Hi everyone, and welcome to our lecture on the fundamental theorem for line integrals. This is a big result and really one of my favorite things to talk about. So let's just jump right to it. Theorem, it says for any smooth curve, now friendly reminder, smooth curve just means like no cusps, no intersections. Just start from A to B. You have a nice smooth curve are defined by a vector function r of t. So we've seen this before, where c is defined by some vector function. Maybe we're in the xy plane, you have an x and a y component, or maybe we're in space, and you have xy and z. It's all fine. It applies for both for the parameter a less than or equal to t less than equal to b. Let f be a differentiable function whose gradients continuous it's all just bookkeeping stuff to make sure everything's defined. And then here's the result. Remember, we're after how to find line integrals. And we want a vector field and we're going to specifically focus on vector fields that are gradient fields. Remember, these are conservative vector fields, where the little f function is the potential function. So it's a very special case. But we're going to say that the line integral of a vector field of a gradient vector field again, this is all like DS Is equal to the potential function evaluated at the endpoint minus the potential function evaluated at the start point. I'm going to leave this on the screen for a second, I'm going to let this sink in. This is probably more amazing than it might realize it has some nice consequences that we're going to talk about. It's called the fundamental theorem. So it should also tell you that this is pretty important and I hope you realize it's pretty close. At least in scope to the fundamental theorem of calculus. I don't know when the last time you remember this guy, but the fundamental theorem of calculus again, if I had a line segment from A to B on the xy plane. And I had some function f, that I wanted to know the definite integral from a to b Have f of x dx. Remember what the fundamental theorem of calculus said, If I know an anti derivatives, capital F, that's when capital S meant anti derivatives and non vector fields. Then this was the antiderivative plugged in at b minus the antiderivative plugged in at a plug did the endpoints. Another way to write this is exactly the same way but I'm going to write it [INAUDIBLE] different version, if I have the derivative of some function, then it becomes the function at b, minus the function at a. Somehow somewhere what happens between the start point and the end point of the curve, doesn't matter. Everything is completely dependent on the value of the function at the start point, and the value at the function at the end point. Let me say one more time. This is pretty important. Everything is determined by the endpoints. Though the particle, the way it's traveling can go bananas, it can go nuts. It can do lose all day, it doesn't matter. But somehow, someway, the endpoints determine the value of this integral. And we've seen before that these integrals have physical significance. They can talk about the work that's done by a force or something like that. Then it's amazing. If you just measure two points, if you have the the data you need for two points. You have everything you need. The one thing you have to be careful for to use this fundamental theorem for line integrals. Sometimes it's abbreviated as FTLI compared to lease the FTC. Fundamental theorem for line integrals is that you have to work with gradient field so this must be a gradient field. This is not true for every single vector field ever. So not every vector field is conservative. It's a gradient field. So just have to check that it is a gradient field or that the vector field is conservative before you apply this property. Let's keep talking about the consequences of this and then we'll actually use this and do some examples. So I want to introduce a definition, if my vector field f is continuous, and the line integral of f ds is independent of path. If the line integral over two paths c1 and c2 are equal for any two smooth curves c1 and c2 with the same initial and terminal points. So one more time a vector field F is continuous, then if it's continuous than the line integral of F ds is independent of path. If for any two smooth curves, you want to see two with the same initial and terminal points, we have the same values of their line integrals. The picture behind this imagine I have two curves Maybe some c 1 that goes from a to b. And imagine I have another curve, perhaps c 2. And it's a little more wavy whatever but it starts and ends at the same point. And I have some vector field will say that the vector field is can is independent a path if it doesn't matter. If I take the vector field and I compute the line integral. Then, I get the same numbers back, I get the same value back. Independent path means it doesn't matter how we get from point A to point B, one could be easier or the other one, it doesn't matter. This is a nice property to have cause it what allows you to do is you take something that's very moving in a very difficult way. Maybe it's lots of loops, the loops are going crazy. And again, you just can replace it with a simpler object. What is the most simplest object that we can use replaces, we'll have a straight line. So it's going to turn every single thing into a straight line which is really nice. And in general, just want you to note that like not every vector field is independent of path. Okay, so this is not true for all of them. However, it is true as we saw by the fundamental theorem for line integrals for conservative vector fields or for gradient vector fields. So if I know that it is built from some potential function, little f. So one of the things you want to get out at the end of this lecture is how do I check vector field is in fact a gradient vector field, another vector field is conservative, same thing. Let's do an example to sort of put some numbers behind this theory. So I have here a vector field F of x, y is x squared y squared. We're in the x y plane, and we're going to let our kirsi be an arc of a parabola defined by y equals to x squared from negative one comma 222 comma eight. So I have just this part of the arc. Then I'm traversing. This is a function. And I'd like to find the work done by this vector field. Remember, this is the same way to say line integral. This is the work done by F. Okay, so we have two ways to do this. So we'll be clever about this and then we'll do brute force as well. So way number one, we're just going to use brute force. And this is the formula that we had before where if I can parameterize the curve, then I can use the parameterization from a to b of f of my parameterization. And then it's a vector. So I want to.it with the derivative vector, this is a function. So I can certainly parameterize the parabola that is not a problem they already do that for us member function is one of those parameterizations we're supposed to know. And I'm just going to go ahead and make up the formula. So let's do that first. So first off I need to get the parametrization r of t. In this case you remember when we have a function your X becomes the T and the widest becomes to set of x squared, it's 2T squared. Okay, and we're going to go from the t is negative 1 to 2 and I'm going to need the derivative of this position. And when I take the derivative, it's just 1 comma 4 t. Let me plug in we get the definite integral from minus one to two of wherever we take the parameterization and we plug it in, so we get t squared the vector t squared. That's coming from x being t, and then the function is x squared comma y square so becomes four t to the fourth. We squared the two t squared. That vector is then dotted with the derivative vector, which is one comma four t. This dot product will return a scalar function. So I have the definite integral from minus one to two of, multiply and add them up as you go t squared plus 16 t to the fifth dt. This is a nice polynomial easy to integrate t squared, of course has an antiderivative of t cubed over three, and then 16 t to the fifth becomes 16 over six, t to the six, and we work that out from minus one to two. And we get when we get here we get eight over three plus 16 over six. Two to the six minus parentheses minus one over three plus 16 over six. Don't be heroes don't do this in your head, grab a calculator, and we get 171. So that's the way number one not using any fancy machinery. But now let me show you why this example is good for this sort of lecture here. It let's do way number two, so note that the gradient vector field working with is actually conservative. You can look at x squared combo y squared, and realize that if I think of x squared as the partial derivative, with respect to x, then that means that in my function, I'm going to have to have x cubed over three for the same reasons as that. If that partial derivative with respect to y is y squared, then I can start to build just by observation by inspection, my function is going to be x cubed over three plus y cubed over three. I can sort of reverse engineer the potential function, little f, that builds this vector field. So this is the thing to note. Now this is just by observation x squared, y squared is an easy function. We can do this if it gets harder. We're going to need a more definite way, but this is where we start. Once I know that then I can use the fundamental theorem for line integrals that I'm integrating the conservative vector field ds. And then I know that this is just the potential function evaluated at the endpoint two comma eight, minus the potential function evaluated at the start point. And I guess I leave it up to you to tell me which is easier now just evaluating the function again, execute over three plus y cubed over 3. This becomes 8 over 3 plus 8 cubed over 3, minus 1 over 3, plus 8 over three, you can check me that once again, you get 171. So the two numbers match. And you can imagine there's a lot less work there's a lot less involved in just realizing I can take advantage of the fact that this is a conservative vector field. And use the fundamental theophylline integrals and the value of this completely depends on the endpoints and the fact that it's traveling in a parabolic arc does not matter. So when you get these kind of questions, the first thing you want to ask is this vector field conservative. And we like to come up with a way to test for that. And I'll show you how to do that. Next, we get into that relationship that we need to introduce a few more definitions. So let's say that a curve C is closed if the starting point is the same as the endpoint. Now you can think of it like a circle, but it certainly doesn't have to be a circle. You can have some bean shaped object or whatever. But we just want a closed curve is one where we start and end at the same place. So we're taking a lap and doing something here. We'll say that a curve is simple. If there are no overlaps, so I can have maybe the infinity sign, this is closed. As if I pick a starting point, my go round, I definitely end where I start, but it's not simple. The two pictures above my circle and my being these are both simple. And so the key takeaway from these two definitions that if C is closed, and the vector field F is a gradient vector field. Then the line integral of F, which is now line integral of the gradient field is some potential function. By the fundamental theorem for line integrals. This says take the says take f of the endpoint minus f of your starting point, but remember C is closed, so that means that the starting point and end point are equal. So when you evaluate the same point of the function and you're going to get two of the same things and it gets zero. So this is a nice way to show that the line integral of the conservative vector field around a closed curve is always true. That's an immediate consequence of the of line integral. So if I give you some crazy function and we're going around the circle. But I tell you conservative, you don't have to do any work walk away. So that's got to be zero. That's a result of the fundamental theorem for line integrals. It turns out the converse is also true. A little trickier to sort of say why but I can at least give you the result. If I know that a vector field is independent, a path, so this is true if that vector field is zero for any closed curve, c. And what's happening is we're getting this relationship between the curve and the function, the vector field that we're over as well as the line in a row. And this is a very, very nice case where things all sort of work out together. This is the nicest case you can talk about. And you get that F. I'll just put a put it over here. So we have it on slide. F is conservative. It's a gradient vector field if and only if you get independent a path, and that's true if and only if, the line integral around any smooth curve, any smooth closed curve is 0. These three things are interconnected, this get into a little bit deeper than we want to go in this class but just realize there's a relationship in terms of what contributes to this line integrals, the function makes a contribution. The curve itself makes a contribution. And when you have the nicest kind of function that is a gradient vector field, so you're built from a potential function. And you have the nicest kind of curve, you have a loop, you have a closed curve, then this has to cancel and you get zero. This leads into the typical iceberg for those conservation laws in physics, the conservation of energy or something like that conservation momentum make. Okay, this is the relationship between them, one and plus and vice versa. All of this depends though on the vector field being conservative, and friendly reminder, a vector field is a vector valued function up and conservative just means. But it's a gradient field of some small f, which we call the potential. And so how do we know this is conservative? Well, we're going to test it in the two variable case we can do this. So if I have a vector field of two variables x comma y, and I have my functions I'm going to call the functions p of x comma y and q of x comma y so some expressions involving lots of x's and lots of lies. Let's go down this this thought process for a second. If the first component is in fact the partial derivative with respect to x, we don't know this, we're just saying what if it's true. So I put a little question mark here. And if the second component is the partial [INAUDIBLE] with receptor y that I know we can get two partial [INAUDIBLE]. That would mean that if I take the mixed partials so f(x, y) by [INAUDIBLE] these [INAUDIBLE] that makes partials commute. I could take f(y, x) and they would have to be the same, this is beanies theorem, that mixed partials commute. If you think about what that means that's a test that's a test to say is this vector field conservative. So what you can look at is the first function which we call p, its derivative with respect to y. So dp d y Is that equal to a second function dq with respect to x. If this is true, then you know you have a confirmed free. So as an example, let's look at F of x common y. And this will be 2x- 3y, -3x + 4y- 8. Again, if I had to do a line integral, I could do a brute force. But let's see, we can be clever. Let's ask ourselves is this like we need to first? Is it conservative? Is it a gradient vector field. So we're going to call this function P. We're going to call this function q, and we're going to run our test. I'm going to look at the partial derivative of p with respect to y, and this turns out to be negative three. And then I'm going to look at and ask myself is it equal to the partial derivative of q with respect to x? Scan this and look at this partial derivative. Have I also got negative three and they match? So that's a yes. So this says that this vector field is in fact, a gradient vector field. You then have to go through once you know that it's true, it's worth your effort to find the potential function find little f of x, y that builds this vector field. You can do this by inspection. You can do this by reverse engineering. In this example I think it's not too difficult. You can look at it and sort of back up and see what it would need to be. Like, if I integrate 2x I'm going to get x squared, and if I integrate my 4y I'm going to need to get minus 3xy + 2y squared- 8y. It's a little bit of guess and check to go through this and to practice this skill, but usually kind of just play around with it until you get the right one. So again, you can check this. If I take the partial with respect to x I get 2x-3y, so that works out, that's my first component. And if I take the partial respect to y, I get -3x+4y-8. Couple things, just one small thing of note here, this is not a unique potential function, I certainly can add a constant, you think about it, I'm integrating so I can always have +c. Because this is for true for any +c, we just pick c to be 0. But there's nothing wrong with saying a potential function is 6 + 6 + 7 + pi, it doesn't matter at the end of the day, it all works out when you solve for these problems. So you can find the potential function, you can then use this in the formula for the fundamental of line integrals. As you go through and answer your question, all right, let's do one last example here find a work done by a vector field x comma y 2y 3 over 2 plus 3x squared y j. Moving an object from the point P 1, 1 to Q 2, 4. So here points are just labeled as PDQ. There's two ways to do this. We could do a brute force method. Parameterize this is a line segment and number so we move the line. Let's do our test first. So test is the vector field conservative. So I take the first component function, and I can just unfortunately call the points being q, but we can also call these, p and q as well as as functions. Hopefully it's clear from context. But let's look at d dy of 2y, 3 over 2. Take a partial derivative here, we get 3 over 2 times 2y to the one half. That's of course 3y to the one half. And I take a partial derivative d, dx of the second components, note the order matters of 3x squared of y, and I get back three square root of y or 3y to the one half. So they are the same they match. This tells me that the function is in fact, the vector field is in fact conservative. So therefore my next step would be to find the potential function that builds this piece. From inspection from playing around with that a little bit, if my partial driven with respect to x is going to be 2y to 3 over 2 that forces my term to exist of 2xy to 3 over 2. And then you could even check if I, what would the partial derivative with respect to y be? If I bring derivative I get three over to the choose cancel the 3 in front. This just works out quite nicely. So I function is actually pretty simple 2 x y to 3 over 2. To find the work number works a fancy way to say the line integral over this curve, the curve connects 112 to 4 of F d s, this is what work is sounds to find. Well, now I know I'm working with a gradient vector field. Build, I mean the nice case, it's not a close. Unfortunately, I don't start at the same point. If that were true, I just walk away and say it's zero. But instead, I will evaluate this at the endpoints. So two comma four, minus the starting point one comma one. And we just start carefully evaluate this in our function, we have two times two, and then four to three over two minus 21123 over two. This turns out to be two times two is 4423 over two is the square root of four cubed. It's two cubed. So times eight minus two, that's 32 minus two that are known as 30. So the work done by this vector field would be 30. And then whatever units are if we cared about you're going to put it here. But again, I care about the number. I'll let you go off and apply this in other places. So keep this in mind is the vector field conservative is a great first question to ask if it is find the potential function, and then use the fundamental theorem for line integrals to make work much, much simpler. All right, great job on this tough concept and examples, do lots of problems and we'll see you next time.