Mathing, by Meher Md Saad, is a channel for the math enthusiasts which aims to provide animated math videos which are not only interesting but also easy to grasp. The videos are largely made using animation, so if you're into these kind of videos, consider subscribing to stay connected for more videos in the future.
I think you'd like Harmony Bloom, a music sequencer plugin that employs similar math/art circular patterns to create complex musical sequences. Its visually striking and quite musical.
If you note you will find pattern in everything like pattern in sorrow pattern in joy pattern in happiness pattern in music pattern in admiration pattern in hateable stuff pattern in pattern which clearly mentions life is too mean choosing over one opinion no time to lose and accept all and much more to learn 😢❤❤❤❤❤
It's brilliant how you decided to explain such a topic that we all have learned and forgotten, but to think twice again in the real world about the importance of spending that time learning it. We should always think about applying math in nature or at least in our imagination
You could make a more beautiful argument for why the slope is -1. Assume some point in time t, we have some intermediate square, but at this point in time the problem is absolutely identical to the starting problem, just renormalize the square lengths. So it means decreasing the length of the square 2 times from the starting point would take the twice the amount of time as decreasing it twice from this position (the velocity is essentially twice as much) - in other words the drop rate is constant - i.e a straight line
10:19 honestly I'm a bit disappointed that the "climax" is just skipped over, so here's my approach to calculating the limit: L = lim_(t->0) (sqrt((S-t)²+t²)-S)/t Let's define f(x)=sqrt((S-x)²+x²) Now, because f(0)=sqrt((S-0)²+0)=sqrt(S²)=S, it becomes clear that L=lim_(t->0) (f(0+t)-f(0))/t. But that's exactly the definition of f'(0). And so, all that's left to get the solution is to keep applying differentiation rules until the answer pops out: f'(x) = (2x - 2(S-x))/(2sqrt((S-x)²+x²)) = (2x - S)/sqrt((S-x)²+x²) And so, f'(0) = (0 - S)/sqrt((S-0)²+0) = -S/sqrt(S²) = -1
There were two questions (the distance and the shape) and only one answer (distance equals 1a). I chose the delay to be a constant fraction of the current square side size. More precisely 1/n of it (1/n of current a). Then it quickly became a convergent infinite geometric series. The sum can be easily calculated using "S=a0/(1-q)" since q is between 0 and 1. The limit of the sum when n approaches plus infinity is 1a. This way you never reach the situation where your current square is of size t units and it no longer decreases. You also don't need to know the solution ahead of proving it. Slope of -1 is a prove not the answer. And I did all that in memory while I paused your video to think about it. Truth be told I think I've seen this problem before and I suspected the answer is 1a about half way into the calculation. The shape of the curve is a different story........
Very engaging, thanks for the video. In my case I use the simmetry between the four dogs, each dog and his "objective" dog forms a rectangular triangle, so the speed is always is at a 45º from the radius to the centre. The speed toward center point is 1/sqrt(2), and because the intial distance from the centre is sqrt(2), the time to reach the center is 1s.
Very late, but hope my comment will help someone. There’s also a solution which requires to use polar coordinates. I remember at physics Olympiad school we pushed this question to the limit and here’s what we’ve got: the path of a dog is a logarithmic spiral (if I remember it correctly) which means that the number of rotations around the centre of a figure (not necessarily a square) is INFINITE!! But using some integrals (or common sense) we can clearly see that the time till the dogs meet is FINITE. This little paradox can be easily explained using the definition of a material point - we neglect the size of a dog. And since we consider the dogs to be infinitely small, it then becomes clear that it would take an infinite number of rotations for them to meet. If anyone interested you can ask for some math behind it in a comment below
My issue with this is it would seem to make the number of spirals and thus the distance each dog traveled dependent on its size. i.e. smaller dogs would make more rotations and thus go farther than larger dogs would before meeting. This is clearly not true from the answer to the problem.
I solved this using scale invariance. First, find that after a time step t, it's the same problem but with a square of length sqrt((1-t)^2+t^2)=(some expession of t, let's call it l). But now there's a problem: with a smaller square, they're moving faster relative to the square size. We fix this by scaling the time steps to the length of the square: t, then t×l, t×l^2... to get squares of length l^n after each time step. The total time is a geometric series with sum t/(1-l). Next, take time limit as t→0 to find that the fraction approaches 1. Coincidentally, it's almost the same limit as in the video.
My intuition was to make each step a function of the side length, and that made a geometric series. Each intermediate square was the same fraction of the previous. Luckily, those infinite series converge. It ended up being a refresher into how geometric series work for me, and covered one of the first 'interesting' math questions I had as a teen: when exactly do they start to converge? My answer was even more inefficient than yours, but a very interesting journey for me.
Relativity: The problem is equivalent of 4 dogs standing at the corners of a shrinking and rotating square. Since rotation doesn't contribute to their distances, each dogs moves the side of the original square. (Assuming that each dog step and rotation is infinitesimal small 🧑🎓).
I found this problem listed as a challenge math problem on a university website a long time ago. They used snails instead of dogs but its the same problem. I wanted to find the equation of the path walked. I used complex analysis to solve it and it took me a VERY long time to figure it out.
While I like seeing the formalism of taking the limit, you can get an initiative feel for why the slope is -1 by noting that at the very start of any step, each dog is moving exactly parallel to the edge and thus reduces the length by the exact amount that it moves forward.
this is so cool. I assume you use Manim for these animations, but manim is always frustrating for me because of the lack of real-time visualization, is there any tool you use for this?
I feel the same when it comes to manim. That's a reason why, manim wasn't my first go-to tool, also another reason being I couldn't master it to my liking (still know nothing of how to do text animations). What I usually did was use a more familiar and easy to visualize thing- Adobe After Effects (Yes a video editor, but I found you can use some programming there called After Effects Expressions which I found easier). At least that's how I made the first dozen of my vids. But the problem with this approach was that it's really limited, and certainly not built for the purpose of math animations. So I'm thinking of trying a new thing called p5.js. This is very easy to visualize and easy to grasp in my opinion. For this video tho, I wanted to do whatever I can using manim, so I generated the figures using After Effects first, then tried doing the same in manim, later added the texts and the Pentagon cartoon whose name I haven't decided in a whole damn year.. using After Effects.
@@Mathing OH COOL i tried processing too but i got demotivated by how high level it was and then got distracted by trying to go super low level and use like opengl which was way too tedious ill look back into it, ty!
My solution to the shape was quite different: I first noticed that regardless of how fast each dog was going, the shape is the same. So dogs moving at 3m/s creates the same path as at 1m/s. Next I noticed that the paths of the dogs must be the same, so you only need to compue the path for one. This also lead me to realize that to get from one dog to the next, you just convert (x,y) to (-y,x). So the dog at (x,y) is always moving towards the dog at (-y,x). This allows you to build some equations: x_n+1 = (end - start)Δt x_n+1 = ((-y_n) - (x_n))Δt and y_n+1 = (end - start)Δt y_n+1 = ((x_n)-(y_n))Δt these nicely convert into differential equations that look like dx = (-x-y)dt dy = (x-y)dt unfortunately from here, you have to be able to actually solve these with a bit of knowledge on the subject, but to work out the less painful steps- d *x* = A *x* dt where *x* = [x,y] and A = [[-1 -1], [1 -1]], then loosely using notation, you can see 1/( *x* ) d *x* = Adt ∫1/( *x* ) d *x* = ∫Adt ln| *x* | = At+C *x* = Ce^(At) Unfortunately, working it out from here is pretty gross if you wanted to prove everything along the way like what it means to take e to some matrix exponent. PS: For those curious about matrix eponentiation (since the internet is a bit lacking in terms of information on the topic). The steps to derive the result are to: - rewrite the function as it's taylor series expansion - diagonalize the matrix to make it easier to handle with higher powers - switch the order to, instead of taking the sum of matrices, have a matrix of the sums - un-taylor series-ify the result since this works from the ground up instead of applying a formula, it also requires you to know about how matrices are diagonalized and why they work like that for maximum understanding
I have another question. I hope my intuition is working correctly but I'd love to get proven wrong, as that's how I learn the fastest Is it true that a ray casted from the center of the square crosses any (and all) dogs path infinitely many times?
I really like efficient solutions, so this video lowkey pains me 😭. The argument that the inefficient solution is beautiful can still be made, but a lot of the time the beauty can be explained by the efficient solution.
11:12 the explanation of this is just take the derivative of the two variables with respect to time, and you get a constant for both, so their relationship is linear.
I remember doing something like this in Minecraft (with command blocks and armor stands). I remember when i made a cardioid and was like: "oh what's that? looks like a heart" (i was ~13 yo)
I viewed the setup as a shrinking, rotating square, so I wrote everything in polar coordinates and set up some differential equations based on the restrictions from the problem.
I think it's easy to prove, that every dog moving always normal to vector of speed of every neighbor. So, every dog is moving with constant speed relate to right neighbor. Sooo path is strictly equal to starting distance. Trajectory is much more interesting task
by changing reference frames it's easy. Take the top two dogs, and fix those on a line (you can always create a line between two points) because the left dog only moves down, and the right dog moves to the left, the distance must be closed by the dog on the right that is moving to the left.
I'd PDE the living heck out of it. Formulation is pretty straightforward: due to the symmetrical nature, take one point and set its moving trajectory dependent on its own coordinates, specifically on a rotation around the center of the square. Since the resulting DE will contain a function like sqrt(x^2 + y^2), polar coordinates is the way to go, and a singularity is expected when points meet (a hint on the behavior of DoF, likely being of a logarithmic nature towards the center). Rest is a routine. Solving analytically, unlike observing patters, includes all possible solutions without guessing work nor heuristic assumptions. The main advantage is that you won't miss anything at all.
