Problem 23 - College Algebra

Question

For the following exercises, write an equation describing the relationship of the given variables. \(y\) varies jointly as \(x\) and \(z\) and inversely as the square root of \(w\) and the square of \(t\). When \(x=3, z=1, w=25\), and \(t=2,\) then \(y=6\).

Step-by-Step Solution

Answer

The equation is \(y = \frac{40xz}{\sqrt{w} \, t^2}\).

Step 1: Understanding Joint Variation

Joint variation implies that a variable, in this case, \(y\), is proportional to the product of two or more other variables. Here, \(y\) varies jointly with \(x\) and \(z\). Thus the initial form of the equation is \(y = kx z\), where \(k\) is a constant.

Step 2: Incorporating Inverse Variation

Inverse variation means the variable \(y\) varies inversely with other variables, which in this situation are \(\sqrt{w}\) and \(t^2\). Hence, the expression \(y\) becomes \(y = \frac{kxz}{\sqrt{w} \, t^2}\).

Step 3: Substitute Known Values to Find Constant

We are given values \(x=3\), \(z=1\), \(w=25\), \(t=2\), and \(y=6\). Substitute these into the equation to determine \(k\): \[6 = \frac{k \times 3 \times 1}{\sqrt{25} \times 2^2}\].

Step 4: Simplify and Solve for Constant

Simplify the equation: \(\sqrt{25} = 5\) and \(2^2 = 4\), allowing the equation \(6 = \frac{3k}{5 \times 4}\). This simplifies to \(6 = \frac{3k}{20}\). Solve for \(k\) by multiplying both sides by 20 and dividing by 3: \(k = \frac{6 \times 20}{3} = 40\).

Step 5: Write the Final Equation

Using the constant \(k\) from Step 4, we substitute back to form the final equation: \[y = \frac{40xz}{\sqrt{w} \, t^2}\].

Key Concepts

Joint Variation

Joint variation occurs when one variable is directly proportional to the product of two or more other variables. In simple terms, this means that as one set of variables increase, the variable that varies jointly will also increase, assuming all other variables remain constant.
For instance, consider the variables in our exercise: if both variables \(x\) and \(z\) influence \(y\) together, we express this relationship as \(y = kx z\). Here, \(k\) is a constant that adjusts the relationship to fit specific values. In essence:

• If \(x\) or \(z\) increases, \(y\) will also increase, assuming everything else stays the same.
• If \(x\) or \(z\) decreases, \(y\) will decrease as well.

This idea of joint variation helps us understand how interconnected multiple variables can be, and it’s a common concept in mathematical modeling and algebraic analysis.

Inverse Variation

Inverse variation describes a situation where one variable increases while another decreases at a consistent rate. The product of the two variables remains constant.
In mathematical terms, if \(y\) varies inversely with another variable, like \(w\) or \(t\) in this instance, \(y\) is inversely proportional to these variables. This relationship can be expressed with an equation such as \(y = \frac{k}{\text{(another variable)}}\).
For our specific problem, \(y\) varies inversely with \(\sqrt{w}\) and \(t^2\). So, the equation becomes \(y = \frac{k}{\sqrt{w} \cdot t^2}\). Here’s what that means:

• If \(w\) or \(t\) increases, \(y\) decreases.
• If \(w\) or \(t\) decreases, \(y\) increases.

Using inverse variation, we can model how one aspect of a system might diminish while another grows, maintaining an underlying balance.

Mathematical Modeling

Mathematical modeling involves creating a mathematical representation of a real-world scenario using mathematical expressions. It's like building a bridge between abstract math and practical reality.
This exercise is a perfect example of mathematical modeling. We take the idea that \(y\) depends on multiple variables, some jointly and some inversely, and build an equation that captures that entire relationship: \(y = \frac{kxz}{\sqrt{w} t^2}\).
Here’s why mathematical modeling is important:

• It allows us to predict outcomes under different conditions.
• It helps in understanding complex systems by simplifying their components into manageable equations.
• It can solve practical problems in engineering, biology, economics, and more.

By substituting known values for \(x\), \(z\), \(w\), \(t\), and \(y\), we determine the constant \(k\), enabling us to use this model in different scenarios.

Algebraic Expressions

An algebraic expression is a combination of variables, numbers, and arithmetic operations that represents a specific relationship or value.
In our case, the resulting expression, \(y = \frac{40xz}{\sqrt{w} \, t^2}\), is not just a random assortment of symbols—it captures the relationship between \(y\), \(x\), \(z\), \(w\), and \(t\) defined by joint and inverse variation principles. Here’s why algebraic expressions are powerful:

• They simplify complex relationships, making them easier to work with and solve.
• They allow us to manipulate equations to find unknown values.
• They form the basis for much of algebra and calculus, where solving for variables is key.

The challenge is often in translating real-world descriptions into algebraic expressions. But once mastered, this skill allows us to work out solutions efficiently, as demonstrated in the exercise.