The Brachistochrone, with Steven Strogatz

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  • Опубликовано: 21 ноя 2024
  • Steven Strogatz and I talk about a famous historical math problem, a clever solution, and a modern twist.
    ------------------
    3blue1brown is a channel about animating math, in all senses of the word animate. And you know the drill with RUclips, if you want to stay posted about new videos, subscribe, and click the bell to receive notifications (if you're into that).
    If you are new to this channel and want to see more, a good place to start is this playlist: • Recommended
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Комментарии • 1,3 тыс.

  • @fenhirbr
    @fenhirbr 8 лет назад +4575

    About your challenge... I do not love to be dunned and teased by foreigners.

    • @300483rahul
      @300483rahul 8 лет назад +68

      haha...that was a good 1

    • @gorkyrojas3446
      @gorkyrojas3446 7 лет назад +258

      Hadn't Newton already solved it when he said that?

    • @zbzb-ic1sr
      @zbzb-ic1sr 7 лет назад +43

      Gorky
      Well, then he must be Newton.

    • @tbssen36
      @tbssen36 6 лет назад +16

      @Gorky OOOOOOOOOOHHHHHHHH SNAP

    • @jmbbao
      @jmbbao 6 лет назад +4

      And also the claw thing

  • @car-keys
    @car-keys 8 лет назад +365

    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.

  • @MindYourDecisions
    @MindYourDecisions 8 лет назад +1316

    Really neat! I wish I had this video when I took physics to learn the connections between different concepts.

  • @MegaAproductions
    @MegaAproductions 8 лет назад +484

    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
      @3blue1brown  8 лет назад +95

      +MegaAproductions Thanks so much!

    • @yonathan4194
      @yonathan4194 8 лет назад +7

      +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

    • @AdityaKumar-ij5ok
      @AdityaKumar-ij5ok 4 года назад +3

      @@3blue1brown hello

    • @cringium
      @cringium 3 года назад +6

      its still 1 comment on this channel. they werent lying

    • @AchillesHeel_1
      @AchillesHeel_1 3 года назад +2

      @@cringium shit yeah

  • @RexGalilae
    @RexGalilae 7 лет назад +2471

    i have an excellent proof for this but this comment section is too small to contain it

    • @Juul32
      @Juul32 7 лет назад +140

      the math geek in me laughed out loud :D genius!

    • @megarafa1199
      @megarafa1199 7 лет назад +158

      Fermat's Last Theorem :)

    • @jamesvangogh4125
      @jamesvangogh4125 7 лет назад +4

      Rafa Martínez-Avial find your owns copycat

    • @0xhba
      @0xhba 7 лет назад +16

      That made my day XD

    • @micayahritchie7158
      @micayahritchie7158 7 лет назад +46

      lol who dragged Fermat into this

  • @unalcachofa
    @unalcachofa 8 лет назад +178

    The best math in youtube by far

    • @daviddonoso6577
      @daviddonoso6577 6 лет назад +5

      Baby you have to google the legendary problem number 6

  • @hreader
    @hreader 8 лет назад +677

    This solution might interest the skateboarding community! Does any skateboard park exploit the solution to the Brachistochrone problem, I wonder?

    • @old-man-two-ears
      @old-man-two-ears 7 лет назад +93

      or rollercoasters for that matter!

    • @shreyashgadute2938
      @shreyashgadute2938 6 лет назад +59

      Yeah,now that I think about it skateboard tracks do look like a cycloid

    • @tald989
      @tald989 6 лет назад +16

      omg I totally thought that while watching the video :)

    • @brainmind4070
      @brainmind4070 6 лет назад +23

      It would definitely be a very smooth, efficient transition. It would give the greatest speed for a given ramp height.

    • @bpark10001
      @bpark10001 5 лет назад +51

      @@brainmind4070 It gives the shortest time, but the speed is independent of the path (being proportional to the square root of the distance dropped).

  • @Larzsolice
    @Larzsolice 8 лет назад +2269

    Do you challenge me because you think you've got an unusually clever solution, or because you think I'm Newton?

  • @maigowang
    @maigowang 6 лет назад +48

    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

    • @medamineelmoussaoui
      @medamineelmoussaoui 3 года назад

      Nice one!!

    • @micknamens8659
      @micknamens8659 3 года назад +18

      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.

    • @masterchiefer25
      @masterchiefer25 7 месяцев назад

      I thought he was asking for another solution to the problem as in another shortest-time curve other than the cycloid.

  • @Avose243
    @Avose243 8 лет назад +20

    This channel has a superb standard of videos and never fails to present new insights into old problems. Thanks for all the effort!

  • @N4w4k
    @N4w4k 8 лет назад +680

    The answer is "no"
    no, I can't :)

    • @N3ONLUV
      @N3ONLUV 7 лет назад +21

      N4w4k You have to *believe*

    • @johannbcmelo
      @johannbcmelo 6 лет назад +3

      If I made it, so can you !! 👊💪 Hahahaha

    • @СимеонДимитров-ц5с
      @СимеонДимитров-ц5с 5 лет назад +2

      why not 'Yes'? "Yes, I'll marry you, Math' xD

    • @ershuanglu3345
      @ershuanglu3345 4 года назад

      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.

  • @DweebsUnited
    @DweebsUnited 5 лет назад +5

    The animation at 4:39 ~ish using a space filling curve is magnificent, and for such a one off display is magnificent

  • @kennethgoodall8980
    @kennethgoodall8980 8 лет назад +29

    This is amazing, honestly keep it up I need more insight into certain math problems and the history that surrounds them

    • @3blue1brown
      @3blue1brown  8 лет назад +15

      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.

    • @kennethgoodall8980
      @kennethgoodall8980 8 лет назад +10

      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
      @3blue1brown  8 лет назад +15

      +Kenneth Goodall Right!?

    • @brevnobia1238
      @brevnobia1238 3 месяца назад

      @@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)

  • @mcdudelydoo3116
    @mcdudelydoo3116 2 года назад +3

    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.

  • @theghostmachine
    @theghostmachine 7 лет назад +68

    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.

    • @Binyamin.Tsadik
      @Binyamin.Tsadik 7 лет назад +1

      Was subscribed here before VSauce too .. I'm a patron !

    • @Binyamin.Tsadik
      @Binyamin.Tsadik 7 лет назад +2

      strontiumXnitrate lol, the worst kind

  • @Harinichandrasekaran-qg5nt
    @Harinichandrasekaran-qg5nt 2 месяца назад

    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.

  • @donaldasayers
    @donaldasayers 8 лет назад +41

    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.

    • @SPACKlick
      @SPACKlick 7 лет назад +5

      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.

    • @ThePotaToh
      @ThePotaToh 7 лет назад +1

      SPACKlick I get the first part of your comment, but what's a cycloidic Taurus? Google doesn't yield any useful results.

    • @RexGalilae
      @RexGalilae 7 лет назад +1

      But Newton was pretty darn savage tho

    • @spur3
      @spur3 7 лет назад +4

      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.

    • @ThePotaToh
      @ThePotaToh 7 лет назад +3

      spur3 oh I see, so the hole would be just a point. I think it's torus rather than Taurus, Google was confused 😝

  • @jesseacummins
    @jesseacummins 8 лет назад +3

    Glad you're posting more videos. These are some of the most beautiful on RUclips.

  • @chetan5848
    @chetan5848 5 лет назад +11

    Strogatz, the chaos man. His Nonlinear Dynamics lectures made life much easier for me.

  • @isaackay5887
    @isaackay5887 2 года назад +2

    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.

    • @nadjlatoual2553
      @nadjlatoual2553 Год назад

      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

  • @brikken1
    @brikken1 7 лет назад +11

    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!

    • @claudecourvoisier1001
      @claudecourvoisier1001 7 лет назад

      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.

  • @c.j.awesome1937
    @c.j.awesome1937 7 лет назад

    easily one of the coolest math ch. vocal , video quality, use of math language... everything real nice

  • @renesax6103
    @renesax6103 8 лет назад +215

    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! :)

    • @3blue1brown
      @3blue1brown  8 лет назад +137

      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.

    • @RalphDratman
      @RalphDratman 8 лет назад +11

      What aspect of MVC do you see as being poorly understood?

    • @renesax6103
      @renesax6103 8 лет назад +27

      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.

    • @renesax6103
      @renesax6103 8 лет назад +20

      yea! I saw those, they are nice, but the most powerful style is 3b1b style ;)

    • @belalsherif553
      @belalsherif553 3 года назад +1

      Hell yeah!!

  • @SliversRebuilt
    @SliversRebuilt 8 лет назад +14

    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.

  • @Math_oma
    @Math_oma 8 лет назад +33

    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.

    • @nadjlatoual2553
      @nadjlatoual2553 Год назад

      If you can help me with a plan for this research as well

  • @StephenRoseDuo
    @StephenRoseDuo 8 лет назад +27

    I love the math and graphics and the cute pi sliding around

  • @GBabuu
    @GBabuu 7 лет назад +8

    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.

  • @elmarkaz
    @elmarkaz Год назад +1

    This mean time is constant, but distance and velocity is variabele

  • @formerlypie8781
    @formerlypie8781 8 лет назад +6

    I must say, I love this channel, you are almost souly responsible for kindling my love of math, thank you

  • @kimchi_taco
    @kimchi_taco Год назад

    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!

  • @oskarlindelof9685
    @oskarlindelof9685 8 лет назад +12

    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?

    • @samuel0100
      @samuel0100 8 лет назад +2

      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.

    • @oskarlindelof9685
      @oskarlindelof9685 8 лет назад +2

      +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.

    • @samuel0100
      @samuel0100 8 лет назад +1

      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)

    • @samuel0100
      @samuel0100 8 лет назад +1

      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

  • @kazul333
    @kazul333 2 года назад

    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!

  • @FlyingSavannahs
    @FlyingSavannahs 4 года назад +21

    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!

  • @mihailazar2487
    @mihailazar2487 7 лет назад +1

    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!

  • @femioyekan8184
    @femioyekan8184 7 лет назад +192

    "Newton stayed up all night....solved it..."

    • @slolilols
      @slolilols 4 года назад +64

      Meanwhile me: *_"I stay up all night and create problems out of thin air"_*

    • @mrroshan1960
      @mrroshan1960 4 года назад +1

      Lol

    • @mann1aswal
      @mann1aswal 4 года назад +20

      Honestly, if the legend is true, then this was absolutely badass.

    • @maxwellsequation4887
      @maxwellsequation4887 4 года назад +7

      He was SIR ISAAC NEWTON

    • @DC-zi6se
      @DC-zi6se 7 месяцев назад

      Leibniz did it as well.

  • @avadakedavra80
    @avadakedavra80 6 лет назад +1

    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*.

  • @aronhegedus
    @aronhegedus 8 лет назад +6

    The animation is so good! his must take you such a long time to do, this was a very fun video to watch

  • @spelunkerd
    @spelunkerd 8 лет назад

    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.....

  • @jagadeeshanms2863
    @jagadeeshanms2863 23 дня назад +3

    Revisiting this after veritasium video😊❤

  • @aniruddhasrivatsav1217
    @aniruddhasrivatsav1217 11 месяцев назад +1

    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

  • @patrickhodson8715
    @patrickhodson8715 8 лет назад +73

    Could you have an intuitive way to see that "rate of change" and "area under the curve" are opposites?

    • @3blue1brown
      @3blue1brown  8 лет назад +57

      I'll do that and much more in the "Essence of calculus" series.

    • @saitaro
      @saitaro 8 лет назад +2

      Can't wait for that. Your Essence of algebra is brilliant! Thanks for what you're doing.

    • @patrickhodson8715
      @patrickhodson8715 8 лет назад +2

      _That_ will be amazing!

    • @wqferr
      @wqferr 8 лет назад +7

      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.

    • @benjwils
      @benjwils 8 лет назад

      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.

  • @anLTproduction
    @anLTproduction 6 лет назад

    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.

  • @eeshan3955
    @eeshan3955 7 лет назад +465

    Who's here after the vsauce video?

    • @zairaner1489
      @zairaner1489 7 лет назад +26

      Yep. But I had already seen it before so does it count?

    • @dhruveshpatel1109
      @dhruveshpatel1109 7 лет назад +3

      I have already seen this channel but not this particular video...

    • @ganaraminukshuk0
      @ganaraminukshuk0 7 лет назад +9

      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.

    • @WilliametcCook
      @WilliametcCook 7 лет назад

      same

    • @InsaneGamesYT
      @InsaneGamesYT 7 лет назад

      yup haha

  • @timetraveller1237
    @timetraveller1237 8 лет назад +1

    one of the best maths/physics videos i have ever seen

  • @wisdom-for-life
    @wisdom-for-life 4 года назад +33

    It's because time and space are an illusion...
    Great visuals/graphs in the video... You are really great at teaching!

    • @david203
      @david203 2 года назад +1

      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.

  • @567secret
    @567secret 6 месяцев назад

    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.

  • @davtor33
    @davtor33 8 лет назад +4

    Awesome. Your videos are fantastically well-made!

  • @dcterr1
    @dcterr1 Год назад

    Wow, very mind-bogglingly elegant solution! I'll have to work on the challenge now!

  • @mikepublic111
    @mikepublic111 8 лет назад +31

    I now need to figure out how to work "brachistochrone" into a conversation.

  • @ShimshonDI
    @ShimshonDI 6 лет назад +2

    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.

  • @FishKungfu
    @FishKungfu 7 лет назад +204

    "Heeyy, Vsauce, Michael here" brought me here.

  • @PurusharthSaxena
    @PurusharthSaxena 6 лет назад

    What I absolutely loved about this video is, light is shown as a particle and as a wave! xD

  • @hectornonayurbusiness2631
    @hectornonayurbusiness2631 8 лет назад +160

    I'd like to see Newtons solution

    • @theflaggeddragon9472
      @theflaggeddragon9472 7 лет назад +17

      It's pretty standard honestly. If you tried to solve the problem elementally , you would most likely wind up with Newton's solution.

    • @eranz1
      @eranz1 6 лет назад +26

      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.

    • @paulovianna1882
      @paulovianna1882 3 года назад +2

      Here, hold my apple...

  • @amoghshukla634
    @amoghshukla634 Год назад +1

    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....

  • @fancmeng7681
    @fancmeng7681 2 года назад +3

    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.

    • @nadjlatoual2553
      @nadjlatoual2553 Год назад

      I have research on this topic. Can you help with university references and notes? Do you have a research plan for this problem?

  • @iamnorwegian
    @iamnorwegian 7 лет назад +1

    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.

  • @Kabitu1
    @Kabitu1 8 лет назад +36

    Are you gonna cover a solution to the challenge question at some point?

  • @alvarol.martinez5230
    @alvarol.martinez5230 8 лет назад

    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!

  • @levitheentity4000
    @levitheentity4000 4 года назад +3

    6:48
    "but for now, all you need to know"
    come ON, i wanna know it all!!!

  • @PauloRicardo-fc4li
    @PauloRicardo-fc4li 5 лет назад

    Your channel is pure culture to enthusiasts of math like myself. Pure gold. Please do not stop.

  • @小刚-b2j
    @小刚-b2j 3 года назад +1

    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.

  • @Palisade5810
    @Palisade5810 7 лет назад +5

    I watched this way before vsauce, man i feel special!!

  • @thearchive26788
    @thearchive26788 4 года назад

    I love your videos. The quality is impeccable and the content is inspiring but i just can't watch them yet.

  • @oaxis8198
    @oaxis8198 4 года назад +73

    3b1b: challenge the audience
    Audiences: meme about newton

  • @cyanide7833
    @cyanide7833 2 года назад

    the challenge at the end is just tooooo good, like its really interesting and exciting just think and ponder on it

  • @dave5194
    @dave5194 7 лет назад +6

    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.

  • @miguelsal3000
    @miguelsal3000 4 года назад +2

    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.

  • @david-paulschulze1957
    @david-paulschulze1957 4 года назад +10

    "By the claw the the lion is revealed"

  • @m_tahseen
    @m_tahseen 4 года назад

    Never knew about the history of the problem .. great job digging into the origin of the problem

  • @Labdominals
    @Labdominals 2 года назад +3

    5:28 I loved your animations of light rays that included the oscillations of the wave itself. Very clever

  • @timothydavis2568
    @timothydavis2568 2 года назад

    3Blue1Browns conversation-with-mathematicians voice is the same as his narration voice. I am stunned and amazed.

  • @li5up6
    @li5up6 3 года назад +3

    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.

  • @donaldwallace9110
    @donaldwallace9110 6 лет назад

    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.

  • @RTPZinana
    @RTPZinana 8 лет назад +4

    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.

  • @tobiasonasch2056
    @tobiasonasch2056 8 лет назад

    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!

  • @theinvisiblearmadilloofdea6204
    @theinvisiblearmadilloofdea6204 8 лет назад +15

    Woohoo, new video!!!!!!!

  • @GaudioWind
    @GaudioWind 8 лет назад +1

    Loved the video. Congratulations. I wish I had every math and physics class like this.
    As for the first part of the challenge, I imagined that we can describe the movement of the particle from two different perspectives. Firstly, from a reference frame that would be moving horizontally along with the center of the circle that rotates under the horizontal line to form the cycloid. In this reference frame, the particle simply describes a portion of circle and the points A and B move horizontally. Well, if we suppose that this circle rotates at a constant speed, then the acceleration of the particle would obviously be pointing towards the center of the circle and constant, mv^2/(D/2), where v is relative to this reference frame. The second frame would be static relative to the cycloid curve. So, in this second frame, the particle would be subjected to its weight (mg) and the normal force (mv^2/r), where now, r=D sin(theta) is the instantaneous curvature radius. Well, it happens, that if we calculate the resulting force, taking the weight and the normal force that we observe in this second frame, considering the geometry and trigonometry, we will see that it is also constant, mg, and all the time pointing towards the circle center too. Since both frames would be inertial and the observed force could be the same if we choose a convenient angular speed omega for the circle, then we can say that the movement of the particle would be the same either if it describes a circle with constant rotational speed equals to sqrt(g/2D) or falls on a cycloid curve under gravitational force, mg.

    • @mcdudelydoo3116
      @mcdudelydoo3116 2 года назад

      I know this is 6 years late but could you explain how exactly you got that the radius of curvature is D sin(theta)?

    • @GaudioWind
      @GaudioWind 2 года назад

      @@mcdudelydoo3116 Thanks for the question. In this reference frame, the instantaneous radius of curvature is no longer the radius of the original circle, but rather the segment connecting P and C in the figure at 10:48 in the video. Note that the yellow line, which is the particle's trajectory, is always perpendicular to this CP line. It is as if the particle moves along a new mutable circle whose center is C, but the position of C changes all the time, as does the size of the radius of this new circle. So you can see that CP relates to the diameter of the original circle as sin(theta).

  • @FunOrange42
    @FunOrange42 8 лет назад +56

    Wait, so what was Newton's solution? Was it the same as Johann's?

    • @rxijin7602
      @rxijin7602 7 лет назад +3

      can't defeat funorange

    • @typo691
      @typo691 7 лет назад +9

      Probably same conclusion but different method(?)

    • @Sivolc11
      @Sivolc11 6 лет назад +11

      en.wikipedia.org/wiki/Brachistochrone_curve

    • @johannbcmelo
      @johannbcmelo 6 лет назад

      I'm with Typo

  • @anthonycannet1305
    @anthonycannet1305 8 лет назад +1

    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.

  • @morelelfrancel6603
    @morelelfrancel6603 8 лет назад +8

    First time I'm seeing a video with no dislike.

  • @김상민-j1z
    @김상민-j1z 5 лет назад

    This explantion is so clear and nice, also good and easy to see.
    Really love it.

  • @codenamelambda
    @codenamelambda 8 лет назад +62

    Is there a way to support you?

    • @aidabit7554
      @aidabit7554 7 лет назад +12

      His patreon link where you can get an early look at videos yet to be released:
      www.patreon.com/3blue1brown

    • @codenamelambda
      @codenamelambda 7 лет назад +9

      Aida Bit I think the patreon account wasn't a thing back then.

    • @krelly90277
      @krelly90277 2 месяца назад

      No. But you can support the channel 3Brown1Yellow.

  • @wecantwaitinpism.2454
    @wecantwaitinpism.2454 4 года назад +1

    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.

  • @benjaminhanson6137
    @benjaminhanson6137 8 лет назад +10

    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

    • @filipsperl
      @filipsperl 8 лет назад

      .

    • @motaaaa
      @motaaaa 7 лет назад

      He said he uses Python, just watch some online course about python and google everything you don't know

    • @HxTurtle
      @HxTurtle 6 лет назад

      or you can just google his github ;)

  • @jkrasney1
    @jkrasney1 3 года назад

    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.

  • @sa8lvi
    @sa8lvi 7 лет назад +3

    I am so happy for you that vsauce discovered your video xD

  • @sirilandgren
    @sirilandgren 4 года назад

    I have only the warmest of feelings for Strogatz after listening to his Great Courses course on chaos.

  • @josevillegas2721
    @josevillegas2721 8 лет назад +3

    Wow, i am amazed by every aspect of this video!!!; what software do you use to make these animations.

  • @richardray7976
    @richardray7976 2 года назад

    Brilliant! These videos of yours are just outstanding!

  • @1ucasvb
    @1ucasvb 8 лет назад +28

    Cool challenge. Now I'm going to be procrastinating my homework, and it's all your fault! D:

    • @ZardoDhieldor
      @ZardoDhieldor 8 лет назад +5

      +1ucasvb
      If it's maths homework, tell the teacher! It should be fine. :D

    • @leelaanandabhavan3083
      @leelaanandabhavan3083 4 года назад +1

      @@ZardoDhieldor I'm not sure about that tho.

    • @elejelly3986
      @elejelly3986 3 года назад +1

      Me too xD

  • @Availablecookie
    @Availablecookie 2 года назад

    I’m not sure if I’m looking at the problem you gave us correctly but I love puzzles. It looks like a straight line because t-0 graph is measuring how long it takes to get into a certain position each starting at zero. If you look at the graph side ways where t=y and 0=x. The end points makes a curve.

  • @ch6339
    @ch6339 8 лет назад +3

    I don't want to bother anyone but the "snell's law" was actually discovered at the same time in France by Descartes and by Snell, that's why in french we often call it the snell-descartes' law, not to be chauvinist or anything, I'm just saying... 😂😂😂😂

    • @sirploom
      @sirploom 4 года назад

      sorry for the 4yrs late responsee, butt, Snell discovered the law in 1621, Descartes was just the first to publish it in 1637, I think Descartes even learnt it from Snell, not sure about that thoughh, that's the reason you guys call it loi de Snell-Descartes

    • @ch6339
      @ch6339 4 года назад

      @@sirploom oh that's good to know ! Thanks 😉

    • @sirploom
      @sirploom 4 года назад

      @@ch6339 no problem! Glad I could help!

  • @jamesraymond1158
    @jamesraymond1158 3 года назад

    You can also put the start and end points anywhere on the surface of the Earth, so that as you slide down on the path, the pull of gravity decreases the deeper you go. In this case, the solution is also a cycloid, but a cycloid in which the circle rolls around the curved surface of the Earth (a circular cycloid). For example, if the two points are on opposite sides of the Earth, then the path traced by a circle rolling around the interior of the Earth would be a straight line, as expected.

  • @NorthernWhisper
    @NorthernWhisper 7 лет назад +4

    So if light works this way and physical objects work this way.. would sound waves follow the same curve under any circumstances?

    • @tempestaspraefert
      @tempestaspraefert 7 лет назад

      Actually, any kind of wave, right? :)

    • @duophile7692
      @duophile7692 7 лет назад +2

      Tempestas Praefert Any kind of wave that refracts due to a change in medium and proportionally to that change in medium, yes.

    • @jackismname
      @jackismname 7 лет назад +1

      Northern whisper ASMR an interesting place where this takes place is during earthquakes, where the waves refraction in a continous manner through the earths layers which are denser near the center

  • @Siryj26
    @Siryj26 8 лет назад

    your videos rule dude, that part on light as a solution was like, genuinely mind blowing

  • @robinsparrow1618
    @robinsparrow1618 7 лет назад +3

    The circle used to graph the brachistochrone is rotating at a constant rate, so it makes sense that it would give a straight line in the angle v time graph.

  • @reptilejesus829
    @reptilejesus829 5 лет назад +1

    Great video - you keep interesting me in things that I had no idea about

  • @wilhelmsarosen4735
    @wilhelmsarosen4735 5 лет назад +7

    Physicist (upon seeing the solid normal line at 7:05): REEEEEEEEE

    • @FlyingSavannahs
      @FlyingSavannahs 4 года назад

      I qualify but don't understand.

    • @aditimuthkhod1252
      @aditimuthkhod1252 4 года назад

      @@FlyingSavannahs Normal lines are generally drawn as dotted lines, but the general may differ from place to place!

    • @FlyingSavannahs
      @FlyingSavannahs 4 года назад

      Ok. That's news to me.

  • @ArjunBahuguna
    @ArjunBahuguna 8 лет назад +2

    I love you. This is brilliant. Going to devour Levi's book asap.

  • @Siryj26
    @Siryj26 8 лет назад +13

    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

    • @3blue1brown
      @3blue1brown  8 лет назад +36

      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.

    • @chapinward
      @chapinward 8 лет назад +3

      visualisation of calculus of variations? please?