Problem 8 - Algebra and Trigonometry with Analytic Geometry

Question

Express the statement as a formula that involves the given variables and a constant of proportionality \(k\), and then determine the value of \(k\) from the given conditions. \(r\) is directly proportional to the product of \(s\) and \(v\) and inversely proportional to the cube of \(p\). If \(s=2, v=3\), and \(p=5\), then \(r=40\).

Step-by-Step Solution

Answer

The constant of proportionality \(k\) is approximately 833.33.

Step 1: Express as a Proportionality Equation

We are told that \( r \) is directly proportional to the product of \( s \) and \( v \), and inversely proportional to the cube of \( p \). This can be expressed as the formula \( r = k \frac{sv}{p^3} \), where \( k \) is the constant of proportionality.

Step 2: Substitute Given Values

To determine the value of \( k \), substitute the given values into the equation. We have \( s = 2 \), \( v = 3 \), \( p = 5 \), and \( r = 40 \). The equation becomes:\[ 40 = k \frac{2 \cdot 3}{5^3} \].

Step 3: Simplify the Expression

Calculate the values inside the fraction:1. The product \( s \times v = 2 \times 3 = 6 \).2. The cube of \( p \) is \( p^3 = 5^3 = 125 \).Substitute these values into the equation to get:\[ 40 = k \frac{6}{125} \].

Step 4: Solve for Constant \(k\)

To find \( k \), multiply both sides of the equation by 125, which gives:\[ 40 \times 125 = 6k \].Calculate the product on the left side:\[ 5000 = 6k \].Finally, divide both sides by 6 to solve for \( k \):\[ k = \frac{5000}{6} \approx 833.33 \].

Key Concepts

Constant of Proportionality

In mathematics, the constant of proportionality is a fundamental concept, especially in relationships involving direct and inverse proportionality. When two quantities are directly proportional, their ratio is a constant, while for inverse proportionality, their product is constant. In this exercise, the constant of proportionality is represented by the variable \( k \). It acts as a scaling factor in the relationship specified by the problem statement.

For example, if we say the speed of a car (\( c \)) is directly proportional to the horsepower of its engine (\( h \)), we can express this relationship as \( c = kh \), where \( k \) would be the constant of proportionality. In the given exercise, the constant \( k \) adjusts the proportional relationship between \( r \), \( s \), \( v \), and \( p \). This factor ensures that the proportional equation holds true when specific values are substituted.

Algebraic Expression

An algebraic expression is a mathematical phrase that can contain numbers, variables, and operators. Such expressions are crucial in formulating relationships between different mathematical entities. In our exercise, we start with creating an expression to capture the relationship given:

• The expression \( r = k \frac{sv}{p^3} \) signifies that \( r \) varies with the product of \( s \) and \( v \) and the inverse of the cube of \( p \).
• This expression is derived from understanding how the variables interact based on the problem's description.

By developing this expression, we are achieving the first step in solving for unknowns within a word problem. Understanding how to arrange and manipulate these expressions is foundational for solving more complex mathematical problems.

Proportional Relationship

A proportional relationship is one where two quantities increase or decrease at the same rate. These relationships can be direct or inverse, which impacts how we construct formulas. In this exercise, the relationship is more nuanced:

• \( r \) is directly proportional to the product \( sv \). This means as \( s \) and \( v \) increase, \( r \) will increase proportionally.
• Conversely, \( r \) is inversely proportional to \( p^3 \). Here, as \( p \) increases, \( r \) will decrease, since it is divided by \( p \) cubed.

Understanding these relationships helps form the expression \( r = k \frac{sv}{p^3} \). This shows both types of proportionality working in tandem to affect \( r \) based on changes in \( s \), \( v \), and \( p \). Acknowledging these intricacies in relationships is essential for accurately solving and predicting outcomes within mathematical equations.

Solving Equations

Solving equations involves finding the unknown variable's value that makes the equation true. This skill is essential for converting word problems into solvable mathematical equations. Let's walk through this process as conducted in the exercise:

• First, recognize the equation from the expression: \( r = k \frac{sv}{p^3} \).
• Plug in the given values: \( s = 2 \), \( v = 3 \), \( p = 5 \), and \( r = 40 \).
• Simplifying gives: \( 40 = k \frac{6}{125} \).
• To isolate \( k \), multiply both sides by 125: \( 5000 = 6k \).
• Finally, divide by 6 to solve: \( k = \frac{5000}{6} \approx 833.33 \).

Finding \( k \) confirms the constant needed to maintain the relationship with given conditions. This clear process demystifies solving equations by showing sequential steps and emphasizing checking your work for consistency.