Problem 34 - Algebra A Combined Function

Question

For each statement, find the constant of variation and the variation equation. \(y\) varies directly as the square root of \(x ; y=2.1\) when \(x=4\) \(x=9\)

Step-by-Step Solution

Answer

The constant of variation is 1.05 and the variation equation is \(y = 1.05 \sqrt{x}\). At \(x=9\), \(y=3.15\).

Step 1: Identify the Formula

Since the problem states that "\(y\) varies directly as the square root of \(x\)", we can express this relationship using the formula \(y = k \sqrt{x}\) where \(k\) is the constant of variation.

Step 2: Substitute Known Values

We are given that \(y = 2.1\) when \(x = 4\). Substituting these values into the formula gives us \(2.1 = k \sqrt{4}\).

Step 3: Solve for Constant of Variation

Since \(\sqrt{4} = 2\), we can rewrite the equation as \(2.1 = k \cdot 2\). Divide both sides by 2 to find \(k\), which gives \(k = \frac{2.1}{2} = 1.05\). So, the constant of variation is 1.05.

Step 4: Write the Variation Equation

With the constant of variation \(k = 1.05\), we can write the variation equation as \(y = 1.05 \sqrt{x}\).

Step 5: Calculate \(y\) for New \(x\) Value

Using the variation equation \(y = 1.05 \sqrt{x}\) and the given \(x = 9\), calculate \(y\) by substituting \(x\) into the equation: \(y = 1.05 \sqrt{9}\).

Step 6: Compute the Result

Since \(\sqrt{9} = 3\), substitute it into the equation: \(y = 1.05 \times 3 = 3.15\). Thus, when \(x = 9\), \(y = 3.15\).

Key Concepts

Constant of Variation

In a direct variation scenario, the constant of variation is a crucial component. It is represented by the letter "k" in the equation. The constant essentially determines the strength and proportionality of the relationship between two variables.

When a quantity varies directly with another, the ratio between them remains constant.To find the constant of variation, you compare given known values of the variables. For instance, in the exercise, when it's stated that "\( y \) varies directly as the square root of \( x \)", the equation becomes \( y = k \sqrt{x} \).

By inserting the specific values into this equation, such as \( y = 2.1 \) when \( x = 4 \), you solve for \( k \), yielding a constant value.Here, using simple algebra to rearrange and solve the equation step by step allows you to find \( k = 1.05 \). This constant is vital as it helps to maintain the direct variation equation's integrity.

Square Root

The concept of a square root is pivotal in understanding this problem. A square root is a number which, when multiplied by itself, gives the original number.

For example, the square root of 4 is 2 because \( 2 \times 2 = 4 \). In mathematical terms, it is often represented by the radical symbol "\( \sqrt{} \)". Square roots are instrumental in variation problems, especially when the relationship involves quadratic terms.

In our problem, the equation \( y = k \sqrt{x} \) implies that \( y \) directly depends on \( \sqrt{x} \), integrating the non-linear element through the square root. To solve the given equation, understanding and calculating the square roots of the given \( x \)-values, like \( \sqrt{4} \) and \( \sqrt{9} \), is key. Using these calculations correctly facilitates deriving the correct \( y \) values in line with the variation equation.

Variation Equation

A variation equation is a mathematical representation of how one variable changes with respect to another. In direct variation, such as in our case, the equation takes the form \( y = k \sqrt{x} \), where the relationship is linear concerning \( \sqrt{x} \).

This relationship ensures that for every unit change in \( \sqrt{x} \), \( y \) changes proportionally based on the constant \( k \). Such equations are instrumental in modelling direct relationships in physics and various applied sciences.

In our example, finding the variation equation involved first identifying \( k \, (1.05) \) using known values. Once determined, this constant allows one to explore how \( y \) will vary for any given \( x \). By substituting new values of \( x \) into the equation, like \( x = 9 \), and using the known \( k \), one confidently calculates new \( y \) outcomes, verifying the proportional changes in a predictable manner.