Problem 38 - Precalculus with Limits

Question

DIRECT VARIATION In Exercises 35-38, assume that is \(y\) directly proportional to \(x\). Use the given \(x\)-value and \(y\)-value to find a linear model that relates \(y\) and \(x\). \(x = 6\), \(y = 580\)

Step-by-Step Solution

Answer

The linear model that relates \(y\) and \(x\) is \(y = 96.67x\)

Step 1: Finding the Constant of Proportionality

Since we know that \(y = kx\), we can find \(k\) by rearranging the equation to \(k = y/x\). Substituting the given values that is \(x = 6\) and \(y = 580\) into the equation we get \(k=580/6\)

Step 2: Calculating the Value of Constant of Proportionality

Having arranged for \(k\), next is to perform the division to find the actual value of \(k\). Doing the division gives \(k\) approximately equal to 96.67

Step 3: Formulating the Linear Model

Now that we have found the constant of proportionality, we can use it to write the linear model that relates \(y\) and \(x\). This is done by substituting \(k\) into the equation \(y = kx\). So the equation of the linear model is \(y = 96.67x\)

Key Concepts

Proportionality Constant

When you hear the term "proportionality constant" in mathematics, it often pops up in discussions involving direct variation. Direct variation happens when two variables change in relation to each other in a consistent manner. In simpler terms, when one goes up, the other goes up at a constant rate, and vice versa.
To find this constant, we refer to it by the symbol \( k \) in the equation \( y = kx \). This means \( y \) is directly proportional to \( x \), and \( k \) is the proportionality constant. It gives us the fixed rate or ratio of change between \( y \) and \( x \).

To find \( k \), you can use division. You simply rearrange the equation to solve for \( k \) as such: \( k = \frac{y}{x} \). Here, using the example where \( x = 6 \) and \( y = 580 \), you substitute these values in and divide to get \( k = \frac{580}{6} \), resulting in a constant of approximately 96.67. This proportion remains consistent as \( x \) and \( y \) change.

Linear Model

Linear models are quite simple in concept but incredibly powerful in application. They depict a straight-line relationship between two variables. In our context of direct variation, the relationship can be expressed as \( y = kx \). This equation represents a straight line on a graph, with \( k \) being the slope of the line.
A linear model helps us predict one variable if we know the other. In the equation \( y = 96.67x \), derived from the given values in the exercise, the 96.67 acts as the link between \( x \) and \( y \). It tells us how much \( y \) will change with every unit change in \( x \).

In practical terms, a linear model like this can assist in numerous ways, such as forecasting expenses, estimating distances, or even predicting outcomes in scientific experiments. Anytime you know one part of a directly proportional relationship, this model helps you find the other part.

Division Operation

The division operation is a fundamental mathematical procedure that is often used to find rates, ratios, or proportions between numbers. In the context of direct variation, division helps us determine the proportionality constant, \( k \).
Consider the general formula for direct variation, \( y = kx \). When rearranged to solve for \( k \), the equation becomes \( k = \frac{y}{x} \). Here, division is crucial to isolate \( k \) and find the exact rate of change from one variable to the other.

The division operation not only calculates this constant but also plays a vital role in verifying the linearity of a relationship. If dividing \( y \) by \( x \) always yields the same \( k \), the relationship is truly linear and consistent. This makes division a powerful tool in unraveling the mysteries of relationships between variables.