Calculus: Eliminating Parameters

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Parametric Equations and Polar Coordinates
Calculus and Parametric Equations
Eliminating Parameters Page [1 of 2]
Looking at the cycloid is sort of fun because we took a look at this interesting curve that sort of naturally appears when we see wheels turning around and moving or tires going through a street. What about the other direction? Suppose that instead of being given the curve and asked for the parametric equations, pretend that someone gave us the parametric equations and said, "Hey, what's the curve?" Well, you can actually figure out what the curve is given some parametric equations. Let's see some examples and see how to actually graph parametric equations if you're given the parameters.
So let's graph the following: How about x = t^3, y = t^2. So those are the parameters given in terms of time. Let me again remind you what that means. If you give me what t is, I will use that to compute the x-location at that time t, and then I'd use that here to find the y-location at that time t, and then I'd know precisely where on the plane the point is located, and as t varies this is going to vary.
So how would you graph that if you're given these parametric equations? One technique is to actually throw them away. That's right, the whole idea is to eliminate them. You spend all of this time understanding what they mean, and now we're going to throw them away. Well, that's one actual method of figuring out what the graph of this looks like; namely, throw away the t. If I just had y equals stuff with x's in it, I have a better grasp of how to graph those things since those are the things that we've been graphing since we were just little, little teeny mathematicians.
So the question is: How can I get rid of the t? One way to do that is to solve for t is the x-equation, and then whatever you get t to equal, plug that into the value of the t you find in the y-equation. So if you eliminate the parameters, all I do in this case is to solve this one for t. So that requires me to take the third root - or cubed root - of both sides, and so I see that = t. So that's what t = - t = . So there you go. Now what I can do is I can insert this new thing for t. This is just another way of saying t, right? That just equals t. So if I replace that little t there by this, then, in fact, I would just have things that have y's and x's in it. Now I insert this for t, and what I would see is the following: I would see that y = - that's what t is; I'm not done yet, I've got to square it -- ^2. Now, what is that? If you think of the cubed root as a power, that would be x^1/3 - that's the power that the cubed root produces - and then if I square that, I would see x^2/3. So, in fact, this path is really sweeping out the graph y = x^2/3.
Now, what is that the graph of? It's the curve that you can see in calculus. Let me draw you a little picture of it. You can actually use calculus to figure it out by finding out where it has a max or a min and where it is increasing or decreasing and find out the concavity. If you did that, you would find out that, in fact, the only thing that is really sort of interesting is that the origin, where in fact the derivative is undefined - the derivative is undefined there - and to the right it's increasing and to the left it's decreasing. So it's going to fall and then go up. If you take a look at the second derivative, you see the second derivative is negative everywhere except where it's undefined at the origin. So, in fact, this thing is always concave down. So it's concave down but decreasing; and then here it's concave down but increasing, and the derivative doesn't exist. But if you use all of this information, you would see this picture. You would see a picture that looks like this - just like that, that's the graph - and notice that there's this very sharp point here. This is the cusp. If you touch that, it won't be very smooth, it would really hurt. There's just one point there, and you'd prick your finger. So it's really a sort of dangerous point there, and that would be the only minimum in this curve, and it's a minimum because the first derivative doesn't exist but the function does exist because with zero, you take the cubed root, you get zero, and square it and you get zero, so the origin is part of this graph.
Anyway, this is actually the path that this parametric equation pair is describing - how the thing is going through on this path. So there's an example of how to graph something using parametric equations.
I thought I'd try one more with you just for fun. Let's try this one. How about x = e^t and at the same time the y-location is given by 3e^2t. So as time varies, the location of x varies like e^t, and y is 3e^2t. So how would I actually figure out what the path of this looks like? How would I graph that? Again, I'll just eliminate parameters. What I want to do here is try to eliminate the t here and replace by stuff with x's. Now, you could actually solve for t here - that's okay -- and then plug it in for t there, and you'll get the right answer, but that's a lot of extra work.
Let's look at this and just see if we can sort of reason out what this might be. This looks almost exactly like this, doesn't it? In fact, there's a burning desire just to replace this whole thing with x and say y = 3x. Well, that's wrong because they're not exactly the same, there's a little two there. So how can I get rid of that little two? Well, I could remember what it means to say e^2t. So I'm going to remind you what that is. In fact, I'll do this in many, many steps just in case. These are things that you don't think about every day. Well, 2t is just t + t, of course. What does it mean if you add exponents? If you're adding exponents that means that you're multiplying the bases. So, in fact, this really is this question in disguise: e^t x e^t because if you have the same base, you just add the exponents, and then I get t + t, which is 2t. So, in fact, this is what y equals, and now I can see that's exactly x, so I can put an x here and an x here. So I see that y = 3 times x times x, so y = 3x^2, and that's a graph that even I know, believe it or not.
That's just a happy-face parabola because it's positive in front. It's going to be centered at the start of the origin, and the three means that instead of being the good, old-fashioned, nice, curvy parabola, it's going to be a little tighter. We're going to move it all in a little teeny bit, so it's going to be a little more dramatic. If you were going to ride on this parabola, it would be a much more interesting ride than the usual parabola. It swoops down and swoops back up. Look at the beautiful job, live, on the fly. Beautiful, that's the graph of y = 3x^2, and that is the path described by these parametric equations.
So you can see that as you get parametric equations and to figure out the graph, all you can do, in fact one nice technique, is to try to eliminate the parameter - eliminate the t - and just get something in terms of x and y, and then graph that.
Anyway, that's all that's really involved in trying to figure out some of these paths. Sometimes they get more elaborate, but you see that the idea is always the same - straight forward - to remove the t, see what you're left with in terms of x and y, and then graph it using regular calculus. We'll take a look at more parametric facts when I talk to you next.