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.
Here is another parametrization of line segment problem that’s a little tougher. Find a parametrization for the line segment AB that uses s. The distance from point A as the parameter. Point A is -3, 8 and point B is 5, 14. And I’ve got those points plotted here, along with the line segment drawn from 1 to the other.
What I want here is a parametrization that uses very specifically the letter s, and s has to be the distance from here, to the point p that’s given.
So first of all I want to find out how long segment AB is. And I’m going to want to find the direction of this vector AB. So I want to find vector AB first, the components. And the components are 5, minus -3 which is 8 and 14 minus 8 which is 6. Now what’s the length of that?
It’s the square root of the first component squared 64, plus the second component squared, 36. So the 64 and 36 is a 100, root 100 is 10. So the total length of this vector is 10. So I want to be able to get point p here at point B when s equals 10.
So I need a direction vector v. It must have, first, the same direction is vector AB and second, it must have length 1. So here’s what I’m going to do. I know that the length of AB is 10. So I’m going to make v equal to 1/10 of vector AB. So that’s 1/10 of 8,6. That gives me .8 and .6. So this vector will have length 1 and points in the same direction as AB.
Now let me observe that vector Ap is going to give me the location of point p on this line segment. So vector Ap. And that’s going to be s times v, let’s say that this is vector v and vector v’s length is 1. If I multiply this vector by the scalar, which is the length of this vector here, from A to p, I would get this vector. So Ap is s times v. So that’s going to be s times this vector; .8, .6.
Now vector Ap in component from is x minus -3, x plus 3, and y minus 8. So if I want to get parametric equations, I just separate these and the components. I get x plus 3 equals .8s. And that means x equals -3 plus .8s. And I get y minus 8 equals .6s. So y equals 8 plus .6s.
So as usual when I’m parametrizing a line segment, I need to have limit on the value of the parameter, in this case s. S starts at 0, and if you go a zero length from point A you will be at point A. And it needs to end at 10 and the total length of the segment, and then you’ll be at point B. So I have to have s going from the 0 to 10. Now let’s just test this out. When I plug in 0, what do I get? I get x equals -3 plus 0.
X equals -3 and I get y equals 8 plus 0, 8. So that’s my left hand point, that’s point A. And when s equals 10,what do I get? I get -3 plus .8 times 10. -3 plus 8 and that’s 5. And y equals 8 plus .6 times 10, 8 plus 6 and that’s 14, and that’s point B.
So our parametrization works. This kind of parametrization that uses the length as the parameter is called an arc length parametrization, that's very important in multi-variable calculus.
Unit
Vectors and Parametric Equations