Problem 32 - Algebra A Combined Function

Question

For each statement, find the constant of variation and the variation equation. \(y\) varies directly as the cube of \(x ; y=32\) when \(x=4\)

Step-by-Step Solution

Answer

The constant of variation is \( \frac{1}{2} \) and the equation is \( y = \frac{1}{2}x^3 \).

Step 1: Identify the Type of Variation

In this problem, the statement 'y varies directly as the cube of x' signifies a direct variation where the formula can be expressed as \( y = kx^3 \), where \( k \) is the constant of variation.

Step 2: Substitute the Given Values

Substitute \( y = 32 \) and \( x = 4 \) into the variation equation \( y = kx^3 \) to find the constant \( k \). This gives us the equation \( 32 = k(4)^3 \).

Step 3: Solve for the Constant of Variation

Calculate \( 4^3 \), which is \( 64 \), and substitute it back into the equation to get \( 32 = 64k \). Solving for \( k \) gives us \( k = \frac{32}{64} = \frac{1}{2} \).

Step 4: Write the Variation Equation

Using the value of \( k \) found, substitute \( k = \frac{1}{2} \) back into the variation equation form \( y = kx^3 \). The variation equation is \( y = \frac{1}{2}x^3 \).

Key Concepts

Direct Variation

In algebra, direct variation is a relationship between two variables where one variable is a constant multiple of the other. When we say "y varies directly as x," it means that increasing x will result in a proportional increase in y. This relationship can be expressed in the form "y = kx," where k is a non-zero constant known as the "constant of variation."

• Direct variation is linear if the power of x is 1.
• The graph of a direct variation equation is a straight line passing through the origin if it is linear.
• It illustrates a proportional relationship.

In our problem, we are dealing with a special kind of direct variation where y varies with the cube of x. This is expressed with the equation "y = kx^3."

Constant of Variation

The "constant of variation" is a vital component in direct variation equations because it determines the ratio between y and x. Knowing this value allows us to understand how much y changes with each unit change in x. When given specific values for x and y, we can solve for this constant using the formula mentioned earlier.
For the exercise, we were given y = 32 when x = 4, and y varies directly as x cubed. By substituting these values into the equation "y = kx^3," we can find:

• First calculate the cube of x: \((4)^3 = 64\).
• Then solve for k using: \(32 = 64k \).
• Thus, \(k = \frac{32}{64} = \frac{1}{2}\).

This tells us that the constant of variation k is \(\frac{1}{2}\).

Cube Function

The cube function in algebra is a function that involves raising a quantity to the power of three, represented as \(x^3\). This function shows how changes in x affect cubic values, which are common in volume calculations. Understanding cube functions is crucial when dealing with higher-degree polynomials.
In the case of direct variation with a cube function, modifying x results in changes to y that are a cubic function of that modification. Thus, when x is doubled, the effect on y is to increase by a factor of \(2^3 = 8\). Understanding the cube function helps visualize how significant even small changes in x can be in affecting y's value.

Variation Equation

The "variation equation" is the form that reveals the relationship between y and x in direct variation problems. It includes both the constant of variation and the variable x raised to some power. Determining this equation is crucial as it allows you to predict y for any given value of x within the context of the problem.
Applying the known constant of variation and the form of variation, we established the equation as "y = \frac{1}{2} x^3" for this particular instance. This equation demonstrates how y changes in relation to the cube of x, scaled by the constant \(\frac{1}{2}\).

• It serves as the mathematical model for making predictions.
• This expression creates a scenario where calculations of y are easily performed for different values of x.

A good grasp of variation equations helps in understanding complex relationships and solving similar algebra problems efficiently.