 Okay, we are, we are studying line integrals. It's very exciting, right? Line integrals, actually today we will, we will prove a theorem, a result about line integrals, which is the first of a series of results that I promised you last time, which relate integration in different dimensions, okay? So let's talk a little more about line integrals. Last time we talked about line integrals of vector fields. Okay, so, but actually this is not the only kind of a line integral that you can have, and so today we'll talk about other kinds as well. Let me remind you what we learned last time. Suppose that you have a vector field on the plane. As always, a vector field can be written in two different ways. You can write it like this, or you can write it like this. And you can choose whatever way you prefer, whichever way you like. So you're given, this is a vector field, okay? And in addition, you are given a curve on the plane. This is a curve with some, in this case, with two boundary points, but in general, we could actually also look at closed curves, which don't have a boundary. So a line integral is a combination, it's a pairing, if you will, it's a marriage of a vector field and a curve, okay? So it takes two objects to produce an integral. You need a vector field, and you need a curve, okay? And so that integral we write like this. We write it F dot dr over C. And we write it also, we write it in the following way. dr here also stands for vector, which has two components, dx and dy, okay? And when you take the dot product of these two vectors, this one having these two components, and that one having those two components, you'll produce p dx plus q dy. And then we can write it down more concretely in a way suitable for calculation by parameterizing the curve. So we write x is a function of some auxiliary variable t and y is a function of some auxiliary variable t, where t is between some values a and b, where say this point will correspond to t equal a, and this point will correspond to t equal p. If you do that, you can actually write it as an ordinary integral, just an ordinary one-dimensional integral from a to b with respect to dt now. But as always, there is sort of a price to pay. And the price to pay is that we have to replace dx by dt, but we can't just replace dx by dt. We write it as dx dt times dt. So this quantity is reminiscent of Jacobians or even earlier in one variable calculus, we always had this factor when we made the change of variables. And this kind of factor show up here as well. So the end result of this is that you get p dx dt plus q dy dt dt. And that's already an ordinary integral in t. So notice the difference between these two. Here I write this integral as an integral over a curve, which is on the plane. But here I rewrite it as an integral over just the basic interval from a to b. Just the basic line segment from a to b. Certainly this is much simpler than this. And we've been able to pass from this form to this form by using this parameterization. This parameterization, as I already explained many times, what it does, it really identifies this complicated curve with this simple line segment. Those are the formulas which describe how they are identified. So this is t equal a again. This is t equal b. And all the values in between correspond to some points on the curve. All right, is that clear? Yes, could I do one simple example? I did an example last time, right? But, right, okay, let's do that. So last time I had an example, which was actually slightly more complicated. In my example last time, I had vector fields defined in three space. But let me just simplify that example a little bit and explain this in two-dimensional case. So example, let's say you have a curve in which x is equal to t squared and y is equal to t cubed, okay? And so you have, that's the curve and t saved from zero to one. And then you have the vector field, which is, let's just say x times i plus x, y squared times j. Doesn't matter. Just to illustrate this, okay? So in this case, what is a line integral? You're going to have an integral from zero to one, which are these limits. And then you simply substitute, you just substitute all of the stuff into this formula. So this is p, this is p, this is q, right? So you get, this is p and p written in terms of the t variable is t squared. Then you have to write dx dt. dx dt is 2t dt plus, well, if I want to write it exactly in that form, I'll just put the dt outside. Plus you put x, y squared, x, y squared is, what, t to the eighth, right? dy dt is 3t squared and then dt. And so then you get some very simple integral, function in t dt from zero to one. Is that clear? Okay, so I will not do this calculation, it's easy. So I'll just stop here. All right, so what do these integrals represent? So last time we discussed, last time we discussed that this integral represents the work done by this vector field viewed as a force vector field, work done by force field F in moving an object or particle along this curve C, right? So, and when we were deriving it, the way we derived it is as usual with integrals, we break the curve into small pieces, right? And break the curve into small pieces. We then approximate each piece by a line segment, right? So let's say this is, I blow up, I blow up one of those segments. So let's say this is, that's this segment, right? Which I blow up on this board to make it easier to keep track of things. At a small scale, because the curve is smooth, we can approximate it well by a line segment, by a linear thing, okay? By straight, by a piece of straight line. And so what, so let's say this is a piece number I. This is a piece number I. There will be, let's say, in this case, maybe a dozen pieces, but eventually we'll make this partition finer and finer, so we'll get thousand, million, billion, so on, right? So let's focus on one of those pieces. And we focus on one of those pieces. We have here, we have, with respect to the x-axis, we have the distances delta x, delta xi, and here it will be delta yi, okay? And then we will have our force vector field. And because we are in a very small scale, in a very small piece of the curve, it will be a good approximation to think that the vector field is just constant along this segment because it's very small, right? So it is going to look like this, and so on. And so you will have some, let's call this vector fi. And so the work done, which as I explained last time, remember I was erasing things, the work done here is easy to find. It's just a dot product of fi and delta r. So the work, let's call it work number i, work done on the i-th segment, is just a dot product of this vector, which is red one, and this one, which is the yellow one. And this yellow vector is, we'll call it delta ri, it has two components, one is delta xi, and the other one is delta yi. So I want to dot this two things, fi dot delta ri, both are vectors, okay? And more concretely then, this is going to be the first component, which will be some pi times delta xi plus the second component qi times delta yi. So that's the work done approximately, approximate works. I have to, I should say this is really approximate because we are making two kinds of approximation here. First of all, we are approximating the actual segment of this curve, it is curvy, even though it is very small, but it's curvy in principle. But we are approximating it by something linear. That's the first approximation we make. And the second approximation we make is that the vector field also varies in general a little bit along this segment, but we are assuming that actually it doesn't vary. So we pick a particular value somewhere in the middle and we just assume that it stays the same along this segment. So this is the kind of approximation that we make at this point. But the idea is that as our partition becomes finer and finer, this approximation, the error of this approximation will go away. So we'll actually get the exact result. So the next step in other words is to take the sum. So take the total work, which is the sum of the work done on each of those segments. And that's approximately now equal to what? To f dot fi dot delta ri, which can also be written as the sum of pi delta xi plus qi delta yi, right? That's the formula we get. I'm essentially repeating what I said last time, but you will see why. You will see in a minute why I want to repeat this. So as always with integrals, this sum as the partition becomes finer actually tends to the integral, what we call the integrals. So that's how we end up with this integral. And what integral? Well, it will have f dot what we call dr. So delta r under this limit becomes differential dr. So we get f dr. That's this line integral, which we are talking about now, precisely as a total work done along the curve. And also you can think in terms of, you can also think in terms of this expression, which is a more detailed expression for this dot product. So that's how you get this second formula on the left board, which is a p dx plus q d1, right? It's the same thing. It's just writing the dot product in components. So that's what we've done so far. And that's called line integral of a vector field. And I think the terminology is a little bit misleading because it suggests as though line suggests that we are integrating over something linear, but we are not. I guess line here means to represent the fact that we are doing a one-dimensional integral. So something which is also line is one-dimensional. So that's why we call it line integral. But in fact, I would rather call it a curve integral or a curve integral because we're really integrating over a curve. So what I would like to say now is that this is not the only type of line integral. Let's follow the standard terminology. This is the only type of line integral that we can have. As you see in this line integral, what we are doing is at the end of the day, we are calculating an expression like this where there are two types of terms. One is, has this form some function times dx. And the second one is some function times dy. So this already makes you wonder what are these dx and dy? That what is it that allows us to insert them under the integral and make the integration? That's the first question. And the second natural question is whether there is something else like dx, dy that we could use potentially also to produce line integrals or integrals over the curves. And the answer is that yes, that there are actually other ways to produce line integrals which are also meaningful from the point of view of applications which have a different interpretation though, not as work done by force. And in trying to generalize it, we simply have to go back and look at this picture one more time. So what exactly, what is the structure of the integral that we obtain? Well, the structure is really, the structure of the integral is really dictated but what we've done at one of the segments at the segment number i. What we've done is we've taken the dot product between this vector fi and the displacement vector along this segment of the curve. But that was, that expression came from physics. That was because we tried to find the work done by this force. But let's look at the result. The result is really the sum of these two expressions, bi delta xi and qi delta yi. These are really the basic expressions that we have written down for this i segment which have eventually led us to consider those integrals. So if we want to generalize these integrals, we should look, we should do it at this step. Once we write the i's term, essentially there is no choice. There will be a particular integral as the limit when we sum up overall i, right? So we see, what are we doing? So we have this segment and let's focus on this term for example. So this term is a product of two factors. The first factor is a value of some function. So it's some number essentially but this number depends on the i's segment. So at the end of the day, when we take the limit, this will become a function. So it's some particular quantity which we attach to this segment. But the second factor is actually has a geometric meaning. The second factor has something to do with the shape of this segment. It actually is the length of the projection of this line segment onto the x coordinate. Of course, somewhere in the background, I imagine the coordinate system, x, y coordinate system. Okay, and so this is a projection of this vector onto the x axis. Whereas delta yi is a projection onto this factor, right? So the way to produce this integral in other words is to combine two things. One is a function like P and the other one is some length. And this length could be the length of the projection onto the x axis or it could be the projection onto the y axis. And if somebody comes and chooses another coordinate system, for example, polar coordinate system which would have, instead of x, y, would have r and theta, they could also take the displacement with respect to one of the coordinates of that system. For example, instead of delta x and delta y, you could have taken delta r or delta theta if you were working with polar coordinates. So in other words, there are many choices here. We choose delta x and delta y for two reasons. First of all, in this case, there is a very nice physical interpretation of the resulting integral. And second of all, our x, y coordinate system is somehow the most basic coordinate system that we can have. There is, however, one quantity, one geometric quantity, one length attached to this picture, which is somehow intrinsic and doesn't depend on any coordinate. And that's just the length of this interval. So in retrospect, even though we've gotten now with very nice integrals which have this very nice interpretation as the work done by force and so on, we can now go back and in retrospect see that actually perhaps the most natural thing to do, now that we're starting to analyze it in more detail, would be to take instead of delta x or delta y, just delta s, namely the length of this segment, right? So this way we get a different type of line integral. Different type of line integral is obtained if we take instead of delta x or delta y, let's even say delta xi or delta yi, we take delta si, which is the length of the segment. Length of the segment. All right, so what do we get in this case? So in this case, let's call this now, let's call this delta si. And of course, delta si is nothing but the square root of delta xi squared plus delta yi squared, right? So how about taking instead of this expression, take the expression, in fact, maybe let me go on this, let me go on this board so that we can follow the analogy more closely. So take the expression, instead of this one, take the expression some, put some fi, where f is some function, some number, well, in this case some number because it's attached to the i-segment, times delta si, which you can also write as fi square root of delta xi squared plus delta yi squared, okay? So if we take the sum now of these expressions, we take the sum of these expressions, this will tend to a different kind of integral, which will still be an integral over c, which we'll have, we'll have f times dx squared plus dy squared. So then, of course, we ask, what does it mean to have such an integral? Well, to do justice to such an integral, we really should parametrize the curve the way we did here. So let's go back to this parametrization. When we parametrize it like this, we get this auxiliary variable t. So now our curve gets replaced for all intents and purposes by a simple line interval, which makes things much easier, right? So we will end up with an integral, again from a to b over this auxiliary variable. We'll have here f, where we substitute x as a function of t and y as a function of t. And then here I will make a trick. I will divide everything by dt and multiply by dt, okay? But if I divide by dt, that means that under the square root I divide by dt squared. So I will end up with dx dt squared dy dt. You had a question. Why did I change it to delta s? Why are we doing all these integrals in the first place? Because we have to, no, because this is what we have to do in this class, but no, because these integrals make sense and then you need it in applications, for example, if you wanna calculate the work done by force. By the way, this is not the only application of line integrals. We'll talk more about them today about the fundamental theory. But for instance, if you look at the end of that section, 16.2, in the exercises, there is discussion of the Ampère's Law. And Ampère's Law is about line integrals of magnetic field caused by current, going, say, you have a current going through a wire and this causes magnetic field. So you would like to know the force of that magnetic field and you can find what the line integral of that magnetic field is by knowing the current along this wire, okay? I'm getting to the answer of your question, okay? This is an introduction to the answer. All right, so this was the reason. I'm explaining now a different type of integral and I will explain now that this too has application to something meaningful, okay? So, but you can think about it. You can think about these integrals from two points of view. One is a kind of utilitarian point of view, which means why do I need this? What is the situation where I actually need these integrals? And this, for this we have answers. For line integrals of vector fields, you have work done by force, for example, but more than that, for instance, Ampère's Law can be understood in terms of line integrals. This type of integrals, as I will explain now, will allow you to calculate things like masses of wires. I'll get to this in a minute, okay? So this is sort of utilitarian point of view. A second point of view is a more abstract point of view. We certainly know the utility of integrals in general, okay? But so far in one variable calculus, we would only calculate the integral of a function over dt. What I try to explain in the last lecture is that you really, you can only understand the most beautiful results in calculus by developing the most general possible integrals that you can take, the theory of most general integrals. So you can stand on the point of view of why are we doing, so I would turn your question around. Your question was why don't we just stay with this? Why do we take the LTSI? Is it just to make our life more complicated? I would turn it around and I say, why only this? Now that we understand this, why not do something more general, you see? And then see if that has some interpretation. So the integrals I'm talking about are the most general integrals over a general curve on the plane. That's the answer, and this is one example, or perhaps, I mean, this example leads to this integral. This is one example, this is an example of an integral of a vector field, and this is another example, okay? You had a question. That's right, that's right, you're right. So the delta Si is approximately equal to the length of the segment. When I wrote length of a segment, I was implicitly referring to this line segment which was approximating my curve. But if you would like to think of delta Si as the original one, then it would be approximate. DS, no, it's always, DS is always the differential, right? So it is a linear increment. It is a linear increment. So the question is about approximation. Well, it's, in some sense, it's true. It depends on how exactly you introduce an notation. But if you introduce in a certain way, you're right, that's what will happen, okay? So, okay, so I've given you this sort of a pep talk about the utility of integrals, but now I have to explain what is this useful for. And this is useful for calculating things like lengths, like the total length, for example. What is the total length of this curve? The total length of this curve is such an integral where the function f is one, you see? So this is the point. So this integral, if f is equal to one, for example. By the way, there is an notation for this which we'll call, let's call it DS. So then if f is one, fDS is just the integral of DS. This is just the length, arc length of C. Which of course something which we discussed before. What we are doing now is we are generalizing that definition of the arc length by allowing an arbitrary function instead of one. And this is something which is useful because let's suppose that you have a wire, a thin wire in the shape of this curve. And let's suppose you want to find a mass of this wire. Okay? So mass of the wire is a funny thing. I guess it was something else which we missed. So we want to find the mass of this wire. Now, but if the wire, if it is homogeneous, that's going to be the same as the length, or it's going to be proportional to the length with some coefficient. But suppose that it's heavier on this side and it's lighter on this side. So it's not homogeneous. What you have is a density function. This is very similar to when we discussed masses of two-dimensional and three-dimensional objects. So you're dealing with some density function. And that will be your function f. And to compute the total mass, you'd have to integrate this function f with respect to this DS, with respect to this measure. In mathematics, this kind of objects are called measures. What I'm trying to explain to you is that the structure of this line integrals, the basic structure of this integrals is as follows. It's always, the integrand is always a product, it could be thought of as a product, of something which is external, like mass density, or vector field. And something which is internal, something which is intrinsic to that curve, and something which has to do with the measurements along the curve. It could be, it could be the length of the projection onto the x-axis, could be the length of the projection onto the y-axis, or it could be the length itself. Each time you get a particular integral, okay? So these are the two types of integral. And I know that it is a little bit confusing because it looks like you have these two different things. And especially if you read in the book, I think it is very confusing. So what I was trying to explain to you is why there is such a variety of integrals and what is the relation between them, okay? Questions? Right. That's right. So in other words what you're saying is, see observe, you can observe the analogy between the following terms, PDX, QDY, or FDS, right? So you have these three different types of quantities which you're allowed to integrate, right? And the first factor here is something which is external. It has to do with vector field or function, like mass density function of the wire and things like that, right? The second one is intrinsic to the curve itself. It has to do with measuring things along this curve. Here you measure something along the X direction, here you measure something along the Y direction, here you just take the length of this little piece for each piece. And these are not all possible ways because, well, this is, in some sense, this is the most intrinsic one because the length is the length, right? It doesn't know about coordinates. This is already a little bit arbitrary because we have chosen a particular coordinate system. And as we know, there are many coordinate systems and essentially there is no preferred one, right? Because, for example, I could rotate the X and Y coordinate system, so that's already a new coordinate system. And we've learned to appreciate other coordinate systems, for example, when we did double integrals. Amongst the coordinate systems, we know it's a polar coordinate system. So, in fact, you could also have expressions like PDR or QD theta. That's also allowed, you see? And each time you have such an integral, the way you compute it is you parameterize your curve and you reduce the integral to the integral over a line segment on the auxiliary t-line, right? So in the case of a line integral of a vector field, you end up with this integral, right? And in the case of a second type of integral, you end up with this type of integral. But in both cases, you get an integral just along the line segment on the t-line. Okay, so this was to explain these different types of integrals because I know that this is something which drives people crazy, because the way it's explained in the book, I think, is not really, how should I say it? There's not enough motivation given. And so what I've tried to do is to motivate them and explain the difference. But I have to say that we mostly will be interested in line integrals of vector fields, not integrals of functions so much with respect to the measure ds, but line integrals of vector fields. This will be our main concern. And here now we will talk about the first example of a kind of a fundamental theorem of calculus applied to this type of integrals. So to explain this, I have to introduce a particular kind of vector field which is actually something we've learned before. So let's remember the notion of gradient, right? Gradient. So if you have a function, again, I work on the plane. So it's a function of two variables. We can take partial derivatives of this function. We have f sub x and f sub y, and we can put them together in what we call the gradient vector, right? This is a gradient vector. When we studied the gradient, we talked about a gradient attached to a particular point, x, y. So it would be a particular point, x, y, it would be a particular point, x, y, and we would evaluate the gradient. So we would have a vector for that particular point. So it was kind of a static object. It was attached to a particular point. But if you look at this formula, it looks like actually we have a vector value function. We actually have a vector attached to each point. So this is precisely a kind of thing that we now call a vector field. So we can now look at the gradient and recognize that the gradient of a function is actually a vector field. This is a vector field. Why? Because at each point, you have a vector. At each point, x, y, you have a vector, namely the two partial, the vector combining the two partial derivatives of this function at this point. And such vector fields are called conservative. And conservative here is not meant to be like conservative liberal, but it's due to a different etymology of the word. Conservative because it conserves something. It has to do with conservation, which is something we'll talk about later. So this is called conservative. Easy to remember. Now at Berkeley, we don't like conservative things, right? But this vector fields are good, so no problems. So what can we say about line integrals of conservative vector fields? So this will be the main subject of the rest of this lecture. It turns out that for such vector fields, we actually have an analog of the fundamental theorem of calculus, of the one variable calculus, which looks as follows. So I preserve, I keep the same picture. I have my curve C, and I consider a line integral of a vector field. It's just that now my vector field is going to be a gradient vector field or conservative vector field. It is a gradient of a function F. And what can we learn about line integrals of such vector fields? So it turns out that we have the following beautiful formula. So on the left-hand side, we have just the line integral of this vector field over this curve. And on the right-hand side, we just evaluate this function at the end points. Okay. So you see, this is the first type of results that I promised last time. I promised last time that we will have results of the following nature. On the left-hand side, we'll have an integral over some domain D of something which is a derivative of another object. And on the right-hand side, we'll have an integral over the boundary of this domain of that object itself. And so you see, in this particular case, perhaps I should explain that this is point A, and this is point D. So in this particular case, your domain D is this curve. And the boundary of this domain consists of two points, B and A. So indeed, on the left, we get an integral over C. And on the right, we just get the values at the two end points. This is the boundary, right? This is the boundary. So what is omega here? Omega is, in this particular case, is just our function F. So this, this integral just means evaluating this function F at the point B minus value at point A. You have a question? Shouldn't be a double integral on the left-hand side? Well, I'm sorry? Here, right. So the question is whether we should put some double integrals on the right. But see, this is a very schematic form. So I'm not keeping track of how many integration signs. In fact, well, up to now, we have an agreement that if it's a double integral, we put two integral signs. If it's a triple integral, we put three integral signs. But in general, you can just say all of them are integrals, right? So we will know what kind of integral it is by looking at the domain of integrations, the two-dimensional, three-dimensional, one-dimensional, right? So this is schematic, but you're right. If it would be consistent with the old notation, the number of integration signs on the left would have to be greater by one than the number of integration signs on the right. So maybe this minus one or something, I don't know. The number minus one and here will be certain number of integration signs, right? But in this particular case, we do have a single integral on the left, right? Because it's an integral over the curve. And on the right, we have an integral over zero-dimensional object. And usually, integral over zero-dimensional object, we never use the integration sign for that. There will be zero integration signs because it's a zero-dimensional object. And the integration over zero-dimensional set is just evaluating your function at the points of that set. With signs, though. So it's important which signs we are taking. And let me explain that. That's actually a very important point. You see, when I parameterize my curve, this gives it orientation because when I identify this curve with this line segment, I'm using, on this line segment, I have a natural orientation. I'm going from lower value to higher value. From lower to greater value, right? So here, A is smaller. A is less than B, of course. Because I write it like this. So when I identify my curve, C was that line interval. This also, it allows me to to move that orientation to the curve. In other words, the curve inherits that orientation of that interval. Which, if this point corresponds to this and this point corresponds to that, the orientation would be like this. But if it was the other way around, the orientation would be in the opposite direction. So in this equation, we're taking the curve oriented from A to B. So it goes like this from A to B. And on the right-hand side, we take the endpoint, evaluate the function at that point, minus the value at the initial point. So it has to be consistent. Because if it's not consistent, you will get an equation with the wrong sign. You'll have an overall sign. Okay? Yes. What does F stand for? This one. It was on this board, but I have closed it. F is a function. F of X, Y. What is a big F? There is no big F here. So the question is, we see big F there, but when we see big F here, but we don't see big F there. So this is what I explained. I said, this is the most general line integral. We take the most general vector field and we take the line integral. Amongst all vector fields, there are special ones which are called conservative. A vector field F is called conservative if it can be written as nabla F, as gradient of some function. Not all vector fields are like this. But from now on, or at this moment, we focus just on the conservative vector fields, vector fields which have this shape, nabla F. In this case, the line integral will look like this because the vector field is nabla F, so we just put this nabla F in the formula. So that's the line integral. The statement of this theorem, which is called a theorem about fundamental theorem for line integrals, line integrals, is that in this particular case, the line integral of this vector field can be written in a much simpler way. It's just the difference of the values of the function at the end points, you see? So this is, first of all, I wanna say that this is a natural generalization of the fundamental theorem of calculus which we learned for functions in one variable. Yes, very good. So the question is, does it mean that the path going from A to B doesn't really matter? This is absolutely correct. And this is one of the most important corollaries of this because on the left-hand side, it looks like the integral depends on the choice of the path. We could go like this, or we could go like this, we could go like this, or maybe just take the most boring one, just a straight line, right? But on the right, nothing tells us, nothing uses the shape of the curve, only the end points. So this formula immediately tells us that actually this line integral, which a priori depends on the choice of the path C, actually does not, only depends on the end points. And that's one of the most beautiful things about it, which you don't see, which without really analyzing this in detail, we wouldn't be able to see that. So it's the first sort of surprising fact about line integrals. And I'm getting to it in a moment. But before I get to this, I wanted to, before I get to this, I want to compare, I want to compare this formula to the formula which we started with last time, which is the fundamental theorem of calculus for functions in one variable, which is that if you have a function in one variable, then we have the formula like this, where you integrate over line segment from A to B and that's F of B minus F of A. So you see there is a clear connection between the two formulas. There is a clear analogy between the two formulas. In both cases on the left, we have a single integral of something which is a derivative. Here's a function in one variable. So there's only one derivative. Here we have a function in two variables. So there isn't a single derivative, but there is a gradient. And what we can do is we can take the line integral of the gradient vector field. And on the right-hand side is exactly the same. It's just evaluating at the end points and taking the difference. So in fact, this is a special case of this more general formula. This is a special case. We have now generalized the fundamental theorem of one-dimensional calculus to the case of line integrals over arbitrary curves. Now, let's look at this in more detail and let's try to figure out some consequences of this. What is this good for? And the first consequence is the one which has already been observed is the independence of the line integral of the path, which is really striking if you think about it because earlier today I wrote some formula for line integral, some specific example. And this example, it was sort of a painstaking process because I had to first of all parameterize my curve. Then I had to express my vector field in terms of this parameterization. Then I had to take the derivatives of X and Y. And I had to put all of this together and then finally integrate. So it's a lot of work. But now what this formula is telling me is that instead of doing all of that, instead of going through this complicated process, what I could do is just take some, there is some function F so that the result is just a difference between the values of this function at the end points. You see? So this is really striking fact. It's a shortcut. It is a shortcut. So now you're happy. Okay, so everybody's happy when things become easier, I guess. But you know, sometimes to get things easier, you have to first make things more complicated to make them easier, right? So I think it was, I forgot who it was who said that I was a physicist. I think it was Edward Teller who said, it's like, what is your goal in life? And he said, my goal in life is to make complicated things easier and easy things complicated. So in a sense, that's not necessarily my goal in this class. But I agree that this is certainly simplification. The right-hand side, we will all agree that the right-hand side is much simpler than the left-hand side, right? But there is a caveat to this, because this does not apply to all vector fields. This only applies to vector fields which are conservative. So if your vector field is not conservative, you're out of luck. There is no simplification. You have to go through this process. You see? But there is a huge class of vector fields to which this applies. And now we will discuss to which class, to which vector field, how to find out, how to see whether your vector field is a good one. Allows for such a simplification. But before I do that, I would like to derive the consequences of this. Or it's called corollaries. So corollary one is that if F is conservative, if F is a conservative vector field, let's just say, is conservative. When I say it's conservative, it means that there is some function F small such that F is equal to nabla of that function, the gradient of that function. Then the line integral of this vector field is independent of the curve. Depends only on the end points, on the end points. In other words, this is equal to the integral over a different curve of any other curve provided that that other curve has the same end points. What I mean by this, of course, is just what I have already drawn here, but let me draw it one more time. So let's say you have this has these two end points, but so does this, this, this, and so on. I don't want to leave the blackboard, but you could, I could walk all over the classroom and then go outside and a sprawl plaza and come back, and as long as I start here and come back here, the line integral, no matter how complicated it would be to calculate it, the result would be the same. You see, so this is really striking, this is a really striking fact, which enables us if the question on the test would be calculate over this curve, it looks complicated, but you know right away that instead you could calculate, say, over the straight line, which is much easier. You see, so this really allows for great simplification. Another way to phrase this, another way to phrase it is the following. Again, we'll be under the assumption that F is a conservative vector field, that if you integrate over a closed curve, a closed curve is one which does not have end points, which does not have a boundary. I'll draw it like this. So this is called, we'll call this such a curve, a closed curve. Closed curve does not have a boundary. We talked about this last time, like a circle, for example. It's boundaries is empty, okay? So if you integrate over this, you will always get zero. So in this case, you don't even need to choose a simplest path, you just know that you get zero. How to see that? Well, let's pick a point here. Okay? So you can think of this point, you can think of this curve as a curve with a boundary. But it's boundary, in its boundary, this one point appears in two different ways. First of all, it appears as the initial point, something that I have previously called A. But second of all, the same point appears also as the end point, which I previously called B, right? It's one and the same point. If you do this integral according to the fundamental theorem for line integrals, and we are allowed to do this because I'm under the assumption that my vector field is the gradient of a function, then we know that this integral is equal to the difference between the values at the point B and the point A. But points B and A are the same in this case. So when I take the difference, I get zero, you see? So this is equal to, this is explanation, f of B minus f of A is zero, right? Because it's the same point. So in fact, these two statements are equivalent to each other for the following reason. If you have two paths, you see, in this statement, which I called corollary one, the emphasis always on the fact that if you have two curves connecting the same points, then the answer is the same, right? But if I, suppose I know that for any closed curve, for any closed curve, the integral is zero, right? This is also implied, this fact will be implied. Why? Because let me draw it on a separate board. So suppose I already know that the integral over close curve, any closed curve of FDR, and I claim that this means that integral over C, and I mean any, and here I mean really any, any closed curve anywhere. If I have an integral over some curve C, this will be equal to the integral of the curve C prime as long as they have the same end points, right? How would I prove this? Well, so here's my curve C. And here's my curve C prime, right? So this is C and this is C prime. So to prove this is the same as to prove that this minus this is zero, right? But what is, what is this minus this? It means that I go like this, but then I change orientation on this one. So I go, so this is minus C prime. It's going like this. So it means I go like this, I go like this. This two curve together, combined into closed curve for which the integral is zero. You see, that's how I get that this is zero. So C minus C prime is a closed curve. Is this clear? Any questions about this? How many people follow this argument, right? How many people have not followed this argument? Don't be afraid. Everybody, you didn't? Not quite, okay. So what's the sticking point? I don't get this argument. Right, so now, so this argument was clear that why this argument was clear, why the integral over the closed curve is zero because you have the same initial and end points. So what I'm trying to explain now is that suppose I know, suppose I have an integral, suppose I have a vector field such that for any closed curve, the integral is zero. I want to say, I want to demonstrate that in that case, the integral over any curve even with boundary, like this, would necessarily be equal to the integral over any other curve with the same boundary. In other words, I would like to derive this formula from this formula. And the way I do it is I say, okay, what is, so I wanted to show that this is equal to this but it's the same as saying that this minus this is zero. Okay, and now I explained that the difference between these two integrals is actually could be thought of as an integral over the curve which is C, the original one, this one, minus C prime, but C minus C prime is actually it's a closed curve. So since I had assumed that for all closed curves it was zero, this implies that this difference is zero. So this was just to say that these two properties are actually equivalent to each other. So you should not think of this corollary one and corollary two as two separate statements. One is path independent, one is path independence. The integral is the same overall path with the same end points. The other one is a statement that the integral over any closed curve is zero. And the point is that these two statements are equivalent to each other. Okay, so where does this leave us? So we see now that if you have a conservative vector field things simplify, let's just put it this way. Things simplify, a lot. So of course then the next question is how do we find out if a given vector field is conservative or not? Because if it is conservative we can apply this and we have all these nice properties and if it's not then we have to use other methods. So this now is really the question. How do you determine whether F is conservative? And if it is, how to find the function, function F such that this vector field is the gradient of that function? That's the question. So as a warmup let's observe one property of conservative vector fields. So we are given, let's suppose that F is conservative. So F is PQ, F is PQ. And let's suppose that it is conservative. So it is nambla F. So let's start with from the other end. Let's suppose it is conservative. What can we say about it? What kind of properties does it have? Say again, if it is conservative then F of B minus F of A is the line integral. Right, that's the formula that we have. But this is still way too abstract. First of all, we talk about some curves and things like that. I would like some very simple algebraic criteria where I don't do any integration but just look at the components PQ and I would like to have a criterion to say whether it's conservative or not. Yes, does this have a proof? Are we going to prove it in this class? Well, I was hoping to prove it in this class but I may run out of time today. So the proof is very simple and it's in the book. The proof is follows almost immediately from the fundamental theorem of one-dimensional calculus when you just spell out what this means. In other words, when you parameterize the curve and substitute everything, you get on the nose the fundamental theorem of one-dimensional calculus. Okay, so I would like an algebraic criterion. You see, I would like an algebraic criterion which does not involve any curves or any integration. So see, what does this mean? This means that P is f sub x and Q is f sub y for some function y. So here's what I can do. I cannot compare these two guys because the two partial derivatives are a priority of nothing to do with each other. But what I can do is I can kind of equalize them in the following way. I can take P sub y. I can take the partial derivative of P with respect to y. If I do that, I get f x y. Right? On the other hand, I can take Q sub x. That's the partial derivative of Q with respect to x. And that's f y x, right? So now remember our good friend, Klerot, told us that if you have a function which has continuous first partial derivatives, then it doesn't matter in which order you take second partial derivatives. That these two things are actually equal to each other. They're equal to each other. So what does it mean? It means that if your vector field is conservative, it will, this property will necessarily have to hold. In other words, P sub y will have to be equal to Q sub x. That's a very simple. Maybe I keep this, but I... So we conclude that in this case, P sub y is equal to Q sub x. Or if you will, d P d y is equal to d Q d x. Here's an example. So let's say the vector field is y e to the x plus sine y times i plus e to the x plus x cosine y plus two y. So this is P and this is Q. So let me ask you this question. Is this vector field conservative? Okay. So in other words, conservative would mean that there exists some function. There is some function above it such that you get this by taking f sub x and you get this by taking f sub y. It's very difficult to guess right away just by looking at those two functions. It's very difficult to guess whether it's such a function f exists. You see, because it's like you should be able to extract in some sense the antiderivative in x and y and see that the result is somehow the same, right? But now we have a very simple criterion as to whether it is conservative or not. Because we can argue as follows. We can say, well, if it is conservative, then d P y would have to be equal to d Q x. Are they equal? So the actual question is are they equal? So let's calculate. What is d P d y? d P d y is e to the x plus cosine y. Everybody agrees with that? Yes? Okay. And now d Q d x is e to the x plus cosine y. It's the same, okay? So there is a chance that it is conservative. I have not yet, so far what I've said is that if the vector field is conservative, then this has to hold. And we see that they are equal. So there is a potential, there is a possibility that this vector field is in fact conservative. But it turns out in fact that in this particular case, this is not only necessary condition, but it is a sufficient condition. And this vector field is in fact conservative. So what is a precise statement and how do we find this function F in this particular case? So by the way, for this particular vector field, then it will be very easy to calculate its line integrals. You don't need to parameterize anything, you know, any curves and so on. You just find this function F and you evaluate that function at the end points. For example, if there are no end points, if it's a closed curve, you know the answer is zero right away. See, so it's a, I'm sorry, how to find the function F? I'm getting to this. No, no, you haven't lost anything. Okay, so what we've established so far is that if F is conservative, then dq, dp, dy is dq dx. In other words, this statement implies this statement. But actually the claim is that the converse is also true. That if this formula is true, dp, dy is equal to dq dx, then there exists a function F such that vector field is a gradient vector. Under one assumption, the converse, so the converse is also true under the assumption that the vector field is defined on a region which is simply connected, a region in R2. One vector field I defined on the plane, right? So it's defined on a region in R2 which is simply connected, simply connected. So I have to explain what simply connected means. Simply connected simply means, it doesn't simply mean that it's connected. It's a different property. So you see the connected, connected is easy. Connected means that this is disconnected, right? This is disconnected, this is connected. So this we know, that's easy, right? Connected means there's only one piece. You cannot separate the object. But simply connected is a slightly more sophisticated property. So well, I drew it in the case of two dimensional regions. But if you want, since we talk about one dimensional objects curves, maybe it's better to just talk about this curve as opposed to the curve which is a union of these two as opposed to curve which is just this one. So this is connected and this is disconnected. But simply connected region, or well, it's true in both settings. It's true for curves or it could be true for these regions like this, as you like. But simply connected is something more complicated. Simply connected means that any curve that you can draw in your region. So let's suppose you have a region like this. Simply connected means that if you draw any curve in this region, like any closed curve like this, you can deform it so as to, under the deformation, it all collapses into one point within the region. So in this particular case, I can deform it. I can make it smaller, smaller, smaller and like this. Now contrast this to the case of a region which has a hole. So it's like an annulus. So in this case, if I take this loop, I cannot deform it continuously to a point. I cannot collapse it into one point because I can make it smaller, all right? I can make it even smaller, but I cannot pass through this hole. You see? So simply connected, put it in the most down to earth's terms is the region which doesn't have any holes. And the hole actually, even if it's just the region without, with one point thrown away, it's already non-simply connected. So this is simply connected, simply connected. And this is not, and this is not. Because even here, because this point is thrown out, I cannot pass by this point. And this turns out to be the only obstacle to the converse statement, which is kind of a mild, it's really a mild restriction because to begin with, most of our vector fields are actually defined on the entire plane, right? For example, this vector field is surely defined on the entire plane. In other words, you can put any X and Y and it makes sense. The formulas make sense, right? So, and the entire plane is certainly simply connected. So, or any disk on the plane is simply connected. It's only when the vector field is not defined at a particular point or is not defined on, there is some hole like this that you cannot apply this result, you see. In fact, in the homework, in one of the exercises of the homework, there is an example of a vector field on, which is defined on a non-simply connected domain for which what we are going to do now does not work, for which this implication this way does not work. So, this is explained in a book, yes. Very good point. So, these are actually connected. These are all connected, right? Because it's only one piece. I cannot separate it into two pieces. It's only one piece, right? So, that's why I talked about connectivity first, because I wanted to emphasize that actually, these are two different properties. You can have a connected region, but non-simply connected. Let us assume though that our region, our vector field is defined on a simply connected domain. For example, this vector field is because it's defined on the entire plane and the plane for sure is simply connected. Then I claim that actually, the converse statement is true, which means that if you know that dp dy is equal to dq dx, which is what we actually found to be the case for our vector field, then there exists a function f so that our vector field is a gradient of that f. So, now the only question that remains is how do we find this f, right? And that's done by a very simple algorithm, by a very simple algorithm, where we simply take antiderivatives of this component of the vector field, an antiderivative of the component of the vector field. So, let's find f in our example. So, what I'm going to do is I'm going to take this, I'm going to take this p and I'm going to take its antiderivative. Because p is, so we're looking for a function f such that this is derivative with respect to x and this is with respect to y. But if so, we should be able to find f by taking antiderivative of this with respect to x, right? So, take antiderivative of p with respect to x. What do we get? Well, so in other words, with respect to x means that we treat y as a constant. Not as a function, not as a variable, but as a constant and x is a variable. So, when we take the antiderivative, we get y e to the x plus x sine y. Don't be surprised that you get x sine y because even though it's a sine y, but it's a sine of y which is now treated as a constant. So, this is treated for all intents and purposes as a constant with respect to x. So, we just put an x in front. That's what gives us the antiderivative. Now, if we were in one variable calculus, we actually wouldn't stop here. We would write a constant, right? This was very important to always remember that you can always write a constant. Why? Because when you take the derivative, constant disappears, right? So, there isn't a unique antiderivative. There is an antiderivative up to addition of a constant. But now, we have two variables. So, this should be a constant with respect to the x variable, but it may very well depend on y. So, what we should do, the only difference from the old calculations is that you should put dependence on y. And now, we find this function of y by comparing to the result for q. I mean, calculating the, so, we sort of go zigzagging like this. We start here, we take antiderivative, but we get some indeterminacy. So, we resolve the indeterminacy by going back to the second derivative, to f of f sub y, right? So, let's take, the first step is like this. Second is differentiate with respect to y. What do you get? You get e to the x plus x cosine y plus c prime of y. That's what we get. But this is what we should get. We should get e to the x plus x cosine y plus 2y. So, we recognize the first two terms have appeared. And now, we want to find what is c prime. Well, comparing to the formula for q, we find that c prime is 2y. And that means that c of y is y squared plus a constant, which is now a known as constant because now we are dealing with a function in one variable, right? So, the answer is this function plus y squared plus a constant. And that's how you find the function f for conservative vector fields. All right, so we'll continue next time. Have a good weekend.