Problem 90 - A Graphical Approach to Precalculus with Limits

Question

Assume that the constant of variation is positive. Let \(y\) vary inversely with the second power of \(x\). If \(x\) doubles, what happens to \(y ?\)

Step-by-Step Solution

Answer

When \(x\) doubles, \(y\) becomes one-fourth of its original value.

Step 1: Understand the inverse variation

In an inverse variation with the square of a variable, if \(y\) varies inversely with \(x^2\), then the relationship can be expressed as \(y = \frac{k}{x^2}\), where \(k\) is the constant of variation.

Step 2: Double the value of x

If \(x\) doubles, then the new value of \(x\) becomes \(2x\). We need to see how this affects \(y\).

Step 3: Substitute the new value of x into the equation

Substitute \(2x\) into the inverse variation equation: \(y = \frac{k}{(2x)^2} = \frac{k}{4x^2}\).

Step 4: Compare new equation with the original equation

Originally, \(y = \frac{k}{x^2}\). With \(x\) doubled, \(y = \frac{k}{4x^2}\). Comparing these, we can see that \(y\) becomes one-fourth of its original value when \(x\) is doubled.

Key Concepts

Constant of Variation

In an inverse variation scenario, the "constant of variation" is a key component that remains unchanged, regardless of how the other variables in the equation shift. When we say that a variable, like \(y\), varies inversely with another, such as \(x^2\), it is effectively maintained by a constant value, often denoted as \(k\). This means the product of \(y\) and \(x^2\) stays the same, or mathematically, \(y \times x^2 = k\).
The constant of variation helps in understanding how changes in one variable cause changes in another, allowing us to predict behavior when variables are altered. It’s like a glue that binds the different elements of the equation, informing how tightly or loosely they are connected. Even if \(x\) changes, the overall structure governed by \(k\) stays intact, providing a reliable measure for calculations. In our example, \(k\) is crucial in determining how \(y\) changes when \(x\) doubles.

Square of a Variable

The term "square of a variable" refers to raising a number or expression to the power of two. When we deal with variables squared, such as \(x^2\), we're essentially multiplying \(x\) by itself, resulting in a new dimension of variability.
In our context of inverse variation, the function \(y = \frac{k}{x^2}\) specifies that \(y\) changes inversely with the square of \(x\). This means that as \(x\) increases, its effect is magnified due to squaring, causing \(y\) to decrease at a rate proportional to the square. Conversely, if \(x\) decreases, \(y\) increases rapidly.
Understanding the impact of squaring \(x\) is critical because it offers insight into nonlinear relationships within a problem, showing that changes are not direct but follow a curving path.

Doubling Effect

When we talk about the "doubling effect" in mathematical contexts, we refer to the impact of multiplying a variable by two on an expression or equation. In the situation where our variable \(x\) doubles,we analyze how the relationship \(y = \frac{k}{x^2}\) is affected.
As seen in the calculations, doubling \(x\) changes the function to \(y = \frac{k}{(2x)^2}\), which is the same as \(y = \frac{k}{4x^2}\). Hence, the original \(y\) is divided by four, demonstrating that \(y\) becomes one-fourth of its initial value. This significant reduction is due to the doubling of \(x\), highlighting how powerful squared relationships are.

• This effect shows how sensitive \(y\) is to changes in \(x\), emphasizing the exponential impact of variable adjustments in inverse variation equations.
• It provides a clear illustration of how interconnected variables can behave in non-linear patterns, which is a core component of understanding inverse variation.