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 tells us her earnings, , are the product of 0.16, the commission rate and the sales price of the vehicle she sells, . Consider if she sells a vehicle worth $9,200, she would then earn 0.16 $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.
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 The value of is a constant greater than zero and is called the constant of variation. In this case, and . These are basically power functions.
Direct Variation
If and are related by an equation of the form:
then we say that the relationship is direct variation and varies directly with, or is proportional to, the nth power of . In direct variation relationships, there is a non-zero constant ratio , where 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 varies directly with the cube of . If when , then find when is 6.
First, we read the first sentence that varies directly with the cube of , this means and we have .
We can now use the and relationship given to find :
So our direct variation formula becomes:
and we can plug in x:
We can see visual this with a graph as well:
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 varies directly with the square of . If when , find when 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 gives us the approximate time it