Problem 20 - Algebra for College Students

Question

If \(y\) is inversely proportional to the square of \(x\), and \(y=\frac{1}{8}\) when \(x=4\), find \(y\) when \(x=8 . \quad y=\frac{1}{32}\)

Step-by-Step Solution

Answer

When \( x = 8 \), \( y = \frac{1}{32} \).

Step 1: Understand the Relationship

Since the problem states that \(y\) is inversely proportional to the square of \(x\), we set up the relationship as \( y = \frac{k}{x^2} \), where \( k \) is the constant of proportionality.

Step 2: Use Given Values

We are given that \( y = \frac{1}{8} \) when \( x = 4 \). Use this information to find the constant \( k \).Substituting into the formula gives:\[ \frac{1}{8} = \frac{k}{4^2} \]\[ \frac{1}{8} = \frac{k}{16} \]

Step 3: Solve for the Constant

Multiply both sides of \( \frac{1}{8} = \frac{k}{16} \) by 16 to solve for \( k \):\[ k = 16 \times \frac{1}{8} \]\[ k = 2 \]

Step 4: Substitute and Solve for New Condition

Now, we use the constant \( k = 2 \) from the previous step to find the new value of \( y \) when \( x = 8 \).Substitute into the equation:\[ y = \frac{k}{x^2} = \frac{2}{8^2} \]\[ y = \frac{2}{64} \]\[ y = \frac{1}{32} \]

Step 5: Conclusion: Verify the Result

The calculated result \( y = \frac{1}{32} \) matches the given answer in the problem. Therefore, when \( x = 8 \), \( y \) is indeed \( \frac{1}{32} \).

Key Concepts

Constant of Proportionality

When we talk about a "constant of proportionality," we're discussing a fixed number that relates two quantities in a proportionality relationship. In the context of inverse proportionality, it serves as the "glue" that connects one variable to another. Without it, you can't accurately express how one variable changes in response to changes in another.

For example, in our exercise, the relationship between variables is given by the formula:

• \( y = \frac{k}{x^2} \)

Here, \( k \) represents the constant of proportionality. It tells us how much wiggle room or freedom one variable has in relation to changes in the other. By knowing \( k \), you get an accurate picture of how \( y \) will change when \( x \) changes. Knowing \( k \) helps us translate the initial conditions (like \( y = \frac{1}{8} \) when \( x = 4 \)) into a usable formula to find new values of \( y \) for other \( x \) values.

Inverse Relationship

In an inverse relationship, as one value increases, the other decreases. This is the opposite of a direct relationship, where both values move in the same direction. When we say "inverse," think of a teeter-totter: if one side goes up, the other must come down.

In mathematical terms, we express an inverse relationship as:

• \( y = \frac{k}{x^n} \)

Here, \( y \) and \( x \) are in an inverse relationship, and \( n \) could be any positive number. The \( k \) acts like a balancing point, ensuring the product of \( y \) and \( x^n \) stays consistent. For our specific exercise, \( n = 2 \) because \( y \) is inversely proportional to \( x^2 \). So as \( x \) increases, \( y \) decreases in a way that their multiplied product remains constant.

Squaring Function

The squaring function is a mathematical operation that takes a number \( x \) and multiplies it by itself, denoted as \( x^2 \). When we apply a squaring function, it impacts the rate and manner of change in relationships like the one in our exercise.

The squaring function plays a crucial role in our inverse proportionality scenario, where \( y = \frac{k}{x^2} \). Here, changes in \( x \) are more pronounced when squared. For instance, if \( x \) is doubled, \( x^2 \) becomes four times larger, not just doubled. This exponential increase in \( x^2 \) ensures that the change in \( y \) is significant, reflecting how powerful the effect of squaring can be. In our problem, even a small increase in \( x \) leads to a notable reduction in \( y \), due to the squaring effect increasing the denominator rapidly.