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Chapter 6: Problem 32

For each statement, find the constant of variation and the variation equation. See Examples 5 and 6. \(y\) varies directly as the cube of \(x ; y=32\) when \(x=4\)

Short Answer

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

Step by step solution

01

Identify the Variation Type

In this problem, we are dealing with direct variation. Specifically, it mentions that \( y \) varies directly as the cube of \( x \). This means that \( y = k \cdot x^3 \), where \( k \) is the constant of variation.
02

Substitute Known Values

We can substitute the given values into the equation to find the constant \( k \). It's given that \( y = 32 \) when \( x = 4 \). Substitute these into the direct variation formula: \( 32 = k \cdot 4^3 \).
03

Solve for the Constant of Variation \( k \)

Simplify and solve for \( k \). First, evaluate \( 4^3 \): \( 4^3 = 64 \).Thus, the equation becomes \( 32 = k \times 64 \).Divide both sides by 64: \( k = \frac{32}{64} = \frac{1}{2} \).
04

Write the Variation Equation

Now that we have found \( k = \frac{1}{2} \), we can write the variation equation. According to the problem statement, \( y \) varies directly as the cube of \( x \). Thus, the equation is: \( y = \frac{1}{2}x^3 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

constant of variation
In direct variation problems, the constant of variation, denoted by \( k \), is a crucial component. This constant is what links two variables together. When \( y \) varies directly as \( x^3 \), it means there is a consistent ratio between \( y \) and \( x^3 \). In simpler terms, for each pair of \( y \) and \( x \), the result of dividing \( y \) by \( x^3 \) is always the same.
  • It tells us how one variable changes in relation to another.
  • It's found by dividing the known outcome (\( y \)) by the substitution result of \( x \) (cubed for our problem).
  • In our example, rather than just quoting \( y \) as \( k \cdot x^3 \), we specifically identified \( k = \frac{1}{2} \).
Once identified, \( k \) can be used to predict other values within the equation. Understanding \( k \) provides a deeper understanding of the relationship between the variables in a variation problem.
variation equation
The variation equation is the mathematical representation of the relationship between variables in a direct variation problem. In these equations, one variable is expressed as a constant multiplied by a function of the other variable. Let's break it down:
  • The equation comes from identifying the type of variation. For example, direct variation, where \( y = kx^3 \), indicates that \( y \) directly depends on the cube of \( x \).
  • By substituting \( k \) (the constant of variation), the equation becomes complete and functional.
  • It allows you to predict the outcome of \( y \) for any value of \( x \) you choose.
In the given problem, after solving for \( k \), the variation equation was determined to be \( y = \frac{1}{2} x^3 \). With this equation, students can substitute different values for \( x \) and calculate the corresponding value of \( y \). This formula highlights the power and simplicity of understanding relationships in algebra.
cube of x
When working with direct variations involving the "cube of \( x \)," it's essential to grasp what cubing a number entails. Cubing \( x \) means multiplying \( x \) by itself twice more, transitioning from 1D to 3D 鈥� basically, calculating \( x \times x \times x \).
  • The operation of cubing is foundational in determining how \( x \) relates to \( y \), especially in higher order variations.
  • The solution's step of cubing \( 4 \) to get \( 64 \) demonstrates this precise calculation: \( 4^3 = 4 \times 4 \times 4 = 64 \).
  • It illustrates how slight changes in \( x \) have a potentially large impact on the value of \( y \), as observed when using the variation equation.
Understanding cubing is powerful, especially in direct variation problems, because it emphasizes not just the magnitude increase with \( x \), but also how it profoundly affects \( y \). Knowing this enables students to anticipate and calculate outcomes effectively in these types of problems.

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Most popular questions from this chapter

Write an equation to describe each variation. Use k for the constant of proportionality. See Examples 1 through 7. \(y\) varies directly as \(x\) and inversely as \(p^{2}\)

The horsepower that can be safely transmitted to a shaft varies jointly as the shaft's angular speed of rotation (in revolutions per minute) and the cube of its diameter. A 2 -inch shaft making 120 revolutions per minute safely transmits 40 horsepower. Find how much horsepower can be safely transmitted by a 3 -inch shaft making 80 revolutions per minute.

Divide. $$ \left(x^{6} y^{6}-x^{3} y^{3} z+7 x^{3} y\right) \div\left(-7 y z^{2}\right) $$

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Find each square root. See Section 1.3. $$ \sqrt{\frac{4}{9}} $$
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