In Exercises 9–20, use point plotting to graph the plane curve de...

Question

In Exercises 9–20, use point plotting to graph the plane curve described by the given parametric equations. Use arrows to show the orientation of the curve corresponding to increasing values of t.x = 2t, y = |t − 1|; −∞ < t < ∞

Accepted Answer

Welcome back. I am so glad you're here. We're asked to draw a graph of the plane curve defined by the following parametric equations. With the help of point plotting indicate the direction of the curve obtained corresponding to increasing values of T using arrows. All right, we are given parametric equations of X equals five ty equals the absolute value of T minus two. And then we are given an interval for T from negative infinity to positive infinity. Then we're given a graph, we have a vertical Y axis, a horizontal xx axis. They come together at the origin. The domain of what's shown for our X values is from negative 20 to positive 15. And the range of what's shown for our Y values is from negative 10 to positive 10. All right, we recall from previous lessons and lessons in the book that the first step when we are graphing with uh parametric equations is that we are going to select T values within our given interval. But within our given interval is from negative infinity to positive infinity, we can never select enough from that interval. So because we're given a graph and we've got that domain from negative 20 to positive 15, we can actually use that. So I'm going to have a start by drawing our table. So the first column of our table is going to be the T values typically what we would start filling in. But this time, we're not going to start filling that in this time, we're going to start filling in our second column, which is the X column and that will be X equals five T. So we're going to start there and then we're going to use that information to get our T values. And then we'll use that to find our Y values. The third column is Y equals the absolute value of T minus two. And then we'll have for our fourth column which will be the XY values. OK. So starting with X, let's start to the left of the origin as far as it goes as negative 20. So we'll have an X value of negative 20 that equals five T. Now solving for T here we divide both sides by five and we get a T value of negative four. But remember that column, what we're finding is the X value. We already assigned it, it's negative 20. Now, we can fill in the T column negative four and wanted to switch colors there. All right, T negative four. Now we can plug T in, in the Y column. Y equals the absolute value of not T but negative four minus two and negative four minus two is negative six. So this is the absolute value of negative six. Well, the absolute value of negative six is positive six. So our Y value here is a positive six. So we can plot this point negative 26 from the origin. We go to the left 20 then we go up six. All right. Now let's choose an X value of negative 10. So we say negative 10 equals five T dividing both sides by five. We're going to get a T value of negative two. But remembering that our X value that we chose is negative 10. Now we can use that T value of negative two to solve for Yy equals the absolute value of negative two minus two, negative two minus two is negative four. So we have the absolute value of negative four. That's a positive four. So this point is negative 10, 4 from the origin, we'll go to the left 10 and we go up four. Alright. Now let's choose an X value of 00 equals five T. And solving for T here dividing both sides by five, we're going to get a T value of zero. And then we can plug that in to solve for Yy equals the absolute value of zero minus two, which is the absolute value of negative two, which is a positive two. So we can plot that point from the origin will go up to for our next X value. Let's choose positive 10. So 10 equals five T dividing both sides by five will get a T value of positive two. For our Y value, we have Y equals the absolute value of two minus two, two minus two is zero. The absolute value of zero is zero. So we have the 0.10 from the origin. We'll go to the right 10 units. Let's have our last X value be positive 15. Solving for T dividing both sides by five, we get a positive three. So plugging that into sulfur, yy equals the absolute value of three minus 23 minus two is one. So the absolute value of one which is positive one. Now plotting the 0.15 1 from the origin, we go to the right 15 and we go up one right now we try to connect that as smoothly as possible. Oh My goodness, this is not very smooth. Well, so that's maybe that's as smoothly as I could possibly do. So the last thing we need to do here is indicate the direction of the curve correspond obtained corresponding to increasing values of T using arrows. So we look at our T column and the first point negative T was negative four. The last 0.15 1 T was positive three. So T was increasing from left to right. We draw arrows along our curve to show that T increases from left to right. And there, we have it, there is our graph well done. We'll catch you on the next one.

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In Exercises 9–20, use point plotting to graph the plane curve described by the given parametric equations. Use arrows to show the orientation of the curve corresponding to increasing values of t.x = 2t, y = |t − 1|; −∞ < t < ∞

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