## Proportional Reasoning

### Linearly Proportional Systems

If \(y\) is linearly proportional to \(x\), we write that as \( y \propto x \). What that means is that \(y\) is related to \(x\) by $$ y = cx $$ where \(c\) is a constant. Notice that the graph of \(y\) vs \(x\) is a straight line that passes through the origin. The slope of this line is \(c\).

Proportional reasoning is more than just some algebra. It is a way of thinking about relationships. Instead of solving equations, you reason, "If \(x\) doubles, \(y\) doubles. Done."

Any two sets of values, \( (x_1,y_1) \) and \( (x_2,y_2) \) (here the subscripts correspond to different values of \(x\) and \(y\). This notation will be used extensively throughout.) are related by $$ \frac{y_1}{y_2} = \frac{x_1}{x_2} $$

### Non-Linearly Proportional Systems

We can expand this line of reasoning to systems that are proportional to powers of each other. If \( y \propto x^n \), then $$ y = cx^n $$ where \(c\) is again a constant and we want to constrain ourselves to \( n>0 \).

We can still use proportional reasoning to think about relationships. If \( n=2 \), we would say, "If \(x\) doubles, \(y\) quadruples". "If \(x\) increase by a factor of 3, \(y\) increases by a factor of 9". Get it?

Any two sets of values, \( (x_1,y_1) \) and \( (x_2,y_2) \) are related by $$ \frac{y_1}{y_2} = \left( \frac{x_1}{x_2} \right)^n $$

### Inversely Proportional Systems

Two quantities are inversely proportional if \( y \propto 1/x^n \), where \( n>0 \). What that means is that $$ y = \frac{c}{x^n} $$ where \(c\) is again a constant. Now, you can argue there is no reason to treat this separately from the previous topic. This is just the same as letting \( y \propto x^n \), where \( n \) can take on positive and negative values. That's true! However, the physical relationship is fundamentally different, thus it is good to look at them separately.

The key difference is that if one increases, then the other *decreases*. If \( n=1 \), we would say, "If \(x\) doubles, \(y\) *decreases* by a factor of 2". "If \(x\) increases by a factor of 3, \(y\) decreases by a factor of 3". Then, for higher powers, the reasoning is the same as before. If \(x\) doubles, \(y\) *decreases* by a factor of 4". Any two sets of values, \( (x_1,y_1) \) and \( (x_2,y_2) \) are related by
$$ \frac{y_1}{y_2} = \left( \frac{x_2}{x_1} \right)^n $$
See the difference? The second ratio is flipped. This corresponds to the inverse relationship between \(x\) and \(y\).

In many of the exercises below, I've simply taken problems like those every book gives for topics later in this course and the next (Electricity, Magnetism, and Thermodynamics). Notice, you don't need to know any physics to solve these problems. They are all solved using proportional reasoning.

### Exercises

*Do Now!* Do these exercises immediately.

*Not Now!* Do these after you start to forget the topic, say in a week.

*More!* More exercises if you want. Maybe review before a test.

### Required Problems

### Solution

1. An object is pushed by a force a certain distance (starting from rest). The kinetic energy acquired is proportional to both the force and the distance. The kinetic energy is proportional to the mass of the object and speed of the object squared. The momentum of the object is linearly proportional to the mass of the object and its speed. So, if the force is doubled with everything else the same, what happens to the momentum?