Parametrics and vector valued functions

[StephenPrivitera]

Considering the problem of finding the area under a parametric curve, I thought,
y=y(t), x=x(t)
A=[inte]ydx=[inte]yx'(t)dt
That result seems straighforward.

I also thought, what if I let the VVF v=<x(t),y(t)> represent the same curve. To find the area, under the curve (I have in the back of my mind the concept of velocity and position), I would solve the integral, r=[inte]vdt.

Should these two results be related? I think they should, but the math shows they aren't. Should the magnitidue of the latter equal the absolute value of the former? Looks like no. Why not?

 

[HallsofIvy]

Thinking of t as time, integrating the velocity vector give the position as a vector with tail at the original point. r= &int;v dt is a vector not a number. Since ||v|| is the speed, &int; ||v||dt gives arc-length, not area under the curve.

 

[M98Ranger]

Thank you for sharing your thoughts on parametrics and vector valued functions. I agree that finding the area under a parametric curve can be straightforward using the method you mentioned. However, when considering a vector valued function, it is important to note that the integral represents the displacement vector rather than the area under the curve. This is because the vector valued function represents both the position and velocity of the curve at different points in time.

In terms of the relationship between the two results, they are related in the sense that they both use the same curve. However, the first integral represents the area under the curve at a specific point in time, while the second integral represents the total displacement of the curve over a given time interval. So while they may not be equal, they are both important in understanding the behavior of the curve.

I hope this helps clarify the differences between the results and why they may not be related in the way that you initially thought. Both approaches have their own significance and can provide valuable insights into the curve.