Hi, I have a question to the equation (14) - the substitution. Where do you get that from, how do you know you had to give y that value? Loved your video tho!!!
Short Answer: you can now ask Chat GPT for the best substitution to use. Longer Answer: the form of the denominator sqrt(c - y) gives us the clue that a trig substitution is the way to go. From there, you can find a table of useful trig substitutions and look for one with the same form as the integral that we have - ie. sqrt(x) / sqrt(a - x). Old School Alternative: Use a table of integrals to integrate this.
Loved how you explained the whole story, and not just the solution. The history behind the math and physics is so much fun to learn ! Amazing work, thank you !
I've only ever seen the hand-wavy version of the solution to the Brachistochrone: "we realize immediately that this is the differential equation for a cycloid...". Thank you for doing the full derivation!
I like this problem. It was worked out in the "classical mechanics" classes I followed. The accompanied story, whether it true or not I don't know, but as educational vehicle it still is in my mind, is that the problem has to do with the question: What is the fastest way to move cannonballs from one place to another. Seemingly, the idea is that one has a higher place and that one digs brachistochrone tracks towards the canons. This in order to react swiftly to enemy attacks and knowing that the number of balls is limited, since these are hard to make in those times. It is a likely story, since a lot of the calculus of that time was related to warfare. Anyway, thanks for the clear presentation and the history behind it.
I hope you haven't stopped making videos, this channel has had some of the best math exposition I've ever seen and seeing the last video was uploaded 7 months ago was a bit deflating.
Thanks for your words. I have been taking a break for a little while in order to pursue some other projects. I will be releasing a few more videos before the end of the year which will cover approximate methods like Ritz & Galerkin.
@good vibrations with Freeball Thank you for your clear explanation, arrived here after reading Newton's quote about solving this in an evening. Love how the mathematics matches my intuitive visualization the Brachistochrone curve. You sir deserve a piece of
I first saw this video when it came out and basically understood nothing except the question that was posed(the nature of the problem), now nearly at the end of my 12th grade, I can understand all the math in the solution and the intuition behind the problem(although not really the understating of why the calculations took the path that they did, I assume it will take a lot more practice to see and execute stuff like that, and why some particular manipulations were made the way they were.)However, I'm extremely content, and excited for the knowledge that uni holds for me, with this video being one of the ones that lit the match in me to delve deeper into this field.
Thank you for your feedback. This is the sort of comment that really warms my heart and makes all of this effort worthwhile. If you are able to follow most of this material in 12th grade (it is really aimed toward graduate students), I'd say you have have a very bright future ahead of you. Wishing you all the best of success at the next level. Have you decided what you'll be studying yet and where?
@@Freeball99thank you for your motivations:) Ive since studied this in more depth and can gladly say that after watching this video thrice over, I can now fully comprehend everything that is going on( except the conception of the beltrami identity, since the video covering lagranges multipliers through the catenary problem really is above my current understanding, every other substitution and method is clear thanks to your phenomenal explanation:). Coincidentally, dr.elliot Schneiders explanation is also fantastically intuitive after understanding yours, and somehow verisatium also came out with an extensive story of the brachistochrone problem right about now(which also covered the Bernoulli solution, mind-blowing). I have applied to institutes across the USA, hoping to pursue an undergrad in cs, or maths(hopefully with electrical engineering). It is too difficult to decide, and my mind flips each day, I just wish that I'm able to learn a lot across all these fields( a lofty goal). My apologies for the extensive message, I would love to know about you and your story too if you ever find the time. Thank you for taking out some of it for this reply sir. Love your videos ❤
Also, @ 1:25 in the problem statement, you that the point mass is acted on only by gravity. But in fact, it is acted on by gravity and also by the reactionary forces of the brachistochrone structure. I recognize this is beside the point of the problem. Nevertheless, at any instant of time, if you draw a free-body diagram of the point mass, we will have the force of gravity and the rreactionary force of the brachistochrone acting on the point mass.
Correct. There are reaction forces from the path itself. I described the problem in the manner that I did because I was looking to present the problem in the context of a optimization problem where our task was to find the path/constraints that minimized our functional. This was an attempt to be consistent with previous videos in the series in which I had described various path optimization problems.
18:22 One can only imagine how much quicker all of this would've gone if these guys could've jumped on a Zoom session and done a screen share with one another. Love your Clemenza accent from the Godfather !
It is amazing how the brilliant minds developed powerful methods at such young ages. And today I'm trying to grasp the meaning of the least action principle and how it relates to Lagrange's equation. To be fair, I'm enjoying studying this topic. The method of calculus of variations is powerful. When I solved the catenary problem I was satisfied with it haha Now I'm looking for more applications. Thank you for your videos! They are really helping me through my course.
Lol when you said that you'd be solving the problem in 10 mins when it took 12 hrs for newton to do so lmao...nice blend of humor, much needed esp near exams
Really nice video. I loved both the maths and the historical background, which gives the problem such a human dimension. I’ve also watched some of your other videos on Lagrange’s equations, these are really great resources to help with the studying. Thanks for the excellent videos, please keep up the good work !
I wonder if you would like to tackle a variation of the brachistochrone in which the path must be of fixed length. I tried to work this out a few years ago and it was very messy. I ended up simulating the solution numerically on python. I had a great deal of fun with this problem but I have never found it worked out analytically anywhere.
A mathematician whom I know in prison studied it a lot, i.e., James' Murray, serving 15 years on the then now closed Virginia State Farm, maximum security prison where I was serving twenty five years for robbery and murder, falsely accused for the murder though at the time I was a murderer; during which time I became a fledgling mathematician, while working in a field bull gang and studying mathematics evenings and non work days as weekends and major holidays. I began my study of mathematics at eighteen. Came home after serving nearly fourteen years and attended VCU after release from prison and minored in mathematics and majored in Mass Communications. I digress. The question is why I never see any video on the witch graphic or problem. I only saw it at a glance in James Murray's book on calculus. Virginia prison facilities or penitentiary had a lot of donated books. I virtually turned the place eventually into a college campus in the late 1960s and early 70s. I have been looking for the witch graphic. James Murray was aka ,"Heart Trouble" for his bad heart. He went home and turned into an alcoholic and died, perhaps, in the mid or late 1970s, having invented a device on paper, which physicists came to look at but left and never gave James Murray their opinion on its merits. He was my inspiration for studying mathematics.
At 17:49, is there a way to calculate the shortest-time path using calculus of variation with the added condition that the curve must never have a smaller value of y than the end point? Or would this require a numerical solution?
Interesting question... I've neither tried it nor have I seen it attempted, but it seems to me that this should be possible by adding a constraint equation and using the Lagrange Multiplier Method. Almost certainly this would require a numerical solution.
Well presented. However, I have few questions: (i) the case is for rolling without slipping. If friction was included what will be the solution? (ii) the effect of gravity is not appearing is the formulation of the problem, was it assumed to be unity for simplicity? Thank you.
If there was frictional loss included, this would have the effect of shortening the path which would flatten out the curve a little. The effect of gravity is just a constant with respect to the integral, so when taking the variation and setting it equal to 0, the magnitude of the gravity has no effect on the shape of the curve.
I'm confused. How can we write dt as ds/v (10:06) when there is acceleration. d = vt when there is NO acceleration, but we have an acceleration of g and an initial velocity of 0, so shouldn't d be 1/2at^2?
So, a few of things... 1. dt = ds/v (time = distance/velocity) holds regardless of whether or not the mass is accelerating. 2. I have assumed that the mass is accelerating, else v would have been a constant and I could have simply moved it outside of the integral. 3. In order to solve this using the path-minimization method shown in previous videos, I chose to rewrite v as a function of the path variable, y, instead of writing it in terms of time (0.5at^2) - we're not trying to solve for the time taken from A to B, but rather we're trying to find an optimal path.
I show this at around the 17:32 mark. A cycloid can be plotted by rolling a wheel along a straight path and by observing a point on the wheel. θ is the angle of rotation of this wheel as shown in the graphic.
i don't see the connection with this experience and theta seems to come out of the blue ! A cycloid is drawn when you roll a ball on a straigth line. here we have a particle "rolling" on a cycloid ! the theta of this "rolling" particle is not the same variable (or is it?)
Rolling wasn't my best choice of words here, but I used it since most demonstrations of this show a sphere rolling down a track, but in fact, the mass is sliding and not rolling since the surface is assumed to be frictionless and the mass is assumed to be a particle with no rotatory inertia. The θ in the solution doesn't directly represent the angle of a rolling wheel. It's a mathematical substitution used to simplify the integration. The connection to a rolling wheel comes after solving the equations. The resulting parametric equations (18 and 19) happen to describe a cycloid, but this wasn't known at the start. You're correct that in the problem, we have a particle moving along the curve, not drawing it. The θ in the solution is not the same as the angle of a rolling wheel. The cycloid emerges as the optimal path, but it's not how we arrived at the solution. It's the result, not the method. The solution method doesn't rely on knowing about cycloids beforehand.
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
Your observation is correct. Stated differently, a change in initial conditions does not change the governing equations of motion. So the general form of the solution remains the same, just the constant of integration change.
I think the conservation of Mechanical energy substitution for the velocity only works for a particle or a puck on a frictionless surface. A ball would have both linear and rotational kinetic energy. The rotational component would take away from the linear.
Yes, you're correct. The original problem statement described the mass as a point-mass so there it has no rotatory inertia. In drawing the point as a circle and then watching several demonstration videos on UA-cam with marbles rolling down a track, I conflated the point mass with marbles rolling, in my description of the problem. Though I did describe the path as being frictionless which means that a mass would not roll along it anyway and so its rotational kinetic energy could be ignored. That said, the mass was modeled correctly as one which was sliding without rolling, so the math is correct with the result being a cycloid. As a side note, however, if this were a marble rolling along the track, then the optimal path would be a shape known as a hypotrochoid.
Very elegent solution. Thankyou. However I have some doubt. The solution looks perfect for sliding particle but for rolling ball to count of conservation of mechanical energy we also need to consider rotational energy (1/2 x moment of inertia x angular velocity 2 - This MOI for a ball being 2/5mr2.) This should slightly modify the curve.
This is great. I stopped short of studying variational calculus at uni, but I've always been interested. Am I correct in understanding that the Beltrami Identity is only valid when F is minimised? or otherwise where is the minimisation happening?
We are always trying to minimize the F in these problems. The Beltrami Identity is what the Euler-Lagrange Equation reduces to when F is not a function of the independent variable, but is a function of the dependent variables only. Solving the Beltrami Identity yields the path that minimizes F.
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?
The trig expression comes from using an integral table at the 14:45 mark. A good mathematician could probably figure this out in their head, but that is above my pay grade. Integral tables will shown how to handle integrands of various forms.
The short answer is that you can now ask Chat GPT for the best substitution to use 😀. The slightly longer answer is that the form of the denominator sqrt(c - y) gives us the clue that a trig substitution is the way to go. From there, you can find a table of useful trig substitutions and look for one with the same form as the integral that we have - ie. sqrt(x) / sqrt(a - x). Old School Alternative: Use a table of integrals to integrate this
Great video. Could you go over quickly how we could apply this program to real world scenario? I'm not sure how to solve for C1, C2, and what my theta value would look like. Thanks
One question: The original problem statement specified a sliding object. Your solution used a rolling object. Therefore, in your example, the conservation of energy equation would include both translational and rotational kinetic energy. Are the solutions for a rolling object versus sliding object the same? I have seen demonstrations of the brachistochrone problem using rolling metal spheres, and the results are similar. Thank you.
Sorry for the delayed response, but I somehow missed this until now... The original problem statement talks about a particle, so by definition contains no rotational inertia. Every demonstration one tends to see involves rolling marbles down a track. In the video, I tend to (incorrectly) refer to the mass a rolling, which it isn't in the problem, it's really sliding.
There has to be an additive coefficient on y, or else there’s no way to have a non-zero y1 boundary condition. What am I missing? Also, given the use of the parameter theta, it seems that given point B(x2,y2), that we would need to solve for theta, but at the same time we need to use point B to solve for some of the coefficients.
If you take the last diagram in the video, Point A is located at (0, h) and Point B at (d, 0). So the height of Point B is always zero by definition based on the choice of axes and the constant c1 is therefore dependent on the height h of Point A (turns out that c1 = 2h and c2 = 0). It can be assumed that θ = 0 at point A and θ = θ_B at Point B. Solving for θ_B will require a numerical solution since θ_B is a function of both h and d. You will find that c1 and c2 depend on the location of point A ONLY. So this alone does not specify the precise cycloid path needed. The parameter, θ_B becomes a necessary parameter for this.
Near the end of the video (17:32), I show a a figure of how the cycloid curve is traced out by a rolling disk. The angle, θ, is the angle of rotation of the disk.
Right at the beginning, you spoke of a marble rolling down a frictionless path. There is no rolling without friction. I understand this is beside the point of the brachistochrone problem, but a frictionless surface implies no rolling.
Yes, you're correct. The original problem statement described the mass as a point-mass so it has no rotatory inertia. In drawing the point as a circle and then watching several demonstration videos on UA-cam with marbles rolling down a track, I conflated the point mass with marbles rolling, in my description of the problem. Though I did describe the path as being frictionless which means that a mass would not roll along it anyway and so its rotational kinetic energy could be ignored. That said, the mass was modeled correctly as one which was sliding without rolling, so the math is correct with the result being a cycloid. As a side note, however, if this were a marble rolling along the track, then the optimal path would be a shape known as a hypotrochoid.
I have taken equations 18 and 19 in order to determine the values of the constants c1 and c2 for the case, let`s say, of the brachistochrone that goes from the point (0,0) to another generical point (1,h), h>0, with the x and y axis taken as you mention in the video. However I can't see there is a solution for any value of h. Could you clarify this point?
what about the rotational motion of the rolling object? Does Euler-Lagrange equation respect for energy conservation from its onset? Wouldn't that Beltrami equation is a consequence of conservation of energy (Noetherm theorem), and what facts has it led to hold a priori conservation of energy if it is going to use same Beltrami equation later?
The original problem statement described the mass as a point-mass so it has no rotatory inertia. In drawing the point as a circle and then watching several demonstration videos on UA-cam with marbles rolling down a track, I conflated the point mass with marbles rolling, in my description of the problem. Though I did describe the path as being frictionless which means that a mass would not roll along it anyway and so its rotational kinetic energy could be ignored. That said, the mass was modeled correctly as one which was sliding without rolling, so the math is correct with the result being a cycloid. As a side note, however, if this were a marble rolling along the track, then the optimal path would be a shape known as a hypotrochoid.
Short Answer: you can now ask Chat GPT for the best substitution to use. Longer Answer: the form of the denominator sqrt(c - y) gives us the clue that a trig substitution is the way to go. From there, you can find a table of useful trig substitutions and look for one with the same form as the integral that we have - ie. sqrt(x) / sqrt(a - x). Old School Alternative: Use a table of integrals to integrate this.
This comes from a table of integrals. We engineers turn to the mathematicians to solves this sort of stuff. Admittedly, integrating functions is not a strong point of mine, but fortunately most of these integrals that we contend with have already been solved by some bright sparks (guys like this ua-cam.com/video/dgm4-3-Iv3s/v-deo.html).
Great explanation! For the last bit, where you find the function of the cycloid, why is it that your original substitution for y can just be written as the equation of a cycloid? Doesnt this technically mean you answered the question when you were substituting?
The substitution for y is what allowed me to arrive at an expression for x. This is a parameterized version of the equation because both x and y are dependent on θ which is why there are two equations. This is not circular reasoning.
@@Freeball99 Ah i see, thanks for the explanation. I was just a bit confused on the parametric stuff. Also, why are the boundary conditions 0 to 90 degrees? Im trying to use the equation to apply it in a real point A and point B scenario
Hey. Great video! However I have a question. How can I plot this cycloid curve in geogebra or any other progamm editor(like Latex) i already tried everything! But somehow this doesnt work
I've never tried it myself, though I'd imaging you could simple generate x and y values for increasing values of θ. What sort of problem are you running in to?
Thank you for sharing this video! I love this channel. Suppose the force is not gravity, but some other constant direction vector field such as pressure from water, and instead of a marble, the interaction is instead water with the solid surface? Accounting for friction, the curve looks like a brachistochrone but appears to flatten a bit. Is it still solvable in a similar method, when such dissipative forces are taken into account?
If you are able to pose the minimization problem in the form of a path-based integral then it will be solvable using Calculus of Variations. But, it's hard to give you a general response without seeing the specifics of the problem. There are certainly methods to include friction (which is path-dependent) into the path minimization problem.
I was bothered by the fact that the faster ball had more kinetic energy though they started from the same height. It took me a while to understand that the ball on the straight line was still accelerating in the end whereas the curved track ball was already slowing down on the level.
Hello, could you maybe explain the step from sqrt (ds^2 = dx^2 +dy^2) to eq. 3 ds = sqrt(1 + ((y')^2))dx ? I dont understand this step. Im sure its not exactly black magic and I just lack practice with calculus but it would still be nice ! Thx
This is just a little algebra... ds^2 = dx^2 +dy^2 = dx^2(1 + (dy^2/dx^2)) = dx^2(1 + (dy/dx)^2) now dy/dx can be written as y', so ds^2 = dx^2(1 + (y')^2) taking the square root of each side gives, ds = dx·SQRT( 1 + (y')^2 ) which is the same as ds = SQRT( 1 + (y')^2 )dx
@@Freeball99 thank you very very much for answering my nooby question. Ive seen it in multiple contexts now without explanation. Also thank you for your videos. Best wishes from Germany!
Nope. Never heard of it until now. If you have some specific links to share on the topic, it would much appreciated. Please forward any thoughts you have to apf999@gmail.com
True. But, to be fair, it's not really a marble, it's a particle so it doesn't have any rotatory inertia - I did not include any rotational effects in the derivation. However, from the point-of-view of a demonstration or experiment, this is typically demonstrated with marbles. In this case, the friction would enforce the no-slip condition, but the rolling friction (which would cause any energy loss) is typically negligible. The result is that the shape of the fastest path turns is the same regardless of whether a particle slides or a marble rolls down the track.
I missed this comment until now. Here is the same response that I posted to Bill Wells... The original problem statement talks about a particle, so by definition contains no rotational inertia. Every demonstration one tends to see involves rolling marbles down a track. In the video, I tend to (incorrectly) refer to the mass as rolling, which it isn't in the problem, it's really sliding. The inclusion of the rotatory kinetic energy would increase the effective mass, but since the mass does not appear in the governing equation, this would have no effect on the optimal path. So we'd still get a cycloid.
Great video, but the cycloid is not the solution for a rolling marble, but rather for a sliding object. A rolling marble converts the potential energy lost to kinetic energy along the curve PLUS the energy of ROTATION of the marble. A physicist's nit.
Yes, you're correct. The original problem statement described the mass as a point-mass so there it has no rotatory inertia. In drawing the point as a circle and then watching several demonstration videos on UA-cam with marbles rolling down a track, I conflated the point mass with marbles rolling, in my description of the problem. Though I did describe the path as being frictionless which means that a mass would not roll along it anyway and so its rotational kinetic energy could be ignored. That said, the mass was modeled correctly as one which was sliding without rolling, so the math is correct with the result being a cycloid. As a side note, however, if this were a marble rolling along the track, then the optimal path would be a shape known as a hypotrochoid.
as a non-mathematician i'm fond of making stupid observations.....the solution is offered but i wonder if experimental observations with a marble on a surface confirm the calculations...what may be true in the world of ideas might not be true in the real world....and it seems to me while the calculations were solving for shortest time, at no point in the pages of calculations was quantity t for time per se actually part of the consideration
Not difficult, just beyond the scope of this video. C2 can be made to be zero based on choice of axes. C1 is linked to the radius of the wheel. This explains how to find it: math.stackexchange.com/questions/889187/finding-the-equation-for-a-inverted-cycloid-given-two-points
@@Freeball99 Beyond the scope usually implies more difficult. That link doesn't show how to find C1. Remember I stated C2=0 was easy to show (I guess you didn't read that part of my post). I'm starting to think it's extremely more difficult to solve since I have yet to find a source that shows how.
@@h1a8 The technique for solving for "a" is set out in the linked article. If you substitute C1 = 2a and θ = t/2 (assuming C2 = 0) then the equations in the video and the article are the same (just the sign on the y-axis is flipped because the cycloid is inverted). Regardless of what you name it, the solution is reasonably easy to find… 1) calculate the slope from A to B and use that to solve for θ 2) use θ and either the x or y equation to solve for C1 - It turns out that C1 (or a) is related to the radius of the circle.
@@Freeball99 What? It is not understood how to express the angle the circle rolls (tracing out the cycloid) as a function of slope of the line segment that connects A to any point on the curve.
Hi, I have a question to the equation (14) - the substitution. Where do you get that from, how do you know you had to give y that value? Loved your video tho!!!
It all makes sense, but that is the only part I dont yet understand:)
Short Answer: you can now ask Chat GPT for the best substitution to use.
Longer Answer: the form of the denominator sqrt(c - y) gives us the clue that a trig substitution is the way to go. From there, you can find a table of useful trig substitutions and look for one with the same form as the integral that we have - ie. sqrt(x) / sqrt(a - x).
Old School Alternative: Use a table of integrals to integrate this.
Newton remains the Don.
Wow I'm so in love in this series a combo of history, motivation and rigorous math...more videos to come ❤
Loved how you explained the whole story, and not just the solution. The history behind the math and physics is so much fun to learn ! Amazing work, thank you !
amazing, also appreciated a lot the historical background/anectodes
I've only ever seen the hand-wavy version of the solution to the Brachistochrone: "we realize immediately that this is the differential equation for a cycloid...". Thank you for doing the full derivation!
Appealing to Newton’s ego to troll him was basically the enlightenment version of trash talking on social media. .
I like this problem. It was worked out in the "classical mechanics" classes I followed. The accompanied story, whether it true or not I don't know, but as educational vehicle it still is in my mind, is that the problem has to do with the question: What is the fastest way to move cannonballs from one place to another. Seemingly, the idea is that one has a higher place and that one digs brachistochrone tracks towards the canons. This in order to react swiftly to enemy attacks and knowing that the number of balls is limited, since these are hard to make in those times. It is a likely story, since a lot of the calculus of that time was related to warfare. Anyway, thanks for the clear presentation and the history behind it.
It´s a pleasure to see your videos! Math is amazing, and so is gossip
You seriously make me laugh, love how you get the viewers excited to learn
I love ur videos, esp the parts when u dicuss the topic historically using timeline and also when u end ur lecture with a quick review...
I hope you haven't stopped making videos, this channel has had some of the best math exposition I've ever seen and seeing the last video was uploaded 7 months ago was a bit deflating.
Thanks for your words. I have been taking a break for a little while in order to pursue some other projects. I will be releasing a few more videos before the end of the year which will cover approximate methods like Ritz & Galerkin.
Well presented! With the most simple and concise solution, that I've ever seen.
too much joy in watching this video. Thank you
I'm very intrigued by this. I wasn't intended to learn but ended up watching the whole thing. Great explanation for a mere mortal like me 👏
Keep up the good work!
The way you tell motivation behind every great problem is fantastic.
@good vibrations with Freeball
Thank you for your clear explanation, arrived here after reading Newton's quote about solving this in an evening. Love how the mathematics matches my intuitive visualization the Brachistochrone curve.
You sir deserve a piece of
π
Really enjoyed this. A nice clear explanation, and good depth of historical context.
Glad you enjoyed it!
Perfect balance between historical background and algebra, great video!
This is a very amazing, inspiring, and exciting video!
Brawesome chistoogood chronenumero explanation. Spasibo senor
I first saw this video when it came out and basically understood nothing except the question that was posed(the nature of the problem), now nearly at the end of my 12th grade, I can understand all the math in the solution and the intuition behind the problem(although not really the understating of why the calculations took the path that they did, I assume it will take a lot more practice to see and execute stuff like that, and why some particular manipulations were made the way they were.)However, I'm extremely content, and excited for the knowledge that uni holds for me, with this video being one of the ones that lit the match in me to delve deeper into this field.
Thank you for your feedback. This is the sort of comment that really warms my heart and makes all of this effort worthwhile. If you are able to follow most of this material in 12th grade (it is really aimed toward graduate students), I'd say you have have a very bright future ahead of you. Wishing you all the best of success at the next level. Have you decided what you'll be studying yet and where?
@@Freeball99thank you for your motivations:) Ive since studied this in more depth and can gladly say that after watching this video thrice over, I can now fully comprehend everything that is going on( except the conception of the beltrami identity, since the video covering lagranges multipliers through the catenary problem really is above my current understanding, every other substitution and method is clear thanks to your phenomenal explanation:). Coincidentally, dr.elliot Schneiders explanation is also fantastically intuitive after understanding yours, and somehow verisatium also came out with an extensive story of the brachistochrone problem right about now(which also covered the Bernoulli solution, mind-blowing).
I have applied to institutes across the USA, hoping to pursue an undergrad in cs, or maths(hopefully with electrical engineering). It is too difficult to decide, and my mind flips each day, I just wish that I'm able to learn a lot across all these fields( a lofty goal).
My apologies for the extensive message, I would love to know about you and your story too if you ever find the time.
Thank you for taking out some of it for this reply sir. Love your videos ❤
www.linkedin.com/in/andrewf9/
Best channel on UA-cam
Its been a long time. Waiting for next video
Amazing video, thank you for this.
Also, @ 1:25 in the problem statement, you that the point mass is acted on only by gravity. But in fact, it is acted on by gravity and also by the reactionary forces of the brachistochrone structure. I recognize this is beside the point of the problem. Nevertheless, at any instant of time, if you draw a free-body diagram of the point mass, we will have the force of gravity and the rreactionary force of the brachistochrone acting on the point mass.
Correct. There are reaction forces from the path itself. I described the problem in the manner that I did because I was looking to present the problem in the context of a optimization problem where our task was to find the path/constraints that minimized our functional. This was an attempt to be consistent with previous videos in the series in which I had described various path optimization problems.
Being a Dutchman, I grant you 100 points for pronouncing "Christiaan Huygens" nearly perfectly!
Höchens.
@@lowersaxon Leider...
18:22 One can only imagine how much quicker all of this would've gone if these guys could've jumped on a Zoom session and done a screen share with one another.
Love your Clemenza accent from the Godfather !
It is amazing how the brilliant minds developed powerful methods at such young ages.
And today I'm trying to grasp the meaning of the least action principle and how it relates to Lagrange's equation.
To be fair, I'm enjoying studying this topic. The method of calculus of variations is powerful. When I solved the catenary problem I was satisfied with it haha
Now I'm looking for more applications.
Thank you for your videos! They are really helping me through my course.
Sir please continue uploading scientific material. You are my favorite physics channel.
After Calculus 3, this was just nice to follow
Lol when you said that you'd be solving the problem in 10 mins when it took 12 hrs for newton to do so lmao...nice blend of humor, much needed esp near exams
Really nice video. I loved both the maths and the historical background, which gives the problem such a human dimension. I’ve also watched some of your other videos on Lagrange’s equations, these are really great resources to help with the studying. Thanks for the excellent videos, please keep up the good work !
Thank you! And you pronounced "Huygens" correctly!
I wonder if you would like to tackle a variation of the brachistochrone in which the path must be of fixed length. I tried to work this out a few years ago and it was very messy. I ended up simulating the solution numerically on python. I had a great deal of fun with this problem but I have never found it worked out analytically anywhere.
Great lecture. Thanks
Amazing explanation .
Thank you so much for this
Brilliant Work!!
Amazing explanation
absolutely wonderful and very interesting.
Excellent job
A mathematician whom I know in prison studied it a lot, i.e., James' Murray, serving 15 years on the then now closed Virginia State Farm, maximum security prison where I was serving twenty five years for robbery and murder, falsely accused for the murder though at the time I was a murderer; during which time I became a fledgling mathematician, while working in a field bull gang and studying mathematics evenings and non work days as weekends and major holidays. I began my study of mathematics at eighteen. Came home after serving nearly fourteen years and attended VCU after release from prison and minored in mathematics and majored in Mass Communications. I digress. The question is why I never see any video on the witch graphic or problem. I only saw it at a glance in James Murray's book on calculus. Virginia prison facilities or penitentiary had a lot of donated books. I virtually turned the place eventually into a college campus in the late 1960s and early 70s. I have been looking for the witch graphic. James Murray was aka ,"Heart Trouble" for his bad heart. He went home and turned into an alcoholic and died, perhaps, in the mid or late 1970s, having invented a device on paper, which physicists came to look at but left and never gave James Murray their opinion on its merits. He was my inspiration for studying mathematics.
Enjoy watching this video. Thanks
Excellent video lecture.
Umakwana iweyo 🔥 grt🤝
haha!!i really appreaciate for the history behind it!! Thanks a lot
Mumakwana boss🔥 grt
At 17:49, is there a way to calculate the shortest-time path using calculus of variation with the added condition that the curve must never have a smaller value of y than the end point? Or would this require a numerical solution?
Interesting question... I've neither tried it nor have I seen it attempted, but it seems to me that this should be possible by adding a constraint equation and using the Lagrange Multiplier Method. Almost certainly this would require a numerical solution.
@@Freeball99 I see, thank you!
So satisfying to watch
Well presented. However, I have few questions: (i) the case is for rolling without slipping. If friction was included what will be the solution? (ii) the effect of gravity is not appearing is the formulation of the problem, was it assumed to be unity for simplicity? Thank you.
If there was frictional loss included, this would have the effect of shortening the path which would flatten out the curve a little. The effect of gravity is just a constant with respect to the integral, so when taking the variation and setting it equal to 0, the magnitude of the gravity has no effect on the shape of the curve.
The math here is so beautiful.
I'm confused. How can we write dt as ds/v (10:06) when there is acceleration. d = vt when there is NO acceleration, but we have an acceleration of g and an initial velocity of 0, so shouldn't d be 1/2at^2?
So, a few of things...
1. dt = ds/v (time = distance/velocity) holds regardless of whether or not the mass is accelerating.
2. I have assumed that the mass is accelerating, else v would have been a constant and I could have simply moved it outside of the integral.
3. In order to solve this using the path-minimization method shown in previous videos, I chose to rewrite v as a function of the path variable, y, instead of writing it in terms of time (0.5at^2) - we're not trying to solve for the time taken from A to B, but rather we're trying to find an optimal path.
what program are you using for the whiteboards?
Great video and presentation!
The app is called "Paper" by WeTransfer. Running on an iPad Pro 13 inch and using an Apple Pencil.
14:48 when substitute y by some sin(theta) what's the meaning of theta in the equation of the rolling marble?
I show this at around the 17:32 mark. A cycloid can be plotted by rolling a wheel along a straight path and by observing a point on the wheel. θ is the angle of rotation of this wheel as shown in the graphic.
i don't see the connection with this experience and theta seems to come out of the blue ! A cycloid is drawn when you roll a ball on a straigth line. here we have a particle "rolling" on a cycloid ! the theta of this "rolling" particle is not the same variable (or is it?)
Rolling wasn't my best choice of words here, but I used it since most demonstrations of this show a sphere rolling down a track, but in fact, the mass is sliding and not rolling since the surface is assumed to be frictionless and the mass is assumed to be a particle with no rotatory inertia.
The θ in the solution doesn't directly represent the angle of a rolling wheel. It's a mathematical substitution used to simplify the integration. The connection to a rolling wheel comes after solving the equations. The resulting parametric equations (18 and 19) happen to describe a cycloid, but this wasn't known at the start.
You're correct that in the problem, we have a particle moving along the curve, not drawing it. The θ in the solution is not the same as the angle of a rolling wheel.
The cycloid emerges as the optimal path, but it's not how we arrived at the solution. It's the result, not the method. The solution method doesn't rely on knowing about cycloids beforehand.
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
Your observation is correct. Stated differently, a change in initial conditions does not change the governing equations of motion. So the general form of the solution remains the same, just the constant of integration change.
thankyou
I think the conservation of Mechanical energy substitution for the velocity only works for a particle or a puck on a frictionless surface. A ball would have both linear and rotational kinetic energy. The rotational component would take away from the linear.
Yes, you're correct. The original problem statement described the mass as a point-mass so there it has no rotatory inertia. In drawing the point as a circle and then watching several demonstration videos on UA-cam with marbles rolling down a track, I conflated the point mass with marbles rolling, in my description of the problem. Though I did describe the path as being frictionless which means that a mass would not roll along it anyway and so its rotational kinetic energy could be ignored. That said, the mass was modeled correctly as one which was sliding without rolling, so the math is correct with the result being a cycloid. As a side note, however, if this were a marble rolling along the track, then the optimal path would be a shape known as a hypotrochoid.
Very elegent solution. Thankyou. However I have some doubt. The solution looks perfect for sliding particle but for rolling ball to count of conservation of mechanical energy we also need to consider rotational energy (1/2 x moment of inertia x angular velocity 2 - This MOI for a ball being 2/5mr2.) This should slightly modify the curve.
This is great. I stopped short of studying variational calculus at uni, but I've always been interested. Am I correct in understanding that the Beltrami Identity is only valid when F is minimised? or otherwise where is the minimisation happening?
We are always trying to minimize the F in these problems. The Beltrami Identity is what the Euler-Lagrange Equation reduces to when F is not a function of the independent variable, but is a function of the dependent variables only. Solving the Beltrami Identity yields the path that minimizes F.
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?
The trig expression comes from using an integral table at the 14:45 mark. A good mathematician could probably figure this out in their head, but that is above my pay grade. Integral tables will shown how to handle integrands of various forms.
at 14:47 why did u choose to substitute y to be c sin squared theta
The short answer is that you can now ask Chat GPT for the best substitution to use 😀. The slightly longer answer is that the form of the denominator sqrt(c - y) gives us the clue that a trig substitution is the way to go. From there, you can find a table of useful trig substitutions and look for one with the same form as the integral that we have - ie. sqrt(x) / sqrt(a - x).
Old School Alternative: Use a table of integrals to integrate this
Great video. Could you go over quickly how we could apply this program to real world scenario? I'm not sure how to solve for C1, C2, and what my theta value would look like. Thanks
math.stackexchange.com/questions/889187/finding-the-equation-for-a-inverted-cycloid-given-two-points
thanks you so much
One question:
The original problem statement specified a sliding object. Your solution used a rolling object. Therefore, in your example, the conservation of energy equation would include both translational and rotational kinetic energy. Are the solutions for a rolling object versus sliding object the same? I have seen demonstrations of the brachistochrone problem using rolling metal spheres, and the results are similar.
Thank you.
Sorry for the delayed response, but I somehow missed this until now... The original problem statement talks about a particle, so by definition contains no rotational inertia. Every demonstration one tends to see involves rolling marbles down a track. In the video, I tend to (incorrectly) refer to the mass a rolling, which it isn't in the problem, it's really sliding.
There has to be an additive coefficient on y, or else there’s no way to have a non-zero y1 boundary condition. What am I missing?
Also, given the use of the parameter theta, it seems that given point B(x2,y2), that we would need to solve for theta, but at the same time we need to use point B to solve for some of the coefficients.
If you take the last diagram in the video, Point A is located at (0, h) and Point B at (d, 0). So the height of Point B is always zero by definition based on the choice of axes and the constant c1 is therefore dependent on the height h of Point A (turns out that c1 = 2h and c2 = 0).
It can be assumed that θ = 0 at point A and θ = θ_B at Point B. Solving for θ_B will require a numerical solution since θ_B is a function of both h and d. You will find that c1 and c2 depend on the location of point A ONLY. So this alone does not specify the precise cycloid path needed. The parameter, θ_B becomes a necessary parameter for this.
i wanna need some more examples and numericals about brachistone
If you have vector from point A to any point on the cycloid, is theta the angle between the x-axis and that vector?
Near the end of the video (17:32), I show a a figure of how the cycloid curve is traced out by a rolling disk. The angle, θ, is the angle of rotation of the disk.
thank you
Right at the beginning, you spoke of a marble rolling down a frictionless path. There is no rolling without friction. I understand this is beside the point of the brachistochrone problem, but a frictionless surface implies no rolling.
Yes, you're correct. The original problem statement described the mass as a point-mass so it has no rotatory inertia. In drawing the point as a circle and then watching several demonstration videos on UA-cam with marbles rolling down a track, I conflated the point mass with marbles rolling, in my description of the problem. Though I did describe the path as being frictionless which means that a mass would not roll along it anyway and so its rotational kinetic energy could be ignored.
That said, the mass was modeled correctly as one which was sliding without rolling, so the math is correct with the result being a cycloid. As a side note, however, if this were a marble rolling along the track, then the optimal path would be a shape known as a hypotrochoid.
How does this curve relate to the isochronous curve. If at all.?
I talk about this at the end of the video at the 19:20 mark. (isochrone & tautochrone are the same thing)
I have taken equations 18 and 19 in order to determine the values of the constants c1 and c2 for the case, let`s say, of the brachistochrone that goes from the point (0,0) to another generical point (1,h), h>0, with the x and y axis taken as you mention in the video. However I can't see there is a solution for any value of h. Could you clarify this point?
"Bernoulli, you're muted!"
what about the rotational motion of the rolling object? Does Euler-Lagrange equation respect for energy conservation from its onset? Wouldn't that Beltrami equation is a consequence of conservation of energy (Noetherm theorem), and what facts has it led to hold a priori conservation of energy if it is going to use same Beltrami equation later?
The original problem statement described the mass as a point-mass so it has no rotatory inertia. In drawing the point as a circle and then watching several demonstration videos on UA-cam with marbles rolling down a track, I conflated the point mass with marbles rolling, in my description of the problem. Though I did describe the path as being frictionless which means that a mass would not roll along it anyway and so its rotational kinetic energy could be ignored.
That said, the mass was modeled correctly as one which was sliding without rolling, so the math is correct with the result being a cycloid. As a side note, however, if this were a marble rolling along the track, then the optimal path would be a shape known as a hypotrochoid.
Amazing!
thank you what is the link of tatuchrone problem? or its not uploaded yet?
I haven't yet made a video on the tautochrone problem. However, there are several of them online.
@@Freeball99 yes, I saw them but your explanation is sth else
That was amazing...
Now please do the solution of the bernoullis "the geometrical way"
a small question: how could you replace y with c1sintheta squared in step 14?
This is just a substitution/change of variable. These sort of tricks are pretty common when performing integrations.
I couldn't understand substitution of y= C_1*sin^2(omega) where it comes from ?
Short Answer: you can now ask Chat GPT for the best substitution to use.
Longer Answer: the form of the denominator sqrt(c - y) gives us the clue that a trig substitution is the way to go. From there, you can find a table of useful trig substitutions and look for one with the same form as the integral that we have - ie. sqrt(x) / sqrt(a - x).
Old School Alternative: Use a table of integrals to integrate this.
Hello, from where did you get that y=c1sin^2(theta)? It showed up out of nowhere.
This comes from a table of integrals. We engineers turn to the mathematicians to solves this sort of stuff. Admittedly, integrating functions is not a strong point of mine, but fortunately most of these integrals that we contend with have already been solved by some bright sparks (guys like this ua-cam.com/video/dgm4-3-Iv3s/v-deo.html).
Great explanation! For the last bit, where you find the function of the cycloid, why is it that your original substitution for y can just be written as the equation of a cycloid? Doesnt this technically mean you answered the question when you were substituting?
The substitution for y is what allowed me to arrive at an expression for x. This is a parameterized version of the equation because both x and y are dependent on θ which is why there are two equations. This is not circular reasoning.
@@Freeball99 Ah i see, thanks for the explanation. I was just a bit confused on the parametric stuff. Also, why are the boundary conditions 0 to 90 degrees? Im trying to use the equation to apply it in a real point A and point B scenario
Hey. Great video! However I have a question. How can I plot this cycloid curve in geogebra or any other progamm editor(like Latex) i already tried everything! But somehow this doesnt work
I've never tried it myself, though I'd imaging you could simple generate x and y values for increasing values of θ. What sort of problem are you running in to?
hello, can someone explain what c1 and c2 means?
c1 and c2 are constants of integration.
Thank you for sharing this video! I love this channel. Suppose the force is not gravity, but some other constant direction vector field such as pressure from water, and instead of a marble, the interaction is instead water with the solid surface? Accounting for friction, the curve looks like a brachistochrone but appears to flatten a bit. Is it still solvable in a similar method, when such dissipative forces are taken into account?
If you are able to pose the minimization problem in the form of a path-based integral then it will be solvable using Calculus of Variations. But, it's hard to give you a general response without seeing the specifics of the problem. There are certainly methods to include friction (which is path-dependent) into the path minimization problem.
@@Freeball99 With friction you have to add the Rayleigh dissipation function to the Euler-Lagrange equation.
I was bothered by the fact that the faster ball had more kinetic energy though they started from the same height. It took me a while to understand that the ball on the straight line was still accelerating in the end whereas the curved track ball was already slowing down on the level.
kinetic energy literally depends upon speed
Hello, could you maybe explain the step from sqrt (ds^2 = dx^2 +dy^2) to eq. 3 ds = sqrt(1 + ((y')^2))dx ? I dont understand this step. Im sure its not exactly black magic and I just lack practice with calculus but it would still be nice !
Thx
This is just a little algebra...
ds^2 = dx^2 +dy^2
= dx^2(1 + (dy^2/dx^2))
= dx^2(1 + (dy/dx)^2)
now dy/dx can be written as y', so
ds^2 = dx^2(1 + (y')^2)
taking the square root of each side gives,
ds = dx·SQRT( 1 + (y')^2 )
which is the same as
ds = SQRT( 1 + (y')^2 )dx
@@Freeball99 thank you very very much for answering my nooby question. Ive seen it in multiple contexts now without explanation. Also thank you for your videos.
Best wishes from Germany!
Bernoulli took 2 weeks to solve this, whereas, it took Newton 12 hours!!
That's crazy!
Ever heard of the witch graphic problem?
Nope. Never heard of it until now. If you have some specific links to share on the topic, it would much appreciated. Please forward any thoughts you have to apf999@gmail.com
Newton was a one badass mathematician.
Huh. How is the marble going to roll down a frictionless path?
True. But, to be fair, it's not really a marble, it's a particle so it doesn't have any rotatory inertia - I did not include any rotational effects in the derivation. However, from the point-of-view of a demonstration or experiment, this is typically demonstrated with marbles. In this case, the friction would enforce the no-slip condition, but the rolling friction (which would cause any energy loss) is typically negligible. The result is that the shape of the fastest path turns is the same regardless of whether a particle slides or a marble rolls down the track.
@@Freeball99 I got it - I was just getting tied in a knot trying to imagine myself rolling down a water slide...
I guess that explains why so many rollercoasters have sections where they go underground.
Well yes, but rollercoasters are influenced by propulsion as well as gravity I’m sure
I love you sir
Redding through comments I see Bill Wells also raised similar query.
I missed this comment until now. Here is the same response that I posted to Bill Wells...
The original problem statement talks about a particle, so by definition contains no rotational inertia. Every demonstration one tends to see involves rolling marbles down a track. In the video, I tend to (incorrectly) refer to the mass as rolling, which it isn't in the problem, it's really sliding.
The inclusion of the rotatory kinetic energy would increase the effective mass, but since the mass does not appear in the governing equation, this would have no effect on the optimal path. So we'd still get a cycloid.
Variational
Great video, but the cycloid is not the solution for a rolling marble, but rather for a sliding object. A rolling marble converts the potential energy lost to kinetic energy along the curve PLUS the energy of ROTATION of the marble. A physicist's nit.
Yes, you're correct. The original problem statement described the mass as a point-mass so there it has no rotatory inertia. In drawing the point as a circle and then watching several demonstration videos on UA-cam with marbles rolling down a track, I conflated the point mass with marbles rolling, in my description of the problem. Though I did describe the path as being frictionless which means that a mass would not roll along it anyway and so its rotational kinetic energy could be ignored. That said, the mass was modeled correctly as one which was sliding without rolling, so the math is correct with the result being a cycloid. As a side note, however, if this were a marble rolling along the track, then the optimal path would be a shape known as a hypotrochoid.
More optimal? I thought optimal meant the best, so what’s better than the best? 😉
Ok, now I see. It's because the curve might not be monotone on y.
as a non-mathematician i'm fond of making stupid observations.....the solution is offered but i wonder if experimental observations with a marble on a surface confirm the calculations...what may be true in the world of ideas might not be true in the real world....and it seems to me while the calculations were solving for shortest time, at no point in the pages of calculations was quantity t for time per se actually part of the consideration
i am from pakistan and my paper held on 13oct
Guess what's the most difficult part? Finding C1 (C2 = 0 is easy). Yet no one can explain how to solve for it. Not even this author of this video.
Not difficult, just beyond the scope of this video. C2 can be made to be zero based on choice of axes. C1 is linked to the radius of the wheel. This explains how to find it: math.stackexchange.com/questions/889187/finding-the-equation-for-a-inverted-cycloid-given-two-points
@@Freeball99 Beyond the scope usually implies more difficult. That link doesn't show how to find C1. Remember I stated C2=0 was easy to show (I guess you didn't read that part of my post).
I'm starting to think it's extremely more difficult to solve since I have yet to find a source that shows how.
@@h1a8 The technique for solving for "a" is set out in the linked article. If you substitute C1 = 2a and θ = t/2 (assuming C2 = 0) then the equations in the video and the article are the same (just the sign on the y-axis is flipped because the cycloid is inverted). Regardless of what you name it, the solution is reasonably easy to find…
1) calculate the slope from A to B and use that to solve for θ
2) use θ and either the x or y equation to solve for C1 - It turns out that C1 (or a) is related to the radius of the circle.
@@Freeball99 What? It is not understood how to express the angle the circle rolls (tracing out the cycloid) as a function of slope of the line segment that connects A to any point on the curve.
Dommage trop de paroles, respire !