00:02
Okay, this question wants us to evaluate this line integral.
00:06
So, to start, we want to check if we have a conservative field, so we're looking for if dqdx is equal to dpdy, and the x derivative of the y component, looking at our vector field, would just be e to the x cosine y, then minus 1.
00:33
Then, for the y derivative of the x component, we get e to the x, the sign turns into a cosine, and then minus 1.
00:45
And i wrote that down.
00:48
This should be a cosine.
00:51
And now we see we do indeed have a conservative field.
00:55
So f is equal to the gradient of some potential function.
01:04
So to go about finding that, we know that f is equal to an x derivative and a y derivative so we're just going to see all the information we can recover so our first contribution to the potential will come from the anti -derivative of the x component so we'll integrate e to the x sine y minus y with respect to x giving us e to the x sine y giving us e to the x sine y minus xy plus some function of y.
01:53
Then for our second contribution to the potential, we integrate the y component.
02:02
So e to the x cosine y minus x minus two.
02:11
And this gives us e to the x sine y minus x we can see here, we'll pick up an e to the x, sine y, a minus xy, and a negative 2y, plus some constant.
02:42
So our potential up to a constant would be phi xy is equal to e to the x times sine y, minus xy, minus 2y.
03:02
So now we can use this to evaluate our line integral because we know the integral over c of f.
03:10
If we have a gradient, it would just be si of the end minus phi of the beginning.
03:23
So we should probably find these ending points and starting points.
03:33
So it actually gives us the curve that we're integrating over.
03:38
So we can just use that...
