Problem 30 - College Algebra

Question

For the following exercises, use the given information to find the unknown value. \(y\) varies inversely with the square of \(x .\) When \(x=4\) then \(y=3\). Find \(y\) when \(x=2\).

Step-by-Step Solution

Answer

When \( x=2 \), \( y=12 \).

Step 1: Understanding Inverse Variation

In problems where a quantity 'y' varies inversely with the square of another quantity 'x', it means that as 'x' increases, 'y' decreases proportional to the square of 'x'. This is mathematically expressed as: \[ y = \frac{k}{x^2} \]where 'k' is the constant of variation.

Step 2: Finding the Constant of Variation

To find the constant of variation 'k', we use the given condition that when \( x = 4 \), \( y = 3 \).Substituting these values into the inverse variation formula, we have:\[ 3 = \frac{k}{4^2} \]\[ 3 = \frac{k}{16} \]Multiplying both sides by 16 gives:\[ k = 3 \times 16 = 48 \]

Step 3: Calculating the Unknown Value of y

Now, with \( k = 48 \), we can find the unknown value of \( y \) when \( x = 2 \).Substitute these values into the equation \( y = \frac{k}{x^2} \):\[ y = \frac{48}{2^2} \]\[ y = \frac{48}{4} \]\[ y = 12 \]

Key Concepts

Constant of Variation

In the realm of inverse variation, understanding the "constant of variation" is crucial. This constant is a fixed number that links the variables involved, enabling us to describe how one variable changes in relation to another. In our exercise, we say that the quantity \( y \) varies inversely with the square of \( x \). This mathematical relationship can be expressed as \( y = \frac{k}{x^2} \), where \( k \) is our constant of variation.

To find \( k \), we substitute known values of \( x \) and \( y \) into the equation. For instance, when \( x = 4 \) and \( y = 3 \), plug these values into the formula: \( 3 = \frac{k}{16} \). By solving this equation, we uncover that \( k = 48 \). This constant tells us exactly how \( y \) adjusts as \( x \) changes its value.

Mathematical Formula

Mathematical formulas are essential tools in solving problems involving variations. Here, the focus is on an inverse variation, which uses a specific formula: \( y = \frac{k}{x^2} \). This formula defines how \( y \) behaves when \( x \) changes. An inverse relationship suggests that as \( x \) gets larger, \( y \) decreases, and vice-versa, proportional to the square of \( x \).

With this understanding, we highlight the utility of substituting given numbers into this formula to solve for unknown values. The clarity of this method is evident when tackling our original exercise - by inputting known \( x \) and \( y \) to determine \( k \), then using this constant to find an unknown \( y \) for a new \( x \). This structured approach ensures reliable and precise results, no matter how \( x \) or \( y \) varies.

Step-by-Step Solution

A structured, step-by-step approach is invaluable when solving mathematical problems, especially those involving variations. Let's walk through the steps to tackle our exercise:

1. **Understanding the Relationship**: Recognize that \( y \) varies inversely with the square of \( x \), captured by the formula \( y = \frac{k}{x^2} \). This means \( y \) and \( x \) are inversely proportional to \( x^2 \).

2. **Finding \( k \)**: Use the known values to find the constant \( k \). With \( x = 4 \) and \( y = 3 \), substitute into the formula to get \( 3 = \frac{k}{16} \). Solving this gives us \( k = 48 \).

3. **Solving for the Unknown \( y \)**: Now, use the constant \( k \) to solve for \( y \) when \( x = 2 \). Substituting \( x = 2 \) into the equation, we find: \( y = \frac{48}{2^2} \), which simplifies to \( y = 12 \).

This step-by-step breakdown not only aids in solving the problem at hand but also reinforces the core concepts of inverse variation.