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United States population growth: In 1960 the population of the United States was about \(180 \mathrm{mil-}\) lion. Since that time the population has increased by approximately \(1.2 \%\) each year. This is a verbal description of the function \(N=N(t)\), where \(N\) is the population, in millions, and \(t\) is the number of years since 1960 . a. Express in functional notation the population of the United States in 1963. Calculate its value. b. Use the verbal description of \(N\) to make a table of values that shows U.S. population in millions from 1960 through \(1965 .\) c. Make a graph of U.S. population versus time. Be sure to label your graph appropriately. d. Verify that the formula \(180 \times 1.012^{t}\) million people, where \(t\) is the number of years since 1960 , gives the same values as those you found in the table in part b. (Note: Because \(t\) is the number of years since 1960, you would use \(t=2\) to get the population in 1962.) e. Assuming that the population has been growing at the same percentage rate since 1960 , what value does the formula above give for the population in 2000? (Note: The actual population in 2000 was about 281 million.)

Step by step solution

01

Express Population in 1963

Write the formula for calculating the population in the year 1963. Since 1963 is 3 years after 1960, we set \(t = 3\). The formula according to the problem is \(N(t) = 180 \times 1.012^t\). Substitute \(t = 3\) into the formula: \(N(3) = 180 \times 1.012^3\). Calculate this using a calculator to find the population in millions.

02

Calculate Population for Each Year (1960-1965)

Using the formula \(N(t) = 180 \times 1.012^t\), calculate the population for each year from 1960 to 1965. Start with \(t = 0\) for 1960, \(t = 1\) for 1961, and so on up to \(t = 5\) for 1965. Record each value to create a table.

03

Create Table of Values

Based on calculations from Step 2, create a table that shows the population in millions for each year from 1960 to 1965. Include all your calculated results from the previous step.

04

Graph Population Over Time

Using the table from Step 3, create a graph where the x-axis represents the time in years (1960 to 1965) and the y-axis represents the population in millions. Ensure the graph has appropriately labeled axes and plot each calculated point.

05

Verify Formula with Table Values

Confirm that the formula \(180 \times 1.012^t\) gives values matching those in the table for each year from 1960 to 1965. Re-check any mismatches by recalculating to ensure accuracy.

06

Predict Population in 2000

Use the formula \(N(t) = 180 \times 1.012^t\) to predict the population for the year 2000. Since 2000 is 40 years after 1960, set \(t = 40\). Calculate \(N(40) = 180 \times 1.012^{40}\) and compare this result with the actual 2000 population, 281 million.

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Key Concepts

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

Population Modeling

Population modeling is a mathematical way to understand how populations change over time. One common model is the exponential growth model, which applies to situations where a population grows by a constant percentage each year. In the given exercise, we're looking at the U.S population modeled with this concept.
Exponential growth implies that the percentage growth rate stays consistent. Here, the U.S. population increases by 1.2% annually. We use the formula: \[N(t) = 180 \times 1.012^t\]where:

• \(N(t)\) is the population at time \(t\), in millions
• 180 is the initial population in 1960
• 1.012 represents a 1.2% annual growth rate
• \(t\) is the number of years after 1960

This approach helps us make future predictions and better understand historical data.

Functional Notation

Functional notation helps us express mathematical relationships clearly. In this context, it shows how the U.S. population changes over the years.When using functional notation for exponential models, like the one in the exercise, you express the model as a function of time, \(N(t)\). Here, \(N\) is our function, and \(t\) represents the number of years since 1960.For 1963, you would plug in \(t = 3\) (since 1963 is 3 years after 1960) as follows:\[N(3) = 180 \times 1.012^3\]This substitute helps us find the population in 1963. This method is handy because it lets us quickly find the population for any year after 1960, providing clarity and precision.

Population Graphing

Graphing population data allows us to visualize trends over time easily. Once you have calculated the population for each year from 1960 to 1965 using the given formula, displaying this data using a graph can provide insight into growth trends.To create a graph:

• Label the x-axis as "Years" (from 1960 to 1965)
• Label the y-axis as "Population in Millions"
• Plot the calculated points for each year (e.g., \((0, 180)\) for 1960)

By connecting these points, you would observe an upward slope, indicating population growth. Such visual representation makes it easier to understand the nature of exponential growth compared to linear growth.

Population Prediction

Predicting future population helps in planning and resource allocation. Using models like the one in this exercise, predictions can be made even for periods well beyond the initial data range.For predicting the U.S. population in 2000:

• Set \(t = 40\) because 2000 is 40 years after 1960
• Use the formula: \[N(40) = 180 \times 1.012^{40}\]

Calculate to find the predicted population. This exercise reveals that according to the model, the population in 2000 would be around the number calculated. Comparing that with the actual 2000 population of 281 million helps us verify the model's accuracy in reflecting real-world data and spotting potential deviations or changes in growth rates.