Problem 106 - College Algebra with Modeling and Visualization

Question

The data in the table satisfy the equation \(y=\frac{k}{x^{n}},\) where \(n\) is a positive integer. Determine \(k\) and \(n\) $$ \begin{array}{ccccc} x & 2 & 4 & 6 & 8 \\ \hline y & 9 & 2.25 & 1 & 0.5625 \end{array} $$

Step-by-Step Solution

Answer

\( k = 36 \), \( n = 2 \).

Step 1: Understand the formula

The equation given is \( y = \frac{k}{x^n} \), where \( k \) is a constant and \( n \) is a positive integer. Our task is to find the values of \( k \) and \( n \) using the data from the table.

Step 2: Write down the equations for each data pair

For each \( (x, y) \) pair, substitute the values of \( x \) and \( y \) into the equation \( y = \frac{k}{x^n} \), and set up an equation.1. For \( x = 2, y = 9 \): \( 9 = \frac{k}{2^n} \)2. For \( x = 4, y = 2.25 \): \( 2.25 = \frac{k}{4^n} \)

Step 3: Express two equations

We now have the two equations from Step 2:1. \( 9 = \frac{k}{2^n} \) can be written as \( k = 9 \times 2^n \).2. \( 2.25 = \frac{k}{4^n} \) can be written as \( k = 2.25 \times 4^n \).

Step 4: Set the expressions for k equal

Since both expressions equal \( k \):\( 9 \times 2^n = 2.25 \times 4^n \).

Step 5: Simplify the equation

Recognize that \( 4^n = (2^2)^n = 2^{2n} \), and substitute into the equation:\( 9 \times 2^n = 2.25 \times 2^{2n} \).

Step 6: Solve for n

Divide both sides by \( 2^n \):\( 9 = 2.25 \times 2^n \).Now, divide both sides by 2.25:\( \frac{9}{2.25} = 2^n \).This simplifies to:\( 4 = 2^n \).Thus, \( n = 2 \).

Step 7: Solve for k using a known pair

Using \( n = 2 \) in one of the original equations, substitute back. For the pair \( (x = 2, y = 9) \):Evaluate \( k = 9 \times 2^2 = 9 \times 4 = 36 \).Thus, \( k = 36 \).

Key Concepts

Exponential Equations

Exponential equations are a type of equation where the variable appears in the exponent. In the given problem, the equation is of the form \( y = \frac{k}{x^n} \), which can be rewritten as \( y = k x^{-n} \). Here, the variable \( x \) is raised to the power of \( n \), and this characteristic makes the equation an exponential equation. An exponential equation expresses how a certain quantity changes rapidly, either by growing or decaying, based on the value of the variable in the exponent. These equations are prevalent in various areas such as population modeling, radioactive decay, and financial calculations.- The role of the constant \( k \) is vital, as it scales the effect of the change in \( x \) on \( y \).- The exponent \( n \) determines how sensitive \( y \) is to changes in \( x \). For larger values of \( n \), \( y \) changes more dramatically with \( x \).The data table demonstrates this behavior of exponential equations through values of \( x \) and corresponding \( y \), showing how \( y \) decreases as \( x \) increases, consistent with inverse variation.

Mathematical Modeling

Mathematical modeling involves constructing mathematical representations of real-world scenarios to predict or understand phenomena. In the exercise, the equation \( y = \frac{k}{x^n} \) models the relationship given in the table.The function captures how \( y \) varies inversely with the square of \( x \). This is illustrated by the process of determining \( n \) and \( k \):- The setup of the two initial equations—one for each data pair—helps to establish a formulaic relationship between \( x \) and \( y \).- We use these equations to isolate \( k \) and subsequently derive a comparison, revealing the value of \( n \), showing how algebraic manipulation models the situation.This kind of modeling is crucial because:

• It helps understand how changes in inputs (here, \( x \)) affect outputs (\( y \)).
• It allows us to predict \( y \) for any value of \( x \), not just those listed in the table.

In applications, mathematical modeling through such equations assists in describing physical phenomena, engineering problems, and statistical relationships, providing a meaningful interpretation of numeric patterns.

Problem Solving

Problem solving involves analyzing the given information, applying appropriate mathematical techniques, and arriving at a solution. The given exercise requires these skills for finding \( k \) and \( n \). Let's go through the methodology:1. **Understanding the Problem:** Initially, comprehend how the table and the equation relate. Recognize \( y = \frac{k}{x^n} \) as a tool to find unknowns based on the effect of \( x \) and \( y \).2. **Formulating Equations:** Using data points leads to forming simultaneous equations. Substituting \( x \) and \( y \) from the table into \( y = \frac{k}{x^n} \) creates solvable expressions.3. **Algebraic Manipulation:** Equations disclose values of \( k \) and \( n \) when manipulated. Equating and simplifying reveal \( n = 2 \) and \( k = 36 \).Here are some steps critical to problem-solving success:

• Always double-check calculations to avoid errors in basic arithmetic or algebra.
• Interpret each solution to confirm it meets the original problem's requirements.
• Reflect on the process to gain insights for similar problems in the future.

Mastering problem solving in mathematics enhances critical thinking and analytical skills applicable beyond mere calculation, fostering a deeper approach to complex challenges.