Problem 30 - College Algebra with Corequisite Support

Question

For the following exercises, use the given information to find the unknown value. \(y\) varies inversely with the square of \(x\). When \(x=4,\) then \(y=3\). Find \(y\) when \(x=2\).

Step-by-Step Solution

Answer

When \(x=2\), \(y=12\).

Step 1: Understand the inverse variation

When we say that "\(y\) varies inversely with the square of \(x\)," it means that as \(x\) increases, \(y\) decreases in a squared way, and vice versa. This relationship can be expressed as: \[ y = \frac{k}{x^2} \] where \(k\) is a constant.

Step 2: Find the constant of variation

We know that when \(x = 4\), \(y = 3\). Plugging these values into the equation, we can find \(k\):\[ 3 = \frac{k}{4^2} \] This simplifies to: \[ k = 3 \times 16 = 48 \] So, the constant \(k\) is 48.

Step 3: Use the constant to find the new value of y

Using the constant \(k = 48\), we can now find \(y\) when \(x = 2\):\[ y = \frac{48}{2^2} \] Simplifying, we get:\[ y = \frac{48}{4} = 12 \] So when \(x = 2\), \(y\) is 12.

Key Concepts

constant of variation

The constant of variation, often represented by the symbol \(k\), plays a crucial role in equations of inverse variation. In an inverse variation, like the one where \(y\) varies inversely with the square of \(x\), the equation is structured as \( y = \frac{k}{x^2} \). Here, \(k\) is a fixed number that represents the constant proportionality in the relationship between \(x\) and \(y\). This constant remains unchanged throughout the different values of \(x\) and \(y\), provided their relationship doesn't change.

• For example, in our problem, when \(x = 4\) and \(y = 3\), we found \(k\) by solving the equation \(3 = \frac{k}{16}\), which gives us \(k = 48\).
• This constant allows us to compute the variable \(y\) for any other given value of \(x\). Knowing \(k\) simplifies the process of finding unknowns in inverse variation problems, as the relationship between the variables remains constant.

Understanding the constant of variation helps solidify your grasp on why these types of relationships behave consistently, making predictions possible as you encounter different values.

variable relationships

Variable relationships in mathematics describe how two or more variables interact with one another in an equation or function. In the context of inverse variation, such as \(y\) varying inversely with the square of \(x\), the relation is expressed through an equation \( y = \frac{k}{x^2} \). Here, \(x\) and \(y\) are variables whose values depend on each other.

• As \(x\) increases, \(y\) decreases because the value of \(x^2\) in the denominator causes the overall value of \(y\) to get smaller. Likewise, when \(x\) decreases, \(y\) increases.
• This reciprocal fluctuation is characteristic of inverse variation relationships, contrasting direct variation where both variables increase or decrease together.

By identifying the type of variation—such as inverse, direct, or even joint—you can predict how altering one variable will affect another.

It's crucial to note the nature of the relationship to correctly adjust the variables and solve equations like \( y = \frac{k}{x^2} \). Understanding this helps in modelling real-world situations where such relationships are present.

algebraic expression

Algebraic expressions are combinations of variables, constants, and operations that together form mathematical statements. In inverse variation problems, an algebraic expression serves as a tool to describe how the variables are linked. For example, the equation \( y = \frac{k}{x^2} \) is an algebraic expression that encompasses an inverse relationship in terms of \(y\) and \(x\).

• Such expressions enable the representation of complex variable interactions in a compact, symbolic manner. They allow you to plug in values easily to evaluate expressions and solve for unknowns.
• In our case, using the expression \( y = \frac{k}{x^2} \), we could substitute values of \(x\) and \(y\) to solve for the constant \(k\), leading us to the solution of the original problem with new values.

Understanding how to manipulate and work with algebraic expressions guides you through solving equations accurately. It forms the backbone of algebra, helping bridge abstract concepts with practical applications.

By practicing these skills, solving real-world mathematical problems becomes more intuitive and systematic.