Review Day: Line Integrals

Today we went over several problems involving Line Integrals. We re-emphasized the need to remember how to solve line integrals through other methods that don't involve the fundamental theorem of line integrals. Tomorrow will be the test so there won't be a blog.

What is path independence?



Today in class we went over the implications set forth by the Fundamental Theorem of Line Integrals. For conservative functions, this theorem actual makes otherwise very long problems very easy by simply finding the potential function and then subtracting the function of the beginning point from the function of the end point. We observed that the following are equivalent (refered as T.F.A.E.)

F is conservative <=> F dot dr is path independent <=> Integral over c of F dot dr = 0 for every closed curve

This gives rise to a new integral: a closed-curve (or closed-loop) integral, which can be seen to the left.

We also tested path independence, which is proved by the Fundamental Theorem of Line Integrals: no matter what path you take, the amount of work done is soley based on the beginning and end points (as long as F is conservative)

What is the fundamental Theorem of Line Integrals?

In today's class we explored the implications set forth by the Theorem of Line Integrals, which states that if F(x,y) is conservative and M, N are continuous, then the Integral over C of F dot dr is f(r(b))-f(r(a)) where a<t<b is the bound. What this means is that if the function given is conservative, you can simply find the potential function, then plug in the extreme ends of the bounds in order to evaluate or find work.

One interesting comment that was brought up by Wilson was, can we reverse the direction of piecewise parametrization? Such as going from (1,0) to (0,0) instead of (0,0) to (1,0) in a triangle that goes (0,0) to (1,0), (1,0) to (1,1) and (1, 1) to (0,0)? We will explore this in the next few days, but this led to a larger observance: the direction doesn't matter if the function is conservative and ends up in the same point, work will always be zero.

Here is the inversion seen in the triangle discussed before:

If the function is conservative, the work will be the same for both paths, as long as they both end up at (1,1)