Problem 39 - College Algebra

Question

For the following exercises, use the given information to find the unknown value. \(y\) varies jointly as the square of \(x\) and the cube of \(z\) and inversely as the square root of \(w\). When \(x=2\), \(z=2\), and \(w=64\), then \(y=12\). Find \(y\) when \(x=1\), \(z=3,\) and \(w=4\).

Step-by-Step Solution

Answer

The value of \( y \) is 40.5.

Step 1: Understanding the Relationship

The problem states that the variable \( y \) varies jointly as the square of \( x \) and the cube of \( z \), and inversely as the square root of \( w \). Therefore, we can represent this relationship using the equation \( y = k \frac{x^2 z^3}{\sqrt{w}} \) where \( k \) is a constant that we need to determine.

Step 2: Finding the Constant of Variation

To find the constant \( k \), we will use the known values: when \( x = 2 \), \( z = 2 \), and \( w = 64 \), \( y = 12 \). Substituting these into the equation gives us:\[12 = k \frac{2^2 \cdot 2^3}{\sqrt{64}}\]Calculate the necessary parts:- \( 2^2 = 4 \)- \( 2^3 = 8 \)- \( \sqrt{64} = 8 \)Substituting these values in, we get:\[12 = k \frac{4 \cdot 8}{8}\]Simplifying the right side yields \( 12 = k \cdot 4 \), thus \( k = 3 \).

Step 3: Solving for the New Scenario

With \( k = 3 \), we now need to find \( y \) for the new values: \( x = 1 \), \( z = 3 \), and \( w = 4 \). Substitute these values into the original equation:\[y = 3 \frac{1^2 \cdot 3^3}{\sqrt{4}}\]Calculate the components:- \( 1^2 = 1 \)- \( 3^3 = 27 \)- \( \sqrt{4} = 2 \)Substitute these into the equation:\[y = 3 \frac{1 \cdot 27}{2} = 3 \cdot \frac{27}{2} = 3 \cdot 13.5 = 40.5\]

Step 4: Final Calculation

After evaluating the equation with the given values, we find that the unknown value of \( y \) is \( 40.5 \).

Key Concepts

Algebraic Equations

Algebraic equations form the backbone of many math problems and are vital in understanding complex relationships between variables. An algebraic equation is an expression that includes numbers, variables, and arithmetic operations, equaling a value. In this exercise, we encountered an algebraic equation that illustrates joint and inverse variation. This type of equation helps describe how different variables are interrelated.

The equation used to solve our problem was given by:

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

This equation tells us that \( y \) varies as a product of the square of \( x \), cube of \( z \), and is divided by the square root of \( w \). Here, \( k \) represents the constant of variation, which stays the same when you keep the relationship among those variables consistent.

Understanding how to set up and manipulate these algebraic equations is key in solving problems involving joint and inverse variation, allowing for predictions and analysis of real-world situations.

Problem Solving

Effective problem solving in mathematics, as demonstrated in our exercise, typically involves a systematic approach. It requires understanding the problem, setting up the correct equations, finding any unknown constants, and applying the calculations to new scenarios. To solve this joint variation problem, one must follow these steps:

• Identify the relationship between variables and express it with an equation.


• Substitute given values to calculate any unknown constants, like the constant of variation \( k \).


• Use the found constant and apply it to a new set of variables to find the unknown value, which in our case was the new \( y \) value.

By following these straightforward steps, problem-solving becomes a structured process. It ensures you have a clear path from the problem to the solution, reducing errors and increasing accuracy.

Inverse Variation

Inverse variation is a significant concept in algebra that describes a relationship where, as one variable increases, another variable decreases. This is the opposite of direct variation, where both variables move in the same direction. Understanding inverse variation helps clarify how some variables depend on the reciprocal behaviors of others. In the exercise provided, inverse variation is demonstrated through the relationship between \( y \) and \( w \). Specifically, \( y \) varies inversely with the square root of \( w \). When \( w \) is larger, \( y \) becomes smaller for the same \( x \) and \( z \). The equation reflecting this is:

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

This shows how an increase in \( w \) (since it is under a square root and in the denominator) decreases \( y \), emphasizing the essence of inverse variation. Recognizing such relationships allows you to adequately answer questions about changing conditions and how they affect outcomes.