Okay. So what I was saying was that on the second exam, it'll be based completely on the homeworks, the fifth homework I have sent and the fourth homework that you finished. Okay. It'll be I mean, I will send you a specific syllabus for it, but that's what it'll be. No, I haven't done any of the proofs of theorems in calculus of variations. So that will not be on the second exam, but I'm going to do a proof on most likely Tuesday next week. And I was saying a moment ago, in calculus of variations, it's rarely that proof, not the proof really. Harvard, I come up with those differential equations. That is really the strength of, I mean, that is the important point I want you to learn from this course. Okay, how did I come up with a differential equation? Because maybe a slightly different version of this problem which is not covered by the theorems. How do you obtain the results for that case? So is this technique that I want to show you, OK, that I will show you on Tuesday, which you can call it a proof of the theorems. Okay? That's really the important idea. I mean, as I said, you can appeal to the theorem then solve the differential equations. Yeah, that's good, that's fine. But what you really need to learn from this course is, how did I get those ODEs so that, you know, you can do this supplies. I'm repeating myself, but how can you get those ODEs for a slightly different problem which we have not covered. All right? Okay, so let's continue with this brackish aquatic problem. I was sort of 1 third of the way through. So I'm going to actually referred to the stuff I wrote before last time. And we'll go over that quickly. Alright. So alright, I've shared this wondering what I was missing. Okay. So we were doing the Saudi I'm a biology if you don't mind if I sniff a little. Ok. So the brackets took a problem just to remind you, said that suppose you have two points. Okay, a and B. And you build a slide. Okay? You build a slide from a to B. And then you put a bead on the slide. So it slides from a to B and it takes a certain amount of time in slides under the influence of gravity. Okay? And the question is, you are allowed to take slides of dif, different curves. You can build any slide you're like. Okay, and the goal is to find the slide so that the bead slides from a to B in the least amount of time under the influence of gravity. Right? As we were talking last time, why isn't just the shortest path? Right? Who wasn't just the shortest path from a to b, right? And we agreed that if you'll make it slide down first, its acceleration will be higher. It might pick up more speed. Even if the path is longer, it might actually take less time to go from a to B in that, right? So a direct line a to B is, you know, is the shortest path, but may not be the fastest way to go. Make it more vertical, longer, but it may go faster. So the correct answer is something in between, like competing these two things. So what I said was that and, you know, who knows actually, it might even be that your shortest path might even go down and then come up. Because if you go down, you know, you are accelerating very fast corners, right? It's not clear at all. So I don't know if you can argue that. What do you think? Could there be such a path? No, because when you make the turn to go back up, that's going to kill lawyer momentum. But still, I mean, you may be picked up so much that he didn't know. Right. Could we argued that? Actually, I don't know the answer. I mean, I I think there are solutions like that, but I'm not a 100% sure. I mean, we have the old later on we'll derive the solution and then we can check it actually. Ok. So I don't know the answer actually. I know how to figure it out, but I don't know the answer. Okay. What what what the way I'm going to do it is I'm going to set up the axes so that the y-axis is pointing down just for convenience. Ok, otherwise you have to write some expressions with a negative. So the y axis is always going to be pointing down. And I'm going to my curve, it's going to be, okay. My curve is going to be y equals y x and it's like this. Okay? And so the problem is find the shape of the slide. Y equals y, x from a to B. So that a bead slides down this slide. There's no friction. In the least time amongst all the possible slides joining a to b, right? Look at all the possible slide joining a to b, k. Find the shape of this slide from a to B, which takes enrich the particle, the bead will take the least tides. Can, there is no friction. So how much, if I, if I pick, pick an arbitrary path, y equals y, x, you know from physics you can work it out. You can use the principle of conservation of mass or many other ways. Once you know the shape of the curve, you know, you can actually predict. You can do other things and you can figure it out from physics, that the time taken is this. If your path is y equals y x, this is the amount of time taken t y. Ok, this can be done from physics. So our goal is the following. Minimize this functional which depends on the curve y equals y x, where v x y x joint X1Y1 to x2y2. That's the problem. Ok? So this is a calculus of variation problem where your function f x, y, y prime. Is this. Okay? That's your F x, y, y prime. So by the way, if you had chosen the y-axis to be pointing up this term, it will be y one minus y. Ok, that's the only difference there and other things would be the same. And then, you know, when you differentiate always you keep getting negative things. So that's why I chose the OK, so that's the y. So now from a theorem on calculus of variations, we notice that this f is independent of x. Okay? So the optimal curve will be the solution of the ODE, this ODE, this ODE. Alright? So you have to differentiate this f with respect to Y prime. And so again, you use this d by d z of one plus z squared with the half, and you cancel out the two. If you do that. Delta f by delta y prime is this quantity. What is this quantity? You can just check that. So if you plug it inside here in this equation, alright? You get this equation for the OD, for the optimal curve. Okay, so this is what I was doing near the end of the class last time. Okay? So now you have to simplify, do some algebra, put it under a common denominator. And you get this, the numerator. And that's minus one. So that's a first order ODE. Okay, and then you square things and bring it in the other side. So you'll get some new constant, C one, sorry, C. Okay? And then you solve this for y prime because I'm going to do is help rebel. So one plus y prime square is this. Okay? And then, by the way here just one more thing I wanted to say. Can you say something about the sign of this quantity is positive or negative? Isn't it negative? Because one plus y prime square has to be positive and then y is always going to be below y1. Okay? But remember my y-axis is pointing down. So the y is greater than L1, right? It's here, sorry, just one minute. So i1 is the highest point in y's here. Of course you could say, you know, maybe the weather, the curve cannot go up, right? Because we know from potential energy it can never go higher than a if it starts with 0 velocity. Correct? So lemme go above a. It could go below B, but lemme go above it. So Just a minor thing, but that's what good is it? He's positive, does something to remember. Okay, So then you've solved for one plus y one square, y prime squared, and then y prime squared. Okay? So then you put it under a common, does put a square root. Now, this should be plus or minus. Right? Here would be, the curve is always going down. Minus means the curve might rise up. Remember I was talking about this situation. Right here is a and here is B. Will you always be like this or will you always could also, this could also happen. Remember the y-axis has pointing down, right? So the negative would be when it is trying to go up. Then y prime, remember y is pointing down. So then y Prime would be negative. All right, so I'm just going to assume that y prime is positive for the moment. And if you read the book, the notes, I've talked about what happens when it's negative. It's, it doesn't change anything. Alright. So both cases going up, going down and then coming up and not it's covered all it's covered in that. I'm just going to take it with that and let's do it like that. Alright, but you read the notes and I talk about plus or minus. So dV by dx is that separable? So you do that. Alright, this is where I was last time. Okay. So how do you integrate this? And I was talking about, well, let's just forget about the Y1, that's some constant. So how do we integrate this? And let's make the C one. Then you would do y equals sine squared theta. Okay? Then it, because it's one minus sine squared cosine square gets rid of the square root there, it gets rid of the square root. Alright? But, so let me put this here a different way on the side. Okay, that works. So I was talking about what's the substitution you try for that. So to make it look like that, it really does take Y minus Y1 because he wanted to make that into a sine squared theta. But then that doesn't take care of the sea. So you have to put a C there. I'd be okay with this. Everyone. All right. So then that takes care of that, and then the C is taken care of here. So D y is 2i sine theta cosine theta d theta. So you plug it into the equation, so you get integral square root of C. And from the top, sine theta d y is 2i sine theta cosine theta d theta. And that is a C minus sine squared theta is cosine theta integral dx. Okay, please, if any state you don't understand something stopped me. Who we get to see. Let me leave this two inside. You'll see why. Equals integral dx. Is that okay? All right. No one's going to tell me how to integrate sine squared theta. You can change sine squared theta two times the quantity one minus sin2 theta. The other double-angle formula, right? Convert powers of trig functions into multiple. That's really how you integrate trig functions. If you have a power of a trig function, sine cubed, sine, sine to the power five. You convert it into multiples of theta. That's how you integrate powers of trig. And there are all kinds of clever ways of doing this. Okay, I'll not go into that, but that's really how you do it. So you will get C. This has one minus cosine two theta. D theta is dx. So x is c times theta minus sine two theta over two plus some constant d. Okay? So we got the solution. So you say, what's the solution? I don't see a y equals f x, right? That is what our goal was. But actually we have it in a slightly strange way. Let me write it a different way. So I'm going to pull the two out. Two theta minus sine two theta plus d. And why is this was a substitution we made. Why is Y1 plus c1 Site C sine squared theta? Okay? So this was a substitution we made and this is the value of x from integration. Ok? So in principle, you could solve the x equation for theta and plug it into this y to get y equals something x, right? That's what I'm going to do because it's very complicated to do that. Okay? So we rewrite this parametric curve to look like something familiar. Okay, this has occurred, which you know, I think you saw this in 242, maybe 243. Okay. Which you will recognize. So I'm going to first write that. So notice this is in a double angle and this is in power. So first thing I'd do is convert this into power into multiples. Okay? So from this you get y is Y1 plus Y2, sorry, y one, C. And this, you write it as a double angle formulas. It's one minus cosine two theta. Okay, so I can thus using the double angle formula. So I can, I'm going to write this again the second. So x is c over two theta minus sine two theta plus d n y is y one. Let D on the left hand side actually. Okay, I'm just rewriting this. I'm not doing anything special. I'm writing it so that it'll be enough home which will recognize in a second. Okay? So I'm going to call this some constant capital T. Okay? And these two thetas, I'm just going to call phi. Okay? Then you will say it looks something like and I'm just going to call little d into capital D. So x is d plus c phi minus sine phi. Y is y one plus c one minus cosine phi. And then here phi is varying. Okay? So this is the parametric equation of the breakfast. Okay? There are some, okay, for some constants. C, d, c, d, and of course we know. Do you recognize this curve? Have you seen this? Did you see this into 4243? It's a popular example. Anyone is Lakewood. And what context do you see it? And I actually don't remember seeing it in T4 A380. I just already knew what the current was. I see. How did you know that? I don't know. I F is some like some math thing I read. A while ago. I don't know. Anyone remembers this hyperboloid for anything. What do you say? I don't like in high school we didn't do the, we talked about how the sigmoid width and you invert it. And if you like, let balls drop from any two points of it, like on the height, they'll get to the bottom at the same time. I tried, that's something different. But that's an interesting thing. You were saying that the two curves joining a to B, which are different and you let the curve's height as thing beat slide. It'll take the same time, right? Yes, since I have the sitewide and if you have like a ball starting here and a ball starting here, and you've got the same time they get to the bottom at the same time. Human directly or by the psych lloyd? Yeah. Yeah, on the segue. Now, the cycler takes the least time. Alice, I don't I guess I might be missing. There is something else do I know I did. There is I mean, the right buried in this. There is another thing, two things we'll take the same time. If what is that? Control member? I can look it up right now. Okay. Look it up. I've I I tell you something else or you can look it up later. Ok. I can look it up and while I'm talking of something else, look it up and tell. Okay. I know there is something about two things had to Curtis taking the same time. What is that? I know. Okay. Anyway, I'll come back to this hike Lloyd, as to what's the significance of this and how it is generated or professor, Yeah, I looked up and it's called the pollen crowding curb, Qur'an, also cycline. Okay. And what does that what problem does it solve? It's that the time taken to fall to the lowest point is independent of its starting point on the curve. Oh, I see. Okay. That's a different question, correct. That's very good. Okay. I'll talk about that in a second. Let me finish this problem and then I'll come back to that. Interesting. That's a different question. Yeah. Okay. So this is a psych lloyd. Now, how has it solved the problem? Let's let's just come back to this. Okay. So I was in the notes, I've solved a specific problem, did I? Right. Okay. So there are three constants to be determined here. Right? There's Y1, Y2, right? So let us write this down here. Okay? So optimal curve is x t plus c three minus. Sine phi Y equals Y1 plus c one minus cosine phi. Ok, now there is a range for which we don't know yet. Okay? So we have C, D phi one, phi two to be determined from the fact the endpoints are X1 and x2, y2. Okay, so these are the two end points. So what is the range of the parameter fee to give you the curve? What are these constants? C and d. Alright? Okay, there's actually some choice here for phi one starting point and that determines phi2 and stuff. So the easiest thing to do is actually, OK, so freedom in choosing phi one, ok, I'll take phi one equal to 0. At phi one equal to 0 curve is that X1Y1 or the point is that X1Y1, right? So that's where. So let's look at this. If you put phi equal to 0, so you put phi equal to 0, then you get one minus one is 0, you get 0 and then you get y ones. But this one is correct. And when you put phi equal to 0 at both these 0's for d should be x1. Okay? So if you plug it in double star, so using this in Douglas part, okay, we get D is x one. So this d is x one. Alright, so let me write the equation again. Lesser. Yep, and you explain why we have freedom in choosing I1. Okay? So it's a little more subtle and I haven't talked through this very deeply. So I think what happens is let me see. I'll come back to that. I haven't really thought it through very carefully. Okay, but I'm thinking that, you know, you can shift your interval little endl, chain these constants a little. But I'm not a 100% sure yet. Alright, so who was that? Who asked me? I did. Who's at a danger? Andrew. Okay. Alright. So let me think about it and then maybe I'll answer ready and I'll send you an email or I'll do something. But there's a little bit of freedom here. I haven't thought through this very carefully. Alright? So I will do that and I will come back to you. But let's see, let's think of another way, right? So there are one to C and D, there are 444 unknown quantities. And how many things are we trying to fit? I guess we're trying to fit four. So there really isn't actually freedom. Rights. Frankly, we don't have freedom. I think phi one equal to 0 comes out automatically. I have to think too that more carefully. All right? I don't think there is any freedom. And the solution actually is phi one equal to 0. But let's come back to that later. So using this, you get d equals x one. So let's plug in the x is x one plus c phi minus sine phi. Y is y one plus c one minus cosine phi. And phi one ranges from, from Saudi fee ranges from phi2, phi2, phi2 and C to be determined from the fact that the right endpoint is x2y2. Alright? So that means, so you plug in phi2 and that's the right endpoints, but therefore you get x2 is x one plus c phi two minus sine phi2. And Y2 is Y1 plus e one minus cosine phi two. Ok? So therefore, phi2 and C are determined by these two equations. Okay? So you have two equations and two unknowns. So someone gives you the right and left endpoints. And then you have to just figure out the c and the Fe2. Okay? So two equations in two unknowns. Okay? You can use Mathematica. By the way, if you forgotten how to use Mathematica, which commands there are? Okay, I update the notes. I don't if I posted them yesterday, but I have them and I've done one example is basically three lines and they all look similar. You just ride on the equation and then just use find root. Okay. So I haven't yet decided about the exam whether you will need Mathematica or not. But I suggest practice it. Alright. It's just three lines. I've, I will post that you will see just three lines of mathmatical commands. One to define, to define the equation. And 1r, you would just say find root. That's it. It's very simple. Okay? There are many more things one can do, but you will see in the notes it's just three lines. You just copy that, use them same time, same thing every time. If your equations are correct, it'll just give you the answer. Alright? Okay, so these are two equations, two unknowns, you solve it. So let me just copy it again. I wanted to point out something else to you. So it's x1 plus c, where c and phi two are determined by these two equations. Do you notice something here about? Notice that when I was doing this problem, the points a and B. Ok, where any two points in space. Then I put my origin and my axes. Right? And you got your curve. Is there's something interesting in this which tells you that your, that your answer doesn't depend on the access chosen, where your origin is and stuff. Do you notice anything in this? Why the answer doesn't depend on the origin chosen. The shape of the curve is not going to change no matter what origin, which is the way it should be. It's a minor thing, but just something interesting. Ok, I will draw something, but then I have to hide this part of the equation. Right? That's her optimal curve. Select look on the right hand side here. What do you see? It's x2 minus x1 and y2 minus y1. So it's really how much a rite you moved and how much down you moved? Your answer depends only on that. Okay. It's the relative distance from a to B, how much right, and how much left doesn't depend on anything else. It doesn't depend on the precise location of a and B. The precise location shows up here. Is that make sense? What I'm saying here is just the relative distance that you travel from a to B. That's Hall Effect set, okay? The shape is determine this shape. And these parameters are determined by just the relative distance. Okay? And then this is the shift based on where you chose your audience. Okay? Alright, so I will stop here. And then if I give you a particular x1, y1, x2, y2, you can just solve this here. Then I'll move on to the next next thing here. Alright, so let me stop this recording.
The brachistochrone problem - II
From Rakesh Rakesh October 20, 2020
5 plays
5
0 comments
0
You unliked the media.