Problem 37 - College Algebra

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

When a variable varies jointly, it means that it is proportional to the product of two or more variables. In this problem, the variable \(y\) varies jointly as the square of \(x\) and the square root of \(z\), which can be written as \(y = kx^2\sqrt{z}\), where \(k\) is the constant of proportionality.

Step 2: Find the Constant of Proportionality (k)

We are given that when \(x = 2\) and \(z = 9\), \(y = 24\). Plug these values into the equation to solve for \(k\):\[24 = k(2)^2\sqrt{9}\]\[24 = k \cdot 4 \cdot 3\]\[24 = 12k\]Now, solve for \(k\):\[k = \frac{24}{12} = 2\].

Step 3: Use the Constant (k) to Find y with New Values

Now that we have \(k = 2\), use it with the new values \(x = 3\) and \(z = 25\):\[y = kx^2\sqrt{z}\]\[y = 2(3)^2\sqrt{25}\]\[y = 2 \cdot 9 \cdot 5\]\[y = 90\].

Key Concepts

Constant of Proportionality

The concept of the constant of proportionality, often denoted as "\(k\)", is a crucial element in understanding joint variation. When we say a variable varies jointly, it means this variable changes concerning the product of two or more other variables, with the constant \(k\) acting as the tie between them.
In the context of our exercise, \(y\) varies jointly as the square of \(x\) and the square root of \(z\). This relationship is expressed by the formula:

• \[ y = k x^2 \sqrt{z} \]

Here, \(k\) is the constant that remains the same across different sets of conditions for \(x\) and \(z\).
When we were given that \(x = 2\), \(z = 9\), and \(y = 24\), we used these values to solve for \(k\). Substituting these into the joint variation equation allowed us to determine that \(k = 2\). All future calculations involving different \(x\) and \(z\) values can use this constant to find the corresponding \(y\). Understanding the role and calculation of this constant ensures consistency and accuracy in solving joint variation problems.

Square Root Function

The square root function is another fundamental concept involved in this joint variation problem. A square root, denoted as \(\sqrt{z}\), represents a value that when multiplied by itself gives the original number \(z\).
In our formula:

• \( y = k x^2 \sqrt{z} \)

The square root function specifically impacts how the variable \(z\) contributes to the joint variation of \(y\).
Unlike a linear function, a square root function increases at a decreasing rate, creating a curve shape rather than a straight line.
This function's outcome becomes significant while solving the exercise, particularly when we substitute different values for \(z\). When we calculated \(y\) for \(x = 3\) and \(z = 25\), notice that from \(\sqrt{z}\) we determine \(\sqrt{25} = 5\), which influences \(y\)'s solution. Understanding and manipulating the square root function helps in providing accurate solutions and grasping the problem's variable interactions.

Quadratic Function

Quadratic functions are expressed in the form of \(x^2\), forming a core element of the joint variation problem at hand. The quadratic component conveys how changes in \(x\) affect \(y\).
Let's revisit the joint variation equation:

• \( y = k x^2 \sqrt{z} \)

The term \(x^2\) implies that \(y\) grows quadratically with respect to \(x\). A quadratic function is nonlinear, generally represented by a parabola when plotted graphically. This means that as \(x\) increases or decreases, the effect on \(y\) intensifies because the squares of numbers grow significantly larger than the numbers themselves.
In the exercise, changing \(x\) from 2 to 3 which influences the value of \(y\) as the square of these values \((2^2 = 4 \text{ and } 3^2 = 9)\) are part of the calculations.
This highlights the potency of the quadratic function in scaling changes to \(y\) within the context of our joint variation problem, offering insights not just through theory but practical application as well.