Assignment 10: The Parametric Time Traveler

Time is Priceless, but it's Free

In high school mathematics, trigonometry and precalculus classes often introduce the concept of parametric curves. In two dimensional graphs, parametric coordinates represent the two coordinates X and Y in terms of a third variable, which we'll call t. The third variable is the "free" one that moves around; X and Y are fixed functions of t. As values change, we form a curve. Most often, we think of the parametric variable as time, which is why we call it t (forming the name of this assignment). It doesn't have to be time, though.

In this assignment, we'll look at two very common parametric curves. First is the parametric line, as we'll build a parametric segment through a point with a given slope. This represents moving across a slider, similar to what we've seen earlier. Second, we'll track the parametric cosine and sine as we travel around a circle. This foreshadows our next assignment on polar coordinates. Hopefully, by the end of these next two assignments, "the vaporous, insubstantial idea [of polar coordinates] becomes a solid there, a thing, a substance in a world of substances."

Parametric Line: Chaos is more freedom ... But no meaning.

One way to start is to plot a set of X and Y. Let's start with something straightforward, like X = t + 1; Y = 2 t - 1 . We need a range for t; let's say t goes from 0 to 5. At t = 0, we start the purple line at X = 0 + 1 = 1; Y = 2 (0) - 1 = -1 . X and Y both change linearly in t. At t = 3, we have X = 4 and Y = 5. At the end, t = 5, we finish at X = 5 + 1 = 6; Y = 2 (5) - 1 = 9 . If we increased the maximum value of t, we'd keep going in the upward direction.

A more meaningful challenge is to find the equation of a parametric line that passes through a given point, with a given slope. Let's try the following example.

Write parametric equations of a line segment through (7, 5) with slope of 3. Graph the line segment using your equations.

We know we must pass through the point where X = 7 and Y = 5. Let's make that where t = 0. Since we need a line, X and Y will be linear equations. So far, we don't know A and B, but we have two parametric lines:

X = 7 + A t ~~~~~ Y = 5 + B t

To find A and B, we need to use the slope. In mathematics, slope can be defined as the change in the vertical component divided by the change in the horizontal component. Since our desired slope is 3, 3 divided by 1, if we increase the horizontal X component by 1 unit we need to increase the vertical Y component by 3 units.

We'd like to set A to a simple value. We can't use A = 0, because that means X would never increase. Let's try A = 1. That makes X = 7 + 1 t . To increase X by 1 unit, we "jump in time" with t, from 0 to 1.

X = 7 + 1 t ~~~~~ Y = 5 + B t

To find B, we know that we have to keep a slope of 3. The value of X changes by 1 when t moves from 0 to 1. The value of Y needs to increase by 3 in the same t interval, so we need B = 3. When t = 0, Y = 5. When t = 1, Y = 5 + 3 (1) = 8 . That works, and we have our equations to graph. The point (7, 5) appears in red, while I let t range from -2 to 2.

X = 7 + 1 t ~~~~~ Y = 5 + 3 t

Parametric Circle: Everything seems simple until you think about it.

Before we reach trigonometry, we've seen the unit circle. It's the set of points where X^2 + Y^2 = 1 . You might have drawn it using X and Y coordinates. How might we graph this in a parametric way?

To draw a parametric circle, we'll need to use some trigonometry. One approach to Assignment 6 involved triangle trigonometry, but we haven't thought much about the circle. Using radian measure, let's look at a few points. We know \cos (0) = 1 ; \sin (0) = 0 . We've also seen some other values, like \cos (\pi / 4) = \sqrt{2} / 2 ; \sin (\pi / 4) = \sqrt{2} / 2 ; \cos (\pi / 3) = 1 / 2; \sin (\pi / 3) = \sqrt{3} / 2 . There are a lot more trig tables if we need them, but let's look at a graph of a few points. I plotted the unit circle in gray, then cosine and sine for radian measures of 0, \frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3}, \frac{\pi}{2} . All the points are on the unit circle.

Though we should be wary of patterns, this promising idea is true. You might have seen the identity \left( \cos t \right)^2 + \left( \sin t \right)^2 = 1 , which can be shown many different ways. This confirms that for any real value of t, the point (\cos t, \sin t) is on the unit circle. We can plot this point for various values of t. If we progress from t = 0 to t = 2 \pi , we travel around the entire circumference and get a complete circle.

What happens if we keep going? Well, we keep going around the circle in revolutions of length 2 \pi . For instance, the value of \cos (3.3 \pi) = \cos (1.3 \pi) = -0.588.

If you have Graphing Calculator, you can experiment with your own circular time travel. In the file asmt10circlerange.gcf, the parametric unit circle is drawn in gray. You can substitute other numbers for the blue and purple points. In the picture below, I used 8 for the blue point and (3.3 \pi) for the purple point. Remember that if the values for cos and sin differ, the point will likely not be on the unit circle.

The title, section headings, and other quotes are taken from Audrey Niffenegger's 2003 book The Time Traveler's Wife.