6.3 Proportion and Variation

A used-car company has just offered their best candidate, Uki, 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 like this one between earnings, sales and commission rate.

Solving Direct Variation Problems

In the example above, Uki’s earnings can be found by multiplying her sales by her commission. The formula e=.16s tells us her earnings, e, are the product of 0.16, the commission rate and the sales price of the vehicle she sells, s.  Consider if she sells a vehicle worth $9,200, she would then earn 0.16 \cdot $9,200 = $1,472 and an $18,400 vehicle would yield her $2,944 in earnings. Doubling the sales price of the vehicle also doubles her earnings. 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 showing Relationship between sales and earnings

The graph shows the data for Uki’s potential earnings. We say that earnings vary directly with the sales price of the car. The general formula for direct variation is y=kx^n The value of k is a constant greater than zero and is called the constant of variation. In this case, k=0.16 and n=1.  These are basically power functions.

Direct Variation

If x and y are related by an equation of the form:


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

Many times we are given a description of a direct variation problem and need to find an unknown.  They key to these problems is to first find the constant of variation using what you know and then you are able to use your new formula to find the unknown quantity.

Example Solving a Direct Variation Problem

The quantity y varies directly with the cube of x. If y=25 when x=2, then find y when x is 6. 

First, we read the first sentence that y varies directly with the cube of x, this means n=3 and we have y=kx^3.

We can now use the x and y relationship given to find k:

So our direct variation formula becomes:

and we can plug in x:


We can see visual this with a graph as well:Graph showing direct variation with a a cube in this example

Note that our direct variation graphs look very different as the first one is linear and the previous example is cubic. These functions can be linear, quadratic, cubic, quartic or even radical, but they all pass through (0,0).

Try it Now 1

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

Solving Inverse Variation Problems

The time is takes to double your money in an continuously compounded investment varies inversely to the interest rate. The formula t=\dfrac{0.7}{r} gives us the approximate time it takes to double the money invested compounding continuously with interest rate, r expressed as a decimal.  If one invests in a continuously compounding account with an interest rate of 5% then it would take them 14 years to double their money. However, if the rate was 10%, they would double their money in only 7 years.

We notice in the relationship between these variables that, as one quantity was doubled, the other was cut in half. The two quantities are said to be inversely proportional and each term varies inversely with the other. Inversely proportion relationships are called inverse variations. See the graph below:

Graph showing Inverse proportion

Inverse Variation

If x and y are related by an equation of the form:


where k is a nonzero constant, then we say that y varies inversely with the nth power of x. In inversely proportional relationships, or inverse variations, there is a constant multiply k=x^ny

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.

Recall that multiplying speed by time gives distance: d=rt. Because the distance is fixed at 100 miles, we have 100=rt, so the time is takes will be given by t=\dfrac{100}{r}

Time varies inversely with rate of speed.

Inverse variation problems are very similar to direct variation. We will first solve for k using k=x^ny and then set up the formula and plug in the new x to find the new y.

Example Solving an Inverse Variation Problem

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

First we want to find k:

k=x^ny = 2^3(25)=8(25)=200

Now we have the formula:


And plug in the new x:


Graph of the inverse variation in this example

Try it Now 2

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

Solving Problems Involving Joint Variation

Many situations are more complicated than a basic direct 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=kyz. If x varies directly with y and inversely with z we have x=\dfrac{ky}{z}.  Notice that we only use one constant in a joint variation equation.

Example Solving a Problem Involving Joint Variation

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.

We start with our general equation based on the description of the variation:


Now we fill in what we know to find k:

\begin{array}{l}  6=\dfrac{k(2)^2}{\sqrt[3]{8}}\\  6=\dfrac{4k}{2}\\  6=2k\\  k=3 \end{array}

Now write our formula:


And now plug in the new y and z:

\begin{array}{l}  \dfrac{3(1)^2}{\sqrt[3]{27}}\\  =\dfrac{3(1)}{3}\\  =1 \end{array}

Try it Now 3

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.

Try it  Now Answers

  1. k=\dfrac{24}{9}y=\dfrac{384}{9}
  2. k=72; y = \dfrac{9}{2}
  3. k=5; x=20


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