chapter

(Requires technology to find a best-fit function.) The accompanying table shows the U.S. international trade in goods and services. $$ \begin{aligned} &\text { U.S }\\\ &\text { International Trade (Billions of Dollars) }\\\ &\begin{array}{crr} \hline & \text { Total } & \text { Total } \\ \text { Year } & \text { Exports } & \text { Imports } \\ \hline 1960 & 25.9 & 22.4 \\ 1965 & 35.3 & 30.6 \\ 1970 & 56.6 & 54.4 \\ 1975 & 132.6 & 120.2 \\ 1980 & 271.8 & 291.2 \\ 1985 & 288.8 & 410.9 \\ 1990 & 537.2 & 618.4 \\ 1995 & 793.5 & 891.0 \\ 2000 & 1070.6 & 1448.2 \\ 2005 & 1275.2 & 1992.0 \\ \hline \end{array} \end{aligned} $$ a. U.S. imports and exports both expanded rapidly between 1960 and \(2005 .\) Use technology to plot the total U.S. exports and total U.S. imports over time on the same graph. b. Now change the vertical axis to a logarithmic scale and generate a semi-log plot of the same data as in part (a). What is the shape of the data now, and what does this suggest would be an appropriate function type to model U.S. exports and imports? c. Construct appropriate function models for total U.S. imports and for total exports. d. The difference between the values of exports and imports is called the trade balance. If the balance is negative, it is called a trade deficit. The balance of trade has been an object of much concern lately. Calculate the trade balance for each year and plot it over time. Describe the overall pattern. e. We have a trade deficit that has been increasing rapidly in recent years. But for quantities that are growing exponentially, the "relative difference" is much more meaningful than the simple difference. In this case the relative difference is \(\frac{\text { exports }-\text { imports }}{\text { exports }}\) This gives the trade balance as a fraction (or if you multiply by 100 , as a percentage) of exports. Calculate the relative difference for each year in the above table and graph it as a function of time. Does this present a more or less worrisome picture? That is, in particular over the last decade, has the relative difference remained stable or is it also rapidly increasing in magnitude?

Blood alcohol content (BAC) is the amount of alcohol present in your blood as you drink. It is calculated by determining how many grams of alcohol are present in 100 milliliters (1 deciliter) of blood. So if a person has 0.08 grams of alcohol per 100 milliliters of blood, the \(\mathrm{BAC}\) is \(0.08 \mathrm{~g} / \mathrm{dl} .\) After a person has stopped drinking, the BAC declines over time as his or her liver metabolizes the alcohol. Metabolism proceeds at a steady rate and is impossible to speed up. For instance, an average ( \(150-1 \mathrm{~b}\) ) male metabolizes about 8 to 12 grams of alcohol an hour (the amount in one bottle of beer^{3} ). The behavioral effects of alcohol are closely related to the blood alcohol content. For example, if an average 150 -lb male drank two bottles of beer within an hour, he would have a BAC level of 0.05 and could suffer from euphoria, inhibition, loss of motor coordination, and overfriendliness. The same male after drinking four bottles of beer in an hour would be legally drunk with a BAC of \(0.10 .\) He would likely suffer from impaired motor function and decision making, drowsiness, and slurred speech. After drinking twelve beers in one hour, he will have attained the dosage for stupor (0.30) and possibly death (0.40) . The following table gives the BAC of an initially legally drunk person over time (assuming he doesn't drink any additional alcohol). $$ \begin{array}{lccccc} \hline \text { Time (hours) } & 0 & 1 & 2 & 3 & 4 \\ \text { BAC (g/dI) } & 0.100 & 0.067 & 0.045 & 0.030 & 0.020 \\ \hline \end{array} $$ a. Graph the data from the table (be sure to carefully label the axes). b. Justify the use of an exponential function to model the data. Then construct the function where \(B(t)\) is the \(\mathrm{BAC}\) for time \(t\) in hours. c. By what percentage does the BAC decrease every hour? d. What would be a reasonable domain for your function? What would be a reasonable range? e. Assuming the person drinks no more alcohol, when does the BAC reach \(0.005 \mathrm{~g} / \mathrm{dl}\) ?

Which function has the steepest graph? $$ \begin{array}{l} F(x)=100(1.2)^{x} \\ G(x)=100(0.8)^{x} \\ H(x)=100(1.2)^{-x} \end{array} $$

Generate equations that represent the pollution levels, \(P(t),\) as a function of time, \(t\) (in years), such that \(P(0)=150\) and: a. \(P(t)\) triples each year. b. \(P(t)\) decreases by twelve units each year. c. \(P(t)\) decreases by \(7 \%\) each year. d. The annual average rate of change of \(P(t)\) with respect to \(t\) is constant at \(1 .\)

Construct both a linear and an exponential function that go through the points (0,200) and (10,500) .
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