 So you have to be a little bit quieter this time because even with the light, I'm not hearing a lot. Sorry. So what we were doing, I remember, is we would have some vector field. We have some vector field. Let me call it GMM, which is a mapping from R in. R in, which means that in, could be in space. We have arrows given by, so at each in R in, we attach an arrow. This is some vector field, which is just an object. Or we can also think of this as measuring a force, sort of more useful as a force. And then we have also maybe some path. Can't see anything on this, but okay. We have some path, yeah. And we can make sense of the line integral, which is the integral of G, which is also mean still. Okay, which also means that what we calculate actually. So suppose Gamma, which has to be a mapping from R to R in, of T. T is on some range. So Gamma actually takes T0, T1 into R in. Right, it's a segment that goes in. And it says that we can calculate this as an ordinary integral from T0 to T1 of G of Gamma of T, which is a vector. And we dot it with, which is also an vector. So both of these are vectors. Let me emphasize that. And then we add the T. This is the line integral, which gives us a measurement. This vector field influences this curve as you traverse the curve, because this is the tangent to Gamma, to vector, because Gamma's a curve. Right? I'm seeing some loose-looking faces. No? All right. This is what we did last time. And we did a couple of examples of this and blah, blah, blah. What I ended with is, so let me remind you before what I did, ended with, is the fundamental theorem calculus in one variable. It says that, well, one version of it says that if I integrate the derivative of some function, those are T's, some function, I take its derivative and I integrate it. Then this is the same, so this is if f is nice. Does it say that? Then this is just evaluate f at each end coin. So this is what enables us to actually do calculus 2, that we can actually compute ended growth, rather than having to take those horrible limits of Riemann songs. And really it just comes down to the mean value theorem suitably interpreted. Right? I assume everybody knows this. If you don't know this and you're definitely in the wrong class and you miss at least a semester of stuff. So we wanted a version of that for these kind of integrals. And the version of that is, so we have some function, let's just continue with g. So I have g taking R into Rn. And suppose that g is the gradient of some function f, where f is a function from R to Rn. And I need some conditions on f. Here I just wrote nice. Here what I need is that f is continuously differentiated from where we're looking. So we have some set U, which is obviously a subset of R. Continuously differentiable on some interval. So we have also if gamma, which takes some other interval, don't care which one, AB into Rn. AB into the smooth curve. And it's contained in g. Then g dotted with the x. So this line interval depends on the input. It actually doesn't have to be differentiable. But it has to be different. The derivative has to be nice. So it has to be continuous. It certainly has to be continuous. So it can be slightly weaker than the differential. I have to have make sense. And the derivative needs to be integral. We can say it's differentiable and that's good enough. We can actually be weaker than that. Because let's say this. Sense then the theorem just makes sense. This is exactly the analogy of that. It's saying I have, let me draw it in the plane this time. I have some vector field in some region. So I have some vector field here. I have some curve that isn't icky. So my vector field has to not just be a random collection of vectors. It has to have some nice continuity. That when I move around here, the vectors don't just suddenly point the other way. They move in a continuous way. And so that's this continuously differentiable. It means that this vector field actually moves in a predictable way. And then I have some curve going from here to here. That's my gamma. And then it doesn't actually matter what the curve is. It just matters that I go from here to here. So this curve has the same line in the group. Or is that curve? It doesn't matter. So something that you're familiar with where these kind of things happen is say in physics, the analogy that it's going to take me to walk from here to the third floor of this building is the same if I go up this stairwell or I go up some other stairwell. I just have to get from here to there. That's assuming that there's no friction. Or the amount of work for me to lift this book from here to here is the same if I do this or if I do this. I still have to fight the same amount of gravity. When I do this, there's not much work. When I do that part, there's all the work. And then it didn't take me. So this condition right here is saying that G has a potential. So in physics terms, that's because it doesn't help you to think in terms of physics, it's fine. But this is saying that G has a potential. That is, there's some function that it represents or there's some sort of quantity here. There's another way to think of it. Another way. Let's call it f here. There's some function. So there exists some function f from R to R in so that G corresponds to downhill. And I have to feel G if I take some surface and get as much as the hills and some valleys and bump there, maybe it goes down here. This is my graph f. This is my R angle here. This is my graph of f. And G corresponds to the gradients. How water would move if I pour water on this surface. And so if I look at this from the top, I know some of this bulk here. Stuff flows down in some way. It corresponds to some flow. And then there's rhyming to roll. It's measuring how much work it is to get from 1.2 and up. And there's lots of paths up the mountain, but they all take the same amount of work if I start in the same place and end in the same place. So that's what this theorem is saying. Another way of saying that is if a conservative vector field, this doesn't mean it's a vector field in a red state that it voted for wrongly, doesn't mean that. Then a line integral. So conservative comes from the fact that there is some quantity which is conserved. So if I were to, instead of looking at the flow lines here, what am I doing wrong here? Instead of looking at the gradient, these gradient curves are so orthogonal to something that has curves that close up. So if I make the level curves of this surface, the gradient flows across them. And so these are sometimes in physics which these are called integral curves. So I have an integral. So sometimes in physics or in other cases, they'll just say this vector field has an integral. That means there's some quantity which is conserved. And I think somebody asked this last time whether this corresponds to, I just forgot what I'm talking about. Good, corresponds to something, but I don't remember what that is. Fun. So are we okay with that? No fundamental theorem. Calculus. One thing that's different here is we have this condition. G has to be the gradient of something. It isn't always true. So there are functions. There are vector fields which are not gradient for which the fundamental theorem doesn't hold. They're really nasty. But it's very easy to construct functions of more variables which are not gradients. So I guess when we prove this, the proof here is pretty easy. So we're assuming that this time, so in some function f, so this is the same as some curve. So I can actually calculate it. It's an integral from T0 to T1 of the gradient of f. Evaluate it again. So this is a vector dotted with the derivative of gamma and this is a one-dimensional integral. So this is, well, maybe it's not clear yet, is the chain rule for this composition. So that means that this is just the integral of d dt gradient of f evaluated again of T. This is just from the chain rule. Put a number T. I do some stuff and I wind up with a number r. So this, this function, let's call that, I don't want the gradient. Otherwise it doesn't make any sense. So this is a function, how about, let me write it this way. This function f composed with gamma is a function from the reals. Remembering the gamma, it gives me a vector. I plug that vector back into f. It gives me a number r. Let's call it h. This is a function, a real-valued function from the reals to the reals. So this is just the ordinary fundamental theorem of calculus. h of T1 minus h of T. Yeah? I can't really see this. Is that d dt of f dot gamma or f of gamma? The composition. So I take gamma, I cram it into f and then I take the derivative of that. Shane Google says that if I have a composition taking r in a gamma, taking r, which is what I have, that if I want to take the derivative, I'm getting confused about composition versus plugging in. Which is a number and gives me a vector. If I start with a number, turning it into a vector and then turning it back into a number. So this guy is a function, takes t to some h of t. I take the derivative of this process. And this is just the gradient of f evaluated at gamma of t. So this is a vector in rn dotted with the tangent vector. That's what the Shane Google says. If I have this kind of a composition and I take the derivative, then it is the gradient vector dotted with the tangent vector. So that's exactly what I'm doing here. So now I'm done because this is just g. No, not g. Yes, it is g. g gamma of b minus g gamma of a. Which I think is what I wrote over there. It's not. It's pretty sure. Did I write f? This is why it's wrong. This is why this is wrong. Sorry. G is the gradient of f. Yes. To find the anti-derivatives you evaluated each end in your final. In terms of this picture, it's just this height. Mine's that height. In terms of who is what. I'm going to split the negative 2. I mean, I would do an example. If you want me to do an example, but the example is stupid because if I just write down here's a function. There's its gradient. Compute, guess what? We just plug in. Yeah. And you say half is 5 to 9. And you say let me do it with an example even though it's stupid. Provided that the eraser erases which it looks like it doesn't. Bose f xy is x squared plus y squared. So the gradient doesn't make it y cubed. Doesn't matter. So the gradient is just 2x 2y. So this is saying if I take the vector field corresponding to 2x, 2y which looks like this. So it's just going outward because I just get longer and longer as you go out. It's 0 here. So I take this vector field. So this is g. And I take some path. Let's take gamma b t squared. So that'll be and let's let it go from 0 and 1. So that's this path here. Compute the integral over gamma. This is my function g. Of g, the x over gamma. Well, I can parameterize everything. So this is, let's just parameterize it. This will be the integral from 0 to 1 to x. Well, I have to plug in 2t, 2t squared. There it is. Gamble prime, 1, 2t, 2t. It'll be that integral which you can do if you want. And the integral from 0 to 1 are 2t plus 4t cubed. I don't actually have to do that integral. This should also be equal to this guy evaluated at 0. I mean, at 1 minus this guy evaluated at 0. So if you do this integral, you should get 2. Did it 2? Okay, good. Thanks. And so this integral is 2. But so is if instead of doing t, t squared I do something that goes like that, the same. So my interpretation is you have this force field which corresponds to the vector field and the line in the rule represents the effect of that force field of moving along the path. So when you're moving in the direction of the force you feel the full force. When the moving perpendicular to the force you don't feel it at all. There's no work. If you're sailing it doesn't have to be straight lines. Any path. So it's calculating work. Ish. And we have to worry about how fast we're going to get things. But in fact, we just show it doesn't matter for a gradient vector field. So it's summing up the effect of this vector field on this curve. I mean, you know, if you're going to track and that's the force you're not going to get pushed along there. So it's just a way to measure what that vector field is doing as you move. Could we simplify most problems then into just straight lines just to make it easier? If, if, if it is a conservative vector field if it's not a conservative vector field this is all not true. In an example where it's not true. Let me give you a very natural example. Let's think of public coordinates. Aren't theta. And let's parameterize the unit circle. The unit circle, right? So I'm just saying start here. Go around. If I want to figure out the length of this curve vector field that coincides with the tangent vector so let's just take a vector field this vector field which is just going around in circles. That's certainly a perfectly good vector field. I'm not calculating anything. I'm just saying we can calculate though. If this were independent of half then the order of the circle should be zero. Why is that a problem? Well, there's something going on right there. Pull the coordinates how a problem at the origin when r is zero theta can be anything. And so as I traverse this I can't just pull this curve back to nothing. There's a problem here at the origin. This curve is very different when this curve goes around and I didn't really think this example through because I wasn't planning to give it. But we can turn this into a real example if you want. But so you can see here I mean I can just take this vector field this vector field so let's write it in x and y so this vector field is the vector field it's not yx it should be only if it's yx so this vector field is y negative x and here's my path just leaving the circle so in x and y this is cosine sine t that is in the gradient vector field apparently so something's going on here. Another easy example is to take this one that's not a gradient vector field. It's not y negative x It's not y negative x It's not y negative x It looks like it to me at the top of the circle sorry just xx or something when x is positive y is positive when y is positive yeah so this is a pseudo example but maybe it gets the point across and it's easy to see that it couldn't be couldn't be just the difference. But as in this example if I take if I take a path it starts here and ends here and has to be zero that's a can to setting here on this surface if I go up and around and back then the amount of work I had to spend to get up here I get back when I come down and this is assuming there's no friction right so if there's no friction there's no perpetual motion because that doesn't cost me anything I earn back just as much as I you know it's perfectly there's no loss so I earn back just as much energy as I spent getting there and you can still calculate work with this sort of stuff even in a non-conservative vector field you can calculate the line integrals and you can calculate sure that we can do the calculations we just can't always simplify them to something easier because we just parameterize that we just do this integral right down in the integral but in this case this integral will be too high vector fields are not conservative so how do we tell if a vector field is conservative saying that this is not a conservative vector field how do we know yeah so did people understand that so let me if this were suppose this guy the claim is conservative so if of x, y changes the gradient of f so that means that fx the x derivative of g of right d d dy of f is x there's just another way of writing that and now if we take the mixed partials d dy of d dx of f then that is and if f or if g were conservative then that should be dx of d dy of f which should be plus 1 but negative 1 is not usually the same as plus 1 so you need to check whether your vector field is conservative you just take the mixed partials and if they're equal everything's great and if they're not equal then it's not take the mixed partials partial of this is 0 for y derivative of this is 0 the x derivative of this is 0 0 is 0 so so energy being conserved how does that relate to the mixed partials means that there is a function there exists a nice function a well behaved function which corresponds to the energy that's your potential energy there is a nice function that doesn't suddenly jump or doesn't behave in weird ways so here this circulatory vector field here has a problem at the origin it can't be reconciled when we bring everything down to 0 there is a singularity here and I can't take away or if we take this guy that one's okay forget that one that one's not a problem anyway so that's the same we have energy and energy varies in a nice way if somehow the energy here we change in some way the mixed partials that wouldn't be a nice function about how the energy varies if we think about the surface corresponding to the energy surface that we're flowing on then if I go this way the tangent plane will vary in a different way then if I go this way and this way so the surface is nasty it means that there exists a function which your vector field is the gradient of there is a conserved quantity so we say a vector field is conservative if it is the gradient of something I mean the word conservative vector field really comes from physics where there is a conserved quantity like energy there's a potential I mean if you want we could say that these are exact forms if you were to see that in life does that sort of imply that there is something outside acting on it like if the energy is jumping around because it's not conservative that would imply there is friction or something I mean it may not have to be energy which is conserved there's lots of conserved quantities that occur in life so there would have to be something from the outside to make it not conservative what outside means but typically in physical systems like if you have friction then if you put in won't be the amount of energy that there is altogether you lose something to friction now if we could crank up a dimension where we see how much is lost to friction then that loss in friction goes somewhere it didn't just go away energy just wasn't burned up it was absorbed by whatever was exerting the friction so if we go up a dimension where we account for that then we have it conserved again so that's where we have potential in kinetic energy you're trading from one to the other if we just look at potential then it's not really but if we have potential plus kinetic then it is okay yeah to see if it's conservative you can use that test with the mixed partial that doesn't work the other way though it's possible to construct none of the other way around if the mixed partials match then for sure the fundamental theorem of the fundamental theorem holds in most cases when the mixed partial matches the fundamental theorem won't hold but it kind of can sort of think about this example this vector field is not a conservative vector field but in fact if I look at this this is okay because it avoids the singularity and so it conserves something so I have to change coordinates this potential function which corresponds to this thing looks like I don't know one over like a specific parameterized path on the gradient field then you might be okay but in this case I think this vector field I drew is that one and this is okay as long as I stay away from both x if I don't go around anyway I don't want to so if I've done any complex analysis I don't think so so there's this well known thing in complex analysis called the Cauchy integral formula which says that you can do path integrals and you calculate around the singularity so you have to add how many poles are inside your curve it's exactly this problem but the thing that you should take away from this is if we can find the potential then everything's great and if we can't find the potential then we have to work hard it was much easier right so in this example I could have just taken this straight line or I could have taken that straight line and it would have been a lot easier to do let's take an easy one g of x, y so if I take the y derivative there I get 2x if I take the derivative of this with respect to x that didn't work here okay let's do that one that one's easy so that one works I really wanted a mixed term but let's do that so suppose I have this this is obviously conservative since no no it's not x squared back here jeez this is not conservative okay let's put x squared here and let's put y cubed there let's do this this one's easy and then we'll do the other one this is obviously conservative because the y derivative of x squared is 0 which is the same as the x derivative because this is supposedly the gradient of sum f time to figure out what f is what I'm thinking you have to do is so if you integrate the x component of it in the x direction what you're going to get is a function or actually I guess the better way to set it up is you're thinking about the nice you're looking for a certain function if you integrate f of x in the x direction what you're going to get is that part so the big function you're looking for, you can put in three pieces a function of only x a function of only y and a function of x and 1 so when you integrate the x component you get the function of x and you get the function of x, y but you don't get the function of only y and when you integrate the y component you get a function of and a function of x, y, and then with those two equations you could put them together in a line. Okay, yeah. So what you said first is also right, and this is what you said. Yeah, there's some f1, f2, and f3 where this has both. Okay, so what you said first was a good way to start. In this case, this one, it'll work because it doesn't exist. So that is I can just integrate x squared dx gives me one third x cubed plus a constant. And if I integrate y cubed dy, this gives me one quarter y to the fourth plus some other constant. And so that means that my function has this part and this is this part. And so that means that my function, any function of the form, one third x cubed plus one fourth y to the fourth plus any constant will work fine. So there's a perfectly valid potential. The thing that you have to keep in mind is that this part might exist. So for example, if what I was trying to, I thought I wrote one down, x squared, x plus xy, does that one work? So that derivative is zero, that one doesn't work. So if I take the x derivative of this and get y, the y derivative of that. Yeah, so let's take this guy, x squared plus y, y cubed plus x. That one will certainly have the right property because the mixed partials are one. And in fact, I can't figure this out. Wait, I'll just use the one you're using before but add it on the left side of it. This one's y. The one that you just did, like the first one you started going up there. You had one that in the one direction derived to be zero and the other direction when you derived it twice was one. So okay, so the derivative of this guy with respect to x is three y squared plus x. Three y squared plus y. So here I shouldn't have x. No. What? I don't know. I shouldn't have y cubed. This is gone. Derivative with respect to x, this is y. And so here I just want an xy here. And now I'm good. No, I know. I have a y squared over two. The derivative of this with respect to y is y. The derivative of this with respect to x is y. So it's certainly answered. I don't need to take the derivative. This is already great. How did we figure out this function? Sorry, I got lost here. How did we figure out this function? So let's take one of the guys and integrate it. So I'm going to integrate x squared plus y squared over two with respect to x. Thinking of y as a constant. And I will get one third x cubed x y squared over two plus some constant. Where this constant is some function of y. And if I integrate the other side, y cubed plus xy dy, then I will get one fourth y to the fourth plus, again, I'm integrating dy. So I will get y squared over two times x plus some other constant depending only on x. And then these guys match. This part's the same. This is my constant at x. This is my constant on y. Yeah. So they're not really constant. They're constant. They're functions which depend only on y. It's constant as far as x is concerned. And this is constant as far as y is concerned. So I'm calling it in constants. But it's really functions only on x. Functions only on y. In terms of this breakdown, when I integrate in terms of x, this is what I see. When I integrate in terms of y, this is what I see. This is the part that is not seen. This is the part that is not seen by the x. And this is the part that is not seen by the y. Those are the constant. So that means that my potential should be one quarter y to the fourth plus x y squared over two plus one third x cubed. And I could throw on any constant if I feel like it. I need another letter n. If one way back to field is conservative, you could just change the path or you could find the potential and the value. You don't have to find the potential. You could just find changed paths or you could do everything that's convenient too. That's perfectly fine. They know why they're messy. That's what I wanted to do. But it seems to me the nature of stuff. So, I'll change topics a little bit. Let's think about what we're apparently going to do. So I want to say some words about that. Gamma of t. And we know that the arc length, well, I don't know if we know this because a lot of people got this wrong on the first exam, but the arc length, does anybody know? Yeah. The integral of the tangent vector. Right? So I just ate the grade. So gamma goes from, I don't know, from t naught to t one. Is the integral from t naught to t one of gamma prime t and I think my curve down here, I take all those tangent vectors, I add their lengths, and it gives me a long, I'm just going to say it because you think about driving on this road and you average your speed over, you know, one second integrals. One second integrals. Add them up and there you go. Yes? Okay. So there we have that. Now we can think of s, so this is almost always the correct, the way that one represents arc length. So almost always s means arc length for a second. Is going to be, well, if I just do this as a function of t, just take my curve and I run it from my start to some time t. This is the distance along gamma. This will now be a function, which is the distance that I travel. So what I'd like to do is I want to rewrite t as some other parameter gamma of u where at every point take the derivative in terms of the u-parameterization. This is one. I take the derivative of gamma, which is now the tangent vectors, that they're always one. At some point in time, I travel one unit of distance. So how would I do that? The tangent vector by the magnitude of the tangent vector? Yeah. That would give you the unit vector in that direction. Right. So now, but I have parameterization. So suppose I hand you, oh, let's do an easy one. So suppose I hand you some curve. Let's use a circle. Let's use a circle radius 2. So this is not parameterized and I'm off by a factor of 2. So this guy has gamma prime of t equal to e everywhere. But I don't want it just to be gamma by 2. What do I want it to be? I want it to be t by 2. So I want this original guy to be the circle of tangent vectors of 2. And then if I go half the speed, if I go around half the speed and my tangent vector will be equal to 1. So I want to divide my speed by 2. One of my speed changes. It's going to have to work a little harder. So would you integrate from t on t? From t on t. I would integrate. I would write that. So, yeah, since I'm almost kind of high. So the idea is I want to solve. I want to define s of t as explicitly as an integral. And then as an n. What do I do here? It just was so easy I didn't have to do it. Right here we have s is to t. So t, I guess it would also call it u. But it's s over 2. U. So that means I want t over 2. Something's wrong with that. Something's wrong here. Oh, yeah. The total distance that I'm when, sorry, is 2. I divide by 2 when I invert. Should I do an example of this? Or should I do an example of this? How about this guy? So gamma of t is, we should use the graph of t to the 3 halves. Wait, t here. Let's start. So this is t goes from, I don't know, 0 to 2. So it's this curve. Now at any point, t, here, s of t, would be that length. So I want this to rewrite this curve. Instead of in this case, I want to rewrite this curve so that at time t, the length is t. Right? I want to rewrite it and in 20 seconds, I run t distance. Right? So at 0, I have nothing and after 1, I will be one unit along this curve. This one is parameterized by the x value. So the length is much greater. So that means I have to figure out that it is how long, maybe I shouldn't have said I'm doing 0, maybe I should be 1. It's this piece of the curve. But how long is this curve after t seconds? So that will be the integral from 1 to 2. The length of the dimension vector. So that will be the length of the vector 1. This is 1t. Square root of the integral from 1 to 2, the square root of 1 plus t. So I make this, I'm su genu equals 1 plus t. So I get 1 plus t. Someone, I'm off by 1. Sorry. 1 plus t to the 3 house 2 thirds should be a minus 1 evaluated from 0 to 1. Sometimes I'm going to 2. I'm going to t. Right? Did I do this in a good way? Yeah. Which is 2 thirds. I'm called this parameter. I don't know. So this is 2 thirds. 1 plus u to the 3 house minus 1 times 2 thirds. My arc length at time u. So at time 0 I have 2 thirds minus 2 thirds, which is 0. And at time 2 I have 2 thirds times 3 to the 3 house times 2 thirds. Okay. So now, I'm out of time, but not much out of time. So now I want to say this the other way around. S equals 2 thirds 1 plus u 3 house minus 2 thirds. I don't want to insult for you. So that's what I'm going to plug in. Now I want to use the curve. Right? This is u. The property that the instant effector is looking for. Modulo aerothetic calculus errors. Which I'm sure I'll rank at. 1 plus 2 thirds. This one minus 1. You get that one there? Alright. So let me stop there. I was hoping to talk about curvature when I won't. So next time we'll talk about curvature.