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 = t − 2, y = 2t + 1; −2 ≤ t ≤ 3

Accepted Answer

Welcome back. I am so glad you're here. We are 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. And our parametric equations are X equals T minus five, Y equals three T plus two. And our T values must be between negative one and four inclusive of negative one and four. Then our, we have a graph with a horizontal X axis, a vertical Y axis. They come together at the origin in the middle, the domain for the given X axis shows it from negative 10 to positive 10. And the domain for the shown part of the Y axis, excuse me, the range for the shown part of the Y axis is from negative 15 to positive 15. All right. So we recall from previous lessons and from the steps in the textbook that step one is to select T values within our given interval. So I'm going to start to draw our table here. The first column is going to be for the T values. The second column we're going to have the work to find our X values where X equals T minus five. The third column is going to be for the work to find our Y values. We're told for the parametric equation Y equals three T plus two. And the fourth column is going to be four hour XY values. All right. So choosing our T values within the given interval, so the lowest we can go is negative one. So I'll start with negative one and then zero, then 123 and four. So we can go all the way up to a positive four inclusive of positive four. And now that we've chosen those, the step two is to compute X and Y. So we've got X equals for A T value of negative one. We'll have negative one minus five and negative one minus five equals negative six. So our first X value is negative six. Now we'll compute our Y value. We have Y equals three multiplied not here by T but by negative one plus two, three, multiplied by negative one is negative three. So we have negative three plus two, which equals negative one. So where X equals negative six, Y equals negative one, plotting that point from the origin will go to the left six and will go down one. All right. Now let's work on A T value of zero. We'll start with X equals not T but zero minus five and zero minus five is negative five. So for A T value of zero, X equals negative five. Now let's work on yy equals three, multiplied not by T, but by zero plus two and three, multiplied by zero is 00 plus two equals two. So where X equals negative five, Y equals two from the origin will go to the left five units and will go up two units. Now let's work on A T value of one X equals not T but one minus five and one minus five is negative four. So we have an X value of negative four. And now let's solve for yy equals three, multiplied not by T but by positive one plus 2, 3, multiplied by one is three. So we have three plus two and three plus two is five. So where X equals negative four, Y equals positive five from the origin, we go to the left four and up five. All right. Now let's work on A T value of two, X equals not T but two minus 52 minus five is negative three. So there's our X value negative three. Now we'll work on yy equals three, multiplied not by T but by two plus 23, multiplied by two is six plus two and six plus two is eight. So where X equals negative three, Y equals eight from the origin, we go to the left three and we go up eight. All right. Next, we will work on a T value of three, X equals not T but three minus 53 minus five is negative two. So our X value is negative two. Now solving for YY equals three, multiplied not by T but by three plus 2, 3, multiplied by three is nine plus +29 plus two is 11. So where X equals negative two, Y equals 11 from the origin, we go to the left two, then up 11. And now our final value for tea is at four, X equals not T but four minus 54 minus five is negative one. So our X value is negative one. Now solving for yy equals three, multiplied not by T but by four plus 2, 3, multiplied by four is 12, 12 plus two equals 14. So when X equals negative one, Y equals 14 from the origin, we go to the left one and up 14. All right. Now step three is to connect these. And then we're asked to indicate the direction of the curve obtained corresponding to the increasing values of T using arrows. Well, at our first point negative six, negative one, T was negative one at our last point negative 1. 14 T was a positive four. So T was increasing as we moved from left to right. So we can draw arrows along our curve from left to right. And there's 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 = t − 2, y = 2t + 1; −2 ≤ t ≤ 3

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