 In this video, we'll investigate curves that are defined parametrically. First, consider the path of an airplane used to write in the sky. Suppose the path of the airplane can be described by the set of parametric equations 2 plus t minus 3 sine of t for x of t and y of t as 4 minus 3 cosine t, where t represents the number of minutes since the pilot began writing. x and y are in units on the coordinate plane, describing the horizontal and vertical location of the airplane in that plane. Using our calculator, we enter our equations into y equals while in parametric mode. We'll set up a table and view the approximate values for x and y. So looking here in our function area, once we're in parametric mode, we see that the display allows us to enter parametric equations. I entered into x1 of t, 2 plus t minus 3 sine of t. Let's remember to be in radian mode. And for y1 of t, I've got 4 minus 3 cosine t. I chose a window that would allow me to see a little bit of the behavior of this curve. So let's go ahead and graph. Okay. I also have an interest in taking a look at the values for each t, what the x and y will be. Now we saw in the calculator that there's a direction associated with the path as t increases. We can depict this on our graph by labeling the path with arrows. What we noticed was that as t increased, the path was moving in this direction and so on. We also noticed that there are a number of interesting characteristics of the graph that we can identify and describe using calculus. We'll be doing a lot of this in the lessons to follow. And this would be things like the slope of the curve at a certain point. The point at which the location of the object moving is the same, but for different t values and so on. So let's take a look at a couple of the values that we see here in the table and identify where they are in our graph. So when t equals 0, we see the x value is 2, the y value is 1. So that's where t equals 0. Let's skip down to, let's say when t is 4, x is 8.27-ish and y is 5.96 or so. So let's see where that is. Each tick mark is one unit, 1, 2, 3, 4, 5, 6, 7, 8 and a little bit and 5 and a bit. So that's where t equals 4. Let's look at t equals 6, 8.8 or so, comma 1.1 or so. So we had 8.8 and 1.1 or so. So here we see that would be t equals 6. So that was just to give a picture of both the direction of the path and some particular values for x and y given a specific t value. Parametric equations also provide us with another way of expressing functions that can be modeled in rectangular form. Consider this set of parametric equations seen here, x of t being t plus 4 and y of t being t squared minus 5. What do we expect the path to look like? Well, the horizontal position, x of t, is described by a linear function of t and the vertical position is described by a quadratic function of the same parameter t. Let's take a look at the graph using our calculator. So here we see our equations, x of t being t plus 4 and y of t being t squared minus 5. We notice here that the path looks parabolic and has a vertex around 4, negative 5 or so. The tick marks are each in increments of 5. So this is vertex of 4, negative 5. We can actually show this algebraically as well by doing something called eliminating the parameter. Eliminating the parameter is a process by which we solve one of the parametric equations for the parameter and then substitute that into the other parametric equation to ultimately obtain a relation involving just x and y. So taking our equation x of t and solving that for t, we get t equals x minus 4. I'm going to then take that and substitute x minus 4 in for t in the equation for y. What we notice is this is actually a quadratic function in vertex form with vertex 4, negative 5, which we see here, 4, negative 5. If I expand this, I get x squared minus 8x plus 16 minus 5, which gives me x squared minus 8x plus 11. So this is a parabola with vertex 4, negative 5 and y intercept at 0, 11, which matches our graph.