by Donny Thurston

In this write-up, I will be exploring parametric equations, largely of the form:

x = mt + b

y = nt + c

Which looks like this, when seen in the graphing calcuator:

As explained on the exploration page, parametric equations are a set of two functions (in two-dimensions, that is) that define points (x,y) on the coordinate plane. The two functions can be represented:

x = f(t)

y = g(t)

We can see here that both x and y are represented by functions of t. Therefore, as differing values of t are used in the equations as input, we can observe two outputs, one each for f(t) and g(t), which also create our x and y for a particular point. As a result, we no longer think of a single input and output on the coordinate plane, such as "plugging in" an x, and getting a y as output. Instead, we "plug in" a t, so to speak, and receive both an x and a y. Let us examine how this process works with graphing calculator technology.

5. Graph several sets of curves for

for selected values of

a,b, andkin an appropriate range fort. (Setaandband then overlay graphs for several values ofk. Repeat for new values ofaandb.)

We are going to select various values for a, b, and k, but first, let's discuss what is meant by an "appropriate range for t." As mentioned, t is the true "input" for the set of equations, and, therefore, the graph. Potentially, we could use all values of t, that is, from negative infinity to positive infinity. However, for the purposes of this exploration, we will see what occurs for set values of t.

So, let us begin by following the instruction here and overlay several graphs for values of k.

Let us assume a, b = 1 and t ranges from 0 to 2.k = 1 in blue, k = 2 in red, k = 3 in green, k = .5 in pink, k = 0 in black, k = -1 in aqua

Here we can see a very reliable pattern occur. It would appear that k represents the slope of each line segment that we observe. Interestingly, however, each line segment differs in what the y-intercept

wouldbe, should the segment be extended to the y-axis. Also observe that in the k = 0 condition, the length of the segment is 2 units, the same range that t covers. What would have hapenned if t had included negative values?Some other interesting features to consider. Each segment intersects at the same segment, (1,1). Could this be related to the fact that a and b are both 1? Let us consider new values for a and b.

In this instance, a = 2 and b = 3. t will still range from 0 to 2.I suggest that the new point of concurrency of my segments will be (2,3).k = 1 in blue, k = 2 in red, k = 3 in green, k = .5 in pink, k = 0 in black, k = -1 in aqua

Looks like it! Is this supported mathematically?

Consider what is happening here. x and y both are the "outputs" that occur for certain values of t, including, in our example, when t = 0. When t = 0, the other values in the parametric equation, including k (and the so-far-unspecified coefficient in front of t in the equation x = a + t), have no influence on x and y since k(0) = 0 for all k. Therefore, for all values of k, (x,y) = (a,b) for t=0.

So, here we have learned that in parametric equations of the form:

k appears to represent the slope of our line segment, and (a,b) is the point of concurrency of the line segments for all values of k. More specifically, and perhaps more importantly, the parametric equation produces (a,b) when t = 0 for any k.

Let us continue.

6. Graph

for some appropriate range for

t.

Interpret. Is there anything to vary to help understand the graph?Let us graph this equation, and then begin to modify it to better understand the functions of the various values within the equation.

t will range from 0 to 3.It appears to fit our previous conjectures. The t = 0 point is (1,-1), and the slope of the line is 2. I am, however, interested in learning what occurs for varying values of m in the equation x = mt + 1, given y = 2t - 1. For convenience, let us use the parametric format I proposed at the beginning of this write-up:

So, consider the graph with varying values of m, when n = 2, b = 1, c = -1, and t ranges from 0 to 3.I suggest that the t = 0 point will remain at (1,-1), independent from m and n, and that n = 2 will have some influence on the slope of the graph. I also suggest that in the m = 0 condition, the length of the segment will be 3, which is the range of t, due to the observation that we had in the previous section.m = 1 in blue, m = 2 in red, m = 3 in green, m = .5 in pink, m = 0 in black, m = -1 in aqua

Here we can observe that (1,-1) was the t = 0 point. Also, the m = 0 condition produced a line of length 3. The slopes, however, varied widely. Upon examination we can see that each slope followed the pattern of slope = n/m. Even in the case of m = 0, the slope was undefined, as if we were presented with n/m = 2/0. I suggest, then, that the slope of each segment produced by parametric equations of the form:

is n/m.

Does this make sense? It does.

Consider the conceptualization of slope as "change in y"/"change in x", or Δy/Δx (or even "rise over run", if that rings a bell). Slope is simply a ratio of how y changes to how x changes. Consider, then, the parametric equation. If (b,c) is our "starting point" for when t = 0, then for each additional unit of t, the x and y value of our point changes by a value equal to m and n, respectively. Therefore, if the x coordinate is adjusted by a value of m, and the y coordinate is simultaneously adjusted by a value of n, then the slope, as we read it on the coordinate plane, will indeed by n/m.

So, we have learned two facts about the parametric equation of the form:

The t = 0 point is (b,c), and the slope of the resulting line segment is n/m. Can we use this information to solve a problem?

7. Write parametric equations of a line segment through (7, 5) with slope of 3. Graph the line segment using your equations. As a line segment, it will have end points. Explore how you would choose endpoints of such that the two distances from (7, 5) are 2 units and 3 units.

The first suggestion in this problem is easy when we apply what we have learned about parametric equations. We want to do two things:

1. Pass through (7,5)

2. have a slope of 3.

So, let us consider the parametric equation of the form:

Since we want to pass through (7,5), it would make sense to use that as our t = 0 point, since we do not have to worry about the interference of m or n. We also want the slope to be 3, so an easy way to produce this would be to set m = 1 and n = 3. This produces these equations, which satisfy the first requirements of the problem:

x = 1t + 7

y = 3t + 5

This can be seen in the following graph, considering t ranges from 0 to 3:

Let us consider the second part of the problem, then. How could we create a segment that had endpoints 2 and 3 units away from the point (7,5)? I suggest that this will explore a final aspect of these parametric equations that we have only touched on:

the range for t. If we can always have a t = 0 point of (7,5) and a slope of 3, given what we know about how parametric equations work, then, holding those constant, we should be able to manipulate our range of t to create a segment of the appropriate length. But how do we calculate the distance from these end points in general terms? It would surely be of no use to simply guess appropriate values of t.Let us attempt to develop a generalized formula for the length of a segment that represents a parametric equation of the form that we have been using for this write-up, namely:

Recall a conclusion we made earlier in the write-up: "If (b,c) is our 'starting point' for when t = 0,

then for each additional unit of t, the x and y value of our point changes by a value equal to m and n, respectively." (emphasis added) For each unit of t, which is a variable we have complete control over, the x and y coordinates adjust by m and n. Let us observe this on a graph!Note that the above graph is consistent with our observations of segment lengths when m or n is zero! If m or n is zero, then that side of the triangle is zero, and so the length of the line segment is all that we have to consider, and is completely dependent on the other (not zero) m or n.

Observe that we can describe the change in x and y as shown above in the legs of the triangle. Notice that b and c have no influence on the length of the segment generated by a parametric equation. The only variables that matter are m, n, and both the starting point (shown above as ) and ending point () of t. Therefore, large values of m and n will create a longer segment, even if given constant values for t. On the other hand, for fixed values for m and n (as seen in the problem we are solving), we should be able to manipulate one of the values of t (we will do both!) to create a segment of the desired length!

So, if we can create the triangle above, we can use the Pythagorean Theorem to determine the length of the segment:

This formula should be able to produce the length of a line segment produced by the parametric equation used for this write-up. Notice, once again, that b and c are absent from the equation. In addition, we can input a desired length and value for one of the points of t, and we should be able to solve for the other end of t! So, let us use the solution to the first part of our equation:

x = 1t + 7

y = 3t + 5

And work to find a segment that is length 3 whose endpoint is 3 units from the point (7,5). Recall that = 0, since that is the our "starting point", and the location of (7,5).

Therefore, if t ranges from 0 to , then we have a segment of length 3, and so the end point is 3 units from (7,5).

So what about the other endpoint being 2 units from (7,5)? Here it is important to remember that t can be negative. If we still allow = 0, then we can simply take whatever value we receive when solving for when d=2, and make it negative. This will cause the segment to extend in the other direction the desired distance, making the segment still 2 units long. If we use the same equation as seen above, we observe that to make d = 2, must be .

So, what if we want both endpoints at the same time? Easy! Just use both computed values of , one negative and one positive. Don't bother with t = 0. So long as 0 is in the domain, the segment will still pass through (7,5), and it will be 2 units in one direction, and 3 units in the other, assuming we used the appropriate endpoints.

So here is the solution to the equation:

x = 1t + 7

y = 3t + 5

For t = [-, ]