First eliminate the parameter, then graph the plane curve of the parametric equations.x(t)=2+costx\left(t\right)=2+\cos tx(t)=2+cost, y(t)=−1+sinty\left(t)=-1+\sin t\right.; 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 x ( t ) = 2 + cos t , y ( t ) = − 1 + sin t y\left(t)=-1+\sin t\right. ; 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 two parametric equations. We've got x equals two 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 one. You're probably more familiar with seeing it like sine squared theta plus cosine squared theta equals one, but you can rewrite it. And, also, it doesn't matter which variable you have. Cosine squared plus sine squared of anything always equals one. 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 here's the idea. So I have x is equal to two plus cosine of t. So what that means is that I'm gonna have to subtract the two to the other side and get cosine by itself. So in other words, x minus two is equal to cosine. Alright? So now what happens is if I wanna 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 two squared equals, and this could become cosine squared. Alright. And we're just going to 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 going to have to move the negative one to the other side. You get y plus one equals sign of t. Therefore, we get y plus one squared equals sign squared t. 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 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 two let me actually write this in purple. So let's x minus two squared plus, and this becomes y plus one squared equals one. 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 two squared plus y plus one squared, and that equals one. 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 one. So it's just got a radius of one because that is just the, because that's the, the right side over there. And the center the center of this is gonna be at this coordinate, which is the the two numbers that are inside the parentheses. Remember, you just always do the opposite of these numbers. So if it's negative two and positive one, the center actually just becomes positive two negative one. Alright? So in the word, the center of our circle is actually gonna be right over here at two comma negative one, and it's gonna have a radius of one. So that means that just means we can sort of draw a radius of one and then connect these pieces over here with just a smooth curve. Alright? So in other words, this is just where r equals one. So r equals one over here. All right. 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 t goes from zero all the way to two 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 going to go counterclockwise around the unit circle like this, because you would start from zero 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 ⁡ t x\left(t\right)=2+\cos t x ( t ) = 2 + cos t , y ( t ) = − 1 + sin ⁡ t y\left(t)=-1+\sin t\right. ; 0 ≤ t ≤ 2 π 0\le t\le2\pi 0 ≤ t ≤ 2 π

( 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

16. Parametric Equations & Polar Coordinates

Parametric Equations

Calculus

16. Parametric Equations & Polar Coordinates

Parametric Equations

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Multiple Choice

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

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

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.

Verified step by step guidance

Step 1: Start with the given parametric equations: x(t) = 2 + cos(t) and y(t) = -1 + sin(t), where 0 ≤ t ≤ 2π.

Step 2: Eliminate the parameter t by using trigonometric identities. Recall that cos²(t) + sin²(t) = 1.

Step 3: Express cos(t) and sin(t) in terms of x and y. From x(t), cos(t) = x - 2. From y(t), sin(t) = y + 1.

Step 4: Substitute cos(t) = x - 2 and sin(t) = y + 1 into the identity cos²(t) + sin²(t) = 1. This gives (x - 2)² + (y + 1)² = 1.

Step 5: Recognize that the resulting equation (x - 2)² + (y + 1)² = 1 represents a circle centered at (2, -1) with a radius of 1. Graph this circle in the Cartesian plane.

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