Unit
Vectors and Parametric Equations
Cornell University
PhD. in Mathematics
Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.
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Cornell University
PhD. in Mathematics
Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.
Finding parametric equations of line segments. Here’s an example; Let point A be -9, 4 and point B be 7, -8. Find parametric equations for segment AB, going from B to A. So this raises the issue that when you are parametrizing a line segment, you can parametrize it so that as t advances you go from one point to the other or vice versa.
So we are going to start form point B. And if B is our starting point, then we need to look at these numbers 7 -8. So remember that our formula is x equals 1 minus t, times the starting x value and that’s 7. Plus t times the finishing x value and that’s -9. And y is 1 minus t times the starting y value, that’s -8, plus t times the ending y value and that’s 4.
So these would be our equations for line segment. Just remember that you have to have the domain restriction; t is between 0 and 1. So when t equals 0, you are going to get the first point 7 -8. When t equals 1, you are going to get the final point -9, 4. Then you get all the points in between when t is between 0 and 1.
Now let’s find parametric equations for AB going from A to B. So it’s going to be very similar, it’s just you are going to switch A and B. So this is going to be our starting point. So this is our starting x value -9, it goes with 1 minus t. And then plus t times, and our finishing x value is 7. Then for y, we have 1 minus t times our starting y value of 4, plus t times our finishing y value of -8. And again, you still need to have the domain restriction t is between 0 and 1.
So you just get the line segment. Otherwise you get a line going through points A and B. Part c is rather interesting. It says find the point Q which is 3/4 the way from A to B. We should use this parametrization of segment AB, because this one goes from A to B. And so if I plug in 3/4 I’ll actually get the point that it's 3/4 the way from A to B.
So that’s what I’m going to do. So I plug in 3/4 here, and I get 1 minus 3/4 which is ¼ times -9. Plus t is 3/4 times 7. And the y value ¼ times 4, plus 3/4 times -8.
So I just have to multiply these out. ¼ times -9, -9/4. Plus 21/4, so 21 minus 9 is 12, 12/4 is 3. And ¼ of 4 is 1, and 3/4 of -8 is -6. So I get -5. And so point Q that I’m looking for, has coordinates 3,-5. And that’s it.
If I had been looking for a point Q which is 3/4 away from B to A, I would have used this parametrization, because this one goes from B to A. But that’s how you find a point that’s in between points A and B.