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)
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)
Mr. Honner also introduced an integral with a little circle symbol drawn in the middle of it. This simply denotes that the integral is taken over a closed path, like a circle. So what would the symbol for an open integral be? Would it be considered open only if the path doesn't end up at the same point that it started with?
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