Problem 8 - Precalculus: Functions and Graphs

Question

Exer. 1-12: Express the statement as a formula that involves the given variables and a constant of proportionallty \(k,\) 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 833.33.

Step 1: Identify the Relationship

The problem states, "\(r\) is directly proportional to the product of \(s\) and \(v\) and inversely proportional to the cube of \(p\)." This means \(r = k \cdot \frac{s \cdot v}{p^3}\), where \(k\) is the constant of proportionality.

Step 2: Substitute Given Values

Plug the given values into the equation: \(r = 40, s = 2, v = 3,\) and \(p = 5\). Substitute these into the proportionality formula: \[ 40 = k \cdot \frac{2 \cdot 3}{5^3} \] This simplifies to: \[ 40 = k \cdot \frac{6}{125} \]

Step 3: Solve for the Constant \(k\)

Rearrange the equation to solve for \(k\): \[ k = 40 \cdot \frac{125}{6} \] Calculate \(k\): \[ k = \frac{5000}{6} \] Which simplifies to \[ k = 833.33\] (rounded to two decimal places).

Key Concepts

Constant of Proportionality

When dealing with direct and inverse variation, the constant of proportionality, represented by \( k \), plays a crucial role. It establishes a specific relationship between the variables involved.
In this exercise, \( r \) is both directly proportional to the product of \( s \) and \( v \) and inversely proportional to the cube of \( p \). This defines the formula:

• \( r = k \cdot \frac{s \cdot v}{p^3} \)

Here, \( k \) is the constant that links these variables proportionally and holds them together within the formula. Unlike variables that may change, \( k \) remains constant for all values in your situation, fixed by the conditions provided in the problem.
Ultimately, finding \( k \) involves substituting the known values into the equation and solving for \( k \), showing just how this constant captures the essence of the relationship between the variables.

Proportional Relationships

Proportional relationships are foundational concepts in mathematics that describe how one quantity varies directly or inversely with another. In this problem, two types of proportional relationships are combined: direct proportionality and inverse proportionality.

• Direct Proportionality: Here, \( r \) varies directly with the product of \( s \) and \( v \). This means as the product of \( s \) and \( v \) increases, \( r \) increases proportionally.
• Inverse Proportionality: Simultaneously, \( r \) is inversely proportional to the cube of \( p \). Hence, as \( p^3 \) increases, \( r \) decreases inversely.

This dual relationship can seem complex, but breaking it down helps. A change in \( s \), \( v \), or \( p \) will affect \( r \). For example, doubling \( s \) or \( v \) doubles their product, directly increasing \( r \) unless offset by changes in \( p \). Understanding these relationships is key to forming and solving the equation \( r = k \cdot \frac{s \cdot v}{p^3} \).

Algebraic Manipulation

Algebraic manipulation involves rearranging and solving equations to find unknown values. In this problem, we use these skills to find the constant \( k \).
The original equation, \( r = k \cdot \frac{s \cdot v}{p^3} \), becomes a puzzle to solve. First, substitute the known values into the formula:

• \( 40 = k \cdot \frac{6}{125} \)

Next, clear the fraction by multiplying both sides by its denominator (125). This isolates \( k \) on one side of the equation:

• \( 40 \times 125 = k \times 6 \)
• Which simplifies to \( 5000 = 6k \)

Finally, solve for \( k \) by dividing both sides by 6:

• \( k = \frac{5000}{6} \)
• Resulting in \( k \approx 833.33 \) when rounded.

These steps demonstrate how algebraic manipulation enables us to deduce unknown values, bringing clarity to complex mathematical relationships.