Problem 38 - Elementary Algebra

Question

Construct a mathematical model given the following. \(y\) varies directly as the square root of \(x\) and inversely as the square of \(z,\) where \(y=15\) when \(x=25\) and \(z=2\).

Step-by-Step Solution

Answer

The model is \(y = 12 \frac{\sqrt{x}}{z^2}\).

Step 1: Identify the Relationship

The problem states that \(y\) varies directly as the square root of \(x\) and inversely as the square of \(z\). This can be expressed mathematically as \(y = k \frac{\sqrt{x}}{z^2}\), where \(k\) is the constant of proportionality.

Step 2: Substitute Given Values to Find \(k\)

We are given \(y=15\), \(x=25\), and \(z=2\). Substitute these values into the equation: \[15 = k \frac{\sqrt{25}}{2^2}\] Simplify to get: \[15 = k \frac{5}{4}\]

Step 3: Solve for \(k\)

To find the value of \(k\), multiply both sides of the equation by \(\frac{4}{5}\): \[k = 15 \times \frac{4}{5} = 12\] So, \(k = 12\).

Step 4: Formulate the Mathematical Model

Now that we know \(k\), substitute it back into the original equation: \[y = 12 \frac{\sqrt{x}}{z^2}\]. This is the mathematical model that describes the relationship between \(y\), \(x\), and \(z\) given the stated conditions.

Key Concepts

Direct Variation

Direct variation is a concept where one quantity increases or decreases in direct proportion to another. When two variables are in direct variation, you can express their relationship with the format \( y = kx \), where \( k \) is the proportionality constant, and \( x \) is the independent variable. In our problem, the case is slightly modified, as \( y \) varies directly with the square root of \( x \).

This means that as the square root of \( x \) increases, \( y \) increases proportionally, provided all other factors remain constant.

This relationship in our exercise is shown as \( y = k \sqrt{x} \), emphasizing that if \( x \) is squared, its impact on \( y \) is determined by rooting the result first.
This direct variation is a fundamental building block in understanding complex relationships in algebraic modeling.

Inverse Variation

In contrast to direct variation, inverse variation describes a situation where one quantity increases as another quantity decreases, and vice versa. The general expression for inverse variation is \( y = \frac{k}{x} \), where \( x \) is the independent variable.

In our particular problem, \( y \) varies inversely with the square of \( z \). This is depicted by \( y = k \frac{1}{z^2} \).

Essentially, this relationship suggests that as \( z \) becomes larger, \( y \) becomes smaller because the square of \( z \) grows at a faster rate.
Inverse variation is crucial for modeling scenarios where a decrease in one parameter leads to an increase in another, a common occurrence in physics and engineering.

Proportionality Constant

The proportionality constant, \( k \), acts as a bridge between direct and inverse variations by quantifying how one variable affects another. It can be termed as a scaling factor derived from known conditions.

In our exercise, determining \( k \) involved substituting the given values of \( y = 15 \), \( x = 25 \), and \( z = 2 \) into the relationship \( y = k \frac{\sqrt{x}}{z^2} \).

This calculation yielded \( k = 12 \). The proportionality constant enables us to accurately describe how alterations in \( x \) and \( z \) will affect \( y \) in the real-world model.
Whenever dealing with mathematical models, the value of \( k \) gives depth and scale to the relationship, dictated by the real-world context.

Algebraic Equations

Algebraic equations serve as the backbone of mathematical modeling. Through them, we can express and resolve complex relationships between variables in a structural form. Algebra, with its symbols and rules, provides a universal language to describe these relationships.

The equation \( y = k \frac{\sqrt{x}}{z^2} \) combines concepts of both direct and inverse variation into a single cohesive model.

• The term \( \sqrt{x} \) accounts for the direct variation.
• While \( \frac{1}{z^2} \) represents the inverse variation component.

Such equations are powerful tools for prediction and understanding, regularly applied in sciences to formulate theories or describe behaviors.
In solving for \( k \) or predicting outcomes in new situations, algebra provides clarity and builds understanding of dynamic systems.