Eliminate the parameter to rewrite the following as a rectangular equation.x(t)=2t−1x\left(t\right)=2t-1x(t)=2t−1y(t)=t5−2y\left(t\right)=t^5-2y(t)=t5−2

y = ( x + 1 2 ) 5 − 2 y=\left(\frac{x+1}{2}\right)^5-2 y = ( 2 x + 1 ) 5 − 2

Eliminate the parameter to rewrite the following as a rectangular equation. x ( t ) = 2 t − 1 x\left(t\right)=2t-1 x ( t ) = 2 t − 1 y ( t ) = t 5 − 2 y\left(t\right)=t^5-2 y ( t ) = t 5 − 2

Everyone. Welcome back. So in this practice problem, we're going to eliminate the parameter from these parametric equations to rewrite this as a rectangular equation. Really, all this means here is that we're gonna get rid of t and end up with an equation that just has x's and y's in it. Let's get started. So remember that basically what happens in this is that you're gonna solve one of the equations for t and then plug it into the other one. There's always two ways you could do this in these problems. You could take the x equation, solve it for t, and then plug that into the y equation. Or you could do the opposite. You can solve the y equation for t and then plug that into the x. There's no strict rules about which one is correct. It really just sort of depends on the on each case. But usually, one is going to lead to a nicer formula that's more familiar. So how do we do this? Well, basically, there's two options here, and the first one is gonna be solving for the y equation and then plugging it into the x. So I wanna walk you through how to do that. So if we take the y equation and then try to solve for t, what you're gonna end up with is that y equals t to the fifth power minus two. I'm gonna add two to both sides and end up with y plus two equals t to the fifth power. Now, in order to get t by itself, because I have t raised to the fifth power, I'm gonna have to basically, take this and I'm gonna have to raise this to the fifth root. So I'm gonna have to raise this whole thing to the fifth root to get those two things to cancel, and this whole thing ends up being the fifth root of y plus two equals t. And now I've solved for t. Now I can plug this into the x of t equation. When I do that, what I'm gonna end up with is that x equals two, and then I'm gonna have to do the fifth roots of y plus two minus one. Alright. So notice how this equation looks nothing like anything that we've ever seen before, but it is a rectangular equation. It does have x's and y's. So usually what happens is that one of these things is gonna be more complicated because you're gonna start off with the more complicated expression and then plug it into the simpler one. But although this is accurate and although this is correct, this is probably the bad choice. And instead, what we're gonna do here is we're gonna do it the other way around. We're gonna solve for this x equation first and then plug it into the y equation, and you'll see that it's a lot nicer. So here's option two. Option two is you basically are gonna solve the x equation for t. So this is gonna be x equals two t minus one. I'm gonna add one to both sides and end up getting x plus one equals two t. Divide by two and divide by two, so you end up getting x plus one over two equals t. Now if I take that and I plug it into the y of t expression, what I'm gonna end up with is I'm gonna end up with y equals this is just gonna be parenthesis x plus one over two because that's what t is, raised to the fifth power minus two. So notice how this is a lot nicer. This is a much better expression for a rectangular equation. It still just has y's and x's, and notice how we have y equals some expression of x raised to the fifth power. Don't actually have to sort of multiply everything out because you could just perfectly leave it fine like this. This would probably look like something like this on a graph, something that looks like to the fifth power, if you were to graph it out, but we don't have to graph it. All we have to do is just get it to a rectangular equation. Alright? So we've eliminated the parameter. That's it for this one, and let me know if you have any questions.