# Parametrics and vector valued functions

1. Oct 17, 2003

### 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?

2. Oct 18, 2003

### 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.