Problem 33 - Precalculus: Functions and Graphs

Question

Exer. 31-34: Examine the expression for the given set of data points of the form \((x, y) .\) Find the constant of variation and a formula that describes how \(y\) varles with respect to \(x .\) $$\begin{aligned} &x^{2} y ;\\{(0.16,-394.53125),(0.8,-15.78125)\\\ &(1.6,-3.9453125),(3.2,-0.986328125)\\} \end{aligned}$$

Step-by-Step Solution

Answer

The constant of variation is \(-10.10625\), and the formula is \(y = \frac{-10.10625}{x^2}\).

Step 1: Identify the Type of Variation

We need to check if this is a direct or inverse variation. The expression given is \(x^2y\). If \(x^2y = k\) (a constant), then it is an inverse variation of the form \(y = \frac{k}{x^2}\).

Step 2: Calculate the Constant of Variation (k)

We can use one data point to find \(k\). Choosing \((0.8, -15.78125)\), substitute into \(x^2y = k \): \[ (0.8)^2(-15.78125) = k \]Calculate this to find \(k\): \[ 0.64 \times (-15.78125) = -10.10625 \]Thus, \(k = -10.10625 \).

Step 3: Verify the Constant with Other Data Points

To ensure the constant \(k\) is correct, substitute other points and verify they give the same \(k\). Using \((1.6, -3.9453125)\):\[ (1.6)^2 \times (-3.9453125) = k \]Calculate:\[ 2.56 \times (-3.9453125) = -10.10625 \]The consistency confirms \(k = -10.10625 \).

Step 4: Establish the Formula for Variation

Since we determined the variation is inverse, and we know \(y\) varies with respect to \(x\) as \(y = \frac{k}{x^2}\), substitute \(k\):\[ y = \frac{-10.10625}{x^2} \]

Key Concepts

Constant of Variation

In problems involving inverse variation, one crucial term to understand is the "constant of variation." This constant, often denoted as \(k\), helps determine the stable factor in inverse relationships. In an inverse variation scenario, if you have an equation like \(x^2y = k\), the term \(k\) is what we refer to as the constant of variation. It represents a fixed value that ties the variables together.

To find this constant, simply choose one data point from your set such as \((x, y) = (0.8, -15.78125)\). Insert these coordinates into the given expression to calculate \(k\):

• Substitute into the expression \((0.8)^2 \times (-15.78125) = k\)
• Calculate to get \(0.64 \times (-15.78125) = -10.10625\)

In this example, the constant of variation \(k\) is \(-10.10625\). This constant remains consistent as shown by substituting other data points, further confirming its value through calculation.

Variation Formulas

Variation formulas are used to express how one variable changes in relation to another. In our exercise, we are dealing with an inverse variation, where the two variables, \(x\) and \(y\), change inversely. This particular relationship can be defined by the formula \(y = \frac{k}{x^2}\).

Here’s how it works in simple terms:

• In an inverse variation, as \(x\) increases, \(y\) decreases and vice versa.
• The inverse relationship is captured by the presence of \(x^2\) in the denominator.
• The constant of variation \(k\) provides the proportion that \(y\) will follow, depending on the value of \(x\).

Knowing the variation formula allows us to predict the behavior of \(y\) for any given \(x\). Substitute our found \(k\) into the formula, thus establishing \(y = \frac{-10.10625}{x^2}\) as the rule governing the behavior of \(y\).

Data Points Analysis

Data points are individual pairs of values that fit into your variation equation. Analyzing these helps confirm the relationship is correctly identified, and the formula is accurate. In this exercise, the points \((0.16, -394.53125)\), \((0.8, -15.78125)\), \((1.6, -3.9453125)\), and \((3.2, -0.986328125)\) need checking for consistency.

• For each point, substitute the \(x\) and \(y\) values into the expression \(x^2y = k\).
• If consistent, each should yield \(k = -10.10625\).
• For example, with \((1.6, -3.9453125)\): \((1.6)^2 \times (-3.9453125) = -10.10625\).
• All data points should validate the same constant \(k\).

Conforming data points confirm the inverse variation and the accuracy of the formula. Through these checks, the integrity of the variation relationship and the constant is maintained, giving clear insight into how \(y\) behaves as \(x\) varies.