Problem 13 - Algebra 2

Question

Suppose that \(x\) and \(y\) vary inversely. Write a function that models each inverse variation and find \(y\) when \(x=10 .\) $$ x=20 \text { when } y=5 $$

Step-by-Step Solution

Answer

The function modelling the inverse variation is \(y=\frac{100}{x}\). The value of \(y\) when \(x=10\) is 10.

Step 1: Identify the constant of variation

Let's find the constant of variation \(k\) using the given values of \(x\) and \(y\). According to the relation \(xy=k\), we substitute \(x=20\) and \(y=5\) which gives us \(k=(20)(5)=100\).

Step 2: Write the inverse variation function

Now that we have calculated the constant \(k\), the inverse variation function can be written as \(y=\frac{k}{x}\), hence \(y=\frac{100}{x}\).

Step 3: Find the value of \(y\) when \(x=10\)

Finally, to find the value of \(y\) when \(x=10\), we substitute \(x=10\) into the function \(y=\frac{100}{x}\), resulting in \(y=\frac{100}{10}=10\).

Key Concepts

Constant of Variation

In problems involving inverse variation, identifying the constant of variation, denoted by \(k\), is crucial. In an inverse relationship between two variables, when one variable increases, the other decreases proportionally, and vice versa. This relationship is expressed as \(xy = k\). Here, \(k\) remains constant for all pairs of \(x\) and \(y\) that satisfy the relationship.
To find \(k\), we substitute the known values of \(x\) and \(y\) into the equation. For example, given \(x = 20\) and \(y = 5\), we calculate \(k = 20 \times 5 = 100\).

• **Directly calculated by multiplying given values of \(x\) and \(y\)
• Essential for forming the inverse function

Understanding the constant \(k\) helps in predicting other values of \(x\) and \(y\) while maintaining their inverse relationship.

Inverse Function

An inverse function in the context of variation is formulated once the constant of variation \(k\) is found. The inverse variation function describes how \(y\) changes with \(x\) according to the formula \(y = \frac{k}{x}\).
This function reflects the inverse relationship where increasing \(x\) leads to a decrease in \(y\), keeping the product \(xy = k\) consistent.

• Transforms known relationships into workable models
• Can be used to compute unknown variable values

The ability to express this relationship as a function is powerful for modeling real-world scenarios where variables interact inversely.

Solve for Variable

Solving for a variable in an inverse variation setup involves isolating the desired variable using the inverse function. For example, finding \(y\) when \(x\) is known involves substituting \(x\) into the inverse function \(y = \frac{k}{x}\).
Let's say \(k = 100\), and we are given \(x = 10\). We substitute to find \(y = \frac{100}{10} = 10\).

• Substitution into the inverse function simplifies the problem
• Useful in determining effects of changes in one variable

This process allows students to determine unknown variables efficiently based on known values and their inverse relationships.

Algebraic Modeling

Algebraic modeling involves creating mathematical expressions or equations to represent real-world problems. In the context of inverse variation, algebraic models like \(y = \frac{k}{x}\) help map out relationships between variables.
These models are extremely helpful because they allow us to predict how changes in one variable affect another, based purely on their mathematical relationship.

• Simplifies complex relationships for better understanding
• Enables practical application in predicting outcomes

Such modeling is a foundational skill that aids students in both academic and real-life problem-solving scenarios. Used correctly, it can show a clear picture of how variables interact through inverse relationships.