Parametric Equations

and Linear Functions

In this assignment, we did explorations using parametric equations. Parametric equations are written in a 2 - vector form, and are a function of t, the time variable. They are an alternative way to write various equations in terms of time. As I was working through the explorations, I got interested in the parametric equations that seemed to represent linear functions. That is, instead of writing a linear equation in the form of y = mx + b, expressing it using a set of parametric equations.


The main part of this investigation began with exploring the equations for various values of a, b and k. What type of graph will equations in this form give us? Let's plug in some values for a, b and k and find out.

These equations appear to represent a linear function. We can change the values of a,b and k to see that any equations in this form will be linear graphs.


If we were to choose an appropriate range for t, and then keep a and b constant while varying k, we would see a family of related graphs. For example, for -100...t...100, where a = 1 and b = 2, we can see the following graphs as k is varied from the integers -3 to 3.

As you can see these parametric equations represent a family of linear equations. In each case, the line contains the point (1, 2), and the slope of the lines vary from -3 to 3.

This exercise led me to think that there must be some relationship between the parametric equation and the resulting linear graph. The question of course is, what is this relationship?

As I pondered this thought, I continued to the next part of the assignment. In it we were told to graph the equation for an appropriate range for t. I used -100...t...100. The graph of this equation was also linear, as seen below.



In need to find the relationship between the two forms of the equations. To do this, I first decided to find the slope and y intercept of the graphed line. In the above case, the slope is 2/1, and the y intercept is -3. I remembered from the last exploration that k controlled the slope of the line. since k=2 in the y equation, perhaps that is where the slope comes from. To check this, I decided to graph and see how the two graphs compared.

In looking art the graphs, we see that thy both pass through (1, -1), However, the slope of the second equation is now become 1/1 instead of 2/1. I'm inclined to believe that the slope of the line will can be found by taking the ratio of the coefficient of t in the y equation over the coefficient of t in the x equation. Let's test this hypothesis. If my idea is correct the graph of the equation should have a slope of 4/3. If we graph this equation as well, we can see if we are correct.

Looks like our hypothesis holds for finding the slope of the line given the parametric equation. One other thing we notice about these three lines is that they all contain (1, -1). We know from our previous exploration that this point is determined by the constant terms of the x and y equations. In order to talk in more detail about the relationships found here, let's define a general form for our set of parametric equations.

Now we can try to find a relationship between the point that we know will always be on the line and the y intercept of the line. I begin this task by keeping the slope of the line a constant and manipulating the m and n variables. The first equation I want to look at is. As you can see from the graph below, the y intercept of this line is -1.4, and this value occurs when t = 0.2.

If we change our equation to , we now see that the y intercept is 3.2 and the t value for this is t = 0.6.

Let's look first at the value of t for each of the equations. For , we see that t = -0.2. This value is related to the value of t if the x equation were solved for zero. Let's check our idea for the second equation . Based on our calculated values, we think t should equal 0.6, and we know from our graph that it does. Excellent. Now, how does the value of t effect the y intercept of the line. We know that the y intercept of a line is the value for y when x = 0. since our y equation does not have an x in it, I believe we can set the y equation equal to zero and Solve for y. In order to do this, we first need to know the value of t.

In the value of t = -0.2. this implies that y = 2(-0.2) - 1 , should be the value of the y intercept. The expression above simplifies to y = -1.4, which is the y intercept for that line.


We can also check in the same manner. We know t = 0.6, so the intercept should equal 2(0.6) + 2, which is 3.2, which is indeed the intercept for that line.


It appears that we have now established formulas for the relationship between parametric and linear forms of the same line. Let's state the relationships formally:


Realize that we get the y intercept formula from solving x = at + m for t when x = 0,which is (-1/a)m, and substituting it for t in the equation y = bt + n.


Now that we have discovered these formulas, we can use them to go in the other direction. That is, given a point on a line and the slope for that line, we can write parametric equations to represent that linear function. For example, if we wanted to find the parametric equations for the line passing through (7,5), and having a slope of 3, we could do the following:

First, we would need to find the y intercept of this line, so we use point slope form to do so:

So, now we know that the y intercept of our line is -16. Since m = 3, we can set 3 = b/a, and substituting -16 for the y intercept, we can write:

Since b/a has to be three, we choose b = 3 and a = 1. The above equation tells us that -3m + n = -16. We know that there are an infinite number of solutions to this equation, and any one of them should work. We choose m = 5 and n = -1. This makes our parametric equation for this line:

Let's graph these equations and make sure we have a line with a slope of 3 and containing (7, 5).


We do! (Celebrate if necessary!)


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