Problem 24 - Calculus and its applications

Question

The consumer price index compares the costs, \(c,\) of goods and services over various years, where 1983 is used as a base \((t=0)\). The same goods and services that cost \(\$ 100\) in 1983 cost \(\$ 226\) in 2012 . a) Model \(c\) as an exponential function rounding the growth rate \(k\) to five decimal places. b) Estimate what the goods and services costing \(\$ 100\) in 1983 will cost in 2020 c) In what year did the same goods and services cost twice the 1983 price?

Step-by-Step Solution

Answer

Cost in 2020: $277; goods doubled in 2008.

Step 1: Determine the exponential model equation

To model an exponential function, given that the cost of goods and services changes from \(c_0\) in 1983 to \(c_f\) in 2012, we use the formula \( c(t) = c_0 \cdot e^{kt} \). Here, \(c_0 = 100\), \(c_f = 226\), and \(t = 29\) (since 2012 is 29 years after 1983).

Step 2: Solve for the growth rate, k

We plug our known values into the formula: \( 226 = 100 \cdot e^{29k} \). Solving for \(k\), we first divide both sides by 100 to get \( 2.26 = e^{29k} \) and then take the natural logarithm: \( \ln(2.26) = 29k \). Therefore, \( k = \frac{\ln(2.26)}{29} \), which calculates to approximately \( k = 0.02754 \).

Step 3: Use the model to estimate the 2020 cost

Now that we have \(k\), substitute \(t = 37\) (since 2020 is 37 years after 1983) into the formula \(c(t) = 100 \cdot e^{0.02754 \times 37}\). This yields \( c(37) = 100 \cdot e^{1.01898} \). Calculating \(e^{1.01898} \) gives approximately 2.77, so \( c(37) \approx 277.00 \).

Step 4: Determine when the price doubled

We need \(c(t) = 200\) (double the cost in 1983). Thus, set \(200 = 100 \cdot e^{0.02754t}\). Simplifying gives \(2 = e^{0.02754t}\). Taking the natural logarithm on both sides, \(\ln(2) = 0.02754t\). Solving for \(t\) yields \(t = \frac{\ln(2)}{0.02754} \approx 25.16\). This means the price doubled around the year 1983 + 25, which is roughly 2008.

Key Concepts

Consumer Price Index

The Consumer Price Index (CPI) is an essential economic indicator. It measures the average change in prices over time, which consumers pay for a market basket of goods and services. Think of it as a thermometer for the economy, telling us how hot or cold it is. The CPI is critical because it helps to understand inflation and the cost of living adjustments.

In real life, the CPI is used to adjust income eligibility levels for government assistance, and it affects the income of almost everyone. With the CPI, you can see how much prices have increased since a base year, such as 1983 where a basket costing $100 then costs $226 in 2012. This increase is known as inflation.

Understanding how to model this with exponential functions gives us insights into past and future price changes. It lets economists predict future costs more accurately and assists in planning both personal finances and national policies.

Exponential Function

An exponential function is a mathematical function of the form \( c(t) = c_0 \cdot e^{kt} \), where \( c_0 \) is the initial value (cost in 1983) and \( t \) represents time. It models situations where growth occurs at a constant rate. This type of growth can result in extremely fast changes over time.

To model the price changes from 1983 to 2012, we used the exponential function to derive the growth rate \( k \). By knowing \( k \), predictions like the 2020 price or when the price doubles can be made by plugging different values of \( t \) into the formula.

Exponential functions are not only used in economics but also in biology, physics, and social sciences. Anytime growth rates themselves grow, the exponential function is a helpful tool.

Natural Logarithm

The natural logarithm, denoted as \( \ln \), is the inverse function of the exponential function with base \( e \). In simple terms, it helps you solve equations in which the exponent is unknown.

In the exercise, natural logarithms help to isolate \( k \) in the equation \( 2.26 = e^{29k} \), by transforming this into \( \ln(2.26) = 29k \). From there, finding \( k \) is straightforward and involves simple algebra.

Natural logarithms are valuable for solving a variety of real-world problems involving exponential growth or decay, such as measuring population growth, radioactive decay, or cooling processes.