Problem 13 - Intermediate Algebra

Question

Find the constant of variation for each of the stated conditions. \(y\) varies directly as the square of \(x\), and \(y=-144\) when \(x=6 .\)

Step-by-Step Solution

Answer

The constant of variation is \(-4\).

Step 1: Understand the Direct Variation Relationship

In mathematics, direct variation means that as one variable changes, the other variable changes in proportion. For this exercise, the problem states that \( y \) varies directly as the square of \( x \). This can be represented with the equation \( y = kx^2 \), where \( k \) is the constant of variation that we need to find.

Step 2: Substitute the Given Values

We are provided with the values \( y = -144 \) and \( x = 6 \). Substitute these values into the direct variation equation: \( -144 = k(6)^2 \).

Step 3: Simplify and Solve for \( k \)

Simplify the equation by calculating \( 6^2 \), which is \( 36 \). Thus, the equation becomes \( -144 = 36k \). Now, solve for \( k \) by dividing both sides of the equation by \( 36 \): \( k = \frac{-144}{36} \).

Step 4: Calculate the Constant of Variation

Perform the division: \( k = \frac{-144}{36} = -4 \). Therefore, the constant of variation \( k \) is \(-4\).

Key Concepts

Constant of Variation

In a direct variation, one variable changes in proportion to another variable. This means if one goes up, the other one may also go up (or down if the constant is negative) in a predictable way. The relationship between these two variables in direct variation is expressed by the equation \( y = kx^n \), where \( n \) determines the nature of the relationship, and \( k \) is the constant of variation. The constant \( k \) tells you how much \( y \) changes for every change in \( x \).

• If \( k \) is positive, \( y \) increases as \( x \) increases.
• If \( k \) is negative, \( y \) decreases as \( x \) increases.

For example, in the formula \( y = kx^2 \), the variable \( y \) varies directly with the square of \( x \). Finding \( k \) involves substituting the given values of \( x \) and \( y \) into the equation and solving for \( k \). This constant helps us predict \( y \) for any value of \( x \).

Square of a Variable

The square of a variable means to multiply the variable by itself. For example, if \( x = 6 \), then the square of \( x \), denoted by \( x^2 \), is \( 6 \times 6 = 36 \). The concept of squaring is important in various mathematical contexts, including geometry, algebra, and calculus.

• When a variable is squared, it grows rapidly. For instance, small increases in \( x \) result in large increases in \( x^2 \).
• Squaring a positive or negative number results in a positive number (e.g., \( (-6)^2 = 36 \)).

In direct variation problems that involve the square of a variable, such as \( y = kx^2 \), the square affects how changes in \( x \) translate to changes in \( y \). It emphasizes the impact of even small changes in \( x \) on \( y \).

Mathematics Problem Solving

Problem-solving in mathematics is a skill that involves understanding the problem, devising a plan, solving the problem, and reviewing the solution. Here's how each step applies to our example.
**Understand the Problem**: We begin by identifying that we have a direct variation problem where \( y \) depends on \( x^2 \). This sets the stage for using the equation \( y = kx^2 \).
**Devise a Plan**: Our goal is to find \( k \). We have specific values for \( y \) and \( x \), so we plan to substitute these into the equation.
**Carry Out the Plan**: Substitute \( y = -144 \) and \( x = 6 \) into the equation, turning it into \( -144 = k(6)^2 \). Simplifying gives us \( -144 = 36k \). Now solve for \( k \) by dividing both sides by 36, which gives \( k = -4 \).
**Review the Solution**: Finally, we check our steps to ensure there are no errors. This means reconsidering our calculations and ensuring the logic aligns — confirming that \( k = -4 \) makes the variation equation consistent with the original problem statement.