00:05
Hold that a particle moves from the origin o to a point q along segments op and pq in figure 19 in the presence of a force field f equals x squared y squared, and we're asked to calculate the work expended when this particle moves, and then we're asked how much work is expended and the particle moves in a complete circuit around the square.
00:36
O .p .q .r.
00:43
So, looking at figure 19 first, first scenario, we can diagram.
00:49
We have our first quadrant in the x, y plane.
00:55
We have the points, the origin o, point p, with coordinates 1 -0, the point q with coordinates 1 -1, and the point r with coordinates 1 -1, and the point r with coordinates 0 -1, and our path starts at the origin, goes to p and then moves vertically to q.
01:42
Now, we have that the partial derivative of f1 with respect to y.
01:53
This is equal to 0, is the same as the partial derivative of f2 with respect to x.
02:07
So, f satisfies the cross -partials condition, and therefore, f is a conservative vector field, so we can find a potential function for f...