Just wanna say that all your videos are beautiful, well edited and animated, insightful, and a joy to watch. Thanks for the awesome content, 3Blue1Brown.
I rarely ever comment on RUclips, but I wanted to say that your content is amazing and most importantly thought provoking! Keep up the good work and thank you for the time you put into your videos!
+3Blue1Brown superb video but one thing i still confused about is why you replace the speed of light in different refractive index with square root of y?because i think this only apply if the particle initial velocity is zero and was accelerated only with gravity whereas light initial velocity in vacuum is about 300,000km/s
I find the final chanllage is actually solved in this video. stop the video at 12:34, note the velocity of piont P = √(2gy)(obtained at 8:00), so the angular velocity of P rotate around C is just √(2gy)/(D Sin(θ)) = √(2g/D). thus the curves in t-θ plane is always stright lines with slope √(2g/D). besides, note θ=1/2 ∠POC (let O be the center of circle), so "P rotate around C with constant angular velocity" is equivalent with "P rotate around O with constant angular velocity", so the wheel rotate with constant rate.
Solution to the challenge: Consider an infinitesimally small time interval dt, when the velocity of the pi-creature is v at angle theta with the vertical line. In this time interval, the pi-creature descends a height of dy = v cos(theta) dt (1). By conservation of kinetic energy, v^2 = 2gy (y being the total height the pi-creature has descended). By Snell's law, sin(theta) = Cv (C is a constant). Take the derivative of both equations above: 2v dv = 2g dy (2) cos(theta) d(theta) = Cdv (3) Putting (1) (2) (3) together, we have d(theta) = Cg dt, so the theta-t function is a straight line :D
A good formal prove. But the teaser asked for an intuitive model for the (t, theta) space. Also Snell's law shouldn't be used as a given, but should be the consequence of the shortest path optimisation in that space.
Thanks! Usually, I don't focus on the history of problems, so I feel Steve was a real addition to the channel for making that part of the focus. I'll probably do more things like this in future videos as well now.
Yeah my mind basically exploded when snells law could be applied, such an amazing use of a particular physical law to reduce a seemingly challenging unrelated problem down to its bare mechanics
@@3blue1brown this legend about Newton solving in one night is pretty ridiculous, no need to divinise a man who was pretty horrible human and pretty stupid when outside pure science (trying all his life to obtain the philosophical stone and making numerology from the bible among other thing)
Makes me feel special that I was subscribed to you before Vsauce. And I love that more people are tuning in, you need more exposure, your videos are simply the best math videos on the net.
If you have a paper, and you try to use the point C like 3b1b does in the video, you will get an equation like v=2cosθ√gR and if you are correctly using point C, you could get that point C is moving in constant speed and thus the whole circle is moving at constant speed.
It is interesting to note that Huygens in 1672 showed by graphical means, that a cycloid was a tautochrone, that is that starting at any point on the curve an object sliding along it will reach the bottom in the same time, thus a cycloid should be the ideal path for a pendulum on a clock. Hugens partly got to the solution as the result of a good guess, the mathematics of a cycloid is pleasingly simple and given that Mersenne had already showed that a circle didn't quite work, the cycloid was the next obvious and easy curve to try. The two problems are clearly linked and I am sure that Newton knew this and proceeded along similar lines.
Interestingly to make a tatochrone pendulum all you need to do is have a flexible string half the arc length of the desired cycloid holding the bob and then cut a solid cycloid of the desired length in half, placing each half either side of the pendulum (curves down small ends meeting at the pivot flat halves away from the pivot). As the pendulum wraps around the cycloid halves the bob traces a tautochronous cycloid beneath. Or, rather than two halves of a cycloid you can use a cycloidic taurus with an internal diameter of 0 for a rather pleasing effect.
Cycloidic taurus: I think, a donut shape, where instead of the cross-section having a circular shape, it has a cycloid. That way, the pendulum line can come through the hole in the middle, and the pendulum can swing in any direction.
Because of the gravitational force , the curve taken by the object is cycloid. U have to consider it. Because of earth is not a perfect sphere , the path taken by the object is cycloid rather than arc of the circle. If we plot our assumption to this conclusion, we can prove the straight lines over the r v/s theta curve
I know it may be late but I was looking for a satisfactory explanation of the warm up challenge in the comments and I could not find any. As so I have tried solving it myself and I finally came to a possible solution. First we must label θ as the angle between the vertical at the center of the circle and the point on the circle's circumference which is tracing the cycloid. We can find the tangential velocity at that infinitesimal point in time if we consider the point of contact C of the circle and the horizontal surface as the instantaneous center of rotation and by using *v = wr* (Eq. 1). If we label the point where the particle is as L, then we can see this becomes *v = wl* (Eq. 2) where l is the length CL. Using the cosine rule we can see that the length l is simply *l = 2Rsin(θ/2)* (Eq. 3). We can also equate the velocity if we consider conservation of mechanical energy since there is no friction. Thus K.E. = P.E. of the particle which is *1/2 mv^2 = mgy* (Eq. 4) where y is the vertical distance from the horizontal surface to the point L. This reduces to *v = sqrt(2gy)* (Eq. 5). Substituting Eq. 3 and Eq. 5 into Eq. 2 yields *sqrt(2gy) = 2wRsin(θ/2)* which can be squared to become *2gy =4 w^2 R^2 sin^2 (θ/2)* (Eq. 6). We can substitute y here with the parametric equations of the cycloid. If we take the y axis downward as positive then it varies slightly that the one in the video as it will be *y = R(1 - cosθ)* (Eq. 7). Substituting into Eq. 6 gives us *2gR(1-cosθ) =4 w^2 R^2 sin^2 (θ/2)* (Eq. 8). Lastly, we can use the trig identity *sin(θ/2) = sqrt((1-cosθ)/2)* (Eq. 9) and transform Eq. 8 into *2gR(1-cosθ) =4 w^2 R^2 ((1-cosθ)/2)* from which we can cancel out the various term to reduce into *g = R w^2* which finally yields *w = sqrt(g/R)* (Eq. 10). This shows that the angular velocity is constant and that the particle's trajectory due to gravity does follow that of the trajectory of a rotating wheel with constant rotation. I hope this helps anyone in the future who might revisit this wonderful video.
You're insanely intelligent, and an inspiration to me. I'm going to try this. If there's one thing I think I've learned from your videos, it's that nothing consistent is off-limits and there is always a different reference frame from which to view your problem.
I heard that you were doing a essence of calculus series. You should consider also doing a essence of multivariable calculus. Multivariable is a class that many people take in all of maths, engineering and science and is surprisingly poorly understood. Thanks! :)
pretty much everything, basic facts about the gradients don't seem to be even remember by lots of people I know and things like Hessian, why its a matrix, jacobians etc seem to be words lots of people don't actually understand.
I was really in doubt about this interview concept when you first presented it, but after just 30 seconds with talking and animation in perfect unison, I was convinced. Great job!
Mads Nielsen Mind washing with pretty pictures...☺️-Don't worry, this kind of phenomenon occurred to me, too, (for another video). Since that, i am more focused on formal audible arguments, than on icons following icons.
0:42 that joke right there earned an instant sub .... also the face that he timed it so it hit exactly at the 42 seconds mark .... this guy is awesome!
Holy crap, Strogatz! You mean the author of the book I loved in undergrad, Nonlinear Dynamics & Chaos! Loved my Dynamical Systems class in undergrad, but damn...it was a beast! Much much much respect for these mathematicians who pioneer the way in these fields, but even more so for those who can beautifully, eloquently, and effectively communicate these advanced concepts in math to plebeians like myself; such is the man you interviewed here: Steven H. Strogatz.
I have research on this topic. Can you help with university references and notes? Do you have a research plan for this problem? I need some references on this topic
7:38 Case Sensitivity: The g in the potential energy term, mgy, is not "the gravitational constant." The gravitational constant, G, is the proportionality constant in Newton's law of universal gravitation that relates the attractive force to the masses involved and their relative position. G = 6.67E-11 m^3/kg•s^2. You are referring to g, which represents the acceleration due to Earth's gravity measured at the surface, 9.8 m/s^2. An innocent slip as I know you know this. Another great video!
When I learn about Brachistochrone problem, I was very surprised and excited that this small question makes a whole new concept "Calculus of Variation." But this video takes me to next level of excitement. Really thought provoking and interesting.
I watch your videos all day, I watch for knowledge and for fun! I can't get enough of this! Your expository is like Nothing I have ever seen before. I'm about to start my PhD in applied Mathematics.... I feel deeply blessed that I came across your channel. You sir, are a breath of fresh air.
One way you could think of it is that the function is the rate of change of the area under itself. Imagine the graph of a function, and pick two points in the X axis. You can see the area under the curve between these two points right? Now what if you nudge one of them just a bit? The area changes just a little, but you can see that, the higher up the function is at the nudged point, the more the area in that interval will change. In other words, how much the area under a curve changes is not only based on how much the length of the interval changes, but also on the value of the function at the end points. Another explanation for those more familiar with discrete mathematics is that an integral is just an infinite sum of differences. If you have something like f(n) = g(n) - g(n-1) What you basically have is that f is the analogous to the "discrete derivative" of g (difference in height divided by difference in length). So what happens when you sum all the consecutive values of f(n) from 1 to, say, m? f(1) + f(2) + f(3) + ... + f(m) (g(1) - g(0)) + (g(2) - g(1)) + (g(3) - g(2)) + ... + (g(m-1) - g(m-2)) + (g(m) - g(m-1)) -g(0) + g(1) - g(1) + g(2) - g(2) + g(3) - g(3) + ... + g(m-1) - g(m-1) + g(m) And you can see the all the terms, except for g(0) and g(m), cancel out, and you end up with g(m) - g(0) This is a little more difficult to imagine on a continuous function, but the principle is the same (taking it to the limit to infinity and yadda yadda yadda). And, of course, the first value of n doesn't have to be 1, that was chosen for simplicity. Out of these two, I prefer the first reasoning just because it's more intuitive, but the second one shows why antiderivatives have indefinite constants: if it's a constant term, it's going to be the same in g(m) and g(0), so they'll cancel out. Therefore, it doesnt matter which constant term you choose, there is no "right" one. Whew, that was a wall of text. Sorry about that, but I wanted to make it as clear as I could in a single comment without edits.
dy / dx = f(x) => dy = f(x) dx The RHS is the area of a tiny sliver of the curve => sum up all the slivers and get integral of dy= integral of f(x) dx => y = F(x) The first step isn't technically mathematically correct but I think it's intuitive enough.
This will be a hand wavy solution to the question but the way I see it, the straight line in the theta-time space indicates a constant rate of change in theta which is the smoothest possible path to the solution as a straight line is always the shortest path: i.e. add any more or less curvature than that exact curve and you'll distort the theta-time curve creating a lower than optimal area under the curve of that graph. Background is only a B.S. in math though so take my intuition with a few grains of salt. Cheers to the work you do, and the tools you've developed - it's all much appreciated!
As a Cornell alumnus, I know Prof. Strogatz is very famous among math and science students for being a great educator, if you could not tell from the video. It's always a pleasure to see/listen to his work.
Saw your video yesterday, really great :) *Here is the answer to your challenge*: particle moves such that speed v is proportional to sin(theta). Differentiate it: rate of change of change of speed is proportional to cos(theta)* rate of change of theta. But rate of change of speed is component of gravity along the curve = g cos(theta). Hence *rate of change of theta = constant*.
8 years later, and that challenge is still on my mind. Not solution, but it got me thinking about all sorts of other variations on the problem for different force terms.
I saw this problem in Physics olympiad national qualifier 25 years ago. Of course, I couldn't even tough the starting point. It has been forgotten for more than 2 decades, and suddenly pop up in front my eyes. One of the biggest mysteries of my life is solved. Thank you!
Here is a solution to your challenge. Acknowledging that the time-minimizing trajectory is linear in the theta-t space, one of the big implications is that angular velocity is constant. BIG. Another property to consider is that, as was mentioned previously in the video, energy is proportionate to the square of velocity, or in most differential cases acceleration. If theta-t is linear, the theta-dot is constant, this means that there is no rotational acceleration being imparted on the system and all of the energy gained from gravity can be focused on tangential velocity, which is what gets us from point a to point b. In this way, like removing friction from a system, we are removing rotational acceleration from the energy equation thus increasing our tangential velocity and, given a path of constant length, this allows us to arrive at the destination the quickest. Note, I think I've been saying tangential through this, I guess what I mean more is cartesian. Tangential implies a circular path.
Really awesome solution, I paused the vide and tried to solve it myself. My first instingt was to use calculus, but I didn´t do very good. However this is how far I came and if someone could help me to continue this calculation it would be really nice. The variable that we want to minimize is the time, and the variable we are looking for is the function between the two points. So we need a relationship between the time and the function. We know that t=s/v, there is a general way to work out the distance between two points on a function. And it says that the distance from the point x=a to x=b on the curve f(x) is the integral from a to b of the function sqr(1+(f´(x))^2). The distance between the points doesn´t really matter so I´m going to say that it is one just to make things easier. This means that s=0integral1(sqr(1+(f´(x))^2))dx We know that v=k*sqr(f(x)) in any given point, but since the speed changes over time we can not plug it in to the formula. Also I´m going to say k=1 just to make things easier. It won´t affect the final curve any way. Correct me if I am wrong but the average speed between x=0 and x=1 is going to be 0integral1(sqr(f(x)))dx. So if we plug this into our formula t=s/v we get t=(0integral1(sqr(1+(f´(x))^2))dx)/(0integral1(sqr(f(x)))dx). So I have a formula for the relationship between the time and the function. Now I just have to find the derivative and set it equal to zero I guess, it will probably end up with i differential equation. There is just a little problem, I have no idea of how to take the derivative of that function, and I don´t even know if it is right for that matter. Is there even an algebraic way to do it? If not, then I have no idea about how to solve this problem algebraically. It would be easier to find the derivative if the variable was x, but it not is f, we are looking for an entire function here! Does anyone know how to continue this solution?
+Samuel Lo ok So if I understod it right then I now have to solve this equation: d/df((0integral1(sqr(1+(f'(x))^2))dx)/(0integral1(sqr(f(x)))))-d/dx(d/df'(x)((0integral1(sqr(1+(f'(x))^2))dx)/(0integral1(sqr(f(x)))))=0 Because I am assuming that what they reference as L is the same as t in my equation. But still I can't solve that. I had propably made a misstanke with the perenthases in the equation but what it says is dt/df-d/dx(dt/df')=0 I just replaced t with what I know that t is equal to in termes of f.
We need to use the instantaneous speed, not average speed. v = sqr(f(x)) ds = sqr(1+(f'(x))^2) dx dt = ds/v = sqr(1+(f'(x))^2) / sqr(f(x)) dx Total time = integral [ sqr(1+(f'(x))^2) / sqr(f(x)) ] dx L = sqr(1+(f'(x))^2) / sqr(f(x)) L is a function of f(x), f'(x)
Great video, I just have one small issue. At 5:55 the video shows a ray of light bouncing off a mirror in an attempt to "minimize" the travel time from A to B, while Steven Strogatz in the interview says "somehow nature is imbued with this property of doing the most efficient thing". But clearly the most efficient thing would be to not bounce off the mirror in the first place and to go directly from A to B. What Fermat's principle states is rays of light traverse the path of *stationary* optical length with respect to variations of the path. In other words, a ray of light prefers the path such that there are other paths, arbitrarily nearby on either side, along which the ray would take almost exactly the same time to traverse. There can be cases where light can be fooled into to taking the longest path!
+Ralph Dratman Great question. Consider a circular mirror. Draw the horizontal and vertical diameters. Consider two points, A and B, on the horizontal diameter equidistant from the centre. Light will travel from A to the top of the circle, call it point O, and reflect to B. Perfectly symmetric. Now draw an ellipse with foci A and B, that goes through O. It will be tangent to the circle at O and will be outside the circle otherwise. Ellipses have the property where the sum of the distances from a point on the ellipse to the two foci is constant. It follows that any point inside the ellipse will have a total distance to the two foci less than that of any point on the ellipse. Since the circle is entirely inside the ellipse it follows that the rays of light took the longest path possible, i.e., any other point on the circle would have been a shorter path. This realization that the light doesn't *choose* the shortest path addresses the religious aspect mentioned in the video (as well as the fact that you can't just use light to solve the travelling salesman problem for example). In the path integral formulation of quantum theory light chooses *all* paths, but most end up cancelling each other out. When a path's nearest neighbours have almost the same length then they add constructively. These paths are called *stationary*, and can occur when they are minimum or maximum, either locally or globally. The reason the light reflects off the point O in the example above is best explained because the circle is tangent to the ellipse (and therefore all the nearest neighbours have nearly constant path lengths).
Chris Shannon I have a quibble with your longest path discussion. If we want to find or verify a putative shortest path, we can test to show that the chosen path is shorter than any other path, including any arbitrary wiggly path. But a putative longest path cannot be longer than every arbitrary wiggly path. It may, of course, as you demonstrate, be longer than any other stationary path. At least, so it seems to me.
What a brilliant story. As you say, awe inspiring. When we learned Snell's law, it was taught without reference to that path being the fastest, hence the path light will naturally take. The facts were taught but not the fundamental principle or reason. Of course that leads to other questions.....
Exactly what is "it" in your comment? How can an illusion predict the mathematics of a curve? Calling time and space an illusion is only true in the philosophy of nondualism, in which pure consciousness is all that truly exists. But such an insight cannot possibly predict any of the laws of nature, including the equations of curves having certain properties. In other words, you have taken a true statement in one model of reality and erroneously applied it to another.
I know nothing about higher math yet I love watching 3Blue1Brown videos. Imaging how these concepts occur in the world is why I watch. This concept reminds me of alpine skiing and skateboarding.
So Newton started solving it grudgingly, knowing he would solve it anyway, and then published it anonymously just to get done with the challenge. It's like a first grader challenging a math major to prove the pythagorean theorem that he just came up with, thinking he came up with something hard.
I read Steven's book 'infinite powers' when I was in high school and you know what... the way he explained it like a story, it felt more like a novel than a book on calculus...he is really an insightful writer....
The answer can be found in the book "Newton's Principia for the Common Reader" by S. Chandrasekhar, pp. 571-578. He points out that Newton had already laid the groundwork for solving this problem in the first part of the Principia.
Strogatz truly is a fantastic communicator of math. Granted, I can only speak from using his book on nonlinear dynamics, but I still got so much enjoyment out of his way of writing. The subject itself is an eye opener and really rewarding, but because I couldn't attend lectures and had to rely on the book, I was very happy to experience such an enjoyable read. He writes to the reader, and it makes all the difference.
Snell's Law is an example that Bartosz Milewski gives of the kind of global thinking that is taken to the extreme in category theory, with its universal constructions. A math instructor of mine also mentions that these kinds of global optimizations (optimizing with the telescope, rather than the microscope, you might say) are done in calculus of variations. I find this duality of local vs global approaches to optimization to be quite interesting.
This is a fantastic explanation. I'm taking precalc this year, and I can intuitively understand this. That is so cool haha. Came from Vsauce. It was an alright video, but he never really explained the reasoning behind the property, which was what I wanted to know. Glad he introduced me to this channel though.
if the shortest path is a curve of a circle, the tangent line of that curve changes angles at a constant rate as you follow each point on the curve. in other words if you look at two points on a circle that are x distance away, the difference in their tangent line angles is the same as any two points separated by x distance. if the formula for slope is (y2 - y1) / (x2 - x1) substituting distance for x and theta(angle) for y the slope is constant throughout. therefore theta changes at a constant rate in respect to time.
This video is really great! Why cannot I watch it when I was an undergraduate! The first time I read the details for the solution of Johann Bernoulli was from Ernst Mach's Science of Mechanics.
I am by no means a mathematician and my understanding of physics is rudimentary, but when you started talking about expressing the problem in term of time over theta, space time diagrams kept popping in my head. every curve or motion is actually a straight line in space time. I don't know maybe i am, reaching but i feel like there is a connection here.
This is not rigorous and I'll probably revisit this later, but I do have a general idea: The lateral force imparted on the object due to gravity is gsin(theta). As the slope is constantly changing, mapping the axis to sin(theta) renormalizes the curve in a way that reduces it to a straight line, as it undoes the encoding of the function. In other words, the function does not look linear on a normal cartesian set of coordinates is because the coordinate system does not factor in the constantly changing lateral force of gravity. By mapping the function into a set of coordinates that cleverly cancels out the constantly changing force of gravity, the optimization problem becomes linear. I also want to thank you for making these videos. I rarely find videos or textbooks that explains things intuitively, and the animations help so much in this regard.
This is specific to the video @ 11:48. There should have been a visualisation of tangent always meeting bottom of circle with changing point of contact C. That took time for me to visualise! Sorry, if that's a comment appears to be rude.
According to the idea of Bernoulli, if an object is tracing a brachistochrone v=k\sin\theta. Since the gravity is vertical, the slope component of the gravity is a=g\cos\theta. This is the acceleration of the object (of unit mass). We know that the acceleration is the derivative of the velocity. Hence d/dt(k\sin\theta)=g\cos\theta. LHS of the above is k\cos\theta d/dt(\theta). Hence d/dt(\theta) = constant function. This is the end of the proof for the challenge 3Blue1Brown proposed, provided that I understand the challenge well.
Your videos are inspirational. They have allowed me to see mathematics, or as Keith Devlin would say, "mathematics has made the invisible visible." And your videos do that for me in a unique way that has not been replicated by any other learning source. I am studying math education, and I would actually like to learn how to make videos like this so that I can make learning more enjoyable for my students. I'm not much of a programmer, but I'd be willing to learn. Is it a trade secret or can you point me in the right direction? Thanks, Grant. -Ben
This was very fun to watch! Actually a nice recap with some additional information about a calculus of variations lecture some time ago. Thanks for that!
Well, I keep a growing list of different interesting topics I think where I think there's room to provide a new, or at least under-discussed, perspective. Unfortunately I can't always spare the time to knock items *off* the list as quickly as they get added.
funny how this problem sneaked into modern society e.g. as a part of scouts training, to find the path from A to B using map, compass, and math of course, that takes the least time and travel it to literally prove it
I've seen several videos on this today (started at Vsauce). An infinite number of cycloids, with varying diameters of original circle, can be fit to two points. Are of of them solutions to the Brachistochrone problem? That cannot be the case because the largest of them is a straight line. So what parameter determines the size of the curve?
It seems from looking further into it that the assumption is the start of the curve must be vertical. I can't find any reason justifying this in the maths.
Recall from the VSauce video that it didn't matter where you start on the curve, it takes the same amount of time to travel to the bottom. This means that larger cycloids don't affect travel time, all you have to do is match the points up.
Travelling from any point between the vertical and your finish point on the same curve takes the same time but travelling on a cycloid with a different curvature takes a different amount of time, so the diameter of the circle which forms your cycloid makes a difference and it can vary infinitely with only two points to define it. There must be a third point or a tangent or another delimiter which defines a fixed cycloid for two given points.
Right, if you start at different points on the SAME cycloid, it takes the same time. But a cycloid formed from a 1cm circle and a cycloid formed from a 1Km circle will have differing travel times. The answer to my original question by the way is that you take the cycloid from the vertical to the required angle because there is no initial hoizontal movement so it falls out of the derivative that the initial motion is vertically down. It was in Bernouli's solution.
Question: even when starting vertically, can there be more than one diameter of a Cycloid to hit the other point? My intuition says no, because of degrees of freedom, but is this correct? (Another thought - if there is more than one, then I imagine there is a fastest one which I would think will be the one where you hit point B where the tangent is most horizontal as if not to waste time going down and then up again...)
The brachistochrome analysis regarding the fastest time from point a to point b is an astonishing problem - not the shortest distance from point a to point b. Both friction and gravity allows the brachistochrome analysis to solve the process. I wish I would have learned such a prospective when I was far younger, from a mental shift.
I would like to answer your question about why the circle rotates at a constant rate while making the cycloid . It is because every point in the the rotating circle has two velocities: one is translational which is constant and the other velocity which is wr where w is rotational velocity and r is radius of the circle . The angle between them changes constantly resulting in changing resultant is velocity . It is zero at the point of contact with the surface and highest at the top of the rolling circle
My educated guess for your challenge is that it minimizes the change in energy. I remember in engineering school we had some examples that were similar where the integral of the work done was minimized when it was a straight line. This also seems to make physical sense in that light will want to minimize its energy state.
About your challenge... I do not love to be dunned and teased by foreigners.
haha...that was a good 1
Hadn't Newton already solved it when he said that?
Gorky
Well, then he must be Newton.
And also the claw thing
The challenge for me is not to find the solution to the question, it’s just to comprehend the question itself.
Really neat! I wish I had this video when I took physics to learn the connections between different concepts.
did you figured that out?
MindYourDecisions hello fresh tall walker
Shit message
hahahahaha!!
Who has the cleverer community? You or 3Blue1Brown?
Just wanna say that all your videos are beautiful, well edited and animated, insightful, and a joy to watch. Thanks for the awesome content, 3Blue1Brown.
I rarely ever comment on RUclips, but I wanted to say that your content is amazing and most importantly thought provoking!
Keep up the good work and thank you for the time you put into your videos!
+MegaAproductions Thanks so much!
+3Blue1Brown superb video but one thing i still confused about is why you replace the speed of light in different refractive index with square root of y?because i think this only apply if the particle initial velocity is zero and was accelerated only with gravity whereas light initial velocity in vacuum is about 300,000km/s
@@3blue1brown hello
its still 1 comment on this channel. they werent lying
@@cringium shit yeah
Do you challenge me because you think you've got an unusually clever solution, or because you think I'm Newton?
+Larzsolice No need to be Newton to do math ;)
+Larzsolice lol
+Gna Nay Depends on the math
None of these both. And that is why you can do better.
haha! Favorite comment.
i have an excellent proof for this but this comment section is too small to contain it
the math geek in me laughed out loud :D genius!
Fermat's Last Theorem :)
Rafa Martínez-Avial find your owns copycat
That made my day XD
lol who dragged Fermat into this
The best math in youtube by far
Baby you have to google the legendary problem number 6
I find the final chanllage is actually solved in this video. stop the video at 12:34, note the velocity of piont P = √(2gy)(obtained at 8:00), so the angular velocity of P rotate around C is just √(2gy)/(D Sin(θ)) = √(2g/D). thus the curves in t-θ plane is always stright lines with slope √(2g/D).
besides, note θ=1/2 ∠POC (let O be the center of circle), so "P rotate around C with constant angular velocity" is equivalent with "P rotate around O with constant angular velocity", so the wheel rotate with constant rate.
This solution might interest the skateboarding community! Does any skateboard park exploit the solution to the Brachistochrone problem, I wonder?
or rollercoasters for that matter!
Yeah,now that I think about it skateboard tracks do look like a cycloid
omg I totally thought that while watching the video :)
It would definitely be a very smooth, efficient transition. It would give the greatest speed for a given ramp height.
@@brainmind4070 It gives the shortest time, but the speed is independent of the path (being proportional to the square root of the distance dropped).
Solution to the challenge:
Consider an infinitesimally small time interval dt, when the velocity of the pi-creature is v at angle theta with the vertical line.
In this time interval, the pi-creature descends a height of dy = v cos(theta) dt (1).
By conservation of kinetic energy, v^2 = 2gy (y being the total height the pi-creature has descended).
By Snell's law, sin(theta) = Cv (C is a constant).
Take the derivative of both equations above:
2v dv = 2g dy (2)
cos(theta) d(theta) = Cdv (3)
Putting (1) (2) (3) together, we have d(theta) = Cg dt, so the theta-t function is a straight line :D
Nice one!!
A good formal prove. But the teaser asked for an intuitive model for the (t, theta) space. Also Snell's law shouldn't be used as a given, but should be the consequence of the shortest path optimisation in that space.
I thought he was asking for another solution to the problem as in another shortest-time curve other than the cycloid.
Thank you for this clear proof!
The animation at 4:39 ~ish using a space filling curve is magnificent, and for such a one off display is magnificent
This channel has a superb standard of videos and never fails to present new insights into old problems. Thanks for all the effort!
This is amazing, honestly keep it up I need more insight into certain math problems and the history that surrounds them
Thanks! Usually, I don't focus on the history of problems, so I feel Steve was a real addition to the channel for making that part of the focus. I'll probably do more things like this in future videos as well now.
Yeah my mind basically exploded when snells law could be applied, such an amazing use of a particular physical law to reduce a seemingly challenging unrelated problem down to its bare mechanics
+Kenneth Goodall Right!?
@@3blue1brown this legend about Newton solving in one night is pretty ridiculous,
no need to divinise a man who was pretty horrible human and pretty stupid when outside pure science (trying all his life to obtain the philosophical stone and making numerology from the bible among other thing)
Makes me feel special that I was subscribed to you before Vsauce. And I love that more people are tuning in, you need more exposure, your videos are simply the best math videos on the net.
Was subscribed here before VSauce too .. I'm a patron !
strontiumXnitrate lol, the worst kind
The answer is "no"
no, I can't :)
N4w4k You have to *believe*
If I made it, so can you !! 👊💪 Hahahaha
why not 'Yes'? "Yes, I'll marry you, Math' xD
If you have a paper, and you try to use the point C like 3b1b does in the video, you will get an equation like v=2cosθ√gR and if you are correctly using point C, you could get that point C is moving in constant speed and thus the whole circle is moving at constant speed.
"Newton stayed up all night....solved it..."
Meanwhile me: *_"I stay up all night and create problems out of thin air"_*
Lol
Honestly, if the legend is true, then this was absolutely badass.
He was SIR ISAAC NEWTON
Leibniz did it as well.
It is interesting to note that Huygens in 1672 showed by graphical means, that a cycloid was a tautochrone, that is that starting at any point on the curve an object sliding along it will reach the bottom in the same time, thus a cycloid should be the ideal path for a pendulum on a clock.
Hugens partly got to the solution as the result of a good guess, the mathematics of a cycloid is pleasingly simple and given that Mersenne had already showed that a circle didn't quite work, the cycloid was the next obvious and easy curve to try.
The two problems are clearly linked and I am sure that Newton knew this and proceeded along similar lines.
Interestingly to make a tatochrone pendulum all you need to do is have a flexible string half the arc length of the desired cycloid holding the bob and then cut a solid cycloid of the desired length in half, placing each half either side of the pendulum (curves down small ends meeting at the pivot flat halves away from the pivot). As the pendulum wraps around the cycloid halves the bob traces a tautochronous cycloid beneath. Or, rather than two halves of a cycloid you can use a cycloidic taurus with an internal diameter of 0 for a rather pleasing effect.
SPACKlick I get the first part of your comment, but what's a cycloidic Taurus? Google doesn't yield any useful results.
But Newton was pretty darn savage tho
Cycloidic taurus: I think, a donut shape, where instead of the cross-section having a circular shape, it has a cycloid. That way, the pendulum line can come through the hole in the middle, and the pendulum can swing in any direction.
spur3 oh I see, so the hole would be just a point. I think it's torus rather than Taurus, Google was confused 😝
Because of the gravitational force , the curve taken by the object is cycloid. U have to consider it. Because of earth is not a perfect sphere , the path taken by the object is cycloid rather than arc of the circle. If we plot our assumption to this conclusion, we can prove the straight lines over the r v/s theta curve
Strogatz, the chaos man. His Nonlinear Dynamics lectures made life much easier for me.
This mean time is constant, but distance and velocity is variabele
I know it may be late but I was looking for a satisfactory explanation of the warm up challenge in the comments and I could not find any. As so I have tried solving it myself and I finally came to a possible solution.
First we must label θ as the angle between the vertical at the center of the circle and the point on the circle's circumference which is tracing the cycloid. We can find the tangential velocity at that infinitesimal point in time if we consider the point of contact C of the circle and the horizontal surface as the instantaneous center of rotation and by using *v = wr* (Eq. 1). If we label the point where the particle is as L, then we can see this becomes *v = wl* (Eq. 2) where l is the length CL. Using the cosine rule we can see that the length l is simply *l = 2Rsin(θ/2)* (Eq. 3).
We can also equate the velocity if we consider conservation of mechanical energy since there is no friction. Thus K.E. = P.E. of the particle which is *1/2 mv^2 = mgy* (Eq. 4) where y is the vertical distance from the horizontal surface to the point L. This reduces to *v = sqrt(2gy)* (Eq. 5).
Substituting Eq. 3 and Eq. 5 into Eq. 2 yields *sqrt(2gy) = 2wRsin(θ/2)* which can be squared to become *2gy =4 w^2 R^2 sin^2 (θ/2)* (Eq. 6). We can substitute y here with the parametric equations of the cycloid. If we take the y axis downward as positive then it varies slightly that the one in the video as it will be *y = R(1 - cosθ)* (Eq. 7). Substituting into Eq. 6 gives us *2gR(1-cosθ) =4 w^2 R^2 sin^2 (θ/2)* (Eq. 8).
Lastly, we can use the trig identity *sin(θ/2) = sqrt((1-cosθ)/2)* (Eq. 9) and transform Eq. 8 into *2gR(1-cosθ) =4 w^2 R^2 ((1-cosθ)/2)* from which we can cancel out the various term to reduce into *g = R w^2* which finally yields *w = sqrt(g/R)* (Eq. 10). This shows that the angular velocity is constant and that the particle's trajectory due to gravity does follow that of the trajectory of a rotating wheel with constant rotation. I hope this helps anyone in the future who might revisit this wonderful video.
I now need to figure out how to work "brachistochrone" into a conversation.
You're insanely intelligent, and an inspiration to me. I'm going to try this.
If there's one thing I think I've learned from your videos, it's that nothing consistent is off-limits and there is always a different reference frame from which to view your problem.
I heard that you were doing a essence of calculus series. You should consider also doing a essence of multivariable calculus. Multivariable is a class that many people take in all of maths, engineering and science and is surprisingly poorly understood. Thanks! :)
Perhaps I will in the future. I did many videos on MVC for Khan Academy (khan style, not 3b1b style), which you may want to check out.
What aspect of MVC do you see as being poorly understood?
pretty much everything, basic facts about the gradients don't seem to be even remember by lots of people I know and things like Hessian, why its a matrix, jacobians etc seem to be words lots of people don't actually understand.
yea! I saw those, they are nice, but the most powerful style is 3b1b style ;)
Hell yeah!!
"Heeyy, Vsauce, Michael here" brought me here.
I wonder when he's gonna kill himself.
Marie Loiseau wat
Same
Same bro
@@___xyz___ whyy??
Glad you're posting more videos. These are some of the most beautiful on RUclips.
I was really in doubt about this interview concept when you first presented it, but after just 30 seconds with talking and animation in perfect unison, I was convinced. Great job!
Mads Nielsen Mind washing with pretty pictures...☺️-Don't worry, this kind of phenomenon occurred to me, too, (for another video). Since that, i am more focused on formal audible arguments, than on icons following icons.
0:42 that joke right there earned an instant sub .... also the face that he timed it so it hit exactly at the 42 seconds mark .... this guy is awesome!
?
I love the math and graphics and the cute pi sliding around
Holy crap, Strogatz! You mean the author of the book I loved in undergrad, Nonlinear Dynamics & Chaos! Loved my Dynamical Systems class in undergrad, but damn...it was a beast! Much much much respect for these mathematicians who pioneer the way in these fields, but even more so for those who can beautifully, eloquently, and effectively communicate these advanced concepts in math to plebeians like myself; such is the man you interviewed here: Steven H. Strogatz.
I have research on this topic. Can you help with university references and notes? Do you have a research plan for this problem? I need some references on this topic
7:38 Case Sensitivity:
The g in the potential energy term, mgy, is not "the gravitational constant." The gravitational constant, G, is the proportionality constant in Newton's law of universal gravitation that relates the attractive force to the masses involved and their relative position. G = 6.67E-11 m^3/kg•s^2. You are referring to g, which represents the acceleration due to Earth's gravity measured at the surface, 9.8 m/s^2.
An innocent slip as I know you know this.
Another great video!
When I learn about Brachistochrone problem, I was very surprised and excited that this small question makes a whole new concept "Calculus of Variation." But this video takes me to next level of excitement. Really thought provoking and interesting.
I watch your videos all day, I watch for knowledge and for fun! I can't get enough of this! Your expository is like Nothing I have ever seen before. I'm about to start my PhD in applied Mathematics.... I feel deeply blessed that I came across your channel. You sir, are a breath of fresh air.
easily one of the coolest math ch. vocal , video quality, use of math language... everything real nice
Could you have an intuitive way to see that "rate of change" and "area under the curve" are opposites?
I'll do that and much more in the "Essence of calculus" series.
Can't wait for that. Your Essence of algebra is brilliant! Thanks for what you're doing.
_That_ will be amazing!
One way you could think of it is that the function is the rate of change of the area under itself. Imagine the graph of a function, and pick two points in the X axis. You can see the area under the curve between these two points right? Now what if you nudge one of them just a bit? The area changes just a little, but you can see that, the higher up the function is at the nudged point, the more the area in that interval will change.
In other words, how much the area under a curve changes is not only based on how much the length of the interval changes, but also on the value of the function at the end points.
Another explanation for those more familiar with discrete mathematics is that an integral is just an infinite sum of differences. If you have something like
f(n) = g(n) - g(n-1)
What you basically have is that f is the analogous to the "discrete derivative" of g (difference in height divided by difference in length). So what happens when you sum all the consecutive values of f(n) from 1 to, say, m?
f(1) + f(2) + f(3) + ... + f(m)
(g(1) - g(0)) + (g(2) - g(1)) + (g(3) - g(2)) + ... + (g(m-1) - g(m-2)) + (g(m) - g(m-1))
-g(0) + g(1) - g(1) + g(2) - g(2) + g(3) - g(3) + ... + g(m-1) - g(m-1) + g(m)
And you can see the all the terms, except for g(0) and g(m), cancel out, and you end up with
g(m) - g(0)
This is a little more difficult to imagine on a continuous function, but the principle is the same (taking it to the limit to infinity and yadda yadda yadda). And, of course, the first value of n doesn't have to be 1, that was chosen for simplicity.
Out of these two, I prefer the first reasoning just because it's more intuitive, but the second one shows why antiderivatives have indefinite constants: if it's a constant term, it's going to be the same in g(m) and g(0), so they'll cancel out. Therefore, it doesnt matter which constant term you choose, there is no "right" one.
Whew, that was a wall of text. Sorry about that, but I wanted to make it as clear as I could in a single comment without edits.
dy / dx = f(x)
=> dy = f(x) dx
The RHS is the area of a tiny sliver of the curve
=> sum up all the slivers and get
integral of dy= integral of f(x) dx
=> y = F(x)
The first step isn't technically mathematically correct but I think it's intuitive enough.
I must say, I love this channel, you are almost souly responsible for kindling my love of math, thank you
The animation is so good! his must take you such a long time to do, this was a very fun video to watch
This will be a hand wavy solution to the question but the way I see it, the straight line in the theta-time space indicates a constant rate of change in theta which is the smoothest possible path to the solution as a straight line is always the shortest path: i.e. add any more or less curvature than that exact curve and you'll distort the theta-time curve creating a lower than optimal area under the curve of that graph. Background is only a B.S. in math though so take my intuition with a few grains of salt. Cheers to the work you do, and the tools you've developed - it's all much appreciated!
As a Cornell alumnus, I know Prof. Strogatz is very famous among math and science students for being a great educator, if you could not tell from the video. It's always a pleasure to see/listen to his work.
If you can help me with a plan for this research as well
Saw your video yesterday, really great :)
*Here is the answer to your challenge*:
particle moves such that speed v is proportional to sin(theta).
Differentiate it: rate of change of change of speed is proportional to cos(theta)* rate of change of theta.
But rate of change of speed is component of gravity along the curve = g cos(theta).
Hence *rate of change of theta = constant*.
Who's here after the vsauce video?
Yep. But I had already seen it before so does it count?
I have already seen this channel but not this particular video...
I actually saw 3Blue1Brown's brachistochrone vid long before Vsauce's take on it. I'm just here to see if any Vsaucers ended up here.
same
yup haha
8 years later, and that challenge is still on my mind. Not solution, but it got me thinking about all sorts of other variations on the problem for different force terms.
3b1b: challenge the audience
Audiences: meme about newton
I saw this problem in Physics olympiad national qualifier 25 years ago. Of course, I couldn't even tough the starting point. It has been forgotten for more than 2 decades, and suddenly pop up in front my eyes. One of the biggest mysteries of my life is solved. Thank you!
Here is a solution to your challenge. Acknowledging that the time-minimizing trajectory is linear in the theta-t space, one of the big implications is that angular velocity is constant. BIG. Another property to consider is that, as was mentioned previously in the video, energy is proportionate to the square of velocity, or in most differential cases acceleration. If theta-t is linear, the theta-dot is constant, this means that there is no rotational acceleration being imparted on the system and all of the energy gained from gravity can be focused on tangential velocity, which is what gets us from point a to point b. In this way, like removing friction from a system, we are removing rotational acceleration from the energy equation thus increasing our tangential velocity and, given a path of constant length, this allows us to arrive at the destination the quickest.
Note, I think I've been saying tangential through this, I guess what I mean more is cartesian. Tangential implies a circular path.
10:24 this is also given in Thomas calculus . I have the Indian edition, page 654
Really awesome solution, I paused the vide and tried to solve it myself. My first instingt was to use calculus, but I didn´t do very good. However this is how far I came and if someone could help me to continue this calculation it would be really nice.
The variable that we want to minimize is the time, and the variable we are looking for is the function between the two points. So we need a relationship between the time and the function. We know that t=s/v, there is a general way to work out the distance between two points on a function. And it says that the distance from the point x=a to x=b on the curve f(x) is the integral from a to b of the function sqr(1+(f´(x))^2). The distance between the points doesn´t really matter so I´m going to say that it is one just to make things easier. This means that s=0integral1(sqr(1+(f´(x))^2))dx
We know that v=k*sqr(f(x)) in any given point, but since the speed changes over time we can not plug it in to the formula. Also I´m going to say k=1 just to make things easier. It won´t affect the final curve any way. Correct me if I am wrong but the average speed between x=0 and x=1 is going to be 0integral1(sqr(f(x)))dx.
So if we plug this into our formula t=s/v we get t=(0integral1(sqr(1+(f´(x))^2))dx)/(0integral1(sqr(f(x)))dx). So I have a formula for the relationship between the time and the function. Now I just have to find the derivative and set it equal to zero I guess, it will probably end up with i differential equation. There is just a little problem, I have no idea of how to take the derivative of that function, and I don´t even know if it is right for that matter. Is there even an algebraic way to do it? If not, then I have no idea about how to solve this problem algebraically. It would be easier to find the derivative if the variable was x, but it not is f, we are looking for an entire function here! Does anyone know how to continue this solution?
en.wikipedia.org/wiki/Calculus_of_variations
The knowledge you need to learn is "calculus of variations". You need to use the Euler-Lagrange equation.
+Samuel Lo ok So if I understod it right then I now have to solve this equation:
d/df((0integral1(sqr(1+(f'(x))^2))dx)/(0integral1(sqr(f(x)))))-d/dx(d/df'(x)((0integral1(sqr(1+(f'(x))^2))dx)/(0integral1(sqr(f(x)))))=0
Because I am assuming that what they reference as L is the same as t in my equation. But still I can't solve that.
I had propably made a misstanke with the perenthases in the equation but what it says is dt/df-d/dx(dt/df')=0 I just replaced t with what I know that t is equal to in termes of f.
We need to use the instantaneous speed, not average speed.
v = sqr(f(x))
ds = sqr(1+(f'(x))^2) dx
dt = ds/v = sqr(1+(f'(x))^2) / sqr(f(x)) dx
Total time = integral [ sqr(1+(f'(x))^2) / sqr(f(x)) ] dx
L = sqr(1+(f'(x))^2) / sqr(f(x))
L is a function of f(x), f'(x)
Then we do the partial derivatives: ∂L/∂(f(x)) and ∂L/∂(f'(x))
Finally, solve the differential equation:
d/dx (∂L/∂(f'(x))) - ∂L/∂(f(x)) = 0
Great video, I just have one small issue.
At 5:55 the video shows a ray of light bouncing off a mirror in an attempt to "minimize" the travel time from A to B, while Steven Strogatz in the interview says "somehow nature is imbued with this property of doing the most efficient thing".
But clearly the most efficient thing would be to not bounce off the mirror in the first place and to go directly from A to B.
What Fermat's principle states is rays of light traverse the path of *stationary* optical length with respect to variations of the path. In other words, a ray of light prefers the path such that there are other paths, arbitrarily nearby on either side, along which the ray would take almost exactly the same time to traverse. There can be cases where light can be fooled into to taking the longest path!
+Chris Shannon Great point! We never emphasized that it's a local minimization.
+Chris Shannon How do you arrange to get the longest path?
+Ralph Dratman Great question.
Consider a circular mirror. Draw the horizontal and vertical diameters. Consider two points, A and B, on the horizontal diameter equidistant from the centre. Light will travel from A to the top of the circle, call it point O, and reflect to B. Perfectly symmetric. Now draw an ellipse with foci A and B, that goes through O. It will be tangent to the circle at O and will be outside the circle otherwise.
Ellipses have the property where the sum of the distances from a point on the ellipse to the two foci is constant. It follows that any point inside the ellipse will have a total distance to the two foci less than that of any point on the ellipse. Since the circle is entirely inside the ellipse it follows that the rays of light took the longest path possible, i.e., any other point on the circle would have been a shorter path.
This realization that the light doesn't *choose* the shortest path addresses the religious aspect mentioned in the video (as well as the fact that you can't just use light to solve the travelling salesman problem for example). In the path integral formulation of quantum theory light chooses *all* paths, but most end up cancelling each other out. When a path's nearest neighbours have almost the same length then they add constructively. These paths are called *stationary*, and can occur when they are minimum or maximum, either locally or globally. The reason the light reflects off the point O in the example above is best explained because the circle is tangent to the ellipse (and therefore all the nearest neighbours have nearly constant path lengths).
+Chris Shannon Really cool. Thanks for the explanation!
Chris Shannon I have a quibble with your longest path discussion. If we want to find or verify a putative shortest path, we can test to show that the chosen path is shorter than any other path, including any arbitrary wiggly path. But a putative longest path cannot be longer than every arbitrary wiggly path. It may, of course, as you demonstrate, be longer than any other stationary path.
At least, so it seems to me.
"By the claw the the lion is revealed"
3Blue1Browns conversation-with-mathematicians voice is the same as his narration voice. I am stunned and amazed.
Awesome. Your videos are fantastically well-made!
What a brilliant story. As you say, awe inspiring. When we learned Snell's law, it was taught without reference to that path being the fastest, hence the path light will naturally take. The facts were taught but not the fundamental principle or reason. Of course that leads to other questions.....
It's because time and space are an illusion...
Great visuals/graphs in the video... You are really great at teaching!
Exactly what is "it" in your comment? How can an illusion predict the mathematics of a curve? Calling time and space an illusion is only true in the philosophy of nondualism, in which pure consciousness is all that truly exists. But such an insight cannot possibly predict any of the laws of nature, including the equations of curves having certain properties. In other words, you have taken a true statement in one model of reality and erroneously applied it to another.
I know nothing about higher math yet I love watching 3Blue1Brown videos. Imaging how these concepts occur in the world is why I watch. This concept reminds me of alpine skiing and skateboarding.
one of the best maths/physics videos i have ever seen
Are you gonna cover a solution to the challenge question at some point?
So Newton started solving it grudgingly, knowing he would solve it anyway, and then published it anonymously just to get done with the challenge. It's like a first grader challenging a math major to prove the pythagorean theorem that he just came up with, thinking he came up with something hard.
6:48
"but for now, all you need to know"
come ON, i wanna know it all!!!
I read Steven's book 'infinite powers' when I was in high school and you know what... the way he explained it like a story, it felt more like a novel than a book on calculus...he is really an insightful writer....
Physicist (upon seeing the solid normal line at 7:05): REEEEEEEEE
I qualify but don't understand.
@@FlyingSavannahs Normal lines are generally drawn as dotted lines, but the general may differ from place to place!
Ok. That's news to me.
I read Strogatz, Nonlinear Dynamics and Chaos.
Revisiting this after veritasium video😊❤
Your channel is pure culture to enthusiasts of math like myself. Pure gold. Please do not stop.
5:28 I loved your animations of light rays that included the oscillations of the wave itself. Very clever
My preferred and the most elegant solution I've seen to the brachistochrone is by that of the calculus of variations in which is kind of pops out.
I'd like to see Newtons solution
It's pretty standard honestly. If you tried to solve the problem elementally , you would most likely wind up with Newton's solution.
The answer can be found in the book "Newton's Principia for the Common Reader" by S. Chandrasekhar, pp. 571-578. He points out that Newton had already laid the groundwork for solving this problem in the first part of the Principia.
Here, hold my apple...
Strogatz truly is a fantastic communicator of math. Granted, I can only speak from using his book on nonlinear dynamics, but I still got so much enjoyment out of his way of writing. The subject itself is an eye opener and really rewarding, but because I couldn't attend lectures and had to rely on the book, I was very happy to experience such an enjoyable read. He writes to the reader, and it makes all the difference.
I watched this way before vsauce, man i feel special!!
Yet another beautiful presentation of a very clever proof. It's been a very nice surprise to hear Steven Strogatz participating on one of your videos!
Wait, so what was Newton's solution? Was it the same as Johann's?
can't defeat funorange
Probably same conclusion but different method(?)
en.wikipedia.org/wiki/Brachistochrone_curve
I'm with Typo
What I absolutely loved about this video is, light is shown as a particle and as a wave! xD
Woohoo, new video!!!!!!!
Snell's Law is an example that Bartosz Milewski gives of the kind of global thinking that is taken to the extreme in category theory, with its universal constructions. A math instructor of mine also mentions that these kinds of global optimizations (optimizing with the telescope, rather than the microscope, you might say) are done in calculus of variations. I find this duality of local vs global approaches to optimization to be quite interesting.
This is a fantastic explanation. I'm taking precalc this year, and I can intuitively understand this. That is so cool haha. Came from Vsauce. It was an alright video, but he never really explained the reasoning behind the property, which was what I wanted to know. Glad he introduced me to this channel though.
if the shortest path is a curve of a circle, the tangent line of that curve changes angles at a constant rate as you follow each point on the curve. in other words if you look at two points on a circle that are x distance away, the difference in their tangent line angles is the same as any two points separated by x distance. if the formula for slope is (y2 - y1) / (x2 - x1) substituting distance for x and theta(angle) for y the slope is constant throughout.
therefore theta changes at a constant rate in respect to time.
This video is really great! Why cannot I watch it when I was an undergraduate! The first time I read the details for the solution of Johann Bernoulli was from Ernst Mach's Science of Mechanics.
I have research on this topic. Can you help with university references and notes? Do you have a research plan for this problem?
I am by no means a mathematician and my understanding of physics is rudimentary, but when you started talking about expressing the problem in term of time over theta, space time diagrams kept popping in my head. every curve or motion is actually a straight line in space time. I don't know maybe i am, reaching but i feel like there is a connection here.
This is not rigorous and I'll probably revisit this later, but I do have a general idea:
The lateral force imparted on the object due to gravity is gsin(theta). As the slope is constantly changing, mapping the axis to sin(theta) renormalizes the curve in a way that reduces it to a straight line, as it undoes the encoding of the function. In other words, the function does not look linear on a normal cartesian set of coordinates is because the coordinate system does not factor in the constantly changing lateral force of gravity. By mapping the function into a set of coordinates that cleverly cancels out the constantly changing force of gravity, the optimization problem becomes linear.
I also want to thank you for making these videos. I rarely find videos or textbooks that explains things intuitively, and the animations help so much in this regard.
This is specific to the video @ 11:48. There should have been a visualisation of tangent always meeting bottom of circle with changing point of contact C. That took time for me to visualise!
Sorry, if that's a comment appears to be rude.
Is there a way to support you?
His patreon link where you can get an early look at videos yet to be released:
www.patreon.com/3blue1brown
Aida Bit I think the patreon account wasn't a thing back then.
No. But you can support the channel 3Brown1Yellow.
Thanks!
Cool challenge. Now I'm going to be procrastinating my homework, and it's all your fault! D:
+1ucasvb
If it's maths homework, tell the teacher! It should be fine. :D
@@ZardoDhieldor I'm not sure about that tho.
Me too xD
According to the idea of Bernoulli, if an object is tracing a brachistochrone
v=k\sin\theta.
Since the gravity is vertical, the slope component of the gravity is
a=g\cos\theta.
This is the acceleration of the object (of unit mass).
We know that the acceleration is the derivative of the velocity. Hence
d/dt(k\sin\theta)=g\cos\theta.
LHS of the above is
k\cos\theta d/dt(\theta).
Hence
d/dt(\theta) = constant function.
This is the end of the proof for the challenge 3Blue1Brown proposed, provided that I understand the challenge well.
Your videos are inspirational. They have allowed me to see mathematics, or as Keith Devlin would say, "mathematics has made the invisible visible." And your videos do that for me in a unique way that has not been replicated by any other learning source. I am studying math education, and I would actually like to learn how to make videos like this so that I can make learning more enjoyable for my students. I'm not much of a programmer, but I'd be willing to learn. Is it a trade secret or can you point me in the right direction? Thanks, Grant.
-Ben
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He said he uses Python, just watch some online course about python and google everything you don't know
or you can just google his github ;)
This was very fun to watch! Actually a nice recap with some additional information about a calculus of variations lecture some time ago. Thanks for that!
You say "challenge", I say "homework"
This explantion is so clear and nice, also good and easy to see.
Really love it.
where do you get ideas for your videos? I imagine it takes awhile to organize your thoughts and results to make something you're content with
Well, I keep a growing list of different interesting topics I think where I think there's room to provide a new, or at least under-discussed, perspective. Unfortunately I can't always spare the time to knock items *off* the list as quickly as they get added.
visualisation of calculus of variations? please?
I have only the warmest of feelings for Strogatz after listening to his Great Courses course on chaos.
8:07 "The velocity in the first one is v1" - me: "rotate"
positive rate gear up
I understand your comment because of Mayday.
@@NoriMori1992 we don't care
windows_x_seven
Did I do something to you? I'm pretty sure I didn't.
funny how this problem sneaked into modern society
e.g. as a part of scouts training, to find the path from A to B using map, compass, and math of course, that takes the least time and travel it to literally prove it
I've seen several videos on this today (started at Vsauce). An infinite number of cycloids, with varying diameters of original circle, can be fit to two points. Are of of them solutions to the Brachistochrone problem? That cannot be the case because the largest of them is a straight line. So what parameter determines the size of the curve?
It seems from looking further into it that the assumption is the start of the curve must be vertical. I can't find any reason justifying this in the maths.
Recall from the VSauce video that it didn't matter where you start on the curve, it takes the same amount of time to travel to the bottom. This means that larger cycloids don't affect travel time, all you have to do is match the points up.
Travelling from any point between the vertical and your finish point on the same curve takes the same time but travelling on a cycloid with a different curvature takes a different amount of time, so the diameter of the circle which forms your cycloid makes a difference and it can vary infinitely with only two points to define it. There must be a third point or a tangent or another delimiter which defines a fixed cycloid for two given points.
Right, if you start at different points on the SAME cycloid, it takes the same time. But a cycloid formed from a 1cm circle and a cycloid formed from a 1Km circle will have differing travel times.
The answer to my original question by the way is that you take the cycloid from the vertical to the required angle because there is no initial hoizontal movement so it falls out of the derivative that the initial motion is vertically down. It was in Bernouli's solution.
Question: even when starting vertically, can there be more than one diameter of a Cycloid to hit the other point? My intuition says no, because of degrees of freedom, but is this correct? (Another thought - if there is more than one, then I imagine there is a fastest one which I would think will be the one where you hit point B where the tangent is most horizontal as if not to waste time going down and then up again...)
The brachistochrome analysis regarding the fastest time from point a to point b is an astonishing problem - not the shortest distance from point a to point b. Both friction and gravity allows the brachistochrome analysis to solve the process. I wish I would have learned such a prospective when I was far younger, from a mental shift.
First time I'm seeing a video with no dislike.
+Morel EBELLE Some idiots ruined it
I ruined it :(
you can always "undislike"
I love your videos. The quality is impeccable and the content is inspiring but i just can't watch them yet.
0:44
Off topic but he looks like J.S Bach
Is it just me?
I saw him and I thought exactly the same thing. Maybe they all looked the same back then
I would like to answer your question about why the circle rotates at a constant rate while making the cycloid .
It is because every point in the the rotating circle has two velocities: one is translational which is constant and the other velocity which is wr where w is rotational velocity and r is radius of the circle . The angle between them changes constantly resulting in changing resultant is velocity . It is zero at the point of contact with the surface and highest at the top of the rolling circle
My educated guess for your challenge is that it minimizes the change in energy. I remember in engineering school we had some examples that were similar where the integral of the work done was minimized when it was a straight line. This also seems to make physical sense in that light will want to minimize its energy state.
your videos rule dude, that part on light as a solution was like, genuinely mind blowing
Bernoulli : *challenge mathematicians*
Newton: so you have chosen death