[Script Info]
Title: 
[Events]
Format: Layer, Start, End, Style, Name, MarginL, MarginR, MarginV, Effect, Text
Dialogue: 0,0:00:00.00,0:00:00.45,Default,,0000,0000,0000,,
Dialogue: 0,0:00:00.45,0:00:04.13,Default,,0000,0000,0000,,In the last video, we saw that\Nif a vector field can be
Dialogue: 0,0:00:04.13,0:00:12.58,Default,,0000,0000,0000,,written as the gradient of a\Nscalar field-- or another way
Dialogue: 0,0:00:12.58,0:00:17.08,Default,,0000,0000,0000,,we could say it: this would be\Nequal to the partial of our big
Dialogue: 0,0:00:17.08,0:00:23.06,Default,,0000,0000,0000,,f with respect to x times i\Nplus the partial of big f, our
Dialogue: 0,0:00:23.06,0:00:28.10,Default,,0000,0000,0000,,scalar field with respect to y\Ntimes j; and I'm just writing
Dialogue: 0,0:00:28.10,0:00:30.69,Default,,0000,0000,0000,,it in multiple ways just so you\Nremember what the gradient is
Dialogue: 0,0:00:30.69,0:00:35.49,Default,,0000,0000,0000,,--but we saw that if our vector\Nfield is the gradient of
Dialogue: 0,0:00:35.49,0:00:38.78,Default,,0000,0000,0000,,a scalar field then we\Ncall it conservative.
Dialogue: 0,0:00:38.78,0:00:47.82,Default,,0000,0000,0000,,So that tells us that f is a\Nconservative vector field.
Dialogue: 0,0:00:47.82,0:00:50.51,Default,,0000,0000,0000,,
Dialogue: 0,0:00:50.51,0:00:54.14,Default,,0000,0000,0000,,And it also tells us, and this\Nwas the big take away from the
Dialogue: 0,0:00:54.14,0:00:58.40,Default,,0000,0000,0000,,last video, that the line\Nintegral of f between two
Dialogue: 0,0:00:58.40,0:01:05.90,Default,,0000,0000,0000,,points-- let me draw two points\Nhere; so let me draw my
Dialogue: 0,0:01:05.90,0:01:09.21,Default,,0000,0000,0000,,coordinates just so we know\Nwe're on the xy plane.
Dialogue: 0,0:01:09.21,0:01:12.78,Default,,0000,0000,0000,,My axes: x-axis, y-axis.
Dialogue: 0,0:01:12.78,0:01:16.82,Default,,0000,0000,0000,,Let's say I have the point, I\Nhave that point and that point,
Dialogue: 0,0:01:16.82,0:01:19.62,Default,,0000,0000,0000,,and I have two different paths\Nbetween those two points.
Dialogue: 0,0:01:19.62,0:01:24.19,Default,,0000,0000,0000,,So I have path 1, that goes\Nsomething like that, so
Dialogue: 0,0:01:24.19,0:01:27.00,Default,,0000,0000,0000,,I'll call that c1 and it\Ngoes in that direction.
Dialogue: 0,0:01:27.00,0:01:30.27,Default,,0000,0000,0000,,
Dialogue: 0,0:01:30.27,0:01:32.38,Default,,0000,0000,0000,,And then I have, maybe in\Na different shades of
Dialogue: 0,0:01:32.38,0:01:38.01,Default,,0000,0000,0000,,green, c2 goes like that.
Dialogue: 0,0:01:38.01,0:01:41.01,Default,,0000,0000,0000,,They both start here\Nand go to there.
Dialogue: 0,0:01:41.01,0:01:43.52,Default,,0000,0000,0000,,We learned in the last video\Nthat the line integral
Dialogue: 0,0:01:43.52,0:01:48.33,Default,,0000,0000,0000,,is path independent\Nbetween any two points.
Dialogue: 0,0:01:48.33,0:01:57.93,Default,,0000,0000,0000,,So in this case the line\Nintegral along c1 of f dot dr
Dialogue: 0,0:01:57.93,0:02:03.56,Default,,0000,0000,0000,,is going to be equal to the\Nline integral of c2, over the
Dialogue: 0,0:02:03.56,0:02:10.87,Default,,0000,0000,0000,,path c2, of f dot dr. The line,\Nif we have a potential in
Dialogue: 0,0:02:10.87,0:02:14.50,Default,,0000,0000,0000,,a region, and we may be\Neverywhere, then the line
Dialogue: 0,0:02:14.50,0:02:17.59,Default,,0000,0000,0000,,integral between any two points\Nis independent of the path.
Dialogue: 0,0:02:17.59,0:02:19.39,Default,,0000,0000,0000,,That's the neat thing about\Na conservative field.
Dialogue: 0,0:02:19.39,0:02:21.19,Default,,0000,0000,0000,,Now what I want to do in this\Nvideo is do a little bit of
Dialogue: 0,0:02:21.19,0:02:23.68,Default,,0000,0000,0000,,an extension of the take\Naway of the last video.
Dialogue: 0,0:02:23.68,0:02:25.71,Default,,0000,0000,0000,,It's actually a pretty\Nimportant extension; it might
Dialogue: 0,0:02:25.71,0:02:27.54,Default,,0000,0000,0000,,already be obvious to you.
Dialogue: 0,0:02:27.54,0:02:29.55,Default,,0000,0000,0000,,I've already written this\Nhere; I could rearrange
Dialogue: 0,0:02:29.55,0:02:31.41,Default,,0000,0000,0000,,this equation a little bit.
Dialogue: 0,0:02:31.41,0:02:32.74,Default,,0000,0000,0000,,So let me do it.
Dialogue: 0,0:02:32.74,0:02:35.31,Default,,0000,0000,0000,,So let me a rearrange this.
Dialogue: 0,0:02:35.31,0:02:36.98,Default,,0000,0000,0000,,I'll just rewrite\Nthis in orange.
Dialogue: 0,0:02:36.98,0:02:42.73,Default,,0000,0000,0000,,So the line integral on path c1\Ndot dr minus-- I'll just go
Dialogue: 0,0:02:42.73,0:02:48.67,Default,,0000,0000,0000,,subtract this from both sides\N--minus the line integral c2 of
Dialogue: 0,0:02:48.67,0:02:53.11,Default,,0000,0000,0000,,f dot dr is going\Nto be equal to 0.
Dialogue: 0,0:02:53.11,0:02:55.61,Default,,0000,0000,0000,,All I did is I took this take\Naway from the last video and
Dialogue: 0,0:02:55.61,0:02:58.43,Default,,0000,0000,0000,,I subtracted this\Nfrom both sides.
Dialogue: 0,0:02:58.43,0:03:04.37,Default,,0000,0000,0000,,Now we learned several videos\Nago that if we're dealing with
Dialogue: 0,0:03:04.37,0:03:08.56,Default,,0000,0000,0000,,a line integral of a vector\Nfield-- not a scalar field
Dialogue: 0,0:03:08.56,0:03:10.69,Default,,0000,0000,0000,,--with a vector field,\Nthe direction of the
Dialogue: 0,0:03:10.69,0:03:12.09,Default,,0000,0000,0000,,path is important.
Dialogue: 0,0:03:12.09,0:03:20.37,Default,,0000,0000,0000,,We learned that the line\Nintegral over, say, c2 of f dot
Dialogue: 0,0:03:20.37,0:03:26.58,Default,,0000,0000,0000,,dr, is equal to the negative of\Nthe line integral of minus c2
Dialogue: 0,0:03:26.58,0:03:32.83,Default,,0000,0000,0000,,of f dot dr where we denoted\Nminus c2 is the same path as
Dialogue: 0,0:03:32.83,0:03:35.74,Default,,0000,0000,0000,,c2, but just in the\Nopposite direction.
Dialogue: 0,0:03:35.74,0:03:39.37,Default,,0000,0000,0000,,So for example, minus c2 I\Nwould write like this-- so let
Dialogue: 0,0:03:39.37,0:03:42.44,Default,,0000,0000,0000,,me do it in a different color\N--so let's say this is minus
Dialogue: 0,0:03:42.44,0:03:46.74,Default,,0000,0000,0000,,c2, it'd be a path just like\Nc2-- I'm going to call this
Dialogue: 0,0:03:46.74,0:03:49.10,Default,,0000,0000,0000,,minus c2 --but instead of going\Nin that direction, I'm now
Dialogue: 0,0:03:49.10,0:03:51.38,Default,,0000,0000,0000,,going to go in that direction.
Dialogue: 0,0:03:51.38,0:03:52.98,Default,,0000,0000,0000,,So ignore the old c2 arrows.
Dialogue: 0,0:03:52.98,0:03:56.03,Default,,0000,0000,0000,,We're now starting from\Nthere and coming back here.
Dialogue: 0,0:03:56.03,0:03:58.10,Default,,0000,0000,0000,,So this is minus c2.
Dialogue: 0,0:03:58.10,0:04:00.22,Default,,0000,0000,0000,,Or we could write, we could\Nput, the minus on the other
Dialogue: 0,0:04:00.22,0:04:06.25,Default,,0000,0000,0000,,side and we could say that\Nthe negative of the c2 line
Dialogue: 0,0:04:06.25,0:04:13.77,Default,,0000,0000,0000,,integral along the path of c2\Nof f dot dr is equal to the
Dialogue: 0,0:04:13.77,0:04:19.51,Default,,0000,0000,0000,,line integral over the reverse\Npath of f dot dr. All I did is
Dialogue: 0,0:04:19.51,0:04:21.63,Default,,0000,0000,0000,,I switched the negative on\Nthe other side; multiplied
Dialogue: 0,0:04:21.63,0:04:23.37,Default,,0000,0000,0000,,both sides by negative 1.
Dialogue: 0,0:04:23.37,0:04:27.90,Default,,0000,0000,0000,,So let's replace-- in this\Nequation we have the minus of
Dialogue: 0,0:04:27.90,0:04:31.48,Default,,0000,0000,0000,,the c2 path; we have that right\Nthere, and we have that right
Dialogue: 0,0:04:31.48,0:04:33.88,Default,,0000,0000,0000,,there --so we could just\Nreplace this with
Dialogue: 0,0:04:33.88,0:04:34.80,Default,,0000,0000,0000,,this right there.
Dialogue: 0,0:04:34.80,0:04:35.67,Default,,0000,0000,0000,,So let me do that.
Dialogue: 0,0:04:35.67,0:04:37.67,Default,,0000,0000,0000,,So I'll write this\Nfirst part first.
Dialogue: 0,0:04:37.67,0:04:43.41,Default,,0000,0000,0000,,So the integral along the curve\Nc1 of f dot dr, instead of
Dialogue: 0,0:04:43.41,0:04:49.20,Default,,0000,0000,0000,,minus the line integral along\Nc2, I'm going to say plus the
Dialogue: 0,0:04:49.20,0:04:51.20,Default,,0000,0000,0000,,integral along minus c2.
Dialogue: 0,0:04:51.20,0:04:55.94,Default,,0000,0000,0000,,This-- let me switch to the\Ngreen --this we've established
Dialogue: 0,0:04:55.94,0:04:57.48,Default,,0000,0000,0000,,is the same thing as this.
Dialogue: 0,0:04:57.48,0:05:00.72,Default,,0000,0000,0000,,The negative of this curve, or\Nthe line integral along this
Dialogue: 0,0:05:00.72,0:05:04.95,Default,,0000,0000,0000,,path, is the same thing as the\Nline integral, the positive of
Dialogue: 0,0:05:04.95,0:05:07.35,Default,,0000,0000,0000,,the line integral along\Nthe reverse path.
Dialogue: 0,0:05:07.35,0:05:13.61,Default,,0000,0000,0000,,So we'll say plus the line\Nintegral of minus c2 of
Dialogue: 0,0:05:13.61,0:05:19.27,Default,,0000,0000,0000,,f dot dr is equal to 0.
Dialogue: 0,0:05:19.27,0:05:20.67,Default,,0000,0000,0000,,Now there's something\Ninteresting.
Dialogue: 0,0:05:20.67,0:05:23.55,Default,,0000,0000,0000,,Let's look at what the\Ncombination of the path
Dialogue: 0,0:05:23.55,0:05:26.56,Default,,0000,0000,0000,,of c1 and minus c2 is.
Dialogue: 0,0:05:26.56,0:05:28.37,Default,,0000,0000,0000,,c1 starts over here.
Dialogue: 0,0:05:28.37,0:05:30.28,Default,,0000,0000,0000,,Let me get a nice,\Nvibrant color.
Dialogue: 0,0:05:30.28,0:05:32.58,Default,,0000,0000,0000,,c1 starts over here\Nat this point.
Dialogue: 0,0:05:32.58,0:05:36.54,Default,,0000,0000,0000,,It moves from this point\Nalong this curve c1 and
Dialogue: 0,0:05:36.54,0:05:38.02,Default,,0000,0000,0000,,ends up at this point.
Dialogue: 0,0:05:38.02,0:05:39.84,Default,,0000,0000,0000,,And then we do the minus c2.
Dialogue: 0,0:05:39.84,0:05:43.59,Default,,0000,0000,0000,,Minus c2 starts at this point\Nand just goes and comes back
Dialogue: 0,0:05:43.59,0:05:45.81,Default,,0000,0000,0000,,to the original point;\Nit completes a loop.
Dialogue: 0,0:05:45.81,0:05:48.27,Default,,0000,0000,0000,,So this is a closed\Nline integral.
Dialogue: 0,0:05:48.27,0:05:52.44,Default,,0000,0000,0000,,So if you combine this,\Nwe could rewrite this.
Dialogue: 0,0:05:52.44,0:05:53.66,Default,,0000,0000,0000,,Remember, this is just a loop.
Dialogue: 0,0:05:53.66,0:05:56.36,Default,,0000,0000,0000,,By reversing this, instead of\Nhaving two guys starting here
Dialogue: 0,0:05:56.36,0:05:58.45,Default,,0000,0000,0000,,and going there, I now can\Nstart here, go all the way
Dialogue: 0,0:05:58.45,0:06:00.66,Default,,0000,0000,0000,,there, and then come all\Nthe way back on this
Dialogue: 0,0:06:00.66,0:06:02.63,Default,,0000,0000,0000,,reverse path of c2.
Dialogue: 0,0:06:02.63,0:06:06.88,Default,,0000,0000,0000,,So this is equivalent to\Na closed line integral.
Dialogue: 0,0:06:06.88,0:06:12.15,Default,,0000,0000,0000,,So that is the same thing as an\Nintegral along a closed path.
Dialogue: 0,0:06:12.15,0:06:15.73,Default,,0000,0000,0000,,I mean, we could call the\Nclosed path, maybe, c1 plus
Dialogue: 0,0:06:15.73,0:06:18.20,Default,,0000,0000,0000,,minus c2, if we wanted to be\Nparticular about
Dialogue: 0,0:06:18.20,0:06:18.90,Default,,0000,0000,0000,,the closed path.
Dialogue: 0,0:06:18.90,0:06:23.39,Default,,0000,0000,0000,,But this could be, I drew c1\Nand c2 or minus c2 arbitrarily;
Dialogue: 0,0:06:23.39,0:06:29.60,Default,,0000,0000,0000,,this could be any closed path\Nwhere our vector field f has a
Dialogue: 0,0:06:29.60,0:06:33.00,Default,,0000,0000,0000,,potential, or where it is the\Ngradient of a scalar field,
Dialogue: 0,0:06:33.00,0:06:34.96,Default,,0000,0000,0000,,or where it is conservative.
Dialogue: 0,0:06:34.96,0:06:38.62,Default,,0000,0000,0000,,And so this can be written as\Na closed path of c1 plus the
Dialogue: 0,0:06:38.62,0:06:45.93,Default,,0000,0000,0000,,reverse of c2 of f dot dr.\NThat's just a rewriting
Dialogue: 0,0:06:45.93,0:06:49.04,Default,,0000,0000,0000,,of that, and so that's\Ngoing to be equal to 0.
Dialogue: 0,0:06:49.04,0:06:53.05,Default,,0000,0000,0000,,And this is our take\Naway for this video.
Dialogue: 0,0:06:53.05,0:06:56.11,Default,,0000,0000,0000,,This is, you can view\Nit as a corollary.
Dialogue: 0,0:06:56.11,0:06:59.16,Default,,0000,0000,0000,,It's kind of a low-hanging\Nconclusion that you can make
Dialogue: 0,0:06:59.16,0:07:01.61,Default,,0000,0000,0000,,after this conclusion.
Dialogue: 0,0:07:01.61,0:07:05.90,Default,,0000,0000,0000,,So now we know that if we have\Na vector field that's the
Dialogue: 0,0:07:05.90,0:07:09.57,Default,,0000,0000,0000,,gradient of a scalar field in\Nsome region, or maybe over the
Dialogue: 0,0:07:09.57,0:07:13.44,Default,,0000,0000,0000,,entire xy plane-- and this is\Ncalled the potential of f;
Dialogue: 0,0:07:13.44,0:07:15.32,Default,,0000,0000,0000,,this is a potential function.
Dialogue: 0,0:07:15.32,0:07:17.37,Default,,0000,0000,0000,,Oftentimes it will be the\Nnegative of it, but it's easy
Dialogue: 0,0:07:17.37,0:07:21.67,Default,,0000,0000,0000,,to mess with negatives --but if\Nwe have a vector field that is
Dialogue: 0,0:07:21.67,0:07:24.65,Default,,0000,0000,0000,,the gradient of a scalar field,\Nwe call that vector
Dialogue: 0,0:07:24.65,0:07:26.10,Default,,0000,0000,0000,,field conservative.
Dialogue: 0,0:07:26.10,0:07:29.88,Default,,0000,0000,0000,,That tells us that at any point\Nin the region where this is
Dialogue: 0,0:07:29.88,0:07:33.86,Default,,0000,0000,0000,,valid, the line integral from\None point to another is
Dialogue: 0,0:07:33.86,0:07:36.15,Default,,0000,0000,0000,,independent of the path; that's\Nwhat we got from
Dialogue: 0,0:07:36.15,0:07:37.13,Default,,0000,0000,0000,,the last video.
Dialogue: 0,0:07:37.13,0:07:42.89,Default,,0000,0000,0000,,And because of that, a closed\Nloop line integral, or a closed
Dialogue: 0,0:07:42.89,0:07:45.57,Default,,0000,0000,0000,,line integral, so if we take\Nsome other place, if we take
Dialogue: 0,0:07:45.57,0:07:52.92,Default,,0000,0000,0000,,any other closed line integral\Nor we take the line integral of
Dialogue: 0,0:07:52.92,0:07:57.49,Default,,0000,0000,0000,,the vector field on any closed\Nloop, it will become 0 because
Dialogue: 0,0:07:57.49,0:07:58.74,Default,,0000,0000,0000,,it is path independent.
Dialogue: 0,0:07:58.74,0:08:02.24,Default,,0000,0000,0000,,So that's the neat take away\Nhere, that if you know that
Dialogue: 0,0:08:02.24,0:08:05.37,Default,,0000,0000,0000,,this is conservative, if you\Never see something like this:
Dialogue: 0,0:08:05.37,0:08:10.58,Default,,0000,0000,0000,,if you see this f dot dr and\Nsomeone asks you to evaluate
Dialogue: 0,0:08:10.58,0:08:13.84,Default,,0000,0000,0000,,this given that f is\Nconservative, or given that f
Dialogue: 0,0:08:13.84,0:08:16.71,Default,,0000,0000,0000,,is the gradient of another\Nfunction, or given that f is
Dialogue: 0,0:08:16.71,0:08:19.96,Default,,0000,0000,0000,,path independent, you can now\Nimmediately say, that is going
Dialogue: 0,0:08:19.96,0:08:24.17,Default,,0000,0000,0000,,to be equal to 0, which\Nsimplifies the math a good bit.
Dialogue: 0,0:08:24.17,0:08:24.47,Default,,0000,0000,0000,,