First eliminate the parameter, then graph the plane curve of the parametric equations.x(t)=2+costx\left(t\right)=2+\cos t, y(t)=−1+sinty\left(t)=-1+\sin t\right.y(t)=−1+sint; 0≤t≤2π0\le t\le2\pi0≤t≤2π

(x−2)2+(y+1)2=1\left(x-2\$$right)^2$$+\left(y+1\$$right)^2=1(x$$−2)2+(y+1)2=1

First eliminate the parameter, then graph the plane curve of the parametric equations. x ( t ) = 2 + cos t x\left(t\right)=2+\cos t , y ( t ) = − 1 + sin t y\left(t)=-1+\sin t\right. y ( t ) = − 1 + sin t ; 0 ≤ t ≤ 2 π 0\le t\le2\pi 0 ≤ t ≤ 2 π

Welcome back, everyone. So hopefully you had a chance to try this practice problem. We're given 2 parametric equations. We've got x equals 2 plus cosine t and y equals negative one plus sine of t. These are parametric equations that involve trig functions. So what we're first gonna do is not graph them. The first thing we're actually gonna do is eliminate the parameter, and then we're gonna get a rectangular equation. And then we'll go ahead and graph the plane curve because it'll be it'll be much easier than just trying to brute force a bunch of t values. Alright? So let's get started here. Remember, the whole entire thing here is we're trying to basically rely on this trigonometric identity that we know, which is that cosine squared t plus sine squared t is equal to 1. You're probably more familiar with seeing it like sine squared theta plus cosine squared theta equals 1, but you can rewrite it. And, also, it doesn't matter which variable you have. Cosine squared plus sine squared of anything always equals 1. Alright? So here's the idea. So I have equations here involving cosine, and I have another equation involving sine. I need to rewrite both of those equations in terms of cosine and sine so I can plug those in and eliminate the t parameter and only just get x's and y's. Okay? So the here's the idea. So I have x is equal to 2 plus cosine of t. So what that means is that I'm gonna have to subtract the 2 to the other side and get cosine by itself. So in the words, x minus 2 is equal to cosine. Alright? So now what happens is if I want to use this identity, I need cosine squared, not cosine, so I just have to square both sides. So in other words, this just becomes x minus 2 squared equals, and this could become cosine squared. Alright? Now we're just gonna do the exact same thing with the sine equation. So with this over here, I'm just going to get that y is equal to negative one plus sine of t. I'm going to isolate sine. So I'm just gonna have to move the negative one to the other side. You get y plus one equals sin of t. Therefore, we get y+2 equals sin2t. Okay. So what does this equation end up becoming or this identity? Well, basically, what happens is when you move it down here and then you sort of combine all of your, and you relate this to the x and y equations that you've just gotten, what we end up getting here is that x minus 2 let me actually write this in purple. So let's x minus 2 squared plus, and this becomes y plus 1 squared equals 1. Right? So cosine squared just ends up being cosine squared ends up being this piece over here, and the sine squared ends up being this piece over here. Alright? So if you just combine them, then you just get this x minus 2 squared plus y plus 1 squared, and that equals 1. So this is a more familiar equation that just involves x's and y's. We've eliminated the parameter. We've actually seen this type of equation before. This is actually just the equation of a circle. So this is an equation of a circle. Radius of 1. So it's just got a radius of 1 because that is just the, because that's the, the right side over there. And the center the center of this is going to be at this coordinate, which is the the two numbers that are inside the parentheses. And remember, you just always do the opposite of these numbers. So if it's negative 2 and positive 1, the center actually just becomes positive 2 negative 1. Alright? So in the word, the center of our circle is actually gonna be right over here at a 2 comma negative one, and it's going to have a radius of 1. So that means that just means we can sort of draw a radius of 1 and then connect these pieces over here with just a smooth curve. Alright. So in other words, this is just where r equals 1. So r equals 1 over here. Alright. And this is the equation of our parametric curve. Now, what happens here is that we're supposed to sort of iterate on values or we're supposed to when you're graphing these things, you're always supposed to pay attention to the t values. And in this case, what we're told here is that 0 that t goes from 0 all the way to 2 pi. So you can kind of actually think about this like going around the unit circle. So what you would do is you would start over here, and you're basically just gonna go counterclockwise around the unit circle like this because you would start from 0 and then go all the way around to pi. Alright. So that's the orientation of that plane curve. That's it for this one, folks. Let me know if you have any questions.

First eliminate the parameter, then graph the plane curve of the parametric equations.x(t)=2+cos⁡tx\left(t\right)=2+\cos t, y(t)=−1+sin⁡ty\left(t)=-1+\sin t\right.y(t)=−1+sint; 0≤t≤2π0\le t\le2\pi0≤t≤2π

(x−2)2+(y+1)2=1\left(x-2\$$right)^2$$+\left(y+1\$$right)^2=1(x$$−2)2+(y+1)2=1

(x+2)2+(y−1)2=1\left(x+2\$$right)^2$$+\left(y-1\$$right)^2=1(x+2)2+(y$$−1)2=1

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32m

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8. Polar Coordinates and Complex Numbers

Eliminate the Parameter

Algebra and Trigonometry

8. Polar Coordinates and Complex Numbers

Eliminate the Parameter

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Multiple Choice

First eliminate the parameter, then graph the plane curve of the parametric equations.
$x (t) = 2 + \cos t$ , $y$\left$(t)=-1+$\sin$ t$\right$.$ ; $0$\le$ t$\le$2$\pi$$

$$\left$(x-2$\right$)^2+$\left$(y+1$\right$)^2=1$

Graph showing a circle centered at (2, -1) in the Cartesian plane.

$$\left$(x-2$\right$)^2+$\left$(y+1$\right$)^2=1$

Graph showing a circle centered at (2, -1) in the Cartesian plane.

$$\left$(x+2$\right$)^2+$\left$(y-1$\right$)^2=1$

Graph showing a circle centered at (2, -1) in the Cartesian plane.

$$\left$(x+2$\right$)^2+$\left$(y-1$\right$)^2=1$

Graph showing a circle centered at (2, -1) in the Cartesian plane.

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First eliminate the parameter, then graph the plane curve of the parametric equations.x(t)=2+costx\(\left\)(t\(\right\))=2+\(\cos\) t, y(t)=−1+sinty\(\left\)(t)=-1+\(\sin\) t\(\right\).y(t)=−1+sint; 0≤t≤2π0\(\le\) t\(\le\)2\(\pi\)0≤t≤2π

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First eliminate the parameter, then graph the plane curve of the parametric equations.
$x (t) = 2 + \cos t$ , $y\(\left\)(t)=-1+\(\sin\) t\(\right\).$ ; $0\(\le\) t\(\le\)2\(\pi\)$