The Brachistochrone Problem and Solution | Calculus of Variations

[@FacultyofKhan]

I've addressed your requests for a classical mechanics series! Here is my playlist, which starts with the Principle of Stationary Action! kzbin.info/aero/PLdgVBOaXkb9DSSqQZWfBrZy_rOljWmA3j

[@ernestschoenmakers8181]

Yeah you have to do a trig substitution indeed but it's not difficult cause if you take the function x=(sqrt(y/(1-y)) which is a bit of the same function and you integrate it over the domain of y then you get the following result: x= arcsin(sqrt(y)) - sqrt(y(1-y)) + a constant and that's it.

[@donaldcasey1080]

not sure if anyone gives a damn but if you guys are bored like me during the covid times then you can watch pretty much all of the new movies and series on instaflixxer. Been streaming with my gf recently :)

[@dangeloamari2444]

@Donald Casey Yup, I've been watching on InstaFlixxer for years myself =)

[@physicsmathsworld2033]

😂😂 last words are so hilarious 🔥👌 "I am the kind of person who would like to fall down from sky to ground rather than ground to hell " btw , your channel is one of the most beautiful channel for Physics & maths ❤ for me . Thank you

[@schatter91]

awesome final comment on sky and hell! great video btw

[@siom7617]

That was actually helpful. An absolutely brilliant explanation, better than the professor at my university!

[@ja8898]

So this is what the textbook meant when they said, "the solution is trivial and we leave it to the reader to solve"

[@I_am_Dipanjan]

Goldstein, huh.

[@michasz4297]

You're the great teacher! Keep making those wonderful videos!

[@marciadysart6368]

no

[@karvakferenc1009]

This are a gift, not a problem

[@RaunySilva]

I hope the best for you. I am really grateful, thanks

[@kacperkazaniecki596]

Amazing! So happy I found your channel (at 3am lol). This is the first video I've watched, but I can tell that the quality is outstanding so really well done. One question though, at 7:50 why would we use trig substitutions? Of course, it works and we get a nice result however if I were to be working this out myself, I'd be comfortable up to that point. Then I wouldn't really think of using trig substitutions. So what's the 'insight' in this that I could learn from?

[@FacultyofKhan]

Thank you for the kind words! Correct me if I'm wrong, but I believe you're asking about what made me decide to use trig substitutions here. It's because when you're faced with integrals involving square roots of rational expressions, a good idea is to use trig substitutions (e.g. replace x by sin(theta) or something similar). This is because the derivatives of the inverse trig functions like arcsin, arccos, etc. often involve square roots of rational expressions. A good thing to do when you're stuck is to look at integration tables to find the closest form corresponding to your integral of interest. If you don't want to rely on external aids, however, your best bet is to keep practicing. Hope that helps!

[@kacperkazaniecki596]

Faculty of Khan ahh right that makes sense, thank you. If you don't mind I'll ask another question. 3Blue1Brown did a similar video on the brachistochrone problem however he explained how Johann Bernoulli solved it. He did so by using Fermats principle of "light always takes the path of least time". The way to connect this idea is to think of the infinite layers of different and increasing velocities as an object is falling down the ramp (and different refraction indices for light). By making this connection you can use Snell's law and arrive at the equation of a cycloid. Since light will take the quickest path which is the brachistochrone curve, if our object follows this path it'll slide down the ramp in the quickest possible time. So I'm happy with this method of deriving it, however I don't really understand why "light would always take the quickest path" obviously it has no prior knowledge of all the routes and then it decides. Tried googling around but couldn't find anything that explains it. Do you perhaps know a more fundamental principle or a point of view that could help me understand this? Doing a school presentation on this on Wednesday and would like to be very well researched for it. Apologies for asking so many questions.

[@FacultyofKhan]

The short answer to your question is that light doesn't actually 'know' beforehand, but that if we express light as a wave, and that wave follows all paths of travel, then most of the paths will cancel out with each other because of destructive interference, leaving only the quickest path behind. If you're interested, you can look up some of the sources mentioned in this stackexchange thread: physics.stackexchange.com/questions/59607/how-does-light-know-which-path-is-fastest Hope that helps!

[@d3x-dt3]

Well, I think I am going to be here for a while. You've hooked me in, sir! Subscribed :D

[@FacultyofKhan]

Appreciate the support!

[@prithvitimalsina399]

Please explain why you have taken (dx) out?? what is the meaning of doing so?? What if i take out (dy) outside? ? and what about the brachiostrochrone problem in curved space...

[@FacultyofKhan]

I took dx out so that I get an integral expression of a function of x, y, and y' (dy/dx). That's ultimately what I want if I want to apply the Euler-Lagrange equation (see: kzbin.info/www/bejne/qXfUoWWieKqpra8). I *could* have taken dy out in which case my integral expression would have been a function of y, x, and x', but I would personally rather choose y as the dependent variable instead of x. Also, if I took the dy out, then I wouldn't be able to use the Beltrami Identity, which makes calculations a lot simpler.

[@prithvitimalsina399]

Would that be incorrect if i take 'dy' out?? It would be the function of x, y, and x' (dx/dy) we can still use the euler lagranges equation. .that would be a solution for x. But it would be great if you could explain what is the "meaning" of taking 'dx' out? Or 'dy' out??

[@FacultyofKhan]

It wouldn't be incorrect; it would just take longer and the calculation would be more annoying. Also, I'm not sure how to explain the meaning of taking 'dx' out; it's a simple factorization.

[@healinghub1112]

I like your lectures......coz its precise and explainatory. Plz make a series on classical mechanics.....you would get loads of subscribers then.....coz that's something people denand

[@healinghub1112]

I meant demand ....in they end

[@FacultyofKhan]

Thank you! Classical Mech is something I intend to work on, so I will get to it eventually!

[@saitaro]

Thank you very much, sir, for this series, it was a pure joy to watch it.

[@diegoortegadod]

Hello Faculty of Khan! You have no idea how helpful this video has been for me... I am currently an IB student doing my Internal Assessment for my Mathematics class exactly on this problem. I have one question, at 9:02, how did you end up with that equation on the left? I understand you replace -C1 for K, but how did (1-cos(x)) turn into (x-sin(x)), and why did you get a value of K2 which is added at the end? This is my only question regarding the mathematics of your video, which is very well explained. Thank you!

[@ellaa7985]

Diego Ortega omg I’m doing the same thing right now

[@edwardzhuang2001]

wtf me too LOL

[@FacultyofKhan]

The (1-cos x) was being integrated, so it changed to (x - sin x) (integral of 1 is x; integral of cosine is sine). The K2 is the constant of integration that gets added. Hope that helps!

[@matiassantacruz5487]

Diego Ortega how did it go??

[@diegoortegadod]

@@FacultyofKhan thank you!

[@nickallbritton3796]

I like your videos, but I'm having trouble putting it all together. I came to this channel having hit a wall in classical mechanics: I want to understand the principle of stationary action and the Euler-Legrange. What I did before was learned differential calculus when I couldn't move ahead in physics without it, but I'm finding it very difficult to pick up variational calculus? Any advice to help me put all the pieces together? Thank you a lot for your videos!

[@FacultyofKhan]

I haven't covered the principle of stationary action, but now that you and a number of other commenters have mentioned it, I'm going to try to push out a video sooner rather than later. Hopefully, that might help you 'put all the pieces together'!

[@KhalidH-o9u]

Many thanks! I just didn’t understand how do you know what do you assume y ?? How do we know it is (h-c1)sin(x/2)^2 ?

[@manassivakumar1349]

at 7:35 while taking the square root why are we not considering the negative value

[@TGRRohit]

How did newton solve it without knowing Lagrange

[@AkshayPatil-of7ii]

hey your way of explanation is awesome.Please make some video on hamiltonian.

[@gatoradeee]

Amazing.

[@FacultyofKhan]

Thank you!

[@saulmartinezcastaneda8507]

Thanks. Would you tell me why TETA(A)=0 & TETA(B)=TETA(L)?

[@jibran8410]

i've got a question , Why do we have to use the beltrami identity? , we didn't use it last time when we had geodesics of speheres and planes and they weren't dependent of x as well , only dependent on theta and y'. So why didn't we use the beltrami identity there, and if you do have a function dependent on x , what do you use?

[@carmencook7026]

My mind was slightly blown and after four minutes had to figure out how or why I was watching this. Lol community college dropout issues. IKR I may not understand it anymore I do miss mathematics

[@zheyushan1472]

Could you do a video on Brachistochrone involving friction please

[@FacultyofKhan]

Well, a problem involving friction would take rather long to do and I don't know if it will provide as much instructional value to my viewers. Nonetheless, I'll put it in my to-do list and see if I can get to it!

[@rodrigogarcia362]

Thank you. You, Euler, and Lagrange saved my life. This helped a lot on my EE.

[@レオ-s9m4o]

Same, I am also writing my EE on physics with Euler Lagrangian equations... while my EOY is postponed to DP2. I am not handing out my EE in a full piece for the first draft. T_T

[@haggaisimon7748]

well, we aren't fallen angels, aren't we? great lecture. like the sequential style of presentation.

[@IMADUSHYANTSVNIT]

thankyou for the clear explanation can we solve this problem in polar coordinates (r,θ) (r is the distance from origin and θ is the angle for direction from the x-axis) because their ds will become r*dθ simply and if yes please help me to solve

[@CE113378]

At 5:00, what do you mean by "stationary"? In what sense is T made to be stationary by finding the right function y(x)? I don't know what "stationary" means in this context. Stationary to me means "not moving." But T is a time interval. Time intervals don't move. Mass moves. Electromagnetic waves move (i.e. the energy of EM waves propagate). In what sense does T move when you don't have the right function y(x)? This is the second video on KZbin. In both videos, the point is made that we need to find the function y(x) that makes T stationary. But I don't understand what "stationary" means.

[@Ki0212]

When we say a function is stationary at a point, it means that a small change in the input changes the function by a negligible amount. In other words, a function is stationary at its local extrema (minima/maxima)

[@م.رضوانحسن]

Great explanation, thank you very much.

[@marcoguedes3038]

The only one thing I cannot make out in all these brachistochrone solutions is how the y substitution works...where did that trigonometric expression come from? How do we know that y is equal to that?... Could someone explain it to me?

[@syamalchattopadhyay2893]

Excellent video lecture

[@pratikprajapati8620]

Which software you are using? Please let me know

[@mikoaj2177]

Hey, great video! I have question: in your functional's denominator you have sqrt(2g(h-y)). But y(0) =h, and then denominator is equal to 0. Then the integral is improper. Does it changes something?

[@FacultyofKhan]

If you substitute the expressions in the integral in terms of theta and change things so that you're integrating with respect to theta, the issues with the improper nature go away, so no, it doesn't really change anything.

[@kimhbryan]

at 6:34, how did you end up with the second term after applying Beltrami's Identity?

[@emir2750]

Yeah that also bugged me but I think it is from -y' × đF/đy' , so I did not calculate it myself but i think if we take the partial derivative of F to the y' and multiply it with y' we can get the second term, what do you think

[@Mysoi123]

The function F is the square root inside the time integral. the partial derivative of F give the y’/(sqrt(2g(h-y)*(1+y’^2) and multiple with the y’ from the Beltrami’s identity you get y’^2 from the second term.

[@jibran8410]

What do you use to make your posts?

[@davidperezhurtado838]

Hi faculty of khan, first congrats for this fantastic videos. I have a homework that says "an airplane is flying on a vertical plane (x,y) with a velocity v=ky. Find the path that minimize the time of travel from point 1 to 2", and the diagrama is the same as yours, except that the points are (x1,y1) and (x2,y2). So since i have the velocity i replace and get the time to minimize as T= integral ((1+y'2)^(1/2))/(ky))dx and at the end i get the path as (y^2)/(2*(A1-1)^1/2)=x+C2 where A1 and C2 are cosntants to find with the boundary conditions. So my question is if the equation to minimize (T) was ok or i make a mistake? Thx

[@andrewli-yangliu9555]

The resultant cycloid concaves downwards when the Brachistochrone is supposed to concave up. Why is this the case?

[@FacultyofKhan]

You might be looking at the wrong portion of the cycloid; what are you plotting? The part of the cycloid corresponding to the brachistochrone is concave up.

[@pratikprajapati8620]

Which software are you using for explanations?

[@elonmusk3003]

i like the way you express things but your videos are not series wise if you can make whole series of classical mechanics that it will cover all topic than it will be surely much helpful for us

[@FacultyofKhan]

I'm planning to start that soon. In fact, the foundation has been laid with this video: kzbin.info/www/bejne/ppWyYmeapqiSpKs

[@mariannecruiser]

Despite reading previous comments, I still don't understand how y = h-c(sin^2 (theta/2). I understand it is for the purpose of convenient integration, but how did y come to equal this? What trig identity was used, etc?

[@dikukhanikar2837]

This can be seen more clearly, just substituting the (h - y) = c1sin^2(theta/2), I think now it will make sense why this substitution is done there. :)

[@info-hub457]

I am a bit confused by the partial derrivative with respect to y' of sq root 1+y'^2. Shouldnt we have y'/sq root(1+y'^2 ??

[@knudvaneeden]

yes, your calculation d/dy' ( 1 + y'^2 ) = y' / squareroot( 1 + y'^2 ) is certainly correct.

[@AmSofiane]

Amazing !

[@smrd0110]

Your site fills a crucial educational need. Can you do videos on pursuit problems?

[@mohdashraf9344]

Is brachistochrone curve ball also has maximum velocity than all other curves?????🤔

[@FacultyofKhan]

Not quite; the curve with the maximum velocity is actually shown earlier in the video, around 1:20 or so.

[@yazanmashal6437]

Hi, could you please explain how you got y = h - csin^2(theta/2) Thank you

[@shivanshpandey3778]

Actually, this is just a substitution, so that our integration gets easier to solve. And substitutions are found through guess or taking the idea from, solving many such integrations earlier.

[@dikukhanikar2837]

This can be seen more clearly, just substituting the (h - y) = c1sin^2(theta/2), I think now it will make sense why this substitution is done there. :)

[@festusudeh9872]

Can you give some worked exercises l?

[@alf3071]

Is there a curve that can have the point B sit above point A? or the particle can never exceed A's height because that's it's max potential energy?

[@FacultyofKhan]

If you're only moving under gravity with no other kinetic energy to propel your particle at the start, then no it cannot exceed A's height.

[@kingofspades8856]

I wonder if there is somewhere explained a purely geometrical solution to this problem in the vein of 17th century masters.

[@alijassim4380]

thank you

[@AdiCherryson]

So what is the answer to the Brachistochrone problem? Is it any cycloid connecting the points A and B? Or just the shortest one? On the Wikipedia (en.wikipedia.org/wiki/Brachistochrone_curve) there's this curve going under the "ground level". Is it the only one correct orientation of the cycloid?

[@FacultyofKhan]

It depends on how far the points A and B are oriented from each other. If they're reasonably close, the brachistochrone curve doesn't have to go below ground level (i.e. if you go from the start of the wikipedia animation to a third of the way through, then that's still a brachistochrone but it stays above ground level). The cycloid connecting the two points A and B, however, is unique.

[@dutchjack]

at 4:38 you have the velocity stated as a function of y. This velocity has no direction, really for the expression dt = ds/v to be true, does v not have to be a vector in the direction of ds? I can see how the answer works because the expression used for velocity is actually the maximum possible magnitude of velocity in the direction of s, but this still bugs me. There might be another subtlety I'm missing which ensures that the velocity vector is always pointing along s, maybe some invocation of Newtons 2nd law?

[@FacultyofKhan]

You're right that v has to be vector in the direction of dS for the expression to be true. The reason this is the case is that the problem we've formulated is a particle falling down a solid set path from A to B (e.g. we're dropping a marble from A to B using a custom wooden curve). The marble must stay in contact with the wooden curve (it can't just float around); therefore, the velocity of the marble is tangent to the wooden curve. In terms of this problem, this means that v must be in the direction of dS. Hope that helps!

[@SkylanderBlazier]

hey just a quick question, why are you making the time stationary? thanks :)

[@FacultyofKhan]

It's because the brachistochrone is the path of least time (i.e. a minimum). To find the local minimum of a function, you have to set its derivative to zero (i.e. make the function stationary). Same idea here. Here's a reference intro video: kzbin.info/www/bejne/bHnIgpZteLiekNU

[@jacobsnoises]

This is actually a fantastic video. One question though, at 9:27, why does it correspond to (theta)=0? If we choose x=x1 when y=h, how are we meant to know that it corresponds to (theta)=0?

[@FacultyofKhan]

Thank you! It's theta = 0 because at theta = 0, y = h (since cos 0 = 1 and y = h + K1/2*(1-cos theta)). Since y = h corresponds to the x-coordinate of 0 ((0,h) is the starting point for our curve), that means the entire point (0,h) therefore corresponds to theta = 0. Let me know if that clarifies things!

[@wenhanzhou5826]

11:40 made my day xD

[@barrichellonoor7766]

Hi, in 7:17, what happened to the 2g and c^2??

[@FacultyofKhan]

They get combined into C1, where C1 = 1/(2g*C^2). Hope that helps!

[@habibali9250]

How to solve Numerically : last two equations. Any MATLAB code with solution.

[@blancocd]

what about with initial velocity??

[@9117mickey]

6:41 you've subbed into the Beltrami identity , how have you got (y')^2 in the numerator ? should it just be y' , cheers in advance everything else makes perfect sense

[@mikoaj2177]

I thought about this for a bit, but there's no mistake. Look, the beltrami identity says: F-y'*(dF/dy')=0 So, dF/dy' really has y' in the numerator, but you forgot to multiply it to y' :)

[@XiaosChannel]

after watching this two times, I'm still gonna say this does not answer my question posted at the Euler-Lagrange video: what would be an *intuitive* meaning for the equation in a real problem?

[@FacultyofKhan]

Ok, I'm actually unsure how to explain the intuition of the Euler-Lagrange equation. It's basically an analog of dy/dx = 0 when we're finding local minima and local maxima. That's pretty much the intuition right there; I can't think of anything else to add, and neither can any of the books I have.

[@XiaosChannel]

I saw a video trying to do this: kzbin.info/www/bejne/gYLJkoSPg8Rpb7s Does what he said makes sense to you in that particular question?

[@FacultyofKhan]

Ok, so I watched the video. He seems to be describing a specific case of the Euler-Lagrange equation that's used in classical mechanics (i.e. the principle of stationary action). I haven't actually made any videos on the Principle of Least Action or on the Euler-Lagrange equation in the context of classical mechanics. It has been requested before and I'm probably going to add a video on it soon. Hopefully that resolves things!

[@XiaosChannel]

Faculty of Khan you are saying that while one can give intuitive examples in that field, you don't find a way to "transfer" it into this problem. is this correct?

[@FacultyofKhan]

Yes, pretty much. The intuition is more obvious when there's actual Physics behind the Euler-Lagrange equation, like in Classical Mechanics.

[@cytyy8531]

Whyy he use half angle instead of angle? can anyone explain it to me?

[@cytyy8531]

or it is still possible to use full angle but half angle is much easier?

[@lyanna518]

Whats K1 stands for ?

[@FacultyofKhan]

It's just a constant of integration, equal to -C1.

[@chrischan8896]

how do we solve for K1 at 10:05? finding the actual numerical value is necessary to plot the brachistochrone curve right? i am trying to model one but i do not know what values to put into the final parametric euqations.

[@FacultyofKhan]

It's rather difficult to solve for K1 given that the equations in 10:05 are rather complex (they require you to solve theta_L as well before you get K1), and involve trig ratios + the variables themselves. I'd recommend first solving for theta_L in the second equation, substituting into the first equation, and then using a numerical approach to solving nonlinear equations (e.g. Newton-Raphson) to get your K1. Hope that helps!

[@chrischan8896]

Thank you for replying, the suggestion for using the Newton-Raphson method is very helpful. However, how do i solve for theta L in the second equation when in the first place, there are still two unknowns?

[@FacultyofKhan]

You solve for theta_L in terms of K1, so it would be something like arccos(1 + 2h/K1), substitute that into the first equation, and then solve for K1 using Newton-Raphson.

[@chrischan8896]

why is it not possible to simply substitute h and L into the two equations?

[@nikhilahilani5809]

Wow.

[@admiralhyperspace0015]

Again how do you know that u subtitution?

[@mr.mottelee4464]

Will you make a video about the second variation? If not, what reference book are you using for this video? (I assume it also contains discussion about second variation)

[@FacultyofKhan]

Perhaps later I'll make a video. The book I'm using right now is one on Mathematical Physics, and although it contains a chapter on Variational Calc, it doesn't go into too much depth so it doesn't talk about the second variation. I'll probably use another resource for the second variation, but I haven't figured that part out yet since there's still some preliminary topics I'd like to cover first.

[@eriknelson2559]

Think you can apply the same (family of) solutions when the particle begins with an initial velocity, the answer being the cycloid which connects the two points and if extrapolated backwards would have imparted the initial velocity at the initial point

[@qurat-ul-ainmughal4003]

Please help me.. mjhay kch problem solve kr k DE dain.. contact no.. DE skty hain?

[@turboleggy]

What a boss

[@mutlucankartal9524]

I would rather fall form the ground to the hell:)))

[@christofelmalik4240]

Anyone can tell me about the trig subs?? Wht the value of y is become like that????

[@FacultyofKhan]

It became like that because that's a relatively convenient way to compute the integral; this is just something that you have to know to use whenever you see an integral involving complicated square root expressions. I encourage you to compute the integral yourself with y set to h-C1*sin^2(theta/2) to see how things simplify. Hope that helps!

[@postyoda1623]

Very interesting; one question I have is that I know Newton solved this problem but obviously that was before Euler-Lagrange equation and Analytical Mechanics. So how did he do it?

[@mrshovelbottom7475]

Geometry and Snell's law

[@abdulsalamone]

11:40 was like thug... 😎

[@danielbarr4354]

When we arrived at the parametric equations at the end, what is the relevance of time when plotting points + graphing the curve? Don't we need time when graphing parametric equations? Thanks

[@agrajyadav2951]

i tried but couldn't find myself :(((

[@Fida1998]

‪Brachistochrone curve is furmulated using a combination of minimizing the the distance and maximizing the speed ‬

[@wajahatali3719]

Define brachistochrone problem with all it's possible result also solve the the problem

[@david-yt4oo]

how the hell is someone supposed to know that they have to make that substitution 7:53

[@FacultyofKhan]

From my answer above: "It's because when you're faced with integrals involving square roots of rational expressions, a good idea is to use trig substitutions (e.g. replace x by sin(theta) or something similar). This is because the derivatives of the inverse trig functions like arcsin, arccos, etc. often involve square roots of rational expressions. A good thing to do when you're stuck is to look at integration tables to find the closest form corresponding to your integral of interest. If you don't want to rely on external aids, however, your best bet is to keep practicing and go back to your Calculus 2 notes if you have them around".

[@debendragurung3033]

Dont be hard on yourself. there are some problems, ull just have to be great on yourself if u just understood it and admit u just couldnt have solved it . Though this video is 13 minute long, Brachistochrone is the problem that plagued for half a century and but fortunately had the greatest of minds working on it, fortunate enough that we could see the lights on it.

[@Nithesh2002]

Alternatively, you can sub u = h - y, and multiply the resulting integral by (sqrt(u))/(sqrt(u)) and your final answer after splitting the integral up so you have elementary integrals is x = sqrt(c1(h-y) - (h-y)^2) - c1/2 * sin^-1(2(h-y)/c1 - 1) + c2 which is easier to do , but obviously not as nice hehe

[@ernestschoenmakers8181]

@@Nithesh2002 Yeah i had the same answer but there's nowhere to see in any Brachistochrone discussion and i don't understand why.

[@Nithesh2002]

​@@ernestschoenmakers8181 I think it's because most solutions prefer to show the more common, parametrised form of a cycloid; which allows for easier recognition. Nevertheless, our approach still gives the correct answer, albeit in a less neat form.

[@KhoderAlshaar]

the joke! hahahahah 11:40

[@furey1]

These videos are beautiful

[@Independent_Man3]

best joke 11:42

[@asifakhan4971]

How you are khan

[@supriya1729]

Op

[@aj-uo3uh]

Vague formulation of the problem, the paths you presented are not possible with just gravity, you need friction. I like the speed and quality of the presentation though.

[@JamesWylde]

Good video, too much mid roll ad spam

[@marciadysart6368]

your mom

[@kennymaccaferri2602]

This synthesised voice .... Aaaaarrrrgggghhhh no me gusta. Para nada. But thanks for the graphics.