Question

The population \(P\) (in thousands) of Las Vegas, Nevada from 1960 through 2005 can be modeled by \(P=68.4 e^{0.0467 t}\), where \(t\) is the time in years, with \(t=0\) corresponding to 1960. (Source: U.S. Census Bureau) (a) Find the populations in \(1960,1970,1980,1990,2000\), and 2005 . (b) Explain why the data do not fit a linear model. (c) Use the model to estimate when the population will exceed 900,000 .

Short answer

The populations in \(1960,1970,1980,1990,2000\), and 2005, calculated using the model, are approximately \(68.4, 97.6, 139.0, 198.1, 282.1\), and \(338.4\) (all in thousands, respectively). A linear model isn't a good fit because it assumes a constant rate of growth, but the data indicates an increasing rate, hence an exponential model is better. Finally, the population of Las Vegas, using the provided model, is estimated to surpass 900,000 or 900 (in thousands) around late 2004 or early 2005.

Step-by-step solution

Calculate Populations in Specific Years

Substitute the respective \(t\) value into the model \(P=68.4 e^{0.0467 t}\). For example, for 1960, \(t\) is 0, so \(P = 68.4 e^{0.0467 \cdot 0} = 68.4\). Repeat this calculation for the other years (remember \(t\) refers to the number of years after 1960).

Explain why Data Doesn't Fit a Linear Model

In a linear model, growth is constant, however, the rate of population increase in the model \(P=68.4 e^{0.0467 t}\) is exponential, not constant. That means, as \(t\) increases, the population doesn't just grow by a fixed amount but grows more and more rapidly. This is a characteristic feature of an exponential model.

Estimation of Time for Population to Exceed 900,000

To estimate when the population will exceed 900,000, create the inequality \(68.4 e^{0.0467 t} > 900\) (since \(P\) is in thousands, this is equivalent to 900,000). Solving for \(t\), we find the time when the population will exceed 900,000. This can be done by isolating the exponential part, taking the natural logarithm (ln) on both sides and then solving for \(t\).

Key concepts

Population Modeling

Population modeling is a powerful tool that helps us understand and predict changes in population over time. In the exercise, the population of Las Vegas from 1960 to 2005 is described using an exponential model. This model is particularly useful when a population is expected to grow at a rate proportional to its current size. This sort of growth is often rapid, making exponential models ideal for forecasting populations that are expanding quickly.

By analyzing previous population data, demographicists and planners can create accurate mathematical descriptors that account for varying rates of growth, fertility, mortality, and migration. These factors collectively influence population trends and are essential for making informed decisions regarding resource allocation, urban planning, and environmental management. A well-designed population model provides significant insight into past trends and future projections.

Exponential Functions

Exponential functions are mathematical expressions where the variable acts as an exponent. This specific function takes the form of \(e^{kx}\), where \(e\) is Euler's number, an irrational number approximately equal to 2.718. It serves as the base of the natural logarithm, making exponential functions crucial in continuous growth models.

The function given in the exercise, \(P = 68.4 e^{0.0467 t}\), represents exponential growth where the population grows by about 4.67% per year. This is fundamentally different from linear growth where increments are constant - in exponential growth, the rate of increase magnifies as the quantity itself increases. Such functions effectively model growth patterns in finance, natural settings, and, as we see here, populations.

U.S. Census Data

U.S. Census Data provides a wealth of information that is vital for numerous analyses, including population modeling. The Census Bureau collects data on various demographic factors like age, sex, race, and housing, offering a comprehensive snapshot of the United States' demographic landscape.

For the Las Vegas population model, historical census data forms the basis for constructing an exponential function that describes how the population changed over time. Utilizing authentic data ensures that models are accurate and reflective of real-world circumstances, which can then be used to predict future demographic trends accurately. Census data is used not only in population studies but also in economic planning, social services provision, and political representation.

Mathematical Modeling

Mathematical modeling involves creating mathematical representations to study complex systems. It helps to simplify and abstract real situations into logical forms that can be analyzed and solved.

In the case of the Las Vegas population, modeling is achieved through an exponential function, a common mathematical tool used for projecting future trends based on established patterns. By substituting specific years into the given function, one could predict population figures for those years. The model also highlights the exponential nature of change, which is a fundamental characteristic in many ecological, financial, and engineering contexts.

• Allows exploration of different scenarios.
• Helps understand relationships within data.
• Supports sound decision-making.

With well-structured mathematical models, we can simulate real-world behaviors with significant accuracy, aiding planners and policymakers in forecasting needs and challenges.

Inequalities in Exponential Equations

Inequalities in exponential equations involve using inequalities with exponential expressions. In the given exercise, students solve the inequality \(68.4 e^{0.0467 t} > 900\) to determine when the population will surpass 900,000.

To solve such inequalities, you typically isolate the exponential component and apply the natural logarithm to both sides in a stepwise manner:

- Start with the inequality: \(68.4 e^{0.0467 t} > 900\)
- Divide by 68.4: \(e^{0.0467 t} > \frac{900}{68.4}\)
- Take the natural log of both sides: \(\ln(e^{0.0467 t}) > \ln\left(\frac{900}{68.4}\right)\)
- Simplify using the property \(\ln(e^x) = x\): \(0.0467 t > \ln\left(\frac{900}{68.4}\right)\)
- Finally, solve for \(t\): \(t > \frac{\ln\left(\frac{900}{68.4}\right)}{0.0467}\)

This process helps identify when the population is projected to reach or exceed a certain milestone, underscoring the practical application of algebra in demographic studies.