Learning Objectives
- Solve direct variation problems.
- Solve inverse variation problems.
A used-car company has just offered their best candidate, Nicole, a position in sales. The position offers 16% commission on her sales. Her earnings depend on the amount of her sales. For instance if she sells a vehicle for $4,600, she will earn $736. She wants to evaluate the offer, but she is not sure how. In this section we will look at relationships, such as this one, between earnings, sales, and commission rate.
Direct Variation
In the example above, Nicole’s earnings can be found by multiplying her sales by her commission. The formula [latex]e = 0.16s[/latex] tells us her earnings, [latex]e[/latex], come from the product of 0.16, her commission, and the sale price of the vehicle, [latex]s[/latex]. If we create a table, we observe that as the sales price increases, the earnings increase as well.
| s, sales prices | e = 0.16s | Interpretation |
|---|---|---|
| $4,600 | e = 0.16(4,600) = 736 | A sale of a $4,600 vehicle results in $736 earnings. |
| $9,200 | e = 0.16(9,200) = 1,472 | A sale of a $9,200 vehicle results in $1472 earnings. |
| $18,400 | e = 0.16(18,400) = 2,944 | A sale of a $18,400 vehicle results in $2944 earnings. |
Notice that earnings are a multiple of sales. As sales increase, earnings increase in a predictable way. Double the sales of the vehicle from $4,600 to $9,200, and we double the earnings from $736 to $1,472. As the input increases, the output increases as a multiple of the input. A relationship in which one quantity is a constant multiplied by another quantity is called direct variation. Each variable in this type of relationship varies directly with the other.
The graph below represents the data for Nicole’s potential earnings. We say that earnings vary directly with the sales price of the car. The formula [latex]y=k{x}[/latex] is used for direct variation. The value [latex]k[/latex] is a nonzero constant greater than zero and is called the constant of variation. In this case, [latex]k=0.16[/latex].
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}[/latex]
then we say that the relationship is direct variation and [latex]y[/latex] varies directly with [latex]x[/latex]. In direct variation relationships, there is a nonzero constant ratio [latex]k=\dfrac{y}{{x}}[/latex], where [latex]k[/latex] is called the constant of variation, which help defines the relationship between the variables.
How to Solve a Direct Variation Problem
First: Find the constant of variation, by writing the model (the equation) for direct variation: [latex]y=k{x}[/latex]
Plug the given [latex]y[/latex] and [latex]x[/latex] into the model.
Solve for [latex]k[/latex], which is the constant of variation.
Second: Write the equation of variation by plugging the constant of variation into the equation [latex]y=k{x}[/latex].
Third: Substitute the remaining [latex]x[/latex] value into the equation of variation.
Simplify to find [latex]y[/latex].
Example: Solving a Direct Variation Problem
The quantity [latex]y[/latex] varies directly with [latex]x[/latex]. If [latex]y=15[/latex] when [latex]x=6[/latex], find [latex]y[/latex] when [latex]x[/latex] is 10.
First: Find the constant of variation, by writing the model (the equation) for direct variation.
[latex]y=k{x}[/latex]
Plug [latex]y=15[/latex] and [latex]x=6[/latex] into the model.
[latex]15=k{(6)}[/latex]
Solve for [latex]k[/latex] by dividing both sides of the equation by 6.
[latex]\dfrac{15}{6}[/latex] = [latex]\dfrac{k(6)}{6}[/latex]
[latex]2.5=k{}[/latex]
So, the constant of variation is [latex]k=2.5[/latex].
Second: Write the equation of variation by plugging the constant of variation (in this case, [latex]k=2.5[/latex] ) into the equation [latex]y=k{x}[/latex].
[latex]y=2.5{x}[/latex]
Third: Substitute the remaining value (in this case, [latex]x=10[/latex]) into the equation of variation.
[latex]y=2.5{(10)}[/latex]
Simplify to find [latex]y[/latex].
[latex]y=25[/latex]
Therefore, if [latex]y[/latex] varies directly with [latex]x[/latex], and if [latex]y=15[/latex] when [latex]x=6[/latex], then [latex]y=25[/latex] when [latex]x[/latex] is 10.
Watch this video to see a quick lesson in direct variation. You will see more worked examples.
TRY IT
1) The quantity [latex]y[/latex] varies directly with [latex]x[/latex]. If [latex]y=16.8[/latex] when [latex]x=2[/latex], find [latex]y[/latex] when [latex]x[/latex] is 3.
1a) Find the constant of variation.
1b) Find the equation of variation.
1c) Find y when [latex]x=3[/latex].
2) The quantity [latex]y[/latex] varies directly with [latex]x[/latex]. If [latex]y=20.7[/latex] when [latex]x=18[/latex], find [latex]y[/latex] when [latex]x[/latex] is 11.
2a) Find the constant of variation.
2b) Find the equation of variation.
2c) Find y when [latex]x=11[/latex].
3) The cost to fill your car’s gas tank varies directly with the number of gallons you put in your tank. Let [latex]x[/latex] be the number of gallons you put in the tank, and let [latex]y[/latex] be the cost, in dollars, of the gasoline. Suppose that the car in front of you pumped 12 gallons of gas. You were able to see that the pump said $36.48. You are planning on pumping 15 gallons of gas into your car.
3a) Find the constant of variation for this scenario.
3b) What does that constant of variation represent in this scenario?
3c) Find the equation of variation for this scenario.
3d) Find the amount you will pay for pumping 15 gallons of gas into your car.
Inverse Variation
Water temperature in an ocean varies inversely to the water’s depth. Between the depths of 250 feet and 500 feet, the formula [latex]T=\dfrac{14,000}{d}[/latex] gives us the temperature in degrees Fahrenheit at a depth in feet below Earth’s surface. Consider the Atlantic Ocean, which covers 22% of Earth’s surface. At a certain location, at the depth of 500 feet, the temperature may be 28°F.
If we create a table we observe that, as the depth increases, the water temperature decreases.
| d, 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. |
We notice in the relationship between these variables that, as one quantity increases, the other decreases. The two quantities are said to be 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]y=\dfrac{k}{{x}}[/latex]
where [latex]k[/latex] is a nonzero constant, then we say that [latex]y[/latex] varies inversely with [latex]x[/latex]. In inversely proportional relationships, or inverse variations, there is a constant multiple [latex]k={x}y[/latex].
How to Solve an Inverse Variation Problem
First: Find the constant of variation, by writing the model (the equation) for inverse variation: [latex]y=\dfrac{k}{{x}}[/latex]
Plug the given [latex]y[/latex] and [latex]x[/latex] into the model.
Solve for [latex]k[/latex], which is the constant of variation.
Second: Write the equation of variation by plugging the constant of variation into the equation [latex]y=\dfrac{k}{{x}}[/latex].
Third: Substitute the remaining [latex]x[/latex] value into the equation of variation.
Simplify to find [latex]y[/latex].
The following video presents a short lesson on inverse variation and includes more worked examples.
Example: Solving an Inverse Variation Problem
A quantity [latex]y[/latex] varies inversely with [latex]x[/latex]. If [latex]y=4.65[/latex] when [latex]x=2[/latex], find [latex]y[/latex] when [latex]x[/latex] is 5.
First: Find the constant of variation, by writing the model (the equation) for inverse variation.
[latex]y=\dfrac{k}{{x}}[/latex]
Plug [latex]y=4.65[/latex] and [latex]x=2[/latex] into the model.
[latex]4.65=\dfrac{k}{2}[/latex]
Solve for [latex]k[/latex] by multiplying both sides of the equation by 2.
[latex]4.65×2=\dfrac{k}{2}[/latex]x2
[latex]9.3=k{}[/latex]
So, the constant of variation is [latex]k=9.3[/latex].
Second: Write the equation of variation by plugging the constant of variation (in this case, [latex]k=9.3[/latex] ) into the equation [latex]y=\dfrac{k}{{x}}[/latex]
[latex]y=\dfrac{9.3}{{x}}[/latex]
Third: Substitute the remaining value (in this case, [latex]x=5[/latex]) into the equation of variation.
[latex]y=\dfrac{9.3}{{5}}[/latex]
Simplify to find [latex]y[/latex].
[latex]y=1.86[/latex]
Therefore, if [latex]y[/latex] varies inversely with [latex]x[/latex], and [latex]y=4.65[/latex] when [latex]x=2[/latex], then [latex]y=1.86[/latex] when [latex]x[/latex] is 5.
Try It
4) The quantity [latex]y[/latex] varies inversely with [latex]x[/latex]. If [latex]y=4.8[/latex] when [latex]x=55[/latex], find [latex]y[/latex] when [latex]x[/latex] is 8.
4a) Find the constant of variation.
4b) Find the equation of variation.
4c) Find y when [latex]x=8[/latex].
5) The quantity [latex]y[/latex] varies inversely with [latex]x[/latex]. If [latex]y=11.25[/latex] when [latex]x=4[/latex], find [latex]y[/latex] when [latex]x[/latex] is 6.
5a) Find the constant of variation.
5b) Find the equation of variation.
5c) Find y when [latex]x=6[/latex].
6) The time it takes to put up a fence varies inversely with the number of people working on the fence. Let [latex]x[/latex] be the number of people working on the fence, and let [latex]y[/latex] time it takes to complete the fence, in hours. Suppose your neighbor’s fence had 5 people working on it and it took 14 hours. You know that you will have 4 people to work on your fence of the same length. You are wondering how long it will take to complete your fence.
6a) Find the constant of variation for this scenario.
6b) What does that constant of variation represent in this scenario?
6c) Find the equation of variation for this scenario.
6d) Find the amount of time that it will take 4 people to finish your fence.
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