Problem 20 - Intermediate Algebra

Question

Find the constant of variation for each of the stated conditions. \(r\) varies inversely as the cube of \(t\), and \(r=\frac{1}{16}\) when \(t=4 .\)

Step-by-Step Solution

Answer

The constant of variation \( k \) is 4.

Step 1: Understand the relationship

An inverse variation between two variables means when one variable increases, the other decreases. For this problem, we have the relationship that \( r \) varies inversely with the cube of \( t \). This can be expressed mathematically as \( r = \frac{k}{t^3} \), where \( k \) is the constant of variation we need to find.

Step 2: Substitute given values into the equation

Given \( r = \frac{1}{16} \) when \( t = 4 \), we substitute these values into the equation: \( \frac{1}{16} = \frac{k}{4^3} \).

Step 3: Simplify the equation

Simplify \( 4^3 \) to get 64. The equation becomes \( \frac{1}{16} = \frac{k}{64} \).

Step 4: Solve for the constant of variation \( k \)

To find \( k \), multiply both sides of the equation by 64: \( k = 64 \times \frac{1}{16} \). Simplify the right side: \( k = \frac{64}{16} = 4 \).

Key Concepts

Understanding the Constant of Variation

In the context of inverse variation, the constant of variation, often denoted as \( k \), is a crucial element. It links the two variables involved. In a direct variation, one variable is equal to \( k \) times the other, showing a straightforward ratio of change. However, things differ in inverse variation. Here, as one variable increases, the other decreases proportionally to maintain the constant \( k \).

In this particular problem, \( r \) varies inversely as the cube of \( t \). This means as \( t \) increases, \( r \) becomes smaller. The equation to express this relationship is \( r = \frac{k}{t^3} \).

By inserting the given values of \( r = \frac{1}{16} \) when \( t = 4 \), solving the equation reveals that \( k = 4 \). This constant signifies the fixed value that determines the correlation between \( r \) and the inverse cube of \( t \).

Exploring the Cube Function

The cube function is pivotal in the relationship described by the exercise. A cube function involves raising a number to the third power. Its formula is expressed as \( t^3 \), where \( t \) is the variable. This function exhibits rapid growth because every increase in \( t \) leads to an exponentially larger increase in \( t^3 \).

In this scenario, since \( r \) varies inversely with \( t^3 \), each alteration in \( t \) results in a significant impact on \( r \). For example, when \( t \) is doubled, \( t^3 \) becomes eight times larger, leading to a drastic reduction in \( r \). Understanding how the cube function scales helps in grasping how negativity impacts inverse variation.

The inverse relationship here is not linear due to the cube function's nature, but rather, it highlights how sensitive \( r \) is to changes in \( t \).

Unveiling Mathematical Relationships

Mathematical relationships, such as the one between \( r \) and \( t^3 \), illustrate how certain variables are interconnected. Recognizing these patterns and functions is essential for solving problems involving inverse variation.

In this problem, the relationship is a classic example of inverse variation, which often manifests in real-world contexts. Important factors include:

• Proportionality: As one variable changes, the other adjusts inversely.
• Dependency: The change in one produces a predictable effect on the other, given the constant \( k \).
• Non-linear dynamics: Especially with non-linear functions like cubes, the relationship shows complex changes.

These points are pivotal in understanding how inverse variation problems are structured and solved, allowing one to predict outcomes for different scenarios.