Problem 86 - A Graphical Approach to Precalculus with Limits

Question

Solve each problem. If \(y\) varies directly with \(x\) and inversely with \(m^{2}\) and \(r^{2},\) and if \(y=\frac{5}{3}\) when \(x=1, m=2,\) and \(r=3,\) find \(y\) if \(x=3\) \(m=1,\) and \(r=8\)

Step-by-Step Solution

Answer

\( y = \frac{45}{16} \)

Step 1: Understanding Direct and Inverse Variation

The problem states that variable \( y \) varies directly with \( x \) and inversely with \( m^2 \) and \( r^2 \). Direct variation means \( y \) is proportional to \( x \), and inverse variation means \( y \) is inversely proportional to the square of \( m \) and \( r \). This can be written as: \( y = k \frac{x}{m^2 r^2} \), where \( k \) is the constant of variation.

Step 2: Finding the Constant of Variation

We use the provided values \( y=\frac{5}{3} \), \( x=1 \), \( m=2 \), and \( r=3 \) to find the constant \( k \). Substitute these into the variation equation: \[\frac{5}{3} = k \frac{1}{2^2 \cdot 3^2}\]Solving for \( k \), we get:\[\frac{5}{3} = k \frac{1}{36} \rightarrow k = \frac{5}{3} \times 36 = 60\]

Step 3: Substituting New Values to Find y

Now that we have \( k = 60 \), use the new values \( x=3 \), \( m=1 \), and \( r=8 \) in the equation: \[y = 60 \frac{3}{1^2 \cdot 8^2}\]Simplify and solve the expression: \[y = 60 \frac{3}{64} = \frac{180}{64} = \frac{45}{16}\]

Step 4: Simplifying the Fraction

The fraction \( \frac{45}{16} \) is already in its simplest form, as 45 and 16 have no common factors other than 1. Therefore, \( y = \frac{45}{16} \).

Key Concepts

Constant of Variation

The concept of a constant of variation is crucial in understanding how variables relate to each other in direct and inverse variation scenarios. When a relationship is described as a direct variation, it implies that one variable increases as the other does, maintaining a constant ratio. In an inverse variation, one variable increases as the other decreases, with again a constant but different ratio.

In mathematical terms, once we determine the constant of variation, denoted by \( k \), we have a foundational tool to relate the variables consistently, no matter their current values. For the given problem, the relation is expressed as:

• Directly with \( x \)
• Inversely with \( m^2 \) and \( r^2 \)
• Equation: \( y = k \frac{x}{m^2 r^2} \)

Finding \( k \) involves taking known values of \( y, x, m, \) and \( r \) and substituting them into the equation to solve for \( k \). If conditions change, we use the same value of \( k \) to find new variable results. In this problem, \( k \) was calculated as 60, acting as the pivotal anchor between the variables.

Proportionality

Proportionality describes a consistent relationship between variables where a change in one leads to a predictable change in another. This is foundational in mathematics, especially with direct and inverse variations, as they are types of proportional relationships. In direct variation, the relationship can be penned down as \( y = kx \), where \( k \) represents a constant and \( y \) increases with \( x \).

Inverse proportionality is a bit different. When \( y \) is inversely proportional to the square of \( m \) and \( r \), represented here as \( y = \frac{k}{m^2 r^2} \), changes in \( m \) or \( r \) critically affect \( y \).
Therefore:

• If \( m \) or \( r \) increase, given \( y \) is inversely proportional to them, \( y \) actually decreases.
• Conversely, if \( m \) or \( r \) decrease, \( y \) increases.

In our exercise, understanding this relationship enabled us to manipulate values of \( m \) and \( r \) to find the appropriate \( y \), reinforcing the interdependence of these variables within their proportionality constraints.

Rational Expressions

Rational expressions involve fractions where the numerator and denominator are polynomial expressions. In our exercise, simplifying rational expressions was essential after determining new values for \( x, m, \) and \( r \). These expressions are key in expressing relationships involving varying quantities that are not linear but whose behavior can be modeled with fractions comprising integers or expressions.

In the given problem, once \( k \) was known,

• The expression for \( y \) becomes \( 60 \frac{3}{1^2 \cdot 8^2} \).
• Simplifying: \( 60 \frac{3}{64} = \frac{180}{64} = \frac{45}{16} \).
• This is a rational expression reduced to its simplest form.

This step ensures that \( y \)'s value is accurately derived and expressed, underscoring the utility and importance of handling rational expressions in variations and proportion problems.