Problem 24 - Intermediate Algebra

Question

Find the constant of variation for each of the stated conditions. \(y\) is directly proportional to the square of \(x\) and inversely proportional to the cube of \(z\), and \(y=4 \frac{1}{2}\) when \(x=6\) and \(z=4\).

Step-by-Step Solution

Answer

The constant of variation is 8.

Step 1: Understand Direct and Inverse Variation

In the problem, it is stated that \( y \) is directly proportional to the square of \( x \), which is represented mathematically as \( y \propto x^2 \). Simultaneously, \( y \) is inversely proportional to the cube of \( z \), represented as \( y \propto \frac{1}{z^3} \). Combining both, we have \( y \propto \frac{x^2}{z^3} \).

Step 2: Set up the Variation Equation

Convert the proportional relationship into an equation by introducing a constant of variation, \( k \). Thus, the equation becomes \( y = k \frac{x^2}{z^3} \).

Step 3: Substitute Given Values into the Equation

Substitute the values \( y = 4.5 \), \( x = 6 \), and \( z = 4 \) into the equation \( y = k \frac{x^2}{z^3} \). This gives us \( 4.5 = k \frac{6^2}{4^3} \).

Step 4: Simplify the Equation

Calculate the powers: \( 6^2 = 36 \) and \( 4^3 = 64 \). Substitute these into the equation, resulting in \( 4.5 = k \frac{36}{64} \).

Step 5: Solve for Constant of Variation \( k \)

Multiply both sides by \( \frac{64}{36} \) to solve for \( k \). Thus, \( k = 4.5 \times \frac{64}{36} \).

Step 6: Calculate the Constant of Variation

Perform the multiplication and division: \( 4.5 \times \frac{64}{36} = \frac{4.5 \times 64}{36} = 8 \). The constant of variation \( k \) is 8.

Key Concepts

Direct Variation

In mathematics, when we say that a variable "varies directly" with another variable or its power, it implies that the two variables are related by a constant multiple. In the exercise, it's noted that \( y \) is directly proportional to the square of \( x \). This establishes that as \( x \) increases or decreases, \( y \) will do the same at a rate that is consistent overall. More formally, if \( y \propto x^2 \), it can be expressed as an equation:

• \( y = kx^2 \)

where \( k \) is the constant of variation. This means that the relationship between \( y \) and \( x^2 \) is consistent, regardless of changes in \( y \) or \( x \), assuming other factors remain constant.

Inverse Variation

Inverse variation, as opposed to direct variation, means that one variable increases while the other decreases proportionally, and vice versa. In this exercise, \( y \) is inversely proportional to the cube of \( z \). A direct representation of this can be understood through the formula:

• \( y \propto \frac{1}{z^3} \)

This suggests that as \( z \) gets larger, \( y \) becomes smaller, all else being consistent. Again, by introducing a constant of variation, \( k \), this relationship transforms into an equation:

• \( y = \frac{k}{z^3} \)

The inverse relationship helps describe how \( y \) behaves in reverse proportion to the changes in \( z \), emphasizing that \( z \) holds an inverse effect on \( y \).

Constant of Variation

The constant of variation is the element that links the relationships described by direct and inverse variation. Using a constant to relate these variables makes it possible to form specific equations that can be solved. In this exercise, we combined direct and inverse variations to model:

• \( y = k \frac{x^2}{z^3} \)

The task was to find \( k \). By substituting in given values for \( y \), \( x \), and \( z \), we determine \( k \). When \( y = 4.5 \), \( x = 6 \), and \( z = 4 \), solving gives:

• \( k = 4.5 \times \frac{64}{36} = 8 \)

Here, \( k = 8 \) is our constant of variation. It effectively "connects" how changes in \( x \) and \( z \) manifest in changes in \( y \). This constant is emblematic of the fixed nature of these relationships.

Mathematical Relationships

Mathematical relationships, like those seen through direct and inverse variations, provide a structured way to express how different variables interact with each other. These relationships are key tasks in math, helping us convert descriptive statements into solvable equations. Simply stated:

• Direct relationships increase or decrease in tandem.
• Inverse relationships move in opposite directions.

The fusion of these in equations like:

• \( y = k \frac{x^2}{z^3} \)

allow for complex interactions between variables to be predicted and calculated. Understanding these relationships fosters more profound insights into how systems behave.