Problem 22 - College Algebra

Question

For the following exercises, write an equation describing the relationship of the given variables. \(y\) varies jointly as the square of \(x\) and the square root of \(z\) and inversely as the cube of \(w\). When \(x=3, z=4,\) and \(w=3,\) then \(y=6\).

Step-by-Step Solution

Answer

The equation is \( y = 9 \cdot \frac{x^2 \sqrt{z}}{w^3} \)."

Step 1: Understand the Type of Variation Involved

The problem states that \(y\) varies jointly as the square of \(x\) and the square root of \(z\), and inversely as the cube of \(w\). This means we have a joint and inverse variation equation.

Step 2: Write the Proportional Relationship

We express the relationship as a proportional equation: \[ y \propto \frac{x^2 \sqrt{z}}{w^3} \]Here, \(\propto\) denotes that \(y\) is proportional to the expression \(\frac{x^2 \sqrt{z}}{w^3}\).

Step 3: Introduce a Constant of Proportionality

Convert the proportionality into an equation by introducing a constant \(k\):\[ y = k \cdot \frac{x^2 \sqrt{z}}{w^3} \]

Step 4: Substitute Given Values to Find \(k\)

Substitute the given values \(x = 3\), \(z = 4\), \(w = 3\), and \(y = 6\) into the equation:\[ 6 = k \cdot \frac{3^2 \sqrt{4}}{3^3} \]

Step 5: Simplify and Solve for \(k\)

Simplify the equation:\[ 6 = k \cdot \frac{9 \times 2}{27} \]\[ 6 = k \cdot \frac{18}{27} \]\[ 6 = k \cdot \frac{2}{3} \]Solve for \(k\):\[ k = 6 \cdot \frac{3}{2} \]\[ k = 9 \]

Step 6: Write the Final Equation

Now that we have found \(k = 9\), substitute it back into the original equation:\[ y = 9 \cdot \frac{x^2 \sqrt{z}}{w^3} \]

Key Concepts

Proportional Equation

A proportional equation is an expression that shows how one quantity depends on others through multiplication by a constant. When dealing with variations, it often uses the symbol \(\propto\) to indicate this dependence.

For joint and inverse variations, the concept involves more than one variable affecting the main variable, which in this case is \(y\). Here, \(y\) varies with the joint action of the square of \(x\) and the square root of \(z\), and the inverse action of the cube of \(w\).

This specific proportional relationship for the exercise is expressed as:

• \( y \propto \frac{x^2 \sqrt{z}}{w^3} \)

This tells us that \(y\) increases with an increase in \(x\) or \(z\), but decreases with an increase in \(w\), adhering to the rules of joint and inverse variation.

Understanding the role each variable plays is key in setting up and solving these kinds of equations.

Constant of Proportionality

The constant of proportionality, often denoted by \(k\), acts as a bridge to convert a proportional relationship into an equation. It quantifies the relationship such that regardless of the proportionality, the equation remains consistent across different values of the variables.

To find this constant, substitute known values of the variables into the proportional equation, which provides a specific value for \(k\). In the given exercise, once we recognize our proportional relationship:

• \( y = k \cdot \frac{x^2 \sqrt{z}}{w^3} \)

Inserting the specific values (\(x = 3\), \(z = 4\), \(w = 3\), and \(y = 6\)), leads us to accurately determine that \(k = 9\).

Therefore, \(k\) gives meaning to the equation beyond variables' abstract symbols, allowing us to predict \(y\) for any values of \(x\), \(z\), and \(w\).

Solve for Constant

Solving for the constant \(k\) involves substituting given values into the equation and simplifying to isolate \(k\). Let's walk through this process step-by-step, as shown in the exercise:

Given the equation:

• \(6 = k \cdot \frac{3^2 \sqrt{4}}{3^3}\)

Simplify inside the fraction:

• \(6 = k \cdot \frac{9 \times 2}{27}\)
• \(6 = k \cdot \frac{18}{27}\)

Reduce the fraction to its simplest form:

• \(6 = k \cdot \frac{2}{3}\)

To isolate \(k\), multiply both sides of the equation by \(\frac{3}{2}\):

• \( k = 6 \cdot \frac{3}{2}\)
• \( k = 9 \)

In conclusion, once \(k\) is known, it solidifies the relationship, and we can rewrite the final equation as \( y = 9 \cdot \frac{x^2 \sqrt{z}}{w^3} \). Solving for the constant is a crucial step in turning theoretical relationships into practical tools for problem-solving.