[Music] okay welcome back again everybody today we're doing section 16.3 which is all about the fundamental theorem of line integrals so let me pull up the slides and we'll get started here we go uh so another little bit of a fair warning this lecture might be a little bit long but at least it's not super technical I would say um there are some new definitions and some new ideas we're going to talk about but it's not as technical as say section 15.9 uh the one about the change of variable formula uh here's the overview so we're going to quickly review the fundamental theorem of calculus part two then we're going to see how we essentially generalize that theorem to get the fundamental theorem for line integrals then we'll talk about independence of path and we're going to spend some time talking about independence of path and how it relates to conservative Vector Fields so then we'll talk about what conservative Vector fields are we'll talk about a specific characteristic of conservative Vector fields and then Computing conservative Vector fields and then finally how this all relates to conservation of energy all right so here's the first bit um you got to kind of dig back into your brain a little bit and think about the fundamental theorem of calculus part two so you should have learned this back in calc 1 this is a actually a statement of it that the book calls the net change theorem but it's an equivalent statement so here it is let fime be continuous on this closed interval from A to B then the integral of fime big fime from A to B is f of B minus F of a so this is like everybody's favorite version or or favorite part of ftc2 but I've always found that the first part is really the more profound part of the fundamental theorem of calculus we'll use that a few times today um generally speaking what this is saying is that the integral of the the rate of change of a function over an interval is equal to the net change of that function over the interval so it's kind of like the way that I think of it is if you add up a bunch of instantaneous rate of change over a period Then you get the net change of the function over that period that's really what it's saying um now what we're going to do is we're actually going to generalize this theorem uh for line integrals and what we're going to do is we're going to interpret the gradient of f as a derivative of the function it's not really much of a stretch because you remember the gradient is the vector of all the partials so if we kind of view that as a quote derivative of the function you're going to see that uh the statement of the fundamental theorem for line integrals is pretty much the same as this but generalized to taking an integral over an arbitrary path that's the idea so uh this is what's called the fundamental theorem for line integrals here it is uh let C be a smooth curve given by a vector function R of T we're going to let F be a differentiable function of n variables whose gradient Vector uh d f is continuous on C then what we have is that the line integral of the gradient of f over that curve is equal to the value of f evaluated at the end points of the curve and then taking the difference of those values so one thing is just take a moment and look at how virtually identical this is to ftc2 right it's like taking the integral of a derivative and then you can evaluate at the end points right here just like this same kind of idea right here so another thing to note here is that since uh having uh dell f it means we have a conservative Vector field so we have a conservative Vector field and so what this theorem says is that we can evaluate the integral of a conservative Vector field over a curve C just by knowing the value of F at the end points of c and that's really what's so pH Al about this theorem um I mean the fact that you only really need info at the end points says a lot about the behavior of the function itself okay uh we're going to prove the case where f is just a function of three variables for clarity but the proof for the invariable case is pretty much the same so you should try it uh it's really not that difficult it's maybe a good chance for you to kind of uh flex your mathematical notation skills the the reason that we get clarity by only using three variables is because I can explicitly write out the sum that we're going to see without having to use an additional Sigma notation so if you want to if you want to try to do this proof on your own for the N variable case you should but you're going to have to use a sigma notation at some point in there to talk about adding up and many terms right here's the proof let C be a smooth curve and let F be a differentiable function of three variables we'll say where the gradient of f is continuous on that curve C then we want to compute this integral right here and by substituting in our parameterization of the curve we can reexpress the integrand like so and then here if we do this dot product you see we got a dotproduct of two vectors right here so compute the dot product and what we're going to get is partial F partial x times partial X partial T that's the product of the first two components right the respective first two components then we're going to add partial F partial y dydt That's the the product of the second components and then partial F partial Z partial Z partial T that's the product of the third components right so this is what we get when we do the dot product now we can reexpress this using a derivative because this is really just an instance of the chain rule the chain rule that we learned back in oh man I think it was section maybe maybe 13 it wasn't the multivariable chain rule somewhere in Chapter 13 I think somewhere in Chapter 13 but this is really the derivative of this composition f of R of T so we can rewrite this whole thing as F of sorry the derivative of f of R of T now we can employ the fundamental theorem of calculus for a single variable function and we get this ftc2 for single variable functions boom and that was the result so we showed that this line integral is equal to the difference of these two values okay now uh next we need to talk about independence of path so first let's define a path I've used the term before but let's explicitly Define it a path is just a piece-wise smooth curve and it's just what we refer to as a path means that it's smooth or uh if it's not uh you you could at least break it up into a finite Union of piecewise smooth curves that's the idea and so what do we mean when we say path Independence so a vector field F if it's a continuous Vector field on a domain D we say that the integral of f over a curve C is path independent or independent of path if the integral of f over C1 is equal to the integral of f over C2 for any two different paths in D that have the same initial point and the same terminal point okay so that's a bit of a mouthful um this is when I want to throw up the Thinker for a minute and I want you to read this definition to yourself a couple of times just to see what it's saying okay and while we're at it since you're thinking about it uh say it out loud too so read this out loud to yourself or to your peer or to your colleague or to your dog or your cat just to give your brain that feedback and think about what it means when you say it out loud okay all right so just to kind of summarize things what we say when we talk about path or what we mean when we talk about path Independence is that the value of this integral doesn't change no matter which path you take between point A and B so for example if I had a point a right here and a point B right here and then I had a path connecting them maybe this one right here we can compute a line integral over that path but there's more than one way to get from A to B we could get from A to B in this way or we could get from A to B in this way maybe I should do one that is actually smooth there we go that way or we could compute or we can compute a line integral from A to B by going you know this way right so the the the the meaning of path Independence is that it doesn't matter which one of those paths you integrate over the integral is going to give you the same result all right that's path Independence so now suppose C1 and C2 are two paths that have the same initial point a and terminal point B and we saw that in general this will not be the case and I think maybe uh maybe if I was drawing this picture hopefully it kind of like made you tip your head a little bit like that seems a little bit specious I don't know if that's a I don't know if that's actually true well remember in general it is not true and in in general it is not true if I want to integrate over this path it'll be a very different result than if I integrate it over this path instead so generally speaking it is not true path Independence is not a universal property of of line integrals and we're going to see exactly when this property is true and we get that from an import a correl of the fundamental theorem for line integrals so here it is uh if your vector function f given by the gradient of f is continuous and C1 and C2 are smooth curves with the same initial and same terminal points then the integrals are equal right so this is this is one of the conditions that we have here in other words this says that the line integral of a conservative Vector field is independent of path this is the big takeaway this is this is a coral are of FTL FTL um essentially if you have a conservative Vector field and you want to compute a line integral between any two points A and B here's a and here's B it actually will not matter which way you go it will not matter which path you use the the value of the line integral will be the same as long as your vector field is conservative that's the thing that's the catch right there all right um now we need a couple more definitions because we're going to start to look at this connection between path Independence and conservative Vector Fields but we need to build up some additional conditions to draw the explicit connection so first we need to Define what a closed curve is so a closed curve is just a a curve whose uh terminal point coincides with it with its initial point so very simply put a closed curve is a curve that goes back to where it started that's a closed curve right the terminal point and the initial Point are the same that's a closed curve now we have this extra theorem here so this one we're actually going to prove um it's a it's a byond conditional so there's actually two directions that we need to prove here and the theorem says this if your integral is path independent then the integral over any closed curve is zero for any possible closed curve in your region D but it also says the other direction sorry the other direction it says if it turns out that your line integral is zero for every closed path then your integral is path independent so this is an if and only of statement path indep dependance is equivalent is equivalence is equivalent to saying the integral is always going to be zero for any closed path that's what it's saying let's prove it so we can see that this is actually true and we're going to do both directions separately so here's the first Direction let C be any closed path in D we can parameterize it by R of T and the first thing we're going to suppose is that our integral is path independent in D so we're assuming path independence first and then we're going to show that that means that if if you integrate over a closed curve you're going to get zero all right so choose any two points A and B on a curve C all right uh C is a closed path maybe I'll draw a little picture here um let me do this there we go so here's our curve C and what you can do is you can pick any two points A and B on that curve that you want maybe I'll pick this as the first point and this as the the second point so this is a this is B all right that's what we're doing so far now we're going to rewrite this curve c as a union of two separate curves that's the next thing we're going to do and we're going to let C1 be the path from A to B so C1 is basically this path right here oops C1 and then C2 is going to be the path from B to a so C2 is going to be this path right here C2 there we go now let's look at our line integral so here it is broken up our line integral over the curve C we can rewrite it as a line integral over the union of these two curves which we can then split up into a sum of two integrals using finite additivity and then what we're going to do is we're going to use them use that result from the previous section about line integrals uh remember if we change the orientation of the path then the value of the line integral is going to change or it's gonna it's going to switch the sign of the the line integral is going to switch so here this is equivalent to this that's what I'm saying so integrating over C2 meaning going from B to a is the same thing as integrating over the opposite of C2 from A to B but you got to change the sign so not exactly the same it's opposite in sign all right however remember that the initial and terminal points of C1 and negative C2 coincide so remember the the initial point of C1 is a right and then that is now going to be the initial point of negative C2 right so here's C1 and here's negative C2 so the initial point and the terminal points coincide on these these two curves right here right but but remember we assumed path Independence so if your path independent that means that the value of the integral is the same so the value of this integral is the same as the value of this integral and if we're taking their difference we're going to get zero there it is so in other words if we assume path Independence then the integral over any closed curve is going to be zero that that's the first direction of implication right there so we just proved that now let's do the other direction so we're going to assume the other direction suppose that this integral is zero for any closed path C in D so maybe even this one there you go kind of the same thing uh we're going to let A and B be any two points in D sorry let me get rid of that maybe I should draw D okay pretend D is like the screen the whole region right uh let's pick any two points A and B and then we're going to create any two paths C1 and C2 connecting A and B so maybe C1 looks like oops where'd you go there you go maybe C1 looks like this right and then we're going to make C2 we go like this here's C2 all right now what we're going to do is we're going to define a curve c as the union of these two curves but we're going to change the orientation of C2 so remember C2 went from A to B so negative C2 the opposite of C2 is going to go from B to a so that's what we're doing right here all right now because we assumed that uh the because we assumed that this integral is going to be zero we can start with that but we can rewrite this integral as the integral over the union of these two curves and then we can split it up into two integrals using finite additivity again and then we get this right here so negative C2 it's the same thing we did before actually right so if we change the orientation of this curve it changes the sign of the integral so we get a negative right here right so that's the piece right there changing the orientation changes the sign of the line integral all right now this is pretty much all we needed because what we have is that this difference is equal to zero so that means that these two integrals are actually equal to each other and that is the definition of path Independence also because uh uh C1 and C2 or sorry um C1 and C2 oh yeah that yeah path Independence these two integrals have yeah these integrals are equal to each other so it's path independent the integral's path independent all right okay so now we've proven this we Prov proven that path Independence is equivalent to that the line integral being zero over any closed path now let's talk about conservative Vector fields for a minute okay so the line integral of any conservative Vector field is independent of path so it follows that the line integral of a conservative Vector field over any closed path is going to be zero that's the combination of what we what we just figured out right now so what does this mean physically physically this means that the net workor done by a conservative force field as it moves some object around a closed path is going to be zero and this makes sense because really the displacement is zero so you have to kind of think of this in terms of a you know imagine there's a conservative force field and you want to push this object around in the force field well if you push an object around and you end up back where you started the displacement is zero and so the net work that you've done is zero that's really what this is saying this is like mathema ially encoding that idea um now what we're going to do is we're going to get the necessary conditions for the converse the other direction of corollary one so let's refresh our memories about corollary one here this one so this one said uh this the line integral of a conservative Vector field is independent of path all right that's kind of that the same thing we were just saying a minute ago the question is what conditions do you need for the other direction like you know is it true that if if your integral is independent of path then you must have a conservative Vector field hm possibly but turns out there's some extra conditions we need not super strong conditions but conditions nonetheless so let's get back to that couple definitions one we need a more rigorous definition of what an open set is so we've been talking about open sets before um you use them back even in Cal 1 and before open intervals things like that so how do we really ously Define what an open set is an open set is an open is called open if for any point p in your set there's a dis centered at P that lies entirely inside the set d right so let me draw a picture of this for a minute let's say here is my okay I'm going to use a dotted line here here's my set d d now the reason we can say d is open is because you can pick any point that you want in D maybe this point right here and there exists an Open disc that lies entirely inside of D or a disc that lies entirely inside just like that and if you can do that for every possible point in your set then your set is considered open right the the key thing is this the lying entirely inside of D so for example uh a set that even includes one of its boundary points cannot be considered open because if you put your if you put your Center at that boundary point and you draw a disc around it the dis will reach inside and outside the set so open sets are are very special they're basically like you know they're they're unions of a bunch of open discs that's the idea okay um now what does it mean to be connected so a set D is called connected if any two points in D can be joined by a path that lies inside of D so let me give you an example here um here this set right here you there we go this set is connected and the reason is because it doesn't matter where you pick two points you can draw a path that connects them just like so or if you've got like a point here you can connect it to these points here so it's connected if you can connect any two points without leaving the set now for comparison R of this one here for comparison think about this set right here the union of these two sets Okay so let's say that our set D maybe I'll call this D1 and this is D2 We can say d is the union of D1 and D2 well if I pick a point over here and a point over here I can't connect them with a path without leaving the set so this is a set that is not connected not connected at all yep okay now open set connected set why do we need this because this is going to give us the conditions for the converse of that theorem so here it is suppose that f is a vector field that is continuous on an open Connected region D now if your integral is path independent in D then F must be a conservative Vector field on D in other words that means there is a potential function f such that the gradient of f is equal to your vector field F on the region d all right so this result combined with the previous result corer 1 really yields that the only Vector fields for which a line integral is path independent are conservative Vector Fields as long as your region is an open Connected region that's the one catch so if you've got an open and connected region then the only Vector fields which are path independent are conservative Vector fields and vice versa all right that's the idea um now this one we're going to going to go through the full proof uh you can also read this proof in the book but this is one that is definitely worth proving because the techniques are really really useful so math uh math Majors make sure you pay attention to the proof techniques that we use here here we go um oh yeah another note we're going to prove it in R2 just because uh it's easy to show in R2 but the technique can be readily generalized to any number of dimensions and again you can try that you should try math majors in particular um follow the technique of the proof and then see if you could understand how to generalize it to any number of Dimensions this one is a little bit trickier than the last one but the idea is basically the same you just think have to think about how you would uh how you would describe the open dis and how you would describe the um the the curves that's basically it so here we go in inr2 so back to the plane let D be an open and connected region in R2 let AB be any fixed point in D okay suppose that our Vector field f is continuous on D then we claim that the potential function f that we seek is given by this integral for any point XY and D all right so this may be a little bit weird at first to wrap your head around but remember we're doing a line integral we're just doing a line integral and we're Computing the integral from this point a to the point x comma y that's basically what this is what this is doing right um now this is kind of like those functions that are introduced back in back in calc one I mean they are very similar where the function itself like the thing that is changing is this upper bound of the integral notice the input of this function is X comma y right it's two variables well the integral here the thing that is changing is this upper point the the terminal point of this curve from a comma B to X comma y That's what's changing so this integral itself is a function of X and Y it's a function of that terminal point that's the idea all right so now that we've got that set up um because we're assuming that this integral is path independent it doesn't matter which path we take from a comma B to X comma y so let me repeat that one more time we we assumed that this is path independent so it doesn't matter which path we're taking from here to here the value of the integral will be the same right the reason that's important is because that that's actually what gives us a well-defined function so one thing your previous math instructors may not have mentioned is that if you do want to Define your own function you need to show you need to actually verify that it's well defined well- defined essentially means that uh it doesn't matter what representation of input you use you still get the same result that's what this is saying and that's exactly what we're what this path Independence gives us so if your path independent it doesn't matter which way you go from a a comma B to X comma y you still get the same result so that means this function that we've defined is actually well defined we can we can we can call it a function and we can investigate it as a function now since D is an open set that means there exists here's d by the way that means there exists a dis entirely contained within the set D just like this now what we're going to do what we're going to do is we're going to we're going to choose any point X1 y in the dis and we're going to choose it so that the coordinate here X1 is strictly less than x so very much the picture let me zoom in for a minute right here's a here's the point x comma y we're just going to choose a point that's directly to the left that's it directly to the left of that point and this the x coordinate of this point is going to be X1 that's what we've got going so far now we're going to Define the path c as the path that goes from a comma B to this point X1 Y and then goes from X1 y to XY so the union of these two paths is what we're calling C all right that's the setup all right now let's investigate um if we look at our potential function f our our potential potential function f uh and we look at the line integral over this uh this curve C we can again split split it into a sum of two line integrals by finite additivity and then what is the curve C1 it's just the curve that goes from Ab to X1 y so right here and then what's the curve C2 it's the curve that goes from X1 y to XY just like so all right okay so I hope I hope everyone's still with me um this is what we just had on the previous slide I'm just copying it again now here's where things get cool uh the first integral right here this integral this one right here is independent of X this integral does not depend on X you'll notice here X1 is a fixed Point X1 is a fixed point so it is not varying this integral does not depend on X just depends on y so consequently when we compute the partial derivative of this integral it's going to be zero so let's go through that we're going to compute the partial derivative of little f with respect to X so we're doing the partial derivative of this sum of integrals now the partial derivative of this integral with respect to X is zero because it doesn't depend on X it's a constant with respect to X right and then what we're what we're left with is the partial derivative with respect to X of the second integral right there okay so I hope that makes sense um if it doesn't please do ask questions in class and I will be happy to go back through the proof let's continue now what we're going to do is let's write our Vector field f as Pi QJ just like the standard structure so that way we can rewrite this remaining integral here like so just like that now here's the thing on this line segment that you see right here remember this is the line segment that goes from X1 y to XY it's this line segment right there on that line segment Y is constant there's no up there's no down Y is constant so that means that Dy equals z this thing equals zero what we're going to do is we're going to parameterize this line segment with a parameter T where T goes from X1 to X and then we can rewrite this as a single integral sorry a sing a variable of a single oh my gosh an integral of a single variable function so try to follow along with this for a minute uh this is going to be zero because Dy is zero and what we can do is we can parameterize X using T right here we can basically parameterize this variable with a T and T is going to go from X1 to X right and we've got DT right there now we can apply ftc1 and when we compute the derivative of this integral we get the original function evaluated at the upper bound which is p of XY by ftc1 okay um this is a good time to pause and just let this marinate don't pause for too long but let this marinate for a minute look at it think about it try to make sense of what we just did because we're about to do it again all right here we go um oh there we go all right so now we're basically going to repeat this process but we're going to repeat it with a different curve where we have a a vertical line segment instead remember we're allowed to do this because the integral is path independent it's path independent so it doesn't matter how we go from here to here the value of the integral will be the same so we're going to do this whole thing again and it's going to be pretty much identical the only difference is we're going to look at things with respect to Y so here's our potential potential function we can split it up into an integral over two curves by finite additivity then here what we're going to do is instead of picking a point that's directly to the left we're going to pick a point that is directly below the point XY let me zoom in so here's XY we're going to pick a point directly below and this point we're going to label as X comma y1 so y1 is a constant right now okay now what is C1 C1 is the path that goes from a b to xy1 and C2 is the path that goes from xy1 to XY right here now same kind of argument as before uh if you look carefully here this integral right here is independent of Y y it's independent of Y this does not depend on y so that means if we differentiate this function with respect to Y we compute the partial with respect to Y this integral is just going to become zero right here and then what we're left with is we're left with the derivative with respect to Y of the second integral right here that's all that's left okay same kind of thing as before here we go there we go now on this line segment X is constant there's no left or right oops I did that backward left or right change there's no change in X so that means that DX is equal to zero then we can use T as our parameter where T goes between y1 and Y and we can rewrite this as an integral of a single variable again so remember since DX is zero this term is going to disappear entirely and then we can reparameterize this in terms of T and then now what we're doing is doing the partial derivative or maybe I'll just say derivative of this integral and by ftc1 again we're going to get Q of XY Q of XY so all together this is what we have just found out this is what all all those steps were for we found out that P is actually equal to the partial derivative of f with respect to X and we found that Q is equal to the partial derivative of f with respect to Y so behold we've written our our vector field f as partial F partial x i plus partial F partial y j or in other words the gradient of f the gradient of f in other words f is a conservative Vector field by definition and a potential function for f is given by this integral and that's the proof hooray so um yeah it's a pretty fun proof I I recommend that you read over it make sense of the structure it's just a very nice proof it's pretty elegant all right so just to recap everything we've done so far we just showed that the only Vector fields for which a line integral is path independent on an open Connected region are conservative Vector Fields that's one thing to take away from this however we still haven't developed a simple way to determine if a given Vector field at least on R2 or R3 is conservative so the question is that like if I give you a vector field can you tell if it's conservative it'd be nice if we had a a simple way to do that um we showed this as true but that can be kind of tedious to have to prove right we would like a quicker way of showing this um what we're going to do is we're actually going to establish another necessary characteristic of conservative Vector fields and then we're going to show that under some additional conditions on the domain like the region that the characteristic is actually sufficient to guarantee a conservative Vector field so there's like a there's a couple more pieces we're missing to like guarantee a conservative Vector field so here it is uh this is this is the the characteristic that we're talking about and we're just going to stick to R2 and R3 for now um things can get a little bit hairier if you go up in the dimensions and that's mainly because of how curl comes into play here which we're going to talk about very briefly in a moment and then you will see it more in the next the next section so um here is theorem 5 if f equals p i QJ is a conservative Vector field and P and Q have continuous first order partial derivatives on D then throughout D this equality has to hold or maybe a little more appropriately we can say that this difference is going to be zero this difference is going to be zero so let's kind of summarize what we're looking at here if you've got a conservative Vector field on some domain D and in R2 right and P and Q have continuous first order partial derivatives so p and Q are C1 then what we know is that this difference is going to be zero right so it's like the the conservativeness of the vector field and the continuity of these partial derivatives give you this result all right now here's a little note from the future future future uh this value right here this expression is actually related to a concept called the curl of the vector field F the curl of of the vector field uh I will discuss curl in detail in the next section and then we'll discuss it again in section um 16.5 all right so more specifically what is this expression well the value here the value of this is basically the K component of the curl of the vector field F and what this theorem is essentially saying is that if your vector field F on R2 is Conservative then it's not pushing anything out of the XY plane either up or down right so if this value is zero then your vector field isn't pushing anything out of the XY plane in any way that's what it's saying um the proof is actually very quick so we'll do this proof right now let F be our uh our Vector field on R2 let's assume that it's a conservative Vector field that has continuous first order partial derivatives uh because it's conservative that means there exists a potential function f so that we can write p as the partial of f with respect to X x and q as the partial of f with respect to Y and then frankly by calling upon close theorem we can show that that difference is zero so what is a partial p with respect to Y partial P partial y well it's taking the derivative with respect to Y of this so using our livets notation this is actually equal to this second derivative but by claro theorem we know that this has to be equal to this the um the mixed second order partial derivatives are equal because they're continuous so because they're continuous we can switch the order of differentiation here and preserve the value of that of that derivative for that that mixed partial derivative okay but this is really just equal to differentiating this with respect to X but this is equal to Q so we get partial Q partial X and we've shown through this long chain of equalities that this is equal to this or in other words this minus this is equal to zero which is what we needed to prove right so if your vector field is Conservative then this has to be true on a domain as long as these are continuous and so on and so on but that's the idea why is this so handy well I think I show it here do I show it here yeah we've almost got yeah yeah we've almost got the other direction I guess the reason this one is so nice is because if this is not true then your vector field can't be conservative that's the contrapositive of this statement the contrapositive is hey if this is not true then your vector field can't be conservative definitely can't be conservative but that's not quite what we wanted not quite what we wanted what we would like to know is is there a way to determine if a vector field is conservative and for that we're going to need a couple more definitions here we go a lot of extra definitions good news is the definitions are simple pun intended here um so now let's talk about simple curves so simple curves are curves that don't intersect themselves anywhere between the end points here's a simple curves curve uh this is not a simple curve here's a simple curve here is not a simple curve so simple curves can't cross themselves that's the idea yeah simple curves don't cross themselves and then this is just a combo of um what a closed curve is this curve is not closed because it doesn't en close a region you can think of it like that it doesn't the terminal and the initial points don't coincide so it's not closed this one they do and it's a simple curve I'm sorry um and it's a simple curve so it's closed and it's simple this one is a closed curve so the initial and terminal points coincide but the curve crosses itself so it's not simple all right now that we have this notion of a simple curve we can talk about a Simply Connected region so this is where things get important a Simply Connected region in the plane is a connected region such that every simple closed curve in D encloses only points that lie inside D so in other words your region D is Simply Connected if it has no holes and if it's not consisting of more than one piece so very these are like the nicest regions you could imagine right Simply Connected they're they're pretty much like the the go-to thing that you imagine when you talk about regions right but not every region is so nice here's an example of a region that's not Simply Connected the reason is that I can pick a curve say um I could pick this curve right here and that curve encloses points that are not in the region D namely these points right same thing over here I can draw a curve right here and this curve encloses points that are not in D so this region cannot be Simply Connected and if it's more than two I mean if it's more than one piece it's definitely not simply connected right because you know like it's just yeah it's going to it's going to be split in half right so really what we're talking about wait do I say it on the next slide I don't it's a little bit down the way um one way you can interpret this is that uh your region is Simply Connected connected if it has no holes it can't have any holes in it that's really what we're trying to go for and as long as we have this condition we get exactly what we were looking for so here is the statement suppose you've got a vector field on an open Simply Connected region D in R2 suppose that these component functions are C1 and suppose that this difference is zero then your vector field must be conservative boom that's the goal right there so that that was like the last little bit that we needed so this is another moment where I want you to take some time just to to think about this particular statement think about this theorem and what it says right and now the good news is that it doesn't really seem like too much of of a condition to need to require right we just need our region to be like a a Simply Connected blob and we need the uh the first order partial derivatives of these component functions to be continuous and then we need this difference to be zero and if those things are all true we're guaranteed that our Vector field is conservative this is great because this gives us a pretty simple way of verifying whether whether or not a vector field is conservative so if all these are true you've got a conservative Vector field all right now here is the cool stuff so the these these results are some of my favorite um yeah the new condition is really just that we need no holes in our region and it turns out that these holes in these regions turn out to be extreme important extremely important and the consequences are studied in depth in a complex analysis course and Beyond like in my own research so in my own research uh looking at fractal Zeta functions um those holes are basically the complex dimensions of this fractal object and those holes correspond to poles of the zeta function itself so you can think of them as asmp tootes you can think of them as ASM tootes that there's basically like a puncture in the region right at that point okay so these holes turn out to be critically important later on and if you want to go down a fun but important Rabbit Hole I encourage you to read up about koshy's integral theorem and the residue theorem the residue theorem um math Majors you will certainly encounter this later on when you take a course in complex analysis uh for the rest of you you will probably end up reaping the benefits of these of these theorems and I hope you get to see them as well at some point um basically it says if you do if you want to compute a line integral you can kind of look at what's left over when you integrate around these holes that's really what it means because everywhere else your integral is going to be zero anyway that's kind of like summing it up very roughly um so we are actually going to prove this theorem but we're going to wait until the next section we're going to wait until the next section when we have a more powerful tool called greens theorem um yeah and then that'll but so for now we're going to accept it as true but it'll be easier to prove in the next section all right now let's do some examples um and these ones I do want you to try I know I've been doing a lot of talking maybe you've had to do some thinking but you haven't had to do any work yet right so I do want you to try this on your own my question is simply this is this Vector field a conservative Vector field justify yes or no right so think about it for a minute okay okay now say something out loud to yourself about how you could get started what would you try first right and now try to work it out on your own um justify your assertion my hint to you is think about theorem six all right and now let's take a look at it together so remember theorem 6 basically says it says that if you're on a Simply Connected region and um oh my jeez I forgot like the statement already uh if you're on a Simply Connected region and these have continuous first order partial derivatives and those that particular difference is going to be zero then your vector field must be conservative so let's check that let's check that um first of all this function this Vector function is defined on all of R2 and all of R2 is a Simply Connected region right uh but check this out if you compute these these first order partials right here partial P partial Y and partial Q partial X they're not equal they're not equal so their difference is not zero and that's all we need to to to assert that the vector field is not conservative here I graphed it using Mathematica so you can you can do that on your own just to get a peek at it um but because this difference is not zero our Vector field can't be conservative yeah that's the contrapositive right that's the contrapositive of theorem six um or that previous theorem even so let's see why for a minute here the reason this Vector field is not conservative is because there's a there's a source right here in the field and the work produced by the field will be positive if you want to push an object around the source this way or it'll be negative if you push it around this way and so what's happening is the the work done is not going to be zero so that's one way to see it now if you did one here everything would be fine you would actually get that the line integral here is going to be zero it's going to be zero there but the moment you do one around this Source right here you're going to get a nonzero line integral that's the idea all right okay let's do another one is this Vector field conservative and this is one that I want you to try too so think about it for a minute say something out loud to your friend or your colleague your peer yourself and then work it out on your own and see if it's conservative all right here we go so we can appeal to theorem six we know the things that we need to verify we just need to take a look at the um at the region itself and make sure that it's uh Simply Connected we need to look at the partial deriva atives uh be sure that they are continuous and then we need to look at that difference right the partial P partial y partial Q partial X and see what we get all right so here we go when you compute the individual partials so if you compute the partial of this with respect to Y this is going to go away and you're going to get 2X and then if you compute the partial of this with respect to X this is going to go away and you're going to get 2x so these partials are equal which means their difference is definitely zero also the domain of this function is all of R2 and all of R2 is definitely Simply Connected there's no holes right um so by the previous theorem theorem six this field is indeed conservative it has to be conservative this is a conservative Vector field now why is that the case kind of using the same intuition as before um you can draw closed curves anywhere you want and you're going to have place is where the force is positively or negatively or orthogonally aligned with the tangent of the curve and so what's going to happen is that's going to that's going to start canceling everything out as long as you have a closed curve right a simple closed curve um in order for the object to move that way it's going to be transferring energy back to the field to maintain its trajectory so the field is conserving energy right like if you try to go this way you're going to have to fight but then when you go this way you're going to get a syst and adding up all of those changes is going to make the work equal to zero it's going to make the work equal to zero so you get this conservative Vector field all right now how do we actually compute these things so we've looked at characteristics of conservative Vector fields we got some necessary conditions to determine if a vector field is conservative we've talked about path Independence and looking at line integrals over closed curves and so on and so on now the question is how do we actually compute these things meaning um how do we yeah how can we actually compute the potential function like how do we compute the potential function to to create a conservative Vector field um we're going to do an example in three dimensions but the process is equivalent for n dimensional Vector Fields it is definitely a process that is automatable so it's a good thing to just make sense of how the process works and then recognize that once you get up to like Dimension maybe four five that it just becomes a bunch of doing the same thing over and over again all right here we go uh let's say we have this Vector field let's find the potential function such that the gradient of f is equal to F right so find the potential function itself um this one you're welcome to try on your own but if you feel like you wouldn't even know where to begin that's okay for those that do want to try it think about it for a moment how would you find the potential function what's necessary all right you can say something out loud about what you think is necessary for this to be or for necessary for your potential function what do you need and then try to work it out on your own and we're going to do it together in a moment all right so here is the beginning if we want to find the potential function for this given Vector field we would need the partial derivatives of these things sorry I'm sorry we would need the partial derivative of our function the partial derivatives of our function to be these things right here in other words we need this to be true so if there is such a potential function if if it even exists right then it must be the case that the derivative of that potential function with respect to X is equal to this component function and the derivative of that potential function with respect to Y is this component function and the derivative of this potential function with respect to Z must be this potential function so it's a lot of requirements right but that's basically what it means because that would mean that the gradient of little f is going to be our Vector field so these are the conditions that must be satisfied in order for the function to be a potential function for bold F all right that's the basic idea um I numbered these here just for ease of reference I'm going to refer back to them a couple of times um I'll try not to flash back and forth too much but I'm going to have to a couple of times so let's start with this if we anti-differentiate that third equation with respect to X then we get this result so let me go back for just one moment here we're going to look at this function right here and we're going to anti-differentiate it with respect to X okay so anti- differentiate this with respect to X and what you get is this wait am I looking at the right one hold on wait wait wait wait wait sorry not three hold on hold on I have a typo here that should be two hold on a second let me be sure I don't have a typo okay sorry about that yeah that was a typo I knew something was off because I was like where did the exponential go um so anti-differentiating equation one with respect to X yields this result so equation one is this one right here there it is so if we anti-differentiate this with respect to X then what we get is we get f is equal to xy^2 plus some other function of Y and Z so this is an important detail to take note of remember when you compute an anti-derivative it's not unique it's Unique up to some quote unquote constant term but that constant term is constant with respect to the derivative that you're taking so here you can consider this like the plus C when you do an anti-derivative but the catch is that this could be more than just a constant it could be an entire function of Y and Z because if you differentiate a function of Y and Z with respect to X you get zero so when you do these anti-derivatives you have to remember that it's not just going to give you an additional constant it's going to give you an additional function a function whose derivative is zero with respect to that derivative right that's what this is saying right here so now we've got this so again we're kind of working through what this potential function would have to be to satisfy those three equations that we saw on the previous slide now what we're going to do is we're going to take this result and we're going to differentiate it with respect to Y so differentiate with respect to Y and you find that the partial derivative of f with respect to Y has to be 2xy plus uh the partial derivative of G with respect to Y so kind of hold this in your mind for a minute and I'm gonna I'm going to flash backward again I mean backward again um this is what F suby has to look like and this is also what F suby has to look like so just comparing the terms here right 2xy 2x y y g y y of z g y y of Z this tells us immediately that the partial derivative of G with respect to Y has to be this there we go so those have to be equal to each other and that's what I'm saying right here comparing the terms in equation two that's correct this time um this has to be equal to e 3 Z now we can repeat this process again we can anti-differentiate this function with respect to y and then what we get is y * e 3 Z plus some function of Z so by the same argument I was making a moment ago this could be any function of Z because if you differentiate it with respect to Y you get zero so now we have that g has to be equal to this and we can substitute that into what we had just a moment ago so now we're slowly building this potential function oops ah I meant this is this all right now we're going to continue the process now we're going to differentiate this with respect to Z and when we do that we get that the partial of f with respect to Z is equal to this now we're going to compare to equation three and when we compare to equation three we're going to see that H Prime of Z has to be zero and so that means h of Z has to be equal to a constant let me go back to the equations for a minute right here see there's no there's nothing here right there's just this piece just that piece so when we go here this has to be equal to zero but if the derivative is equal to zero then that means that your function has to be constant and therefore we have our potential function there it is so we kind of did this reverse engineering process right where we're like okay if we wanted it to be the potential function these partial derivatives need to be equal to these component functions okay if those are equal what happens when we anti-differentiate and then what is the what do the results give us and then we did all these comparisons right so in the end we found this and it's always fun just to verify that this is correct so just take this function and compute the partial derivatives and you should get these component functions back let's see that we do differentiate this with respect to X gone gone and you get y^2 check uh differentiate this with respect to Y and you're going to get what 2xy plus e to 3 Z and that disappears check and then differentiate this with respect to Z and what are we going to get that's gone that's gone this three is going to come down from the chain Rule and we get 3 y e to 3 Z check so indeed this is our potential function as required and that's how you do it right so now we know how to compute conservative Vector Fields now let's finish up by talking about conservation of energy so this is where the physicists will probably get really excited I hope you do it's it's it's very it's a very elegant result so maybe you've guessed from the names that have been floating around this whole time that all of this is strongly related to conservation of energy and now we're going to make that connection explicit we're going to make it explicit so here it is suppose that our Vector field f is a Contin is force field and it moves an object along a path c c is given by this parameterization R of t t goes between a and b and we're going to label R of a as capital A that's the initial point and R of B is B that's the terminal point of C so C goes from A to B just like so now by Newton's second law of motion f equals ma we know that the force at a point on C remember this is the force at a point on the curve it's related to the acceleration by fals Ma so we get fals Ma and remember that the acceleration vector is just the second derivative of the position Vector so acceleration is the second derivative so we know this f equals m * R Prime of T now we're going to reexpress the work done by the force on this object through this calculation so what we're going to do is we're going to calculate the work that the force field does in moving an object along the curve C and we're just going to plug in stuff that we already know and then use um uh one of those one of those properties of um of the dot product from way back in chapter 12 I mean way back in chapter 12 so here we go line integral over C we can substitute in our parameterization F of R of t. RP Prime of T DT so we've changed our our variable of integration to the parameter T now remember that F of R of T by Newton's second law is ma a so M * R Prime now we can actually Factor the M out of the integral by linearity and we can compute this dotproduct right here so going from here to here I'll say um is actually a result from chapter 12 so if you look back in section oh geez is it chapter 12 man I can't remember if it's 12 or 13 um but if you compute this remember there's a there's a simple product rule for the dot product right so if you want to compute the derivative of a DOT product like so then you're basically going to use an analogy of the dot product for I'm sorry an analogy of the product rule for a scalar product you're going to get this Prime Times this plus this times this Prime that's essentially the idea but then because you're going to end up getting two of the same terms you can write it as two times this right here that's where this two is coming from so maybe I can make it more explicit sorry maybe I'll just write it out um what am I saying D DT of RP Prime of t. RP Prime of T is equal to R Prime t r Prime t plus r Prime t r Prime T but that's just equal to two R Prime um double Prime where did you go t r Prime T there it is oh Dots Dots sorry these should be dots it's like something feels off here those should be dots okay so if you differentiate this dot product right here you end up getting two of these right here so that way if you divide by two you get what we had before so if we divide both sides by two you get this and this is exactly what's going on here so this is right here so I hope that makes it a little more clear um I know my handwriting is a little messy but that's how we're going from here to here just a clever clever use of that product rule for for the dot product all right now another thing we're going to use another property of the dot product remember if you take a vector and you dot it with itself then you get the length squared so RP prime. RP Prime is the length of RP Prime squared cool stuff now what we've got is the derivative of this length with respect to T but we're also integrating with respect to T so by the fundamental theorem of calculus we can just evaluate this at the two end points and take the difference so ftc2 coming into play here then do the evaluation you get R Prime b^ squ minus r prime a oops typo squared oh another typo right there that should be a squared and I fix it at the end all right now last little notation change here remember that the derivative of the position Vector is the velocity Vector so R Prime is v r Prime is V and then now we can split this up and we can write it as 12 mv^ 2 minus 12 m v^ 2 or I should say the size of of V evaluated at B here evaluated at a here there we go all right now what is this we just showed that the work done is equal to this value here and the quantity that you're seeing in these two terms half the mass times the square of the speed that's what's known as the kinetic energy K of the object so physicists and Engineers even maybe you recognize this before we even got there that's the kinetic energy we can we can denote it by K and so what we can do is we can rewrite this work equation as the work done is the kinetic energy at B minus the kinetic energy at a there we go so that's the work done by the force field in moving an object along c so it's the change in the kinetic energy at the end points of C all right so hold on to this for a minute now we're going to look at it from a different perspective here we go if we further assume that our force field is conservative if it's a conservative force field then that means there's some potential function little F or if we take the gradient of f we get bold F now we Define the potential energy P of an object at any point as the opposite of f so this is what we call the potential energy this is why I've been referring to this as a potential function like a potential function um so we're going to Define p as the opposite of f so that way capital F or bold f is equal to the opposite of the gradient of P um I note it here but the reason that we choose the negative sign for this is just to agree with the intuition that work done against a force field increases the potential energy right so it's like um if you're you know if you're pressing against a wall and the wall is not moving you're increasing your potential energy right and then if the wall were to disappear you would go like that so that's what we're talking about the reason this is negative is because uh if you're working against the force field you should be increasing your potential energy all right now by the fundamental theorem for line integrals remember because this is a conservative Vector field the line integral here is really just negative this line integral by substitution right and then we can use FTL to just evaluate this at the end points of the curve but this is B and this is a so we get that the work done is also equal to the difference in the potential energy at the end points all right now here's the magical part actually I feel like all that was the magical part but I guess here's the punch line here's the punch line if we put these two work equations together we have that this is to this so the change in the kinetic energy is equal to the change in the potential energy moving some things around and we get that the potential energy at a plus the kinetic energy at a has to be equal to the potential energy at B plus the kinetic energy at B that's the total energy so total energy at a is equal to the total energy at B this is the conservation that you learn about in the law of conservation of energy so in other words if you're going from point A to point B under the influence of this conservative force field the total energy remains constant and that's the law of conservation of energy and this is why we call it conservative and potential so I hope you like that result very cute very clean very nice all right and on that note let's play so um I'm excited to work with you all about this uh when we do an activity related to it um thank you for hanging in there for a slightly for for a long lecture um if you have questions please ask in class class uh I do recommend that you review those proofs because some of those techniques are really handy um but I'll stop rambling here and I will see you all next time thank you again for watching [Music]