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 the 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 2 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 gonna lead to a nicer formula that's more familiar. So how do we do this? Well, basically, there's 2 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 5th power minus 2. I'm gonna add 2 to both sides and end up with y plus 2 equals t to the 5th power. Now in order to get t by itself, because I have t raised to the 5th power, I'm gonna have to basically, take this, and I'm gonna have to raise this to the 5th root. So I'm gonna have to raise this whole thing to the 5th root to get those two things to cancel, and this whole thing ends up being the 5th root of y+2 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 2, and then I'm gonna have to do the 5th roots of y plus 2 minus 1. 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 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 2. Option 2 is you basically are gonna solve the x equation for t. So this is gonna be x equals 2 t minus 1. I'm gonna add 1 to both sides and end up getting x plus 1 equals 2 t. Divide by 2 and divide by 2, so you end up getting x plus 1 over 2 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 1 over 2 because that's what t is, raised to the 5th power minus 2. 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 5th 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 5th 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.