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5.7 Variation

Learning Objectives

By the end of this section, you will be able to:

  • Solve direct variation problems.
  • Solve inverse variation problems.
  • Solve problems involving joint variation.

Solve direct variation problems

When two quantities are related by a proportion, we say they are proportional to each other. Another way to express this relation is to talk about the variation of the two quantities. We will discuss direct variation and inverse variation in this section.

Lindsay gets paid \( \$ 15 \) per hour at her job. If we let \( S \)  be her salary and [latex]h[/latex] be the number of hours she has worked, we could model this situation with the equation \( S = 15 h \).

Lindsay’s salary is the product of a constant, [latex]15[/latex], and the number of hours she works. We say that Lindsay’s salary varies directly with the number of hours she works. Two variables vary directly if one is the product of a constant and the other.

Example

Which graph represents direct variation and why?

Graph of a straight line through the origin.
Graph A.
Graph of a straight line through (2,0).
Graph B.
Show Solution

Graph A represents direct variation.  While both graphs show a function with a constant rate of change, only the graph of a straight line with a [latex]y[/latex]-intercept of zero shows a constant ratio between [latex]x[/latex] and [latex]y.[/latex]

Direct Variation

For any two variables \( x \) and \( y \), \( y \)  varies directly with \( x \)  if

\( y = k x , \ \ \text{where} \ k \neq 0 \).

The constant  \( k \) is called the constant of variation.

How To

Solve direct variation problems.

  1. Step 1. Write the formula for direct variation.
  2. Step 2. Substitute the given values for the variables.
  3. Step 3. Solve for the constant of variation.
  4. Step 4. Write the equation that relates [latex]x[/latex] and [latex]y[/latex] using the constant of variation.

Example

When Raoul runs on the treadmill at the gym, the number of calories, [latex]c[/latex], he burns varies directly with the number of minutes, [latex]m[/latex], he uses the treadmill. He burned 315 calories when he used the treadmill for 18 minutes.

  1. Write the equation that relates [latex]c[/latex] and [latex]m[/latex].
  2. How many calories would he burn if he ran on the treadmill for 25 minutes?
Show Solution

(a) The number of calories, [latex]c[/latex], varies directly with the number of minutes, [latex]m[/latex], on the treadmill, and  \( c = 315 \) when \( m =18 \).

Explanation Steps
Write the formula for direct variation. [latex]y = kx[/latex]
We will use [latex]c[/latex] in place of [latex]y[/latex] and [latex]m[/latex] in place of [latex]x[/latex]. [latex]c= km[/latex]
Substitute the given values for the variables. [latex]315 = k\cdot 18[/latex]
Solve for the constant of variation. [latex]\frac{315}{18} = \frac{k \cdot 18 }{18}[/latex]

[latex]17.5 = k[/latex]
Write the equation that relates [latex]c[/latex] and [latex]m[/latex]. [latex]c = k m[/latex]
Substitute in the constant of variation. [latex]c = 17.5 m[/latex]

(b) Find [latex]c[/latex] when [latex]m=25[/latex].

Explanation Steps
Write the equation that relates [latex]c[/latex] and [latex]m[/latex].

[latex]c = 17.5 m[/latex]
Substitute the given value for [latex]m[/latex]. [latex]c = 17.5  ( 25)[/latex]
Simplify. [latex]c = 437.5[/latex]
Raoul would burn 437.5 calories if he used the treadmill for 25 minutes.

 

Solving Direct Variation Problems

A pre-owned car dealer 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. As she considers the offer, she takes into account the typical price of the dealer's cars, the overall market, and how many she can reasonably expect to sell. In this section, we will look at relationships, such as this one, between earnings, sales, and commission rate.

In the example above, Nicole’s earnings can be found by multiplying her sales by her commission. The formula \( e = 0.16s \) tells us her earnings \( e \) come from the product of \( 0.16 \), her commission, and the sale price [latex]s[/latex] of the vehicle. If we create a table, we observe that as the sales price increases, the earnings increase as well, which should be intuitive. See Table 1 below:

Table 1
[latex]s[/latex], sales price [latex]e=0.16s[/latex] Interpretation
[latex]\$4{,}600[/latex] [latex]e=0.16(4{,}600)=736[/latex] A sale of a $4,600 vehicle results in $736 earnings.
[latex]\$9{,}200[/latex] [latex]e=0.16(9{,}200)=1{,}472[/latex] A sale of a $9,200 vehicle results in $1472 earnings.
[latex]\$18{,}400[/latex] [latex]e=0.16(18{400})=2{,}944[/latex] 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.
Graph of y=(0.16)x where the horizontal axis is labeled, “s, Sales Price in Dollars”, and the vertical axis is labeled, “e, Earnings, $”.
Figure 1

Figure 1 represents the data for Nicole’s potential earnings. We say that earnings vary directly with the sales price of the car. The formula \( y = k x^n \) is used for direct variation.  The value of \( k \) is a nonzero constant greater than zero and is called the constant of variation. In this case, \( k = 0.16 \) and \( n =1 \).  We saw functions like this one when we discussed power functions.

Direct Variation

If \( x \) and \( y \)  are related by an equation of the form

\( y = k x^n  \), where \( k=0\),

then we say that the relationship is direct variation, and \( y \) varies directly with or is proportional to the [latex]n[/latex]th power of \( x \).  In direct variation relationships,  there is  a non zero constant ratio  \( k = \frac{y}{x^n} \), where \( k \) is called the constant of variation, which help defines the relationship between the variables.

 

How To

Given a description of a direct variation problem, solve for an unknown.

  1. Identify the input, \( x \), and the output, \( y \).
  2. Determine the constant of variation. You may need to divide \( y \) by the specified power of \( x \) to determine the constant of variation.
  3. Use the constant of variation to write an equation for the relationship.
  4. Substitute known values into the equation to find the unknown.

Example

The quantity [latex]y[/latex] varies directly with the cube of \( x \). If \( y = 25 \) when \( x = 2 \), find \( y \) when \( x \) is \( 6 \). 

Show Solution

The general formula for direct variation with a cube is \( y = k x^3 \). The constant can be found by dividing \( y \)  by the cube of \( x \).

[latex]\begin{array}{rl}\hfill k &= \frac{y}{x^3}\hfill \\ \hfill &= \frac{25}{2^3}\hfill \\ \hfill &=\frac{25}{8}\hfill \end{array}[/latex]

Now use the constant to write an equation that represents this relationship.

[latex]y=\frac{25}{8}x^3[/latex]

Substitute \( x = 6 \)  and solve for \( y \). 

[latex]\begin{array}{rl}\hfill y&=\frac{25}{8}6^3\hfill \\ \hfill &=675\hfill \end{array}[/latex]

Analysis

The graph of this equation is a simple cubic, as shown in Figure 2 below.

Graph of y=25/8(x^3) with the labeled points (2, 25) and (6, 675).
Figure 2

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 (0, 0).  

Try It

The quantity \( y \) varies directly with the square of \( x \). If \( y = 24 \) when \( x = 3 \), find \( y \) when \( x \) is \( 4 \).

Show Solution

[latex]\frac{128}{3}[/latex]

Access the online resource below for additional instruction and practice with direct variation problems:

Mathispower4u: Direct Variation. YouTube. https://www.youtube.com/watch?v=WGqmAmzUODM

Solving Inverse Variation Problems

Water temperature in an ocean varies inversely to the water’s depth. The formula \( T = \frac{14,000}{d} \) 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.

In Table below, we observe that, as the depth increases, the water temperature decreases.

Table 2
[latex]d[/latex], depth (ft) [latex]T=\frac{14{,}000}{d}[/latex] Interpretation
[latex]500[/latex] [latex]\frac{14{,}000}{500}=28[/latex] At a depth of 500 ft, the water temperature is 28° F.
[latex]1000[/latex] [latex]\frac{14{,}000}{1000}=14[/latex] At a depth of 1,000 ft, the water temperature is 14° F.
[latex]2000[/latex] [latex]\frac{14{,}000}{2000}=7[/latex] At a depth of 2,000 ft, the water temperature is 7° 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 Figure below 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 \( y = \frac{k}{x} \) for inverse variation in this case uses \( k = 14,000 \).

Graph of y=(14000)/x where the horizontal axis is labeled, “Depth, d (ft)”, and the vertical axis is labeled, “Temperature, T (Degrees Fahrenheit)”.
Figure 3

Inverse Variation

If \( x \) and \( y \) are related by an equation of the form

\( y = \frac{k}{x^n} \)

where \( k \) is a nonzero constant, then we say that \( y \) varies inversely with the [latex]n[/latex]th power of \( x \). In inversely proportional relationships, or inverse variations, there is a constant multiple \( k = x^n y \).

Example

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.

Show Solution

Recall that multiplying speed by time gives the distance. If we let \( t \) represent the drive time in hours and \( v \) represent the velocity ( speed or rate) at which the tourist drives, that \( vt = \text{distance} \). Because the distance is fixed at \( 100 \) miles. \( vt = 100\) so \( t = \frac{100}{v} \). Because time is a function of velocity, we can write \( t(v) \).

[latex]\begin{array}{rl}\hfill t(v )&=\frac{100}{v}\hfill \\ \hfill &=100v^{-1}\hfill \end{array}[/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. We say that time varies inversely with velocity.

How To

Given a description of an indirect variation problem, solve for an unknown.

  1. Identify the input, \( x \), and the output \( y \).
  2. Determine the constant of variation. You may need to multiply \( y \) by the specified power of \( x \) to determine the constant of variation.
  3. Use the constant of variation to write an equation for the relationship.
  4. Substitute known values into the equation to find the unknown.

Example

A quantity \( y \) varies inversely with the cube of \( x \). If \( y = 25 \) when \( x = 2\), find \( y \) when  \( x \) is \( 6 \).

Show Solution

The general formula for inverse variation with a cube is \( y = \frac{k}{x^3} \). The constant can be found by multiplying \( y \) by the cube of \( x \).

[latex]\begin{array}{rl}\hfill k&=x^3y\hfill \\ \hfill &= 2^3\cdot 25\hfill \\ \hfill &=200\hfill \end{array}[/latex]

Now we use the constant to write an equation that represents this relationship.

[latex]y=\frac{200}{x^3}[/latex]

Substitute \( x = 6 \) and solve for \( y \).

[latex]\begin{array}{rl}\hfill y&=\frac{200}{6^3}\hfill \\ \hfill &=\frac{25}{27}\hfill \end{array}[/latex]

Analysis

The graph of this equation is a rational function, as shown in Figure below.

Graph of y=25/(x^3) with the labeled points (2, 25) and (6, 25/27).
Figure 4

Try It

A quantity \( y \) varies inversely with the square of \( x \). If \( y = 8 \) when \( x = 3 \), find \( y \) when \( x \) is \( 4 \).

Show Solution

[latex]\frac{9}{2}[/latex]

 

Mathispower4u: Inverse Variation. YouTube. https://www.youtube.com/watch?v=awp2vxqd-l4

Solving Problems Involving 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 \( c \), cost, varies jointly with the number of students, \( n \), and the distance, \( d \).

Joint Variation

Joint variation occurs when a variable varies directly or inversely with multiple variables. For instance, if \( x \) varies directly with both \( y \) and \( z \), we have \( x = k y z \). If \( x \) varies directly with \( y \) and inversely with \( z \), we have \( x = \frac{ky}{z} \). Notice that we only use one constant in a joint variation equation.

Example

A quantity \( x \) varies directly with the square of \( y \) and inversely with the cube root of \( z \).
If \( x = 6 \) when \( y = 2 \) and \( z = 8\), find \( x \) when \( y =1 \) and \( z = 27 \).
Show Solution

Begin by writing an equation to show the relationship between the variables.

[latex]x=\frac{ky^2}{\sqrt[3]{z}}[/latex]

Substitute \( x = 6 \), \( y = 2 \), and \( z = 8 \) to find the value of the constant \( k \).

[latex]\begin{array}{rl}\hfill 6&=\frac{k\cdot 2^2}{\sqrt[3]{8}}\hfill \\ \hfill 6&=\frac{4k}{2} \hfill \\ \hfill 3&=k\hfill \end{array}[/latex]

Now we can substitute the value of the constant into the equation for the relationship.

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

To find \( x \) when \( y =1 \) and \( z = 27 \), we will substitute values for \( y \) and \( z \) into our equation.

[latex]x=\frac{3\cdot 1^2}{\sqrt[3]{27}}=1[/latex]

Try It

A quantity \( x \) varies directly with the square of \( y \) and inversely with \( z \). If \( x = 40 \) when \( y = 4 \) and \( z = 2 \), find \( x \) when \( y= 10 \) and   \( z =  25 \).

Show Solution

[latex] 20  [/latex]

 

Mathispower4u: Direct and Inverse Variation. YouTube. https://www.youtube.com/watch?v=is07Wg_0DiY

Key Concepts

  • A relationship where one quantity is a constant multiplied by another quantity is called direct variation.
  • Direct Variation : \( y = k x^n \) , \( k \) is a non zero constant
  • Two variables that are directly proportional to one another will have a constant ratio.
  • A relationship where one quantity is a constant divided by another quantity is called inverse variation.
  • Inverse Variation : \( y = \frac{k}{ x^n} \) , \( k \) is a non zero constant
  • Two variables that are inversely proportional to one another will have a constant multiple.
  • In many problems, a variable varies directly or inversely with multiple variables. We call this type of relationship joint variation.

Section Exercises

Verbal

1. What is true of the appearance of graphs that reflect a direct variation between two variables?

Show Solution

The graph will have the appearance of a power function.

2. If two variables vary inversely, what will an equation representing their relationship look like?

3. Is there a limit to the number of variables that can vary jointly? Explain.

Show Solution

No. Multiple variables may jointly vary.

Algebraic

For the following exercises, write an equation describing the relationship of the given variables.

4. \( y \) varies directly as \( x \) and when \( x = 6 \), \( y =12\). 

5. \( y \) varies directly as the square of \( x \) and when \( x = 4 \), \( y =80\).

Show Solution

\( y = 5 x^2 \)

6. \( y \) varies directly as the square root of \( x \) and when \( x = 36 \), \( y =24\).
7. \( y \) varies directly as the cube of \( x \) and when \( x = 36 \), \( y =24\).
Show Solution

\( y = \frac{1}{1944} x^3 \)

8. \( y \) varies directly as the cube root of \( x \) and when \( x = 27 \), \( y =15\).
9. \( y \) varies directly as the fourth power of \( x \) and when \( x = 1 \), \( y = 6\).

Show Solution

\( y = 6 x^4 \)

10. \( y \) varies inversely as \( x \) and when \( x = 4 \), \( y = 2 \). 

11. \( y \) varies inversely as the square of \( x \) and when \( x = 3 \), \( y = 2 \). 

Show Solution

\( y = \frac{18} {x^2 } \)

12. \( y \) varies inversely as the cube of \( x \) and when \( x = 2 \), \( y = 5 \). 

13. \( y \) varies inversely as the fourth power of \( x \) and when \( x = 3 \), \( y = 1 \).

Show Solution

\( y = \frac{81} {x^4 } \)

14. \( y \) varies inversely as the square root of \( x \) and when \( x = 25 \), \( y = 3 \).

15. \( y \) varies inversely as the cube root of \( x \) and when \( x = 64 \), \( y = 5 \).

Show Solution

\( y = \frac{20}{ \sqrt[3]{x}}\)

16. \( y \) varies jointly with \( x \) and \( z \) and when \( x = 2 \) and \( z = 3 \), \( y = 36 \).

17. \( y \) varies jointly with \( x \), \( z \) and \( w \) and when \( x = 1 \), \( z = 2 \), \( w = 5 \) then  \( y = 100 \).

Show Solution

\( y = 1000 x w z\)

18. \( y \) varies jointly as the square of \( x \) and the square of  \( z \) and when \( x = 3 \) and \( z = 4 \) then  \( y = 72 \).
19. \( y \) varies jointly as  \( x \) and the square root of \( z \) and when \( x = 2 \) and \( z = 25 \) then  \( y = 100 \).

Show Solution

\( y = 10 x \sqrt{ z} \)

20. \( y \) varies jointly as the square of \( x \) and the cube  of  \( z \) and the square root of \( w \). When \( x = 1 \), \( z = 2 \), and \( w = 36 \), then \( y = 48 \).

21. \( y \) varies jointly as \( x \) and \( z \) and inversely as \( w \). When \( x = 3 \),  \( z = 5 \), and \( w = 6 \), then \( y = 10\).

Show Solution

\( y = 4 \frac{xz}{w}  \)

22. \( y \) varies jointly as the square of \( x \) and the square root of \( z \) and inversely  as the cube of \( w \). When \( x = 3 \), \( z = 4 \), and \( w = 3 \), then \( y = 6\).

23. \( y \) varies jointly as \( x \) and \( z \) and inversely as the square root of \( w \) and the square of \( t \). When \( x = 3 \), \( z = 1 \), \( w = 25 \) and \( t = 2 \), then \( y = 6\).

Show Solution

\( y = 40 \frac{xz}{\sqrt{w} t^2}  \)

Numeric

For the following exercises, use the given information to find the unknown value.

24.  \( y \) varies directly as \( x \). When \( x = 3 \), then \( y = 12 \). Find \( y \) when \( x = 20 \). 

25. \( y \) varies directly as the square of \( x \). When \( x = 2 \), then \( y = 16 \). Find \( y \) when \( x = 8 \). 

Show Solution

\( y =  256 \)

26. \( y \) varies directly as the cube of \( x \). When \( x = 3 \), then \( y = 5\). Find \( y \) when \( x = 4 \). 

27. \( y \) varies directly as the square root of \( x \). When \( x = 16 \), then \( y = 4\). Find \( y \) when \( x = 36 \). 

Show Solution

\( y =  6 \)

28. \( y \) varies directly as the cube root of \( x \). When \( x = 125 \), then \( y = 15\). Find \( y \) when \( x = 1{,}000 \). 

29. \( y \) varies inversely with  \( x \). When \( x = 3 \), then \( y = 2\). Find \( y \) when \( x = 1 \). 

Show Solution

\( y =  6 \)

30. \( y \) varies inversely with the square of  \( x \). When \( x = 4 \), then \( y = 3 \). Find \( y \) when \( x = 2 \). 

31. \( y \) varies inversely with  the cube of \( x \). When \( x = 3 \), then \( y = 1 \). Find \( y \) when \( x = 1 \). 

Show Solution

\( y =  27 \)

32. \( y \) varies inversely with the square root of \( x \). When \( x = 64 \), then \( y = 12 \). Find \( y \) when \( x = 36 \). 

33. \( y \) varies inversely with the cube root of \( x \). When \( x = 27 \), then \( y = 5 \). Find \( y \) when \( x = 125 \). 

Show Solution

\( y =  3 \)

34. \( y \) varies jointly as \( x \) and \( z \). When \( x = 4 \) and \( z = 2 \), then \( y = 16 \). Find \( y \) when \( x = 3 \) and \( z = 3 \).

35. \( y \) varies jointly as  \( x \), \( z \), and \( w \). When \( x = 2 \), \( z = 1\), and \( w = 2 \), then \( y = 72 \). Find \( y \) when \( x = 1 \), \( z = 2 \), and \( w = 3 \).

Show Solution

\( y =  18 \)

36. \( y \) varies jointly as \( x \) and the square root of \( z \). When \( x = 2 \) and \( z = 4\), then \( y = 144 \). Find \( y \) when \( x = 4 \) and \( z = 5 \).
37. \( y \) varies jointly as the square of  \( x \) and the square root of \( z \). When \( x = 2 \) and \( z = 9\), then \( y = 24 \). Find \( y \) when \( x = 3 \) and \( z =2 5 \).

Show Solution

\( y =  90 \)

38. \( y \) varies jointly as  \( x \) and \( z \) and inversely as \( w \). When \( x = 5 \), \( z = 2\), and \( w = 20\),  then \( y = 4 \). Find \( y \) when \( x = 3 \), \( z = 8 \), and \( w = 48 \). 

39. \( y \) varies jointly as  the square of \( x \) and the cube of \( z \) and inversely as the square root of \( w \). When \( x = 2 \), \( z = 2\), and \( w = 64\),  then \( y = 12 \). Find \( y \) when \( x = 1 \), \( z = 3 \), and \( w = 4 \). 

Show Solution

\( y =  \frac{81}{2} \)

40. \( y \) varies jointly as  the square of \( x \) and  of \( z \) and inversely as the square root of \( w \) and of \( t \). When \( x = 2 \), \( z = 3\), \( w = 16\), and \( t = 3 \),  then \( y = 1 \). Find \( y \) when \( x = 3 \), \( z = 2 \), \( w =36 \) and \( t= 3\). 

Technology

For the following exercises, use a calculator to graph the equation implied by the given variation.

41. \( y \) varies directly with the square of \( x \) and when \( x =2 \), \( y = 3 \). 

Show Solution

\( y =  \frac{3}{4} x^2 \)

Graph of y = 3/4 x^2.

42. \( y \) varies directly with the cube of \( x \) and when \( x =2 \), \( y = 4 \). 

43. \( y \) varies directly with the square root of \( x \) and when \( x =36 \), \( y = 2 \).

Show Solution

\( y =  \frac{1}{3} \sqrt{x} \)

Grapgh of y = 1/3 sqare root of x

44. \( y \) varies inversely with \( x \) and when \( x =6 \), \( y = 2 \).

45. \( y \) varies inversely as the square of \( x \) and when \( x =1 \), \( y = 4 \).

Show Solution

\( y =  \frac{4}{x^2}  \)

Grapgh of y = 4/x^2.

Glossary

Constant of Variation

the non-zero value \( k \) that helps define the relationship between variables in direct or inverse variation.

Direct Variation

the relationship between two variables that are a constant multiple of each other; as one quantity increases, so does the other

Inverse Variation

the relationship between two variables in which the product of the variables is a constant

Joint Variation

a relationship where a variable varies directly or inversely with multiple variables

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