# Inverse and Joint Variation

### Learning Outcomes

- Solve an Inverse variation problem.
- Write a formula for an inversely proportional relationship.

[latex]d[/latex], depth | [latex]T=\frac{\text{14,000}}{d}[/latex] | Interpretation |
---|---|---|

500 ft | [latex]\frac{14,000}{500}=28[/latex] | At a depth of 500 ft, the water temperature is 28° F. |

350 ft | [latex]\frac{14,000}{350}=40[/latex] | At a depth of 350 ft, the water temperature is 40° F. |

250 ft | [latex]\frac{14,000}{250}=56[/latex] | At a depth of 250 ft, the water temperature is 56° F. |

**inversely proportional**and each term

**varies inversely**with the other. Inversely proportional relationships are also called

**inverse variations**. For our example, the graph depicts the

**inverse variation**. We say the water temperature varies inversely with the depth of the water because, as the depth increases, the temperature decreases. The formula [latex]y=\dfrac{k}{x}[/latex] for inverse variation in this case uses [latex]k=14,000[/latex].

### A General Note: Inverse Variation

If [latex]x[/latex] and [latex]y[/latex] are related by an equation of the form [latex-display]y=\dfrac{k}{{x}^{n}}[/latex-display] where [latex]k[/latex] is a nonzero constant, then we say that [latex]y[/latex]**varies inversely**with the [latex]n[/latex]th power of [latex]x[/latex]. In

**inversely proportional**relationships, or

**inverse variations**, there is a constant multiple [latex]k={x}^{n}y[/latex].

### isolating the constant of variation

To isolate the constant of variation in an inverse variation, use the properties of equality to solve the equation for [latex]k[/latex]. [latex-display]y=\dfrac{k}{{x}^{n}}[/latex-display] Isolate [latex]k[/latex] using algebra. [latex-display]yx^n=k[/latex-display]### Example: Writing a Formula for an Inversely Proportional Relationship

A tourist plans to drive 100 miles. Find a formula for the time the trip will take as a function of the speed the tourist drives.Answer: Recall that multiplying speed by time gives distance. If we let [latex]t[/latex] represent the drive time in hours, and [latex]v[/latex] represent the velocity (speed or rate) at which the tourist drives, then [latex]vt=[/latex] distance. Because the distance is fixed at 100 miles, [latex]vt=100[/latex]. Solving this relationship for the time gives us our function.

[latex]\begin{align}t\left(v\right)&=\dfrac{100}{v} \\[1mm] &=100{v}^{-1} \end{align}[/latex]

We can see that the constant of variation is 100 and, although we can write the relationship using the negative exponent, it is more common to see it written as a fraction.### How To: Given a description of an inverse 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 multiply [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 an Inverse Variation Problem

A quantity [latex]y[/latex] varies inversely 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 inverse variation with a cube is [latex]y=\dfrac{k}{{x}^{3}}[/latex]. The constant can be found by multiplying [latex]y[/latex] by the cube of [latex]x[/latex].

[latex]\begin{align}k&={x}^{3}y \\[1mm] &={2}^{3}\cdot 25 \\[1mm] &=200 \end{align}[/latex]

Now we use the constant to write an equation that represents this relationship.[latex]\begin{align}y&=\dfrac{k}{{x}^{3}},\hspace{2mm}k=200 \\[1mm] y&=\dfrac{200}{{x}^{3}} \end{align}[/latex]

Substitute [latex]x=6[/latex] and solve for [latex]y[/latex].[latex]\begin{align}y&=\dfrac{200}{{6}^{3}} \\[1mm] &=\dfrac{25}{27} \end{align}[/latex]

#### Analysis of the Solution

The graph of this equation is a rational function.### Try It

A quantity [latex]y[/latex] varies inversely with the square of [latex]x[/latex]. If [latex]y=8[/latex] when [latex]x=3[/latex], find [latex]y[/latex] when [latex]x[/latex] is 4.Answer: [latex-display]\dfrac{9}{2}[/latex-display]

[embed]## Joint Variation

Many situations are more complicated than a basic direct variation or inverse variation model. One variable often depends on multiple other variables. When a variable is dependent on the product or quotient of two or more variables, this is called**joint variation**. For example, the cost of busing students for each school trip varies with the number of students attending and the distance from the school. The variable [latex]c[/latex], cost, varies jointly with the number of students, [latex]n[/latex], and the distance, [latex]d[/latex].

### A General Note: Joint Variation

Joint variation occurs when a variable varies directly or inversely with multiple variables. For instance, if [latex]x[/latex] varies directly with both [latex]y[/latex] and [latex]z[/latex], we have [latex]x=kyz[/latex]. If [latex]x[/latex] varies directly with [latex]y[/latex] and inversely with [latex]z[/latex], we have [latex]x=\dfrac{ky}{z}[/latex]. Notice that we only use one constant in a joint variation equation.### isolating the constant of variation

To isolate the constant of variation in a joint variation, use the properties of equality to solve the equation for [latex]k[/latex]. [latex-display]x=kyz[/latex-display] Isolate [latex]k[/latex] using algebra. [latex-display]\dfrac{x}{yz}=k[/latex-display]### Example: Solving Problems Involving Joint Variation

A quantity [latex]x[/latex] varies directly with the square of [latex]y[/latex] and inversely with the cube root of [latex]z[/latex]. If [latex]x=6[/latex] when [latex]y=2[/latex] and [latex]z=8[/latex], find [latex]x[/latex] when [latex]y=1[/latex] and [latex]z=27[/latex].Answer: Begin by writing an equation to show the relationship between the variables.

[latex]x=\dfrac{k{y}^{2}}{\sqrt[3]{z}}[/latex]

Substitute [latex]x=6[/latex], [latex]y=2[/latex], and [latex]z=8[/latex] to find the value of the constant [latex]k[/latex].[latex]\begin{align}6&=\dfrac{k{2}^{2}}{\sqrt[3]{8}} \\[1mm] 6&=\dfrac{4k}{2} \\[1mm] 3&=k \end{align}[/latex]

Now we can substitute the value of the constant into the equation for the relationship.[latex]x=\dfrac{3{y}^{2}}{\sqrt[3]{z}}[/latex]

To find [latex]x[/latex] when [latex]y=1[/latex] and [latex]z=27[/latex], we will substitute values for [latex]y[/latex] and [latex]z[/latex] into our equation.[latex]\begin{align}x&=\dfrac{3{\left(1\right)}^{2}}{\sqrt[3]{27}} \\[1mm] &=1 \end{align}[/latex]

### Try It

[latex]x[/latex] varies directly with the square of [latex]y[/latex] and inversely with [latex]z[/latex]. If [latex]x=40[/latex] when [latex]y=4[/latex] and [latex]z=2[/latex], find [latex]x[/latex] when [latex]y=10[/latex] and [latex]z=25[/latex].Answer: [latex-display]x=20[/latex-display]

[embed]## Licenses & Attributions

### CC licensed content, Original

- Revision and Adaptation.
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### CC licensed content, Shared previously

- Question ID 91393,91394.
**Authored by:**Jenck,Michael (for Lumen Learning).**License:**CC BY: Attribution.**License terms:**IMathAS Community License CC-BY + GPL. - College Algebra.
**Provided by:**OpenStax**Authored by:**Abramson, Jay et al..**Located at:**https://openstax.org/books/college-algebra/pages/1-introduction-to-prerequisites.**License:**CC BY: Attribution.**License terms:**Download for free at http://cnx.org/contents/[email protected]. - Inverse Variation.
**Authored by:**James Sousa (Mathispower4u.com) for Lumen Learning.**License:**CC BY: Attribution. - Joint Variation: Determine the Variation Constant (Volume of a Cone).
**Provided by:**Joint Variation: Determine the Variation Constant (Volume of a Cone)**Authored by:**James Sousa (Mathispower4u.com) for Lumen Learning.**License:**CC BY: Attribution.