Good afternoon everyone. My name is Rakesh Nelson is sick today, so I'm teaching math to 43 for him today. And so why don't we you see me looking sideways. The cameras on my laptop and I have a monitor and your pictures are on the sides. So that's why I'm looking there. And I will request can you all please turn your cameras on? Otherwise, it's like talking to a wall. I don't think anyone likes talking to a wall. So that way it can get interactive class, right? I'd like you to ask me questions. Are He's having students have seen before. Okay. So feel free to interrupt me, ask me questions and I will ask you questions. It's a planter class that way. Alright. So please, if you don't mind, turn your cameras on. I think you'll be more engaged. I'll be more engaged that way. Alright. Thank you. Almost everyone. Okay. And I recognize many students, 242-4200 m. So we're going to come back. Okay? So what we're going to do today is something called the fundamental theorem of line integrals. Okay, but that's just a name. What are we going to be doing for it? So that the cost of being pedantic. I will to show you how you were taught in the last section, perhaps our last class, how to compute an integral on a curve. And then I will show you that in some cases there is a shortcut. Alright? And that's what the fundamental theorem of line integrals is. So let's just look at this example which I'll go through fast. Because I think you have done this before. So this is an example. Suppose you have your vector field F is given by y is x squared j plus x plus z k. Right? I'm trying to find where is it that one can reduce the size of V here. Okay, That's cool. Okay? So there's a vector field and you're given a curve. And you're given the curve C and I give you its parametric equation. X equals t square, y equals t cubed, z equals t plus one, where t ranges 1-3. Okay? You're given a vector field, a vector which changes from point-to-point. Sorry, someone is wearing the lawn in my backyard, so okay. So you're given a curve C and the goal is okay. That's my goal. So is it very nicely for you or is it okay? Is it okay? Okay. It's noisy for me. So, can someone tell me how to compute this integral? How do you go about computing this integral? Anyone, anyone wants to recommend a way to compute this integral? Would you start by finding the derivatives of x, y, and t as a vector, right? So you'll find define the r vector, the manual, okay? Yeah, you define the outer vector. Compute d r by d t and then it puts you right? Yes. Okay, So then, Okay, I get d r by d t. So how about computing the integral though? What do I need to do? I think you would plug in the x, y, and z values into the f function. And then you would do the dot-product in India. Okay, So let's do this factory. Alright? So you define the our function. Our function is the x i, y, j, z k, which on the curve is e square i t cubed plus t plus one k. So, okay, before I go into clear, let's, let's just do this. I should have written this before. What is the strategy to compute this? And this is what value was saying. The way you do it is you take F dot DR, see, and you replace it by a d t and then we don t know how to do a DR integral. But if you replace it by a d t integral, okay? And t will be 1-3. So you convert it into a regular interval which you've done in calc one, okay, and that's how we compute it. As Emanuel said, I need to compute the DRY T3. And then I'll write the F on the curve and then take the dot product. I'll get a function of t and I integrate. Okay, so let's just do that. So R is on the currency, r is x, y, j, z k, which is square, The cubed j plus e plus one k. So d by d t is t squared j plus one k, right? So that's the d r by d t. Now I need to get the F on the curve C. F is a vector field. It's YZ, X squared j and x plus z k. That's the vector field. But all the curves C on the curve C, this has x, y, and z is, I. Plug it in. X is t squared, y is t cubed and z is t plus one. So t cube t plus one. X square. X is t squared. And x plus z would be t squared plus t plus one. Okay? So this says that I need to now take the dot product. F dot d by d t is two t times cubed times t plus one. And then three t squared times t four plus one times that. Okay? Which becomes to T5. Let's do T4, T6. Okay, That's F dot DR. Vitamin D. By the way. Should be recording this a lot. Oh, it's already being recorded. Okay. Okay. So then if I wanted to compute F dot DR over c, that's the same thing as the interval was 123 f.drd, t, d t. So that's 123 of this whole thing. Or plus three keys, cx plus d squared plus d plus one d t. I'll leave it for you to finish that. Okay? So this has probably do this line integral on a curve, F dot DR, integrate on the curve. Okay, so that's good and fine. So why am I repeating a calculation you just did have like in the last lecture. So I wonder to show you that in some special situations, you don't have to integrate at all. There is a quick way to get this answer. And that has various other implications. It's I think multiplications in physics and staff. And what does that situation? Okay, so I'll leave that for you to finish. So in some situations, there is notice the word sum. There is a quicker way. Compute. Okay. Dr. That is not the only reason I'm doing this. It has many other implications. So what is this quicker way is basically says when your vector field F, right? If you take a function little f, then you take a function and you take its gradient, you get a vector. Gradient of f is a vector which changes from point to point. So it's a vector field. So it says when this vector field F is in the form of gradient of f, then it's very easy to compute this integral. It's actually the fundamental theorem of calculus. That's what really does solve, and that is what is called the fundamental theorem of line integrals. But when the word line integral means integrate on a curve, line, curve, that's the idea. So it says, basically says this. Let C be a curve. We went starting point. And the ending point. I'm trying to find a better. Let's see. Let's make it uppercase B as the ending point. But guess what the picture looks like this. I have some curve, doesn't matter what it starts to pay. Ends at B and this is my curve C. I'm going to be integrating on this curve. And I'm going to compute the line integral. Okay? Don't write this yet because I'm going to change it. Now this is the general line integral. I wanted to compute, right? There is no shortcut for a general line integral. But if I replace this f by a different vector field, a special vector field gradient of f. Then this would just equal to the value of f at the, right, at the ending point minus F at the beginning point. Don't have to do any integral. So if your vector field had this very special form, instead of capital F, You have a gradient of little f. Okay? So let us do a computation like this. Yeah, let's just do a simple computation and then I'll show you some other implications of this. Okay, so the theorem says that if you integrate gradient of F dot DR over any curve C, then the value of this integral is the value at the end points. The value of f at the ending point minus the value of f at p beginning point. And if it reminds you of the fundamental theorem of calculus for one variable, it really, the proof is that it comes from that actually. So let's do an example. Okay? So given F is little f with this function, x squared y, z. By the way, anyone has any question on this statement? You all okay with that statement. Okay? Alright. So given f is this function and that curve, C. Okay? Make up any carpenters made up this one in between -1.1. Okay? So at some curve, okay? And my goal is compute integral of C over c, the gradient of F dot DR. My goal. Right? I have this function. I can compute this gradient. And then I can compute that, right? So that is how I would do it if I was using the technique from the first example I showed you. Okay, I will compute gradient of f. I'd get the capital S and then compute d r by d t, computed f naught d by d t and do all that. That's what I would do it. But this fundamental theorem of line integrals tells me that this is equal to, you just need to take the starting point and the ending point. Correct? So from fundamental theorem for line integrals, right? It said this was equal to the value of f at b minus the value of f at p. Don't have to do any integration. Alright? But this is notice it's very special. I keep emphasizing this. I'm sorry for repeating myself. It's not a general vector field, is the gradient of F there. Alright? Okay? So how do we find? So I know my F, F is right here. I need to plug in the value of the beginning and the end points into f. So what is the beginning and then what are the beginning of the end points? How do I find my beginning point for discussing? Anyone wants to tell me? How would that be the inequalities? One? Yeah. Could you say that again? Good, funny thing is I will just looking at you asked your picture and useful task. Okay. So tell me again, how did you find, how would you find the coordinates for ANSD? You use the bounds for TEA. Okay, Good, Good. Okay, so which point? How would you get a lower bounds? So that would be negative one file. Okay? So exactly. So it's the value at d plus two, d square -2.2 t at t equals minus one, the lower bound. So a is one, minus one and minus two. Okay? So I guess you can see ending point v is this equals one, right? The endpoints, so this is -1.1. So ending point is b minus one. And. Okay? So therefore, BR si is f b minus f, which is the value of f at e minus 12 minus one minus one minus two. Okay? Now, F was given to be x squared y, z. Therefore gradient of f or DRC is to plug it in, you get what, nine minus one minus one square, minus one minus two, which is -18 -20. Okay? So we could compute this without doing any integration. But again, it's a very special vector field. All right? Okay. Again, I, sorry, go ahead and write it in before I say probably writing this. Okay. So I just want to go back here. Now, I did not tell you why this theorem is true. It really is a cons. If you just write down this definition and use the chain rule, it comes from the fundamental theorem of calculus for one variable. What you did in Calc one. Okay? So consequence of that, I'm not going to prove it's more important for us learn how to use it. And what are some of the other implications. Okay? So implications of the fundamental theorem for line integrals. Okay? So let me write this down here again, just what the theorem says. And then we'll look at some of the implications. Okay? It says integral of gradient of F dot DR. Oversee is the value of f b minus f. That's what this theorem says. So it says look at this integral. You integrate on any curve C. To compute the value of this integral, what do I need? I need the, of course, the function f original. What do I need to know about the curves C to compute this integral component has to be continuous, right? Okay, yeah, correct. There was an older hypothesis. We're assuming all those things are true. It's continuous, is it that needs to be differentiable and all that, right? But what I'm saying is to compute this integral, whose value is this? What do I need to know about the curve? What information do I need about the curve? To compute the right-hand side? Or else my question is not clear enough or is it too trivial what I'm asking? My question? Yeah. What? I'm trying to compute this integral, right? So it says if you want to compute the gradient of F dot DR integrate on the curve C, you'd say, well, I need to know the curve every Maryland was, how am I going to compute it? That's what we say. But then you look at the answer on the right-hand side. It says f b minus f a. So compute the answer on the right-hand side. What is it based on? What do you need to know about c there? You won't even say something. Same need to know the bounds. Human, you need the beginning and the end point. That's all you need to know. Yeah. So it doesn't matter what shape the curve has, what the curve was doing between the beginning and the end. This integral, which supposedly depends on the whole curve, really just depends on the endpoints. Correct? So that's very bizarre. So in particular, you could have another curve, right? You could have another curve, C1, with the same end points. Okay? And the theorem says that if you compute C1, you'll get the same answer. This is a very bizarre. So it says if the vector field at this very special form, it's a gradient of f, rather a general F. Then in this case, it doesn't matter what the curved shape, what path the curve, just beginning and the end points. Okay? So that's what this fundamental theorem of line integrals, this. Now, do you recall, I mean, some of you've taken your physics towards seven metaphysics to them, kinematics, right? So you have look at the motion of the particle under a conservative force field. A particle unit goes from, you've seen this already. You know, you have physics. We have point a and point B. And there is one particle which goes along C1. Then you ask how much work was done. If you recall, if you've done work done, then the work done is the difference of the potentials. Okay? So it doesn't depend on what path you've chosen. If you go from this point, if you lift a stone from this point to this point, then the work done as the difference of the potential and it doesn't matter what path you took. Okay? That is because your force field, this is the integral for the work. You don't need to know. This enters reminding him this is the work done integral, a much work you did. And this is the force acting on it. So gradient of F is the force F has the potential. So this is the theorem says that if you are in moving a particle in a force field, then the work done depends only on the end points. It doesn't depend on what path we chose. This as. If you have a conservative vector field, meaning the vector force field F comes from the gradient. If it doesn't make sense, don't worry about it if you haven't seen this before. But this is actually, this is from physics. That in physics, if your force field F and B written as the gradient of f optic, normally you write a minus the gradient of f. This is your force and this is the potential. Okay? I don't know if you've done this. So if you had to know the force due to gravity particle moving up and down than the potential is mg z or minus mg. So it's really coming from here. Anyway. I'm not going to go more into that us because not everyone is interested in this, but this is a consequence of this fundamental theorem of calculus. Intelligence. That if you integrate gradient of f on a curve, then the value of that integral is just based on the endpoints, doesn't depend on the curve. Okay, That's one implication of that. Air is a second implication of you. Okay. With this, anyone has a question. Okay? Okay. So here's the second implication. Second implication for fundamental theorem of line integrals. So let's, let's look at a very special curve. Okay? If the curve starts at some, let's put, let's call this point. R. Starts from r, does whatever it likes. It comes back. So this was my curve C. And I'm asking, I give you any function f and I say compute this integral. So what do you think this will be? Your wants to guess? Or a equals zero. Sorry, Let's see who's speaking. Sorry, could you say your name again? I didn't see the move too quickly. It's a manual. I said, why, why is it zero? Because since the line integral is it's used to measure work. Since you started in start and end at the same point, the amount of work done in total would be equal to zero, right? So actually this is a consequence of the fundamental theorem of line integrals, but I want to apply it. So you're telling me the physics point of view which is an application that a consequence of the fundamental theorem of line integrals y is the zero. You're correct, completely correct. But do you mind can you send me a call? This is a consequence of the fundamental theorem of line integrals. Would it be since the starting and end points are the same, that means that f of b and f of a, or the same thing. So when you subtract the one from the other, it would be equal to zero. Exactly. Write this as f of n point minus f of start point. So it's far minus F are not zero. Right? That's exactly how the physics consequences which you were telling me, right? So, so if C is a closed curve, meaning it has the beginning same end points, then the integral of F dot DR over c is. So that's the second consequence. I just said exactly if it has consequences in physics. If you're moving under conservative vector field. And you start from some point, you go all over the place, travel the world, and come back to where you started. Total work done is zero. Alright? But only in a conservative vector field, meaning it's like that. Okay? If I just did this on a closed curve with same n-bit medium, point out it's not zero. It's only even this very special. So the question that then begs the question, you're given a problem like this. Okay? And you say, You know, I want to compute this integral quickly. So you'll say, well, let's see if it is the gradient of something or not. If it is a gradient of something, I'll be able to compute this integral. Very fatigue. So the question is, okay, so the question is how do I figure it out whether a vector field is a gradient of something or not. Okay? So let me write this down or whatever I just said really quickly. If this vector field F is the gradient of f or some function f. Question. I'll do read decide whether F can be written as a gradient of f. And how do we find, okay? Not every F can be written in this form. How much time do I have? I have 16 min. Okay? So I'm going to answer this question in two variables. Variable there is a similar story, but let's just first 22 variables is a vector field in the x, y plane. Okay? So suppose your vector field F is some px y, u, x y j. Okay? So it's a vector field in the x, y plane. So it just has I and j components and everything is dependent on x and y. And we write f as the gradient of f or some function f of x, y, right? Okay? Now gradient of f is just f sub x dy plus f sub y g. So is there and F such that this P x y I plus Q x y j is equal to Fxi plus f y j. Okay, So let me rephrase this question. Given the vector field f, is there, f such that E is F sub X? Okay? Let me put an X, Y. The question. Okay? So I'll give you some function p. We'll do an example in a second. Another function Q. Can you find one function f, so that p is f x and q is FY. Right? You're trying to address that question. So suppose there was, what does it force? So you know, which, for which p and q can you find an F so that this happens? That's my question. If for this p and q there is an f, what condition does it impose on capital P and capital Q? That's what I want to know. Their condition. You see? Suppose this p and q have such an f. What does it force on P and Q? Or something on the derivatives of P and Q? It forces a condition. Does that mean p and q must be continuous, okay? Right. We always assume that everything is continuous and differentiable and everything. All right, you're right, very important patients there, but in this course we don't emphasize that so much. Okay? So I have some functions, P and Q. And for this P and Q pair, there is a single little f for which these two things forward. What properties does it force on P and Q? Because then that's where I can check before I tried to look for f. But actually all the derivatives are you encouraging poses or property? Something we don't pay attention to so much, but actually it's an important property. And you think what conditions if for this pair of functions p and q, which are completely different functions, there is one little f or which P is the x derivative and q is the y derivative. When can that happen? What conditions must be enqueue satisfied? Rather is a condition on the derivatives of p and q. It's not something which is very obvious in the sense we don't think about it so much because we just assume it. Anyone has any idea. Let's wait another minute. Did we already assumed that it's a function of both variables, x and y, correct? Yeah. So it's going to be a relationship between P and Q, but actually derivatives of p and q. So, okay, I could write it but less than. So look at these quantities, px, UX, and QY. There are four quantities here. I claim two of them are equal. Which to? If this is true, then two of these are equal. Would they have to correspond with Clairaut's theorem of like partial derivatives. Yeah, What does it say? Give me once. Suppose f is defined on a disk D that contains the point a, b. And if the functions f of x, y and f of y, x, or the partial derivatives are both continuous on D, Then they would be equal, right? So just to, as I said, I'm not going to emphasize, you know, all the conditions here. Basically says part of most nice functions. You take fx and then you differentiate with respect to y. Or you take first y derivative and then differentiate with respect to x, right? Reverse order. Most nice function, these two are always equal. That's what it says Clairaut's theorem. So if you keep that in mind, which of these four are equal? Which two of them are equal to each other, which to y1 and Q x d y, right? If you take, this is p. So p y is f x y, and QX is f y x. You won't see them. Right? So therefore, ok. So let us go back and backtrack. So my question was, I give you a vector field F. And I ask, can you write this as a gradient of f or some little f? Which means the p will be the x derivative of f and q will be the y derivative. There is such an F. Then I'm going to erase this. Now. If there is such an F, Then BY is f x y q x is f phi x. Therefore q p y is equal to q x. So let me write this as a proposition. Okay? So let f be yeah, I plus Qj variables. Let's put all the variables there. There is a similar theorem in three variables, but we'll skip that for the moment. Then F is equal to gradient of f for some function f. There is some hypothesis here on the domain and stuff which I'm just skipping. Domain cannot have wounds and things. There is a condition there if and only if BY is UX. What's behind you? I like kale, they're all forth. Okay. Alright. Thank guys. They read my class before, so okay. All right. So, okay, so there's this proposition here and we have five-minutes. Let's try to do an example. Okay. I don't think I'll be does anyone have any question on this? Anyone? No. Okay. So we're actually going to take a capital F, the vector field and tried to figure out. And we find that is it from the form group of ethanol. Okay? I'm giving you this vector field. Okay? So show that this and be written as gradient of f. And find the f. Okay? So then once I have that, then I can compute my integrals very easily for this vector field. So if you're thinking in terms of physics, this is my force field. I'm trying to prove that it's a conservative field. And then F is, find its potential. The potential is the function f or negative or five depending on how that's the physics interpretation. But you don't have to worry about it. Okay? So in this case now you know the P is this function and the Q is the other one. Okay? So let's compute the y partial derivative with respect to y. So it's two x. And if you differentiate that, you get minus two over y cubed. You okay with this, everyone. U sub x. That's, I'm trying to verify this condition, sorry, 1 min. So given F is the gradient of f if and only if p is Q x. So I'm trying to verify that condition. So this is p q x is two x minus two over y cubed plus zero. You okay with this? So therefore, EY is QX. Hence, the vector field F is equal to gradient of f or some F. Okay? By equals q, x or the vector field F can be written as the gradient of something. Okay? Now the gradient of f is f sub X. Okay? Am I going too fast as at all? Okay. Molly is wrong, okay? Yes. Any questions? Okay, Alright, so I'm trying to write my vector field F, right? So my vector field F is this, this is the I component, this is the j component. So this has to be matched with f sub x. Then this has to be matched with f sub y. Right? So I want to just copy that down actually. Xy. Just give me 2 min. I'm not going to finish it, but I wanted to write something. Okay? So, so f sub x is equal to that. And f sub y is equal to. That's the question. Okay, Another one thing, I'm going to continue for 1 min on, but before that, I want to tell you, I believe you all get a cold do for your attendance. And the code is zero today. That's what he told me. Okay. The code is zero. But let me just ask you one question. I'm not going to solve this. How would you find an F which satisfies this property? I want a method that F sub X and F sub Y. Is that how do I find the hallway even beginning? You can take the integral with respect to x or y. Okay? That's Tucker, right? Yes. So you said, let's say I start with this first equation and what do I do? You can take an integral with respect to x there. Correct? So when I integrate that with respect to x, I'll get my F. Okay, I'll just do this quickly. Okay. And then you'll do something, but then how are you sure that the second condition is verified? So I'll leave it there, but that's good. That's a very good start and it's almost, you're halfway through there then. Alright. Okay, thanks very much guys. Steel. When to stop.
Math 243 April 21 2:30 PM Substitute lecture
From Rakesh Rakesh April 21, 2023
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