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Removing the parameter in parametric equations (example 2)

Video transcript
Let's see if we can remove the parameter t from a slightly more interesting example. So let's say that x is equal to 3 times the cosine of t. And y is equal to 2 times the sine of t. We can try to remove the parameter the same way we did in the previous video, where we can solve for t in terms of either x or y and then substitute back in. And I'll do that. But I want to do that first, just to show you that it kind of leads to a hairy or an unintuitive answer. So if we solve for-- let's solve for t here. We could do it either one, they're equally complex. So if we solve for t here, we would say divide both sides by 2. You'd get y over 2 is equal to sine of t. And then you would take the arcsine of both sides, or the inverse sine of both sides, and you would get-- I like writing arcsine, because inverse sine, people often confuse it with an exponent, taking it to the negative 1 power. Arcsine of y over 2 is equal to t. Actually, let me do that little aside there. I should probably do it at the trigonometry playlist, but it's a good thing to hit home. Because I think people get confused. So arcsine of anything, let's say, y. The other way of writing that is sine minus 1 of y. These two things are equivalent, when they're normally used. But I don't like using this notation most of the time, because it can be ambiguous. This could mean sine of y to the negative 1 power, which equals 1 over sine of y. And arcsine and this are definitely not the same thing. So you want to be very careful there to make sure that you don't get confused when someone writes an inverse sine like this. But they're not actually taking sine of y to the negative 1 power. On the other hand, if someone were to write sine squared of y, this is unambiguously the same thing as sine of y squared. In fact, I wish this was the more conventional notation because it wouldn't make people think, oh, 2 and minus 1 there, and of course, that's just sine of y squared. So it can be very ambiguous. And of course, if this was a negative, this would be a minus 2, and then this really would be 1 over sine of y squared. That's why, just a long-winded way of explaining why I wrote arcsine, instead of inverse sine right there. Needless to say, let's get back to the problem. So we've solved for t in terms of y. Now we can substitute back here. And we've got an expression for x in terms of y. So we get x is equal to 3 times the cosine of t. But we just solved for t. t is this thing right here. So it's the cosine of arcsine of y over 2. And we have eliminated the parameter, but this is a very non-intuitive equation. We could have done the other way. We could have solved for y in terms of x and we would have gotten the sine of the arccosine. It would have been equally hairy or non-intuitive. But either way, we did remove the parameters so I guess we could mildly pat ourselves on the back. But that's not the purpose of this video. The purpose of this video is to see if there's any way we can remove the parameter that leads to a more intuitive equation involving x and y. And what we're going to do is, I guess you can call it a bit of a trick, but it's something that shows up a lot. Especially when you deal with polar coordinates. And you might want to watch my polar coordinate videos, because this essentially touches on that. But if I said-- let me rewrite them. x is equal to 3 cosine of t and y is equal to 2 sine of t. So what we can do is just think, well, how can we write this? And you know, cosine of t and [? kind ?] of t, how can we relate them? And the first thing that comes to my mind is just the unit circle, or to some degree, the most basic of all of the trigonometric identities. And that is that the cosine squared of t plus the sine squared of t is equal to 1. This comes from the unit circle. I explained it in the unit circle video, and that's because the equation for the unit circle is x squared plus y squared is equal to 1. The cosine of the angle is the x coordinate, the sine of the angle is the y coordinate, and so on and so forth. But this is our trig identity. You don't have to think about it too much right now. Just, I guess, know that it's true and watch some of the other videos if you want it proven that it's true. But if we can somehow replace this cosine squared with some expression in x, and replace the sine or the sine squared with some expression of y, we'd be done, right? And then we would have it equaling 1. And that shouldn't be too hard. We can rewrite this. We can set cosine of t equal to something in x, and we can set sine of t equal in something in y. So let's do that. We divide both sides of the equation by 3. You get x over 3 is equal to cosine of t. And if you divide both sides of this equation by 2, you get y over 2 is equal to sine of t. And then we can use this trigonometric identity. Instead of the cosine of t, we can substitute x over 3. Instead of the sine of t, we can substitute y over 2. And you get x over 3 squared-- that's that, right there, that's just cosine of t squared-- plus y over 2 squared-- that's just sine of t squared-- is equal to 1. And now this is starting to look a lot better than this. This, I have no idea what this is. But hopefully if you've watched the conic section videos, you can already recognize that this is starting to look like an ellipse. We can simplify it a little bit. We could say this is equal to x squared over 9 plus y squared over 4 is equal to 1. And if we were to graph this ellipse-- we will actually graph it-- we get-- let me draw my axis. I'm using this blue color a little bit too much, it's getting monotonous. OK, let me use the purple. So that's our x-axis. That's our y-axis. The major axis is in the x direction because the denominator here is larger than that one. And it's the semi-major radius-- this is going to be the square root of this, it's 3. 1, 2, 3 in that direction. 1, 2, 3. I know I'm centered in 0, because neither of these are shifted. You should watch the conic section videos if this sounds unfamiliar to you. And the semi-minor radius is the square root of 4, so that's 2. So they get 1, 2. And 1, 2. Let me see if I can draw this ellipse. So it looks something like that. There you go. So just like that, by eliminating the parameter t, we got this equation in a form that we immediately were able to recognize as ellipse. When I just look at that, unless you deal with parametric equations, or maybe polar coordinates a lot, it's not obvious that this is the parametric equation for an ellipse. But this, once you learn about conic sections, is pretty clear. It's an ellipse. And it's easy to draw that ellipse. But in removing the t and from going from these equations up here, and from going from that to that, like in the last video, we lost information. We lost, one, what is the direction that we move in as t increases? And we also don't know what point on this ellipse we are at any given time, t. So to do that, let's make our little table. So let's take some values of t. So we'll make a little table. t, x, and y. It's good to pick values of t. Remember-- let me rewrite the equations again, so we didn't lose it-- x was equal to 3 cosine of t, and y is equal to 2 sine of t. It's good to take values of t where it's easy to figure out what the cosine and sine are, and without using a calculator. We're assuming the t is in radiance, just for simplicity. So let's pick t is equal to 0. t is equal to pi over 2. That's 90 degrees in degrees. And t is equal to pi. And so what is x when t is equal to 0? Well, cosine of 0 is 1 times 3, that's 3. What's x, when t is just pi over 2? Cosine of pi over 2 is 0. 0 times 3 is 0. And what's x equal when t is equal to pi? Cosine of pi is minus 1. Minus 1 times 3 is minus 3. Fair enough. Now let's do the y's. When t is 0 what is y? Sine is 0, 0. So 2 times 0 is 0. When t is pi over 2, sine of pi over 2 is 1. 1 times 2 is 2. And when t is pi, sine of pi-- that's sine of 180 degrees-- that's 0. 2 times 0 is 0. So let's plot these points. When time is 0, we're at the point 3, 0. So 3, 0-- 3, 0 is right there. This is t equals 0. When t increases by pi over 2, or if this was seconds, pi over 2 seconds is like 1.7 something seconds. So at t equals pi over 2, we're at the point 0, 2. We're right over here. So this is at t is equal to pi over 2. And then when t increases a little bit more-- when we're at t is equal to pi-- we're at the point minus 3, 0. We're here. So this is t is equal to pi or, you know, we could write 3.14159 seconds. 3.14 seconds. And actually, you know, I want to make the point, t does not have to be time, and we don't have to be dealing with seconds. But I like to think about it that way. I like to think about, maybe this is describing some object in orbit around, I don't know, something else. So now we know the direction. As t increased from 0 to pi over 2 to pi, we went this way. We went counterclockwise. So the direction of t's parametric equations is in that direction. And you might be saying, Sal, you know, why'd we have to do 3 points? We could have just done 2, and made a line. If we just had that point and that point, you might have immediately said, oh, we went from there to there. But that really wouldn't have been enough. Because maybe we got from here to there by going the other way around. So giving that third point lets us know that the direction is definitely counterclockwise. And so what happens if we just take t from 0 to infinity? What happens if we bound t? t is greater than 0 and less than infinity. Well, we're just going to keep going around this ellipse forever. Multiple times. Keep writing over and over, infinite times. If we went from minus infinity to infinity, then we would have always been doing it, I guess is the way to put it. Or if we just wanted to trace this out once, we could go from t is less than or equal to-- or t is greater than or equal to 0. All the way to t is less than or equal to 2 pi. And in this situation, t really is the angle that we're tracing out. If we were to think of this in polar coordinates, this is t at any given time. And I just thought I would throw that out there. It isn't always, but in this case it really is. When you go from 0 to 2 pi radius, you've made 1 circle. But this is about parametric equations and not trigonometry. So I don't want to focus too much on that. But anyway, that was neat. When we started with this, if I just showed you those parametric equations, you'd have no idea what that looks like. But by recognizing the trig identity, we were able to simplify it to an ellipse, draw the ellipse. And then by plotting a couple of points, we were able to figure out the direction at which, if this was describing a particle in motion, the direction in which that particle was actually moving. Anyway, hope you enjoyed that.