This section addresses the difference between relationships that are proportional and those that are not. In so doing, the process introduces the concept called “constant of proportionality”, which in algebra courses is also called the “constant of variation”. A key idea to get across is that the constant is sometimes the same thing as the unit rate.
The unit rate has been used in a previous lesson to solve problems. We revisit that in this lesson, and then show how the unit rate can be used to derive a formula that is in the form of y= kx, where k is the constant of proportionality, or unit rate. As stated previously, I don’t obsess over this unit, but rather take a straightforward procedural approach.
Another key concept in this section is what values are dependent on each other. For example, if gasoline costs $3 per gallon, the cost is dependent on the number of gallons. Students might get the two values confused; a problem stating that 4 gallons of gasoline cost $3 might be expressed as 4 gal/$3. The result would be a unit rate of 4/3 gallons/dollar. To avoid such confusion, it is important to go over what is the more logical choice given the nature of the problem; that is, what the dependent and independent variables are.
In this lesson such concepts are presented informally and based on an intuitive sense; the terms “dependent and independent variable” is not defined here. It is defined formally in a subsequent chapter. This avoids overloading the student with too much information. Also, giving formality to something they have seen before and understood intuitively creates a stronger connection—like travelling on a street whose name is unknown and revealed when seeing it on a map..
Warm-Ups. Two of the problems here are used in the discussion on the difference between proportional and non-proportional relationships. There’s also another problem about averages.
1. If 2 pounds of meat cost $7, how much will 4 pounds cost? (Use proportions to solve) Answer: 7/2 = x/4; 2x =28, x = $14.
2. A copier can print 12 copies in 48 seconds. At this rate, how many copies can it print in one minute? Answer: 12/48 = x/60; ¼ = x/60; 4x = 60; x = 15 copies
3. Ace Skating Rink charges $2 for each hour of skating. If you owe $6, how many hours did you skate? Answer: 3 hours.
4. Diamond Skating Rink charges $2 for each hour of skating, but there is a $5 fee for renting bowling shoes. If you owe $13, how many hours did you skate? Answer: $13 - $5 = $8. Since this doesn’t include the rental fee, divide by 2 to find total hours: 4 hours
5. Lou has an average of 95 on four tests. On the fifth test he scores 100. What is his average over the five tests? Answer: Total of scores for four tests: 95 × 4 = 380. Total for five tests: 380 + 100 = 480. Average over five tests: 480 ÷ 5 = 96.
Constant of Proportionality. It is tempting to start off with “If I have two ratios that are 3/2 and 6/4, what are these called?” with the expectation that you will hear “Proportional”. So just remind them and then move on to show a bunch of equivalent ratios: 3/2, 6/4, 9/6, 12/8.
“Since they all represent the same ratio, then their quotient should be the same, and in fact it is. What is it?”
It is 3/2 or 1 ½ . Since we said that equivalent ratios are proportional, we can also say that a proportion has a constant ratio. This constant has a name: The constant of proportionality.
The constant of proportionality is the ratio in simplest (lowest) terms
Looking at Problems 3 and 4 of the warm-ups (or similar problems), the first problem shows a ratio (rate in this case) of $2/hr. If students calculate costs for 2, 3, and 4 hours they will get $9, $11, and $13. 9/2, 11/3 and 13/4 are clearly not equivalent.
When the quantities are not constant, the quantities are non-proportional.
Examples:
Proportional or non-proportional and why? If proportional what is the constant of proportionality? 4/5, 8/10, 12/16 Answer: Non-proportional 4/5 ≠12/16
6/5, 12/10, 42/35 Answer: Proportional; 6/5 or 1 1/5 is constant of proportionality.
Rates. A rate is a ratio in which the units differ. You might ask the students what a rate is and see if they can cite the definition. Then again, sometimes checking for understanding can be time-consuming and it’s best to just bite the bullet and get on with things.
Let’s say we have some distances for various times travelled:
“Are the ratios equivalent?” Yes. “What is the constant of proportionality?” Easy to see that it’s 50/1 or 50 mph. And in fact when we deal with rates, the constant of proportionality is the same thing as the unit rate.
“Let’s do another: the relationship: Cost of gasoline and number of gallons.” In this case, I want the students to construct a table, given the following information. 2 gallons for $7, 3 gallons for $10.50, 4 gallons for $14.
“I want to find the unit rate so I can figure out the cost of gasoline for some amount of gallons. What is that unit rate?”
The consensus will be $3.50 per gallon, but there might be some who did it in reverse and divided 2 by 7, resulting in a unit rate of 2/7 gallons for $1. If someone did, go with it, but if no one did, go for it anyway! What do I mean? Just this:
We said that we want to find the cost of gasoline for some amount of gallons. Can this be done easily using a unit rate of 2/7 gallon/$1? Or it is easier using $3.50/gallon. Let’s say I want to find the cost of 5 gallons.
It should be obvious to most that the easiest way is to express the unit rate as $3.50/gallon. Someone who is clever might point out that you can set up a proportion:
This is not as straightforward since it involves fractions and results in the equation (2/7)x = 5, for which the solution is x = 5 ∙ 7/2, x = 35/2, or $17.50. Doubtful that anyone will get that far, given that it’s a seventh grade class, equations are new to them, and they tend to avoid equations with fractions. In all likelihood, the notion of 2/7 gallon/$1 is probably enough to cast doubts and convince them that $3.50/gal is the way to go.
I’ll ask how we solve this, and students should readily see that $3.50 is multiplied by 5—which is similar to problems they had to do in a previous lesson in which unit rates are used to solve a problem. Specifically it was about face painting. A face painter at a carnival can paint 2 faces in 8 minutes; how many can be painted in 40 minutes? The unit rate is 0.25 face/minute or ¼ face/minute. The unit rate is multiplied by 40.
In using this approach it’s important to know three things, which I have them write down in their notebooks:
1. What quantity are we trying to find?
2. What is this quantity dependent on?
3. Are the two quantities proportional?
In the gasoline example, we are trying to find the cost of the gasoline, and it is dependent on the number of gallons. In the face painting example, the number of faces painted is the quantity we are trying to find. It is dependent on the number of minutes.
Writing Formulas or Equations Using Unit Rates. This brings us to how to solve problems using equations, a topic that seventh grade students find confusing. I break it down in as procedural a manner as possible.
Proportional relationships can be described by equations of the form y = kx, where k is the constant of proportionality or, when rates are involved, the unit rate. The y variable is what we are trying to find. It is dependent on what x equals.
We can use any letters we want; we are not beholden to x and y. Looking at the examples we just did, the cost of gasoline relating to the number of gallons, let c represent the cost of gasoline, and g the number of gallons. Then we have cost = k times the number of gallons, or c = kg. The constant of proportionality or unit rate for the problem is $3.50 which we’ll write as 3.5 and substitute it into the equation.
We now have c = 3.5g. This is a formula that can be used to calculate the cost of gasoline for any number of gallons, represented by the variable g. I have students use it to calculate the cost of 20 gallons. “It’s a plug in” I’ll remind them. We then have c = 20 ∙ 3.5, and c = $70.
Examples:
Let’s try it for face painting. We had the unit rate = 0.25 min/face. Let f = number of faces painted and m = number of minutes. I have the students write the formulas in their notebooks as I work through it with them. Answer: f = 0.25m
Now I try one they haven’t seen before. The number of legs for a collection of spiders is given: no. of legs/no. of spiders: 8/1, 16/2, 24/3, 32/4
Are the two quantities proportional? Yes. What is the constant of proportionality or unit rate? 8 legs/spider
I’ll have them name the two variables. Let’s say it’s L for number of legs and S for number of spiders. “What is dependent on what?” I’ll ask. The number of legs is dependent on the number of spiders.
Write down L = kS.
What did we say k equals? 8
Write the equation: L = 8S.
I do two or three more examples so they feel comfortable with the procedure.
Example: The cost for 3 pounds of meat is $7.50. Write an equation relating cost to pounds. We’re given the ratio, therefore ratios of cost/weights will be equivalent to the unit rate which is $2.50/lb.
We are trying to find cost (c ) and the cost is dependent on number of pounds (p).
c = 2.5p
Homework. Homework should be a mix of determining whether relationships are proportional or not, finding constants of proportionality, unit rates, and writing equations