# Video: Parameterizing Curves in a Plane

Let 𝐴 = (1, 1) and 𝐵 = (1, 3). Which of the following is a parameterization of the line segment 𝐴𝐵 over 0 ≤ 𝑡 ≤ 1 that starts at 𝐴 and ends at 𝐵? [A] 𝑥 = 1, 𝑦 = 𝑡 + 1 [B] 𝑥 = 𝑡 + 1, 𝑦 = 1 [C] 𝑥 = 1, 𝑦 = 2(𝑡 + 1) [D] 𝑥 = 1, 𝑦 = 2𝑡 + 1 [E] 𝑥 = 2𝑡 + 1, 𝑦 = 1

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### Video Transcript

Let 𝐴 be equal to one, one and 𝐵 be equal to one, three. Which of the following is a parameterization of the line segment 𝐴𝐵 over the closed interval zero to one that starts at 𝐴 and ends up 𝐵? Is it 𝑥 equals one, 𝑦 equals 𝑡 plus one? B) 𝑥 equals 𝑡 plus one, 𝑦 equals one. C) 𝑥 equals one, 𝑦 equals two times 𝑡 plus one. D) 𝑥 equals one, 𝑦 equals two 𝑡 plus one. And E) 𝑥 equals two 𝑡 plus one, 𝑦 equals one.

We begin by recalling that when Cartesian coordinates of a curve are represented as functions of the same variable, usually 𝑡, they’re called parametric equations. And so parametric equations in the 𝑥𝑦-plane, 𝑥 equals some function of 𝑡 and 𝑦 equals some other function of 𝑡, denote the 𝑥- and 𝑦-coordinate of the graph of a curve in the plane. In this case, the points 𝐴 and 𝐵 are as shown. And we want the parameterization of the line segment 𝐴𝐵. Now, in order to derive the parametric equations, we’ll begin by recalling how we find the vector equation of a straight line, as these are intrinsically linked. The vector equation of a line that passes through a point with position back to 𝑟 naught and direction vector 𝐴 is 𝑟 equals 𝑟 naught plus 𝑎𝑡, where 𝑡 which is sometimes is shown as 𝑘 or 𝜇 is an unknown constant.

This essentially tells us how far away from the original point 𝑟 naught we need to move. In this case, the position vector of the point is one, one. Remember, this is just the vector that describes how we get from the origin zero, zero to a point on the line. And we’re using 𝐴 as direction matters. That’s where our line segment begins. The direction vector 𝐴𝐵 is the vector 𝑂𝐵 minus the vector 𝑂𝐴. That’s one, three minus one, one which is zero two. So in vector form, our line is 𝑟 equals one, one plus zero, two 𝑡. Now that we have the vector equation of the line, we’re going to write 𝑟 as a vector itself. We’re going to write it as 𝑥, 𝑦. And of course, we could use 𝑥, 𝑦, 𝑧, if we were worrying about three dimensions. But that’s unnecessary this time.

Then, on the right-hand side, we’re going to add the 𝑥 and 𝑦 components. So we get one plus zero 𝑡, one plus two 𝑡, which of course simplifies to one, one plus two 𝑡. Now, we know that, for the vector on the left to be equal to the one on the right, their individual components must themselves be equal. So 𝑥 is equal to one. And 𝑦 is equal to one plus two 𝑡. And that’s it. We’ve created a pair of parametric equations that describe our line. We’ve got the parameterization of the line segment 𝐴𝐵. It’s 𝑥 equals one, 𝑦 equals one plus two 𝑡. And in this case, that’s option D.