Problem 40 - College Algebra with Corequisite Support

Question

For the following exercises, use the given information to find the unknown value. \(y\) varies jointly as the square of \(x\) and of \(z\) and inversely as the square root of \(w\) and of \(t\). When \(x=2, z=3, w=16,\) and \(t=3,\) then \(y=1 .\) Find \(y\) when \(x=3, z=2, w=36,\) and \(t=5\).

Step-by-Step Solution

Answer

When \( x=3, z=2, w=36, \) and \( t=5, y=\frac{2 \sqrt{15}}{15} \).

Step 1: Understand the Relationship

The problem states that \( y \) varies jointly as the square of \( x \) and of \( z \), and inversely as the square root of \( w \) and of \( t \). This relationship can be expressed by the equation: \( y = k \cdot \frac{x^2 z^2}{\sqrt{w} \sqrt{t}} \), where \( k \) is the constant of variation.

Step 2: Solve for the Constant of Variation \( k \)

Using the provided values when \( x=2, z=3, w=16, t=3, \text{ and } y=1 \), substitute these into the equation to solve for \( k \):\[ 1 = k \cdot \frac{2^2 \cdot 3^2}{\sqrt{16} \cdot \sqrt{3}} \]Simplifying gives:\[ 1 = k \cdot \frac{4 \cdot 9}{4 \cdot \sqrt{3}} = k \cdot \frac{36}{4 \cdot \sqrt{3}} \]\[ 1 = k \cdot \frac{9}{\sqrt{3}} \]Solving for \( k \),\[ k = \frac{\sqrt{3}}{9} \]

Step 3: Calculate \( y \) with New Variables

Using the expression found for \( k \), substitute the new values \( x=3, z=2, w=36, \text{ and } t=5 \) into the equation:\[ y = \frac{\sqrt{3}}{9} \cdot \frac{3^2 \cdot 2^2}{\sqrt{36} \cdot \sqrt{5}} \]Simplify the expression:\[ y = \frac{\sqrt{3}}{9} \cdot \frac{9 \cdot 4}{6 \cdot \sqrt{5}} \]\[ y = \frac{\sqrt{3}}{9} \cdot \frac{36}{6 \cdot \sqrt{5}} \]\[ y = \frac{\sqrt{3} \cdot 6}{9 \cdot \sqrt{5}} \]\[ y = \frac{2 \cdot \sqrt{3}}{3 \cdot \sqrt{5}} \]\[ y = \frac{2 \sqrt{15}}{15} \]

Step 4: Simplify the Result

The expression \( \frac{2 \sqrt{15}}{15} \) is already in its simplest form. Therefore, this is the final answer for the value of \( y \) when \( x=3, z=2, w=36, \text{ and } t=5 \).

Key Concepts

Inverse Variation

Inverse variation is an intriguing mathematical concept. It explains a relationship where one value increases as another value decreases. This is opposite to direct variation, where both values increase or decrease together.

In our exercise, the variable \( y \) varies inversely with the square root of both \( w \) and \( t \). This means as either \( \sqrt{w} \) or \( \sqrt{t} \) increases, \( y \) decreases, given other factors constant. This inverse relationship models situations where a balance or trade-off is required.

• The formula representing inverse variation is: \( y = k \cdot \frac{x^2 z^2}{\sqrt{w} \sqrt{t}} \)
• The inverse elements \( \sqrt{w} \) and \( \sqrt{t} \) appear in the denominator, highlighting the inverse relationship.

Understanding inverse variation helps in scenarios that entail optimizing output by analyzing contributing factors. Its practical applications are widespread, including engineering and physics, where resources tend to be inversely related to outcomes.

Constant of Variation

In mathematics, the constant of variation is a crucial concept. It represents a fixed number that relates two quantities inversely or directly proportional. In our current problem, this constant is denoted by \( k \).

To compute \( k \), you need specific values of the variables involved. You start by substituting known values into the equation for \( y \), which allows you to isolate and solve for \( k \). Here's a brief look:

• Start with the equation: \( y = k \cdot \frac{x^2 z^2}{\sqrt{w} \sqrt{t}} \)
• Use known values to find \( k \): \( 1 = k \cdot \frac{2^2 \cdot 3^2}{\sqrt{16} \cdot \sqrt{3}} \)
• Solve, leading to \( k = \frac{\sqrt{3}}{9} \)

Once calculated, \( k \) becomes a constant that can be used to compute unknown values of \( y \) for new conditions, ensuring the variation equation remains accurate. Understanding \( k \) not only completes the equation but also connects these mathematical variables meaningfully in practical scenarios.

Square Root

The square root is a fundamental mathematical operation that finds the original number whose square provides a given product. It's visually represented by the radical symbol \( \sqrt{} \).

In problems involving variation, the square root often influences how variables change relative to each other. In our exercise:

• Both \( w \) and \( t \) incorporate square roots as \( \sqrt{w} \) and \( \sqrt{t} \).
• These terms in the denominator of our joint variation problem imply that as the value of the square root component increases, the overall function's outcome decreases.

The presence of the square root in equations usually affects how quickly or slowly the dependent variable reacts to changes. For students tackling joint variation problems, mastering square roots can provide insights into complex real-world relationships.