In Exercises 71–76, eliminate the parameter and graph the plane c...

Question

In Exercises 71–76, eliminate the parameter and graph the plane curve represented by the parametric equations. Use arrows to show the orientation of each plane curve.

x = 2t − 1, y = 1 − t; −∞ < t < ∞

Accepted Answer

Welcome back. I am so glad you're here. We are asked to write the corresponding rectangular equation for the following parametric equation by eliminating T draw a graph of the plane curve using the rectangular equation and then indicate the direction of the curve using arrows. Our parametric equations are X equals five T plus one and Y equals seven minus T. And we're told that T is between negative infinity and positive infinity. Then we have a graph, we have a vertical Y axis, a horizontal X axis. They come together at the origin in the middle, the domain for what's shown for our X axis is from negative 15 to positive 15. And the range for what's shown for our Y axis is also from negative 15 to positive 15. All right, let's start off by getting our rectangular equation. So our parametric equations, X equals five T plus one and Y equals seven minus T. What we need to do is eliminate T. So we'll set one of these equal to T it's a little bit easier if we end up with a rectangular equation where Y is by itself on one side of the equation. So I'm going to work with the X equals equation and get T by itself. So we'll subtract one from both sides of the equation and we'll have X minus one equals five T and then dividing both sides by five. And we get that X minus one in the numerator, all divided by five is equal to T. Now, we can use substitution and put X minus one divided by five, substituting that in four T in our equation of Y equals seven minus T. So we'll have Y equals seven minus the quantity of, in the numerator, X minus one. That's all divided by five. And now what we do is we'll set up an XY table. We will choose values for X and plug those in to our equation. Our rectangular equation here and then get the corresponding values for Y and then plot those points. So for X I am going to strategically choose values for X that when we plug them in, they'll divide evenly by five. So for X, I'm choosing negative 14, negative nine, negative four, one six and 11. Now plugging those values in for X. So I'll just demonstrate one of them. And so we've got Y equals seven minus in the numerator. We'll have negative 14 instead of X minus one, all divided by five, negative 14 minus one is negative 15. We're dividing that by five and that will be subtracted from seven, negative 15, divided by five is negative three. So we have seven minus negative three, which is seven plus three, which is a positive 10. So our Y value when X equals negative 14, the Y value is 10. Now you can go ahead and pause the video if you need to plug all of those X values in. But here's what we get for the corresponding Y values when X equals negative nine, Y equals positive nine. When X equals negative four, Y equals positive eight, when X equals one, Y equals seven, when X equals six, Y equals six, when X equals 11, Y equals five. Now we can plot these points the first one negative 14 10 from the origin, we go to the left 14 units and up 10, negative 99 from the origin, we go to the left nine units and up nine negative 48 from the origin, we go to the left four units and up eight 17 from the origin, we go to the right one and up seven, 66 from the origin, we go to the rate six and then up six, 11, 5 from the origin, we go to the right 11 and up five. Now we can connect these for our plane curve. You can imagine that that's straighter and smoother than it is. And the last thing we're asked to do is indicate the direction of the curve using arrows. But this is actually where the T values are most helpful. So if we expand our chart and we figure out the corresponding T values, we just need to know where they're increasing and decreasing. So we can do just the top values for X and Y and the bottom values for X and Y we just need to solve for T. And we can do that using our equation that we have set where T is equal to X minus one divided by five. So we can use our X value of negative 14. So we have negative 14 minus one divided by five. That's going to be a negative 15 divided by five. So that's negative three. So T equals negative three on the left most point that we drew and now we can substitute 11 for X and sulfur T. So we have 11 minus one divided by 5, 11 minus one is 10, 10 divided by five is a positive two. So from left to right, the T values are increasing. So we can draw our arrows pointing from left to right along our plane curve. And there is our graph well done. We'll catch you on the next one.

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In Exercises 71–76, eliminate the parameter and graph the plane curve represented by the parametric equations. Use arrows to show the orientation of each plane curve. x = 2t − 1, y = 1 − t; −∞ < t < ∞