Problem 64 - Applied Calculus

Question

Starting salaries in the United States for associates with master's degrees in financial engineering (FE) have been rising approximately linearly, from \(\$ 74,800\) in 2009 to \(\$ 89,800\) in \(2013 .\) a. Use the two (year, salary) data points (0,74.8) and (4,89.8) to find the linear relationship \(y=m x+b\) between \(x=\) years since 2009 and \(y=\) salary in thousands of dollars.

Step-by-Step Solution

Answer

The linear relationship is \( y = 3.75x + 74.8 \).

Step 1: Identify the two data points

The two data points to use are (0, 74.8) representing the year 2009 with a salary of $74,800, and (4, 89.8) representing 2013 with a salary of $89,800. The salaries are in thousands of dollars.

Step 2: Determine the slope (m)

Use the formula for slope between two points, \( m = \frac{y_2 - y_1}{x_2 - x_1} \). Substitute the given values: \( m = \frac{89.8 - 74.8}{4 - 0} = \frac{15}{4} = 3.75 \). This slope represents the annual salary increase in thousands of dollars.

Step 3: Use the slope-intercept form to find the y-intercept (b)

The slope-intercept form of a line is \( y = mx + b \). Use one of the points, say (0, 74.8), which simplifies finding \( b \) because \( x = 0 \): \( 74.8 = 3.75 \cdot 0 + b \). Thus, \( b = 74.8 \).

Step 4: Write the equation of the line

Substitute the slope \( m \) and the y-intercept \( b \) into the formula \( y = mx + b \). This gives \( y = 3.75x + 74.8 \), which is the linear relationship between the years since 2009 and the salary in thousands of dollars.

Key Concepts

Slope Calculation

When we talk about the slope in the context of linear equations, we're discussing how much the dependent variable (in this case, salary) changes for each unit change in the independent variable (years since 2009). Think of the slope as the 'steepness' of the line or how quickly salaries are rising over the years.
To calculate the slope, we use the formula \[m = \frac{y_2 - y_1}{x_2 - x_1}\]Here

• \( (x_1, y_1) \) is the first data point: (0, 74.8)
• \( (x_2, y_2) \) is the second data point: (4, 89.8)

By substituting these points into the formula, we find the slope, \( m = \frac{15}{4} = 3.75 \).
This means that each year, the salary increases by $3,750. Understanding the slope helps predict future salaries based on past increases.

Y-Intercept

The y-intercept is the point where the line crosses the y-axis. In practical terms, it represents the value of the dependent variable when the independent variable is zero. In our salary example, the y-intercept tells us what the starting salary was in 2009.
When using the equation of a line, \( y = mx + b \), solving for \( b \) when \( x = 0 \) directly gives us the y-intercept.
From the data point (0, 74.8), substituting into the equation, the y-intercept \( b \) is immediately identified as 74.8.
So, the starting salary in 2009 was $74,800. Knowing the y-intercept helps us understand the initial conditions before any changes occurred.

Linear Relationships

Linear relationships describe how two variables relate to each other in a straight line when graphed. The key feature of a linear relationship is that the rate of change is constant. In our exercise, this is shown as a steady increase in salary over the years.
By finding the slope and y-intercept, we established a linear equation \( y = 3.75x + 74.8 \). This equation means that for every year past 2009, the salary increases by \(3,750, starting from an initial salary of \)74,800.
This linear model is beneficial in predicting future data points. Simply plug in the number of years after 2009 to predict the reaching salary.
Understanding how linear relationships work provides insight into patterns and trends that might not be immediately visible. They can be incredibly useful for making predictions in fields like economics, engineering, and various sciences. By knowing the linear relationship, you are essentially unlocking the ability to forecast future outcomes or understand past trends.