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Empty
 one year ago
Time to derive the EulerLagrange equation! @Astrophysics
Empty
 one year ago
Time to derive the EulerLagrange equation! @Astrophysics

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Empty
 one year ago
Best ResponseYou've already chosen the best response.6Alright so first off, I'll throw this up here: \[\int_a^b \sqrt{1+(y')^2}dx\] What's his integral? What function y minimizes this integral, any ideas?

Empty
 one year ago
Best ResponseYou've already chosen the best response.6c'mon gogogooogogoogoogoogoogogoggo none of my questions should take longer than 5 minutes, if you don't know you don't know, just say something and let's move forward I don't want this to drag on cause this is quite long.

Empty
 one year ago
Best ResponseYou've already chosen the best response.6Ahhh no I have to have participation for this one every step of the way, otherwise once you're lost it's hopeless since this is quite difficult. If you gotta do your homework though we can do this some other time if you like.

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Naw it's cool, lets do it!

Empty
 one year ago
Best ResponseYou've already chosen the best response.6Ok! Also anyone else who wants to chime in and try to answer my questions feel free! Ok so back to what's up: That integral is arc length, what function y(x) minimizes the value of that integral given an interval [a,b]? A straight line, y=mx+b as they say! (At least in Euclidean space, let's not worry about nonEuclidean space for now hehe ;P ) So we can rewrite this integral as: \[J[y] = \int_a^b \sqrt{1+(y')^2}dx\] The [ ] square brackets mean that the input is a function and the output is a number. This is a functional! So in our case we know that when y=mx+b this will make \(J[y]\) to be a minimum, as in some sense the derivative will be equal to zero here in some sense! We'll get into this soon!

Empty
 one year ago
Best ResponseYou've already chosen the best response.6Now we can look at a more general case, \[J[y] = \int_a^b F(x,y,y')dx\] Where now F(x,y,y') could be a bunch of other stuff you'd like to integrate such as some stuff about a curve that will let you travel under the force of gravity in the shortest amount of time or it might represent a hanging chain. We'll get more into the details of this later, the key thing is that our goal here is to find in general: What function y minimizes the integral J[y]? Any ideas or questions before I get into it?

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0The idea is to get a function that minimizes/ maximizes the integral, so in a nutshell we're looking for the equation of a line correct?

Mimi_x3
 one year ago
Best ResponseYou've already chosen the best response.0you guys don't have a life.

Empty
 one year ago
Best ResponseYou've already chosen the best response.6Only when the integral has \[F(x,y,y')=\sqrt{1+(y')^2}\] For instance F=TV could be minimizing the action along a path.

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Right right, so this is the principle of least action, so if we have a path, dw:1443264864345:dw quick sketch, it could be any path

Empty
 one year ago
Best ResponseYou've already chosen the best response.6So if I give you say \[F(x,y,y') = \frac{1}{2}m(y')^2  \frac{1}{2}ky^2\] then what y minimizes this? the path of a pendulum or a spring.

Empty
 one year ago
Best ResponseYou've already chosen the best response.6You have to be sharp on this one because we're looking at x as being the independent variable an y as the dependent variable. You would probably feel much more comfortable if I had written the exact same thing as this: \[F(t,x,v)=\frac{1}{2}mv^2  \frac{1}{2}kx^2\] So don't let the dummy names fool you, it's the relationships that matter don't focus on names, they're meaningless symbols.

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Haha, no I thought it was spring as soon as I saw it

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.0The dummy names are fooling me _, it wasnt until i saw the other form i recognized it.

Empty
 one year ago
Best ResponseYou've already chosen the best response.6Ok cool, well let's jump in then. First we need to set the value of y at the boundaries, \[y(a)=A\]\[y(b)=B\] and we want to somehow minimize this J[y] so in order to do that what the hell do we do? Let's pretend we already have the min, and call it y bar: \(\bar y\) Now on top of \(\bar y\) I'll throw another almost entirely arbitrary function eta multiplied by some scaling value and add it to it. dw:1443265381728:dw So here I've drawn eta \(\eta(x)\) as a continuous differentiable function that obeys these boundary conditions which will become important and more clear later: \[\eta(a)=\eta(b)=0\] So now we have this function: \[y(x)=\bar y(x)+k \eta(x)\] So when k=0 we have \[y=\bar y\] which means we're at our minimum possible value. Remember eta is a completely arbitrary function as long as it obeys those rules. So give this a couple read overs, and try to anticipate what we will try to do next and tell me. I'm making some coffee so I'll be back shortly! :D

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0I think that's the best explanation I've seen of eta yet

Empty
 one year ago
Best ResponseYou've already chosen the best response.6I want both of you to give me one possible function eta can be before we move forward.

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0I don't know, it's an arbitrary function right xD

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0I believe k is the constant multiply of eta, so you can pretty much use any function f(x)=kx^2

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0long as it's between the two points

Empty
 one year ago
Best ResponseYou've already chosen the best response.6Ok so two things, the multiple of k won't affect it, but it's not exactly part of eta even though it could have been, this isn't very important. And you're right it needs to connect up, so now this is kind of a tricky and a side tangent but I think it's worth trying to do once. Make sure that your function obeys this: \[\eta(a)=\eta(b)=0\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0I should have set up an actual function for y(x) though

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Ok so we can use equation of a straight line then put parabolas

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0long as it's between the intervals

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.0It took me a freaking minute to understand the relation between k and \(\eta\) and \(\bar y\)

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.0and why \(y=\bar y\). dammit lol

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0I don't I sort of related it to the equation of a line haha

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.0Yeah same. i related to coordinate points along both functions..

Empty
 one year ago
Best ResponseYou've already chosen the best response.6So here's are a couple of ways we could have done it: \[\eta(x)=(xa)(xb)\] \[\eta(x)=\sin[(xa)(xb)]\] But of course these are just completely general functions and it's not important that we have these, just as long as you understand what they would look like. So here I have used f(x) to play the role of y bar, and g(x) is playing the role of eta. Look at how when k=0 we get the graph we want. https://www.desmos.com/calculator/dnhlnprvs6

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Smart idea to use desmos, ok yeah I think I understand it

Empty
 one year ago
Best ResponseYou've already chosen the best response.6Take note, that the animation is pivoting around the points x=0 and x=4 since at these points eta is zero, so it doesn't change the function we have there.

Empty
 one year ago
Best ResponseYou've already chosen the best response.6After seeing this are there any more questions about this? This is really the conceptual bulk of the problem, the next piece is not so much conceptual as it is just going through some highlyadultlevel calculus (well like as tough as calculus 3 derivatives can get).

Empty
 one year ago
Best ResponseYou've already chosen the best response.6There is one last major thing I should lay out before we start pluggin' and chuggin' so the picture isn't quite complete yet but almost and I've laid the groundwork hopefully.

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0We always assumed the f(x) value is the min right? I was never too sure how to know it was ACTUALLY the min, like even with that drawing I made earlier (principle of least action) I assumed that was the min haha.

Empty
 one year ago
Best ResponseYou've already chosen the best response.6Yes we have assumed f(x) is the min. Right now we don't actually know what it is, the picture I've made that one is just some arbitrary one, so f(x)=mx+b would be the minimizing function for \[F(x,y,y')=\sqrt{1+(y')^2}\] however what I've chosen here \[f(x)=\cosh(x2)\] will be some minimizing function for some other choice of \[F(x,y,y')\] Continuing on a little bit with my explanation to try to make this more clear and give us an analytical expression to evaluate: dw:1443267473954:dw So imagine perhaps that J[y] is the arc length integral which gives a minimum value when y=mx+b. and so if we had thrown some eta function multiplied by a constant we'd have like: \[y=mx+b+k\sin x\] if say \(a=0\) and \(b=2 \pi\)

Empty
 one year ago
Best ResponseYou've already chosen the best response.6https://www.desmos.com/calculator/kgjhjslvf7 Imagine integrating arc length. for the curve on the interval \([0,2 \pi]\) here. When is this integral going to give you the smallest value? When k=0. I think this should make this completely clear now.

Empty
 one year ago
Best ResponseYou've already chosen the best response.6I'll let you think about this, ask me any questions, cause this is really the crucial part, once you've connected this picture to the concept I think you'll have it. And like I said earlier, anyone can ask a question any time, it lets me know you're thinking about it and I have purposefully left out things for the sake of trying to make this as easy as possible but if something bugs you, say it!!! :D

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Looks good so far!

Empty
 one year ago
Best ResponseYou've already chosen the best response.6Alright then! Calculus time. We have just by creation made this true: \[\frac{d J[y]}{dk} _{k=0} = 0\] All this says is the derivative of our integral with respect to k is 0 when k=0. Remember, \[y=\bar y + k \eta\] has to be plugged in to evaluate in a second, I'll go ahead and start it out, then you will have to struggle through what I think is the most difficult part, and after you give up I'll give you the answer ;P \[\frac{d J[y]}{dk} = \frac{d}{dk} \int_a^b F(x,y,y')dx = \int_a^b \frac{\partial}{ \partial k} F(x,y,y')dx =?\] So you'll have to use the chain rule, because remember F depends on y and y depends on k. Good luck! I'll be back in 5. :D

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.0I'll come back to look at this later today, I've got to be up in 2 hours and heading to take a nap atleast before leaving. I'll look at all this then :)

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0I think we should finish this off tomorrow, it's 6 am here, I'll work on this tomorrow when I'm not so tired XD, this is fun

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{a}^{b} [\frac{ \partial F }{ \partial y } \frac{ \partial y }{ \partial k }+\frac{ \partial F }{ \partial x }\frac{ \partial x }{ \partial k }]dx\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0That's my set up, now I'm going to sleep xD

Empty
 one year ago
Best ResponseYou've already chosen the best response.6Close, but wrong in a weird way. I think you're cheating :)

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0I used chain rule

Empty
 one year ago
Best ResponseYou've already chosen the best response.6Yes but we also have a third term in F(x,y,y') there's no \[\frac{\partial F}{\partial y'} \frac{\partial y'}{\partial k}\] Also, this term is weird and not correct! \[\frac{ \partial F }{ \partial k} \frac{ \partial k }{ \partial y } = \frac{\partial F}{\partial y}\]

Empty
 one year ago
Best ResponseYou've already chosen the best response.6So now you've got two steps left to do, plug in y and then evaluate at k=0.

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0There should be a partial k in the second term not x

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0and I didn't cheat!

Empty
 one year ago
Best ResponseYou've already chosen the best response.6Awesome! This was pretty much the hardest part so from here it's all mostly down hill.

Empty
 one year ago
Best ResponseYou've already chosen the best response.6Right now evaluate \[\frac{\partial y}{\partial k}\] using \[y=\bar y + k \eta\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ \partial y }{ \partial k } = \eta \]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Now we can use by parts

Empty
 one year ago
Best ResponseYou've already chosen the best response.6Wait, only one of those three derivatives is eta. The other two are different. And you're anticipating we will be using by parts but there's actually no need to if this was the real answer!

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Yeah I see, but w/e I'll do this later tomorrow, going to bed, thanks
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