So in recent videos, we've learned how to eliminate the t parameter from parametric equations to get back to equations just involving x and y. Well, some questions will actually have you do the opposite. They will give you a rectangular equation, and they'll ask you to find parametric equations for it. In other words, they'll ask you to write an xt or a yt. This is also sometimes called parameterizing an equation. It's really just the opposite of eliminating the parameter. So I'm going to walk you through a step-by-step process of how to do this because there are a couple of things you want to keep in mind as you are doing this. Let's just jump right in. Alright. So remember how when we eliminated the parameter, the basic idea was that we would solve one of these equations for t, then plug it into the other one and get rid of the t variable. Alright. Now we're doing the opposite. I'm going to go from an equation that involves just y and x, and I'm going to end up with an xt and a yt. So the first thing that you have to do here is you're going to have to choose an expression for t. You're going to have to pick something, an expression for t, either involving x or y. So what you're going to do here is if this is your equation, the first thing I can do for choosing t is I could just choose t to be x itself. It sounds kind of silly because I'm just choosing one variable to be another, but what happens? Well, what I'm going to do here is once I've chosen my t, is I'm going to solve for xt. I'm going to get back to this in just a second, by the way. So I'm going to solve for xt. If t is equal to x, that means that xt is just t itself. Right? xt is just t. All I've done is swap the equation. So now that I've gotten one of my parametric equations, xt is t, how do I get the second one, the yt? All you're going to do is you're just going to substitute the xt expression into the original equation and then solve for your yt. So in other words, if I pop this into my original equation and every time I see x, I just replace it with a t, what I end up with for y is I just end up with 4(t+1). Alright. So all I've done here in this equation is I ended up with the same exact y equation. I've just traded the x variable for a t. It seems kind of silly because I just pick one variable to be another. But, in fact, if you actually were to plot out a bunch of t values, you would get the same exact x and y pairs as this equation. They actually just describe the same exact line. Alright? So it turns out that setting t equal to x is pretty much almost always going to work in your problems. Alright? Now most problems, what they'll have you do is they'll have you sort of they'll prevent you from from choosing xt equals t because they want you to be a little bit more creative. Alright? Now this only worked because I chose t to be x, but I also could have chosen t to be something else. So let's come up with a different set of parametric equations for this. Alright? So, again, I'm going to have to choose t, and it's going to usually involve some expression of x. So if I look through this equation over here, one of the things I can do is I could set t to be everything that's inside of this parenthesis over here. So I can set t not to just be x, but x+1. Again, I'm just going to solve my x expression for t. So all I'm going to do is just subtract t-1. And then if I pop this into the original expression, what I'll end up with is that 4=y equals 4(t-1+1). The minus 1 plus 1 will cancel and leave me with just 4×t. So, again, this is another perfectly valid set of parametric equations. If you were to plot out a bunch of values, you'll get the same x y pairs as this equation and also your first set as well. So it just depends on what you chose t to be. There's an infinite number of possibilities, but usually, you're going to want to choose something really simple. Alright? That's really all there is to it. There's one thing that I haven't mentioned yet for choosing t, which is that you're going to want to choose a t that avoids domain restrictions. So, for example, one of the best things to do here is either to choose x, a t equals x, or some kind of multiple or addition to x. You're really just going to want to pick variables that are odd powers of x because t can be any very t could be any number and then x could be any number There's no domain restrictions there What you want to avoid is you want to avoid even powers of x like x^2 or the square root of x. Because what's going to happen is when I solve this expression for x, I'm going to get the square root of t and then t can't be any negative numbers because then I'll get imaginaries. You usually gonna want to avoid even powers of x. Alright? So that's all really all there is to it. Let's go ahead and take a look at some examples because I want to walk through a step-by-step process of how to do this. Alright. So let's take a look. So without choosing xt to be t, we're going to have to find parametric equations for these rectangular equations. Alright? Let's take a look at the first one, example a. We've got that y is equal to 2x+5, and, actually, what's given to us already is what t equals. So they've already actually chosen t for us. So the first step in these problems is you're going to define t unless you're already given that what that t expression is going to be. So in this case, we don't have to worry about anything. So if t is equal to x+1, let's move on to the second step. What you're going to do here is you're going to solve that expression of t for xt. So if t is equal to x+1, that means that xt is just going to be I'm just going to have to subtract 1 from both sides. I'm just going to be t-1. So just imagine that you had this equation here, and you were just solving for x. You'd have to move this to the other side, and you just get xt. So that's what happens is you'll often see a lot of times these parentheses get added in once you get x as a function or an expression of t. Okay. The third step is we're now going to plug that xt into our original y of x equation to get y to get yt. So what happens here is if I plug this into my 2,x+5, every time I see x, I replace it with t-1. So this is going to be t-1 with a parenthesis, then +5. Alright? Now what happens here is I get y as a function of t. So this is just going to be 2t-2+5, and we can simplify this minus 2 plus 5 ends up being 2t+3. Alright? So that's my set of parametric equations. I ended up getting that xt is t-1 and yt is equal to 2t+3. And that's all there is to it. That's the 4th step. So let's go ahead and now take a look at this example b over here. I've got y equals, it's a little bit more complicated, x+2 squared minus 3. Let's go through the steps. We're going to have to define t unless it's given to us. Now there's some guidance here. You can always try t equals x. However, the problem told us we actually can't do that. We can't just choose xt to be t. But another perfectly valid option that you'll often see a lot is you can choose t to be whatever expression is inside of the parentheses. Not the not the power, just whatever is in the parentheses. So in this case, what I'm going to do here is I'm just going to set t to be equal to x+2. That's the thing that's inside the parentheses over here. Alright? It's another perfectly valid thing that you can do. Let's just see how it works. The second step is I'm going to solve this expression for xt. If I just flip this around, what I'm what I'm going to get here is that xt is equal to t-2. Alright? Now, Now let's move on to the 3rd step. I'm just going to plug this xt into my original y of x expression. So this ends up being y equals. And now what I have here is parenthesis squared, and the thing that goes in this parenthesis here is going to be this t-2. Right? Because every
Writing Parametric Equations - Online Tutor, Practice Problems & Exam Prep
Created using AITo convert a rectangular equation into parametric equations, start by defining a parameter \( t \) based on either \( x \) or \( y \). For example, if \( t = x + 1 \), then \( x(t) = t - 1 \). Substitute \( x(t) \) into the original equation to find \( y(t) \). This process can also apply to equations like \( x^2 + y^2 = 1 \), using trigonometric identities to derive \( x(t) = 2 \cos(t) \) and \( y(t) = 3 \sin(t) \). Always ensure your choices avoid domain restrictions, particularly with even powers.
Parameterizing Equations
Video transcript
Parameterizing Equations Example 1
Video transcript
Welcome back, everyone. So in this example, we're gonna take a look at this equation, y=2x3, and we're gonna write 2 sets of parametric equations that describe this rectangular equation. So in other words, they want us to parameterize this y=2x3 equation. So let's take a look at the steps of how to do this. The first thing you're gonna do in these problems is you're gonna have to define t unless it's given to you already. In this case, it's not. So in this case, what we're gonna have to do is choose a t to parameterize our equation, and there's always some guidance to this. In this case, you can always just try t equals x or t equals whatever the thing in the parentheses is if you're given some kind of a parentheses in the equation. And in this problem, we're not told that we can't use x=t or t=x. So this is always gonna be a viable solution. So option 1 is that we just choose x=t or t=x. Right? So if we let t=x, then the second step is we're gonna have to solve for x(t). So in other words, we're gonna have to get x in terms of t. And if t=x, so you just flip the equation around. In other words, x(t)=t itself. That's the first equation, the first parameterized equation. Now to get the y equation, we're just gonna have to plug that x(t) into the original equation for x. In other words, the y=2x3. And let's go ahead and do that. So if x(t)=t itself, that means that y is equal to 2. And instead of now plugging in x cubes, we're really just gonna replace this with t3 because that's what x is equal to. Right? So in other words, this really just becomes 2t3. Alright? So this is your x(t) equation, and this is your y(t) equation. So again, it's always kind of silly how these problems work out because if you're not told that you can't use this option or this definition of t, then this is always a viable set or a solution to your parametric equations. You can always just basically replace x with t, and then your y of equation just becomes the original equation, but you've replaced x with t. Alright? So that's always one of your options. So that's one of your sets of parametric equations. Or what we can also do is we're gonna have to pick another sort of definition for t and sort of parameterize the equation a different way. Remember, there's always a ton of different ways that you can do this. There's no really correct answer, but there are some easier ones, for defining t. Alright? So let's go ahead and take a look at the second one here. You're going to define t unless it's given to us. In this case, we can't use t=x, obviously, because we just use that in the first one. So let's just try to pick a different definition for t. What we can do here is whenever you're just exhausted with you know, t equals x or whatever the parenthesis is, you can start to just try to set t in terms of powers of x if you have t if you have powers of x. And so, again, so what you can do here is you can set t=x3. You always wanna avoid even powers of x, so like x2 or the square root of x or something like that, because, remember that you have to sort of avoid those domain restrictions. But t=x3 is a perfectly valid solution because t and x are both defined for all numbers, positive and negative. So t=x3 is perfectly fine here. So now what we do is we've defined t, and we just go ahead and solve for x(t). So now in this case, if we want x in terms of t, I'm going to have to get x by itself. And in this case, what happens is if t=x3, that means that x is actually just going to be the cube root of t. I'm just solving for the other variable, the x. Alright? And, again, this is perfectly fine here because I'm gonna have I'm gonna get negative and positive numbers, and those are perfectly fine. So now, I'm just gonna plug that x(t) into the original equation for y. So in other words, I'm gonna get y=2. And instead of x, now I'm gonna replace it with the function that I have, which is just cube root of t. So the original equation was 2x3. So now if I plug in the x of t expression into this, I'm gonna have the cube root of t, but then I'm gonna have to cube it. So if we simplify this, what we're gonna see here is that this is really just becomes 2, and then the cube root of y cubed, that actually just sort of undoes the cubing. So this actually just ends up becoming 2t. Alright? So your parameterized equations, your second set of parametric equations ends up being x(t)=t and y=2t itself. If you picked a different set of equations, you may have gotten different answers than me, but this is one possible solution to this parameterized equations. Alright. That's it for this one, folks. Let me know if you have any questions.
Equations of Circles & Ellipses
Video transcript
So in the last couple of videos, we saw how to take parametric equations involving trig functions and, using this Pythagorean identity, rewrite them to have only just x's and y's. Well, just as we've done with previous videos, lots of problems will ask you to do the opposite. They'll give you these equations involving x's and y's and ask you to write parametric equations for them. What I'm going to show you is a step-by-step process of how to do this, and it's really just the opposite of what we did with eliminating the parameter. We're still just going to use this Pythagorean identity to write 2 parametric equations for this equation involving x's and y's. Let's go ahead and get started here. We're just gonna jump right into this example. Alright?
So whenever you have an equation that contains something like x2+y2, remember that's an equation of a circle or an ellipse. Right? Obviously, there are other numbers involved here, but the basic idea here is that you have x2+y2=1. So what you're gonna do in these problems is, for example, if you have something like x22+y2, we have obviously some other numbers, but we have x2+y2 in there. So we can parameterize this. The key idea here is we're gonna rewrite this equation to be in the form we have some kind of function of x2 plus some kind of function of y2, and that equals 1 on the right side. So if you take a look at this equation here, we've got something of x2 plus something of y2 equals 1. And if you sort of just sort of match these things, you'll end up seeing that this is actually a function of x that's squared plus some kind of function of y that's squared, and that equals 1. So this is in this form so we can parameterize the equations using this step.
Alright, these steps. And so here's how this was gonna work. Alright? How does this how do we use this to get back to parameterized equations involving just t? Well, the key idea is we're still just gonna use this Pythagorean theorem or, sorry, this Pythagorean identity. So we have cos2t+sin2t=1. And in this case, that something is just the parameter t. Alright. So what you're gonna do here is notice how these equations actually look kind of similar. We have something that's squared plus something that's squared equals 1. So the key idea here is that you're just going to set that equation involving x to be cos2. You're gonna set that function involving x to equal cos2, and you're gonna set that function involving y to be sin2. And then what you're gonna do here is you're just gonna go ahead and try to isolate x and y. Okay?
So here's how this works. I'm gonna take my expression involving x, which is gonna be x22, and I'm gonna set that equal to cos2t. Now, what I have to do is just have to get x by itself. Well, if I take the square root of both sides, I just get that x over 2 is equal to cosine of t, and if I get x by itself just by moving the 2 over to the other side, I end up with 2 cosine of t. Notice how I have just x by itself equal to some kind of expression involving t. That's why you'll see this written as x of t. We're gonna do the exact same thing with the y equation. So, in this case, what happens is I got y2=sin2t, and I'm just gonna go ahead and isolate y and get it by itself. Take the square root of both sides. You'll end up with that y is equal to sine of t, and you don't actually have to do any other steps here, so you could just repeat because y of t is equal to sine of t. Notice how we ended up with 2 parameterized equations, x of t and y of t, that both involve the parameter. This actually makes sense that we kinda ended up back at the 2 equations that we started out with over here on the left side, because all we did was we just did the reverse process of eliminating the parameter. Remember, if you're ever unsure of how you've parameterized your equations, you can always just eliminate the parameter to get back to the original equation, and you should get back to where you started from.
Alright? Now problems would always be this straightforward, so I want to go ahead and walk you through a step-by-step process of how to do that using our next example. Alright? Let's take a look. So we're gonna write parametric equations for this example over here. We've got 9x2+y2=9. So let's take a look here. This is an equation just like the previous one that involves x2+y2. There are other numbers, but their key thing is that there's something like x2+y2 in there. Alright?
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Write parametric equations for the rectangular equation below.
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