Problem 37 - College Algebra with Corequisite Support

Question

For the following exercises, use the given information to find the unknown value. \(y\) varies jointly as the square of \(x\) and the square root of \(z\). When \(x=2\) and \(z=9,\) then \(y=24\). Find \(y\) when \(x=3\) and \(z=25\).

Step-by-Step Solution

Answer

When \( x = 3 \) and \( z = 25 \), \( y = 90 \).

Step 1: Understand Joint Variation

The problem states that y varies jointly as the square of x and the square root of z. This means we can express y in the form: \( y = k \cdot x^2 \cdot \sqrt{z} \), where k is the constant of variation.

Step 2: Find Constant of Variation (k)

We use the given values \( x = 2 \), \( z = 9 \), and \( y = 24 \) to find the constant k.Substitute into the equation: \( 24 = k \cdot 2^2 \cdot \sqrt{9} \).Simplify:- \( 2^2 = 4 \)- \( \sqrt{9} = 3 \).Thus, \( 24 = k \cdot 4 \cdot 3 \).So, \( 24 = 12k \).Solve for k: \( k = \frac{24}{12} = 2 \).

Step 3: Substitute New Values and Solve for y

Now, use the constant \( k = 2 \) to find y when \( x = 3 \) and \( z = 25 \).Substitute into the joint variation equation: \( y = 2 \cdot 3^2 \cdot \sqrt{25} \).Simplify:- \( 3^2 = 9 \)- \( \sqrt{25} = 5 \).Thus, \( y = 2 \cdot 9 \cdot 5 \).Calculate the result: \( y = 90 \).

Key Concepts

Constant of Variation

In mathematics, when we talk about joint variation, we are often referring to how one quantity changes in relation to others. The constant of variation, denoted by \( k \), is pivotal in these equations. It acts as a constant multiplier that scales the relationship between variables. In the equation \( y = k \cdot x^2 \cdot \sqrt{z} \), the term \( k \) remains unchanged when you know the original set of values. Once identified with known values, \( k \) will help find unknown values in similar scenarios.To determine the constant \( k \), you simply use the known values and solve the equation. For example, with the provided exercise, substituting \( x = 2 \), \( z = 9 \), and \( y = 24 \), you get:

• \( 24 = k \cdot 4 \cdot 3 \)
• \( 24 = 12k \)
• Solving gives \( k = \frac{24}{12} = 2 \)

This constant helps simplify the original equation for future calculations.

Square Root

The square root is a fundamental math concept represented by the symbol \( \sqrt{} \). It is the value that, when multiplied by itself, gives the original number. For instance, \( \sqrt{9} = 3 \), as \( 3 \times 3 = 9 \). In variations like this exercise, the square root might show relationships between quantities. When \( y \) varies as the square root of another variable \( z \), it highlights an interesting dynamic where growth may be slower due to the squaring effect on \( x \) but moderated by taking the square root of \( z \). For the given exercises, calculating the square root helps in simplifying the joint variation formula:

• \( \sqrt{9} \) simplified to 3
• For \( z = 25 \), \( \sqrt{25} = 5 \)

Algebraic Expressions

Algebraic expressions are combinations of constants, variables, and operations. In this exercise, the algebraic expression \( y = k \cdot x^2 \cdot \sqrt{z} \) plays a key role. This expression ties together the relationship between \( y \), \( x \), and \( z \).Understanding algebraic expressions is critical: they allow us to represent relationships clearly and perform calculations. Solving these requires understanding operations like multiplication and applying functions such as squares and square roots.Breaking down the given expression:

• \( x^2 \) indicates \( x \) is squared, affecting \( y \) exponentially depending on \( x \)'s value.
• The \( \sqrt{z} \) modulates this effect, influencing \( y \) differently as \( z \) changes.

This understanding assists in predicting how \( y \) will vary, given changes in \( x \) and \( z \). Simplified algebraic manipulation like this makes joint variation problems easier to handle.