Direct Variation
Learning Outcomes
- Solve a direct variation problem
- Use a constant of variation to describe the relationship between two variables
[latex]s[/latex], sales prices | [latex]e = 0.16s[/latex] | Interpretation |
---|---|---|
$4,600 | [latex]e=0.16(4,600)=736[/latex] | A sale of a $4,600 vehicle results in $736 earnings. |
$9,200 | [latex]e=0.16(9,200)=1,472[/latex] | A sale of a $9,200 vehicle results in $1472 earnings. |
$18,400 | [latex]e=0.16(18,400)=2,944[/latex] | A sale of a $18,400 vehicle results in $2944 earnings. |
A General Note: Direct Variation
If [latex]x[/latex] and [latex]y[/latex] are related by an equation of the form[latex]y=k{x}^{n}[/latex] then we say that the relationship is direct variation and [latex]y[/latex] varies directly with the [latex]n[/latex]th power of [latex]x[/latex]. In direct variation relationships, there is a nonzero constant ratio [latex]k=\dfrac{y}{{x}^{n}}[/latex], where [latex]k[/latex] is called the constant of variation, which help defines the relationship between the variables.
recall isolating a variable in a formula
We've learned to solve certain formulas for one of the variables. For example, in the formula that relates distance, rate, and time, [latex]d=rt[/latex], we can solve the equation for rate, [latex]r=\dfrac{d}{t}[/latex] or for time, [latex]t=\dfrac{d}{r}[/latex]. We say we are isolating the variable of interest in these cases. The same idea applies when solving a direct variation problem for the constant of variation, [latex]k[/latex]. Given a direct variation such as [latex]y=kx^n[/latex], we can isolate the constant of variation using the properties of equality to obtain [latex]\dfrac{y}{x^n}=k[/latex].How To: Given a description of a direct variation problem, solve for an unknown.
- Identify the input, [latex]x[/latex], and the output, [latex]y[/latex].
- Determine the constant of variation. You may need to divide [latex]y[/latex] by the specified power of [latex]x[/latex] to determine the constant of variation.
- Use the constant of variation to write an equation for the relationship.
- Substitute known values into the equation to find the unknown.
Example: Solving a Direct Variation Problem
The quantity [latex]y[/latex] varies directly with the cube of [latex]x[/latex]. If [latex]y=25[/latex] when [latex]x=2[/latex], find [latex]y[/latex] when [latex]x[/latex] is 6.Answer: The general formula for direct variation with a cube is [latex]y=k{x}^{3}[/latex]. The constant can be found by dividing [latex]y[/latex] by the cube of [latex]x[/latex].
[latex]\begin{align} k&=\dfrac{y}{{x}^{3}} \\[1mm] &=\dfrac{25}{{2}^{3}}\\[1mm] &=\dfrac{25}{8}\end{align}[/latex] Now use the constant to write an equation that represents this relationship.
[latex]y=\dfrac{25}{8}{x}^{3}[/latex] Substitute [latex]x=6[/latex] and solve for [latex]y[/latex].
[latex]\begin{align}y&=\dfrac{25}{8}{\left(6\right)}^{3} \\[1mm] &=675\hfill \end{align}[/latex]
Analysis of the Solution
The graph of this equation is a simple cubic, as shown below.Q & A
Do the graphs of all direct variation equations look like the example above? No. Direct variation equations are power functions—they may be linear, quadratic, cubic, quartic, radical, etc. But all of the graphs pass through [latex](0, 0)[/latex].Try It
The quantity [latex]y[/latex] varies directly with the square of [latex]y[/latex]. If [latex]y=24[/latex] when [latex]x=3[/latex], find [latex]y[/latex] when [latex]x[/latex] is 4.Answer: [latex-display]\dfrac{128}{3}[/latex-display]
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- Direct Variation Applications . Authored by: James Sousa (Mathispower4u.com) . License: CC BY: Attribution.