Problem 68 - Contemporary Precalculus

Question

Between 1790 and \(1860,\) the population y of the United States (in millions) in year x was given by \(y=3.9572\left(1.0299^{\circ}\right),\) where \(x=0\) corresponds to \(1790 .\)F ind the U.S. population in the given year. $$1859$$

Step-by-Step Solution

Answer

Answer: The population of the United States in 1859 was approximately 31.2 million people.

Step 1: Determine the value of x representing the year 1859

We are given that x=0 corresponds to the year 1790. To find the value of x representing the year 1859, subtract 1790 from 1859: $$x = 1859 - 1790$$ $$x = 69$$

Step 2: Use the value of x to find the population y

Now that we know x=69, we can use the formula given for the population y to find the U.S. population in 1859: $$y = 3.9572(1.0299^x)$$ Substitute x = 69 into the formula: $$y = 3.9572(1.0299^{69})$$

Step 3: Calculate the U.S. population in 1859

Now we can simply calculate the population y using the given value of x: $$y = 3.9572(1.0299^{69})$$ $$y \approx 31.199$$ Since the population is given in millions, the U.S. population in 1859 was approximately 31.2 million people.

Key Concepts

Population Modeling

Population modeling is a useful mathematical tool to understand how populations change over time. It helps predict future population sizes and trends. In this case, we look at the United States population between 1790 and 1860. When building a population model, it's crucial to identify several core factors:

• Base Population: This is the initial population size. In this exercise, it's approximately 3.9572 million in 1790.
• Growth Rate: The rate at which the population grows each year. Here, it represents a growth factor of 1.0299.

These elements are critical as they form the basis for forecasting population trends. A well-constructed model can illustrate how populations expand under given conditions, allowing for strategic planning or historical analysis.

Exponential Functions

Exponential functions describe processes that change at a constant multiplicative rate. They are powerful in modeling phenomena like population growth due to their capacity to reflect steady increases over time.An exponential function takes the form:\[ y = a imes b^x \]

• a: Initial value (e.g., the starting population).
• b: Base or growth factor (greater than 1 indicates growth).

In our exercise, the formula used is \( y = 3.9572(1.0299^x) \), which demonstrates how the U.S. population grew over years from the base year 1790. The exponent \( x \) represents the number of years since 1790, and each increment results in a multiplication by the base growth factor.Exponential growth starts slowly but accelerates as time progresses. This is why exponential functions are particularly effective for modeling populations, which often grow rapidly under favorable conditions.

Mathematical Applications

Mathematical applications are abundant in everyday life, and modeling populations with exponential functions is a great example. By understanding how to apply these functions, students can gain insights into historical trends and future projections.Through the exercise of calculating the U.S. population in 1859, we learn to:

• Identify and manipulate variables to fit an exponential model for historical data analysis.
• Utilize step-by-step problem-solving skills to resolve real-world mathematical problems.

Consider the specific application:- Determining the year corresponds to finding the correct power \( x \), ensuring accurate input for the function.- Calculating the population using the formula translates abstract concepts into tangible outcomes. Such applications strengthen mathematical proficiency and enhance comprehension, demonstrating how theories translate into practical uses across various fields.