Problem 38 - A Graphical Approach to Precalculus with Limits

Question

find the constant of variation \(k\) and the undetermined value in the table if \(y\) is directly proportional to \(x\). $$\begin{array}{|c|c|c|c|c|} \hline x & 1.2 & 4.3 & 5.7 & ? \\ \hline y & 3.96 & 14.19 & 18.81 & 23.43 \end{array}$$

Step-by-Step Solution

Answer

The constant of variation \( k \) is 3.3 and the missing \( x \) is 7.1.

Step 1: Understand Direct Variation

Direct variation means that two variables, \( y \) and \( x \), have a relationship such that \( y = kx \) for some constant \( k \). Our task is to find this constant \( k \) using the values provided.

Step 2: Calculate Constant of Variation \( k \)

Choose any pair of \( x \) and \( y \) from the table. Using \( x = 1.2 \) and \( y = 3.96 \), the formula \( y = kx \) gives \( 3.96 = k \times 1.2 \). Solve for \( k \): \[ k = \frac{3.96}{1.2} = 3.3 \]

Step 3: Check Constant of Variation \( k \)

Verify \( k \) by using another pair \( x = 4.3 \), \( y = 14.19 \): \( 14.19 = 3.3 \times 4.3 \). This holds true, so \( k = 3.3 \) is correct.

Step 4: Find the Undetermined \( x \) Value

Use \( y = kx \) with the undetermined \( y = 23.43 \): \( 23.43 = 3.3 \times x \). Solve for \( x \): \[ x = \frac{23.43}{3.3} \approx 7.1 \]

Key Concepts

Constant of Variation

The constant of variation, often denoted as \( k \), is a fundamental component in understanding direct variation problems. Imagine the constant of variation as the speed in a car journey—it tells us how fast changes in \( x \) leads to changes in \( y \), where \( y \) is directly proportional to \( x \).

In this scenario, the relationship between \( x \) and \( y \) is described by the equation \( y = kx \). This means for every change in \( x \), \( y \) changes by a factor of \( k \). To discover what \( k \) is, you simply use the values given for \( x \) and \( y \) in the problem, like so: if \( y = 3.96 \) when \( x = 1.2 \), substitute these into the equation to get \( 3.96 = k imes 1.2 \).

To solve for \( k \), rearrange the equation:

• Divide both sides by \( x \): \( k = \frac{y}{x} = \frac{3.96}{1.2} \)
• Simplify the fraction: \( k = 3.3 \)

This tells us the rate at which \( y \) increases as \( x \) increases is 3.3.

Proportional Relationships

Proportional relationships are vital in understanding how two variables relate to each other in direct variation. When we say two variables are directly proportional, it implies that as one variable increases, the other does as well, and they maintain a constant ratio.

In this exercise, the relationship between \( y \) and \( x \) has a proportionality characterized by the constant \( k = 3.3 \). This means that for every 1 unit increase in \( x \), \( y \) increases by 3.3 units.

Checking the consistency of this relationship is essential to confirm the value of \( k \). By using different pairs of \( x \) and \( y \) in the table:

• For \( x = 4.3 \) and \( y = 14.19 \), check: \( 14.19 = 3.3 imes 4.3 \) is true, confirming \( k = 3.3 \)
• Any discrepancy might indicate an error in calculations or data.

Once verified, this consistent ratio characteristic confirms the proportional relationship.

Solving for Variables

Solving for variables in a direct variation scenario involves using the constant of variation to find missing values in the relationship between \( x \) and \( y \). This exercise requires finding an unknown \( x \) when \( y \) is given.

Upon finding \( k = 3.3 \), use the direct variation formula \( y = kx \) to solve for the unknown \( x \). When \( y = 23.43 \), substitute back into the equation:

• Set up the equation: \( 23.43 = 3.3 imes x \)
• Isolate \( x \) by dividing both sides by 3.3: \( x = \frac{23.43}{3.3} \)
• Perform the division to find \( x \approx 7.1 \)

This method of plugging in known values and performing basic algebraic manipulations is key in solving for unknown variables in direct variation problems.