Part IV – Straighten Up!
In this activity, you analyzed data and discovered several inverse relationships. Inverse relationships are not that hard to explain … as one variable increases, the other decreases in a predictable manner. However, they aren’t as simple as direct relationships.
Questions
Why are direct relationships more simple and useful than indirect relationships?
We can transform an indirect or inverse relationship into a direct relationship by mathematically altering one of our columns of data. As an example, let’s take the cost versus mpg graph (the last inverse graph) and turn it into a direct relationship (or straight line) graph.
Using your graphing software, create a new column in the data table and call it “gallons per mile.” Have the graphing software calculate the data for this column by taking the reciprocal of the mpg data (i.e., 1/mpg). This will transform your data into a direct relationship. Now graph the transformed data with “gallons per mile” on the x-axis and “cost to drive 100 miles” on the y-axis.
Look at this new graph on the screen. Make sure to print it or show it to the teacher.
Write the equation in y = mx + b format. Use correct y and x variable names that match our data.
How is this equivalent to the equation of the inverse function?
What takes the place now of k, the constant of proportionality, that we saw in the inverse function (refer to Question 13 in Part III)?
Will the graph go through the origin? Please explain.
Now use your graphing software and data table to make one last graph. This time plot “gallons to go 100 miles” (which you calculated for Table 4 in Part III) on the x-axis and “cost to drive 100 miles” on the y-axis (still assuming gas costs $4 per gallon).
Write the equation in y = mx + b format. Use correct y and x variable names that match our data.
What takes the place now of the constant of proportionality? Where does this number come from?
This final graph makes the x-axis easier to follow because it is round numbers instead of decimals. The graph shows a direct relationship between the cost and the number of gallons of gas used to drive 100 miles.
This final graph also shows that small increases in mileage for the least fuel efficient vehicles results in large fuel and cost savings (see graph below). For example, going from 11 gallons per 100 miles (which corresponds to a mileage of 9 mpg) to 9 gallons per 100 miles (which corresponds to 11 mpg) gives large fuel and cost savings ($8). The same 2 mpg increase in fuel efficiency at the higher mileage rates gives smaller and smaller fuel and cost savings. For example, increasing from 26 mpg to 28 mpg corresponds to only saving 0.28 gallons of gas (3.85–3.57 gallons per 100 miles) and saving only $1 on gas costs. Obviously, the best scenario would be to only drive the most fuel efficient vehicles. Yet, small increases in fuel efficiency for the low mileage vehicles can still make a big difference in saving gas and money.
Originally published at http://www.sciencecases.org/gas_mileage/case4.asp
Copyright © 1999–2010 by the National Center for Case Study Teaching in Science. Please see our usage guidelines, which outline our policy concerning permissible reproduction of this work.
