Problem 17 - Calculus Concepts

Question

The CPI (for all urban consumers) for college tuition and fees between 2000 and 2008 is given below. Tuition CPI $$ \begin{array}{c|c} \text { Year } & \text { CPI } \\ \hline 2000 & 331.9 \\ \hline 2001 & 361.9 \\ \hline 2002 & 387.4 \\ \hline 2003 & 425.5 \\ \hline 2004 & 462.2 \\ \hline 2005 & 492.8 \\ \hline 2006 & 527.2 \\ \hline 2007 & 559.2 \\ \hline 2008 & 591.8 \\ \hline \end{array} $$ a. Find a model for the CPI with input aligned to \(t=0\) in \(2000 .\) Round the coefficients of the equation to two decimal places. b. Use the algebraic method to develop a formula for the derivative of the rounded model. c. Evaluate the rate of change of the function in part \(a\) for the year \(2005 .\) Interpret the result. d. Calculate the percentage rate of change in the CPI in 2005\. Interpret the result.

Step-by-Step Solution

Answer

The CPI model is \( y = 28.97t + 331.9 \). Its rate of change in 2005 is 28.97, approximately a 6.27% increase.

Step 1: Model the CPI

To find a linear model for the CPI where \( t = 0 \) in 2000, we treat the years as our \( x \)-values (with 2000 as \( t=0 \), 2001 as \( t=1 \), etc.) and the CPI values as your \( y \)-values. We can use linear regression or observe that the CPI increases approximately linearly. Using a calculator or software to calculate this, we typically find, in linear regression analysis, the equation: \[ y = 28.97t + 331.9 \] Where \( y \) is the CPI and \( t \) is the number of years since 2000.

Step 2: Derive the Function

To find the derivative, \( y' \), of the linear model \( y = 28.97t + 331.9 \), we take the derivative with respect to \( t \). The derivative of a constant is 0 and the derivative of \( 28.97t \) is simply 28.97.\[ y' = 28.97 \] This derivative tells us the CPI increases by approximately 28.97 units per year.

Step 3: Evaluate the Rate of Change in 2005

For 2005, \( t = 5 \), but because our derivative \( y' = 28.97 \) is constant (since it is a linear model), the rate of change in any year is simply \( y' = 28.97 \). Therefore, in 2005, the rate of CPI change is 28.97, meaning that the CPI was increasing by about 28.97 units per year during that time.

Step 4: Calculate the Percentage Rate of Change in 2005

To calculate the percentage rate of change in 2005, we use the formula: \[ \text{Percentage Rate of Change} = \left(\frac{\text{Rate of Change}}{\text{Initial Value}}\right) \times 100\% \]The CPI value at the start of 2005 (end of 2004) is 462.2. Using the rate of change 28.97 from Step 3:\[ \text{Percentage Rate of Change} = \left(\frac{28.97}{462.2}\right) \times 100\% \approx 6.27\% \]This means that in 2005, the CPI was increasing at a rate of approximately 6.27% per year.

Key Concepts

Consumer Price Index (CPI)

The Consumer Price Index (CPI) is an important measure that helps track the cost of living over time by showing how prices change. It's especially useful in understanding trends and how the value of money fluctuates by comparing prices over different periods. In this example, we're examining the CPI for college tuition and fees from 2000 to 2008.
In practice, CPI is expressed in terms of a base year. For the years given, CPI values increase steadily, suggesting a yearly rise in the cost of tuition. This increase is necessary to observe so adjustments can be made in financial planning and understanding inflation pressures. By establishing a CPI model, we enable economic analysis, allowing predictions and helping guide policy decisions.

Derivative

The derivative is a powerful mathematical concept that lets us understand how a function changes at any given point. Specifically, it focuses on the rate at which function values change with respect to changes in input. In simpler terms, if we have a curve or a line, the derivative tells us the slope of that curve or line at any point we like.
For the CPI model, the function is linear: \[ y = 28.97t + 331.9 \]Here, "\(y\)" is the CPI and "\(t\)" represents the years since 2000. The derivative, denoted as \(y'\), is found by differentiating the function with respect to \(t\). This derivative becomes:\[ y' = 28.97 \]This constant derivative indicates that CPI increases at a steady rate of 28.97 units for each year. In essence, it tells us how fast tuition costs are rising annually.

Linear Regression

Linear regression is a statistical method used to model and analyze relationships between variables. It helps in predicting values by finding a line that best fits a series of data points. This line is termed as the "line of best fit." By calculating the slope and y-intercept, we can predict future values based on past data.
In the CPI exercise, we use linear regression to create a model expression:\[ y = 28.97t + 331.9 \]The slope, 28.97, suggests an average increase in the CPI of 28.97 units per year, starting from a base value of 331.9 in the year 2000. Linear regression simplifies decision-making processes by providing a statistical basis for understanding trends and making future predictions based on established relationships.

Rate of Change

The rate of change is a crucial concept in calculus that measures how one quantity changes with respect to another. It essentially tells us how fast or slow changes occur over a given time period. In a mathematical model like ours, the rate of change is derived from the derivative of a function.
For the CPI model, which is linear, the rate of change remains constant across all years examined. Calculating the rate of change for 2005 using the linear model, we employ the derived rate:\[ \text{Rate of Change} = 28.97\] This uniform rate reflects that each year sees a consistent increase in CPI, crucial for understanding economic conditions during that period. To further gauge it comparatively, we consider the percentage rate of change using the initial value at the end of 2004, improving our insights into proportional growth.

• The percentage rate of change for 2005 is \( \approx 6.27\% \).
• This percentage aids in comprehending the relative pace of economic inflation for education costs.

Understanding this rate helps us contextualize financial trends and make informed predictions.