I'm really glad to see this "classical" problem getting solved once again =) The last time I saw it was in my 9th grade at a certain physics olympiad school, when the solution didn't invoke any limits or line integrals, since we didn't even know what a derivative was (and weren't supposed to). The proposed solution was based upon noticing the fact that at any moment any two adjacent dogs travel perpendicular to one another, hence the distance between them shrinks at a speed equal to one of the dogs' velocity, which is 1 unit/s and is constant. Therefore the time it takes before all the dogs meet is equal to 1 unit (the initial distance between 2 adjacent dogs) divided by this speed, which amounts to 1 second. And since the dog's velocity is constant, the required distance is just time times velocity, in this case the answer is 1 unit. I think this might be one of the solutions you found online =) There was also a slightly more complicated version of this problem, with the room being in the shape of an equilateral triangle (or in general an n-sided regular polygon) and with 3 (or n) dogs in the corners respectively. Maybe you'd want to cover this one in one of your next videos =)
Yep, this was the solution I referred to as being too clever & short hehe. About the triangular case, you can actually solve this puzzle for any regular n-gon using the very method in this video! This will include the same steps for example creating a graph of sidelength vs distance, proving that the graph is a straight line, finding relation between two consecutive sidelengths to get the slope.. But in this particular case, the slope won't be equal to -1, it'll just turn out to be a different value instead.
@@Mathing Yeah, I thought so as well. Then here's a couple of different pursuit-type problems that might catch your interest. They're fascinating in their own way, but the soution isn't as straightforward. 1. A rabbit is running along a straight road with a constant speed of 10 m/s. A fox is trying to catch the rabbit by running with a constant speed of 20 m/s, but its velocity vector is always pointing at the rabbit. Initially the fox is standing 300 m from the road and then starts running when the rabbit is 300 m away from the fox. Find the time it takes for the fox to catch the rabbit. 2. Two ships - A and B - are departing simultaneously from two ports located on a straight coastline at a distance L from one another. Ship A is moving perpendicular to the coast with a constant speed, whereas ship B has the same constant speed as ship A, but its direction is always towards ship A. Find the distance between the ships after a long time. As a bonus you can try to derive the trajectories in both problems, but the solutions don't require that. In fact, the answers actually look pretty nice =) Also in the n-gon problem there can be an additional question about the initial accelerations of the dogs and the initial curvatures of their trajectories, so that's that for a change =P
About the triangular case: That is the form of the problem I have first seen also in preparation for the Physics Olympiad. The way I solved it: All velocity vectors summed up at any instance will be zero (for all constellations not only triangular obviously) hence the "center of mass" of the system will not move. Therefore we know that our dogs or whatever will meet in the center (if they ever do). Then we notice that the situation is the same as perceived from any of the dogs. That is to say all connecting lines between the dogs (how many there might be) must remain in the starting configuration bc we can just copy what one dog sees to the other rotating according to the direction it's heading(they might be scaled as the problem does not have any dependency on the dogs distances, but only directions). Now it is rather straightforward: We may take the rotating frame of reference, then all dogs are moving straight to the center with constant velocity (one decomposes the initial velocity into radial and tangential components) and finds the distance to the center. It is nice to see such a seemingly complex problem to have such an easy (to calculate) solution. No saying that your way is not cool as well especially the bit finding the geometric manifestation of the derivative ; ) just probably unlikely to pull off in an exam.
I made a video that generalizes this problem and solve in detail the system of differential equations tha arises to derive the exact equation of the trajectory. I also show where those spirals appear. Look it up on my channel, it is called : Zeno's Mice (Ants) Problem and the Logarithmic Spirals.
I think there is a more intuitive solution. Observe that: 1. Each dog is moving towards another dog, which is moving perpendicular to it. (for example, at the start it is easy to see the velocity vectors are perpendicular and stay this way due to symmetry) 2. Perpendicular motion is well-approximated by a circular motion. Thus, dog A is moving towards dog B, while dog B is moving along a circular arc around it (for small time intervals). Thus the distance between A and B gets shorter by 1 unit/sec as A comes closer to B, while B is trying to stay at the same distance. Since they approach each other at 1 unit/sec, they will meet in 1 second. (the approximation works well on a thin segment of a disk)
Congratulations that you got it! In fact, this was the very solution I referred to as 'short' and 'elegant'. A more general version of the same solution can be found here: www.theguardian.com/science/2019/jun/03/did-you-solve-it-dogs-in-pursuit
now I feel clever, this is what I came up as well! Though I will say, the revelation that the distance between the dogs closes at a constant rate only gave me the distance traveled and the general sense that the radius of the curve must linearly decrease, not an actual shape. I had no idea how to construct that curve, which the video's method gets for free...though as I type this I realize I'm dumb and you just plot the curve in the polar plane
@@grantofat6438 from symmetry we know that the dogs form a square. From the statement of the problem we know that their velocities are pointed toward each other. So we have 4 velocities pointed along the sides of the square. This implies the velocities are perpendicular.
this is not an approximation, 1 is an exact answer, and to simplify your explaination, i can provide a more intuitive explaination imagine you are one of the dogs, and you think that you are stationary (even when the dog is moving), another dog is approaching you directly in a straight path, meaning that the reletive velocitty of the dog wrt to you is always v towards you, and distance between you and the dog is d, t = d/v
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.
Two additional thoughts: 1. If t is negative, we'll get the curve in the opposite direction, i.e. a spiral outwards, continuing the logarithmic spiral pattern outwards, with the center missing. 2. Wikipedia's article (en.wikipedia.org/wiki/Mice_problem ) has mistakenly stated that it will take "infinite time" to approach the center. This is impossible because the distance is finite, and we're moving at constant speed.
1) The value of 't' being negative never occured to me, it would surely look cool. Thanks for pointing it out! 2) I found the same thing, but yes they were wrong.
That mistake had been introduced by someone on August 2 and, fortunately, been corrected again by August 28. en.wikipedia.org/w/index.php?title=Mice_problem&action=history
Hi, just tried out what the graph looks like when t = - 0.2 and yes, it is as you said, a spiral outwards for infinity: drive.google.com/file/d/1GsIMW-I-PUjiBRjOXvcfzrWSuGZSBSp-/view?usp=sharing
The correct thing here is that the spirals will wind an infinite number of times around the center. This confuses some people, since it does so in finite time and distance.
Complex numbers make this easier. By symmetry, each dog's motion is one 90 degree rotation with respect to the previous one. If we place the center of the square at the origin of the complex plane, then b = ia, c = ib, d = ic, a = id, where a, b, c, d are complex-valued functions. Hence we have a' = b - a = ia - a = (-1+i)a so a = e^((-1+i)t), i.e. an exponential spiral Note that I'm removing the "constant speed" constraint here, in order to find the paths more easily.
I've seen this puzzle before. The result is not surprising at all when you consider this: Let's say dog A is walking towards dog B. Since B is always moving at a right angle to A's direction of travel, it is neither getting closer nor farther from A. So, for A, it's just as if B doesn't move; and so A walks 1 unit to get to B. That's not a proof but it's sure compelling, I think.
Very nice video! I'll note that the limit at 10:20 can found in a few ways; it's a nice exercise for L'Hopital's rule or, you can do it with elementary methods with the difference of two squares.
I feel like there should be some nice argument to prove the linearity of the distance vs time using the temporal symmetry of the problem (the system looks very alike at any point in time up to rescalling)
@@Mathing The problem is the constant speed, which gets ever bigger relative to the square as the square shrinks. You can drive this by considering smaller time steps.
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.
There's a very nice physics-y way to get the answer. We already know by symmetry the answer must be in the center of all 4 of them (otherwise rotating 90° would change the answer but otherwise leave the problem unchanged) so we can just think about how long it takes to reach the center. To do that we find the radial velocity of any dog by drawing a radial line and projecting the velocity on to it (fortunately since the shape stays the same, the radial velocity does too). The radial velocity is cos(45) = sqrt(2)/2 and the radial distance is sqrt(2(l/2)^2) = lsqrt(2)/2. In terms of theta that's l/(2cos(theta/2). Dividing the gives the time required to get to the center
If you rotate the figure by 45 deg witt origin at the center of the square, a very straight forward differential equation is easy to derive and the result is a well known curve.
if you compute it with radial coordinates it's even easier because you prove that n^r=θ and that the derivative at the point (0;1) is equal to -1 without using differential equations.
@@lorenzodiambra5210 Nice. But do you get the length of the curve? There is also a simple orthogonality argument that says the length of the path = one side of the square. No equations or arithmetic.
My solution/proof for this would be that every dog is moving towards a dog that is moving perpendicular to it at all times, therefore the distance only changes by the amount the first one is moving. The target dog doesn't ever change it's distance from the follower, only the follower changes distance, therefore the distance would be exactly the same as at start, 1 unit. Edit to add, I hadn't read any other comments or watched the whole video when I wrote this. I see others came up with essentially the same solution.
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.
Solving a system of differential equations (made fairly simple in polar coordinates) allows one to calculate the equations for the dog's path as r(t) = (1-t)/sqrt(2) θ(t) = π/4 - log(1-t).
Beautiful proof! I solved this many year ago, when there was no Internet... I used polar coordinates and was quite simple when you realize the 45° constant!
you can write a differential equation In the coordinates of a particular dog using the fact that the next dog is always ninety degrees rotated from the current dog because this whole system has to keep fourfold. then take derivative of the radius using the chain role and observed that is constant. unlike the solution using perpendicularity this will easily generalize to higher polygon.
After 3:04 I just saw there triangles and pythagorean theorem, so my solution was just Lim [t->0] { √( t² + (1-t)² ) } (what would happen if dog's brain lag was getting shorter and shorter, so it creates new triangles and calculates new distance faster and faster). And it just goes to 1 straight. upd: probably i messed something up, but can't understand that
when you think about it, it is actually quite obvious why the length has to be 1 unit. Since every dog is moving straight towards another dog and the other dog is only moving perpendicular to that direction so the distance traveled by the dog has to be equal to 1 minus the distance between the dogs (only true in the limit because then for every single instance in time your dog is moving on a circle (or rather on the tangent of a circle at the point where tangent and circle meet) around another dog while moving towards a third dog. so if every dog is moving with a slope equal to the tangent of a circle around another dog you can center your point of view on one of the dogs (dog1) and it would look like another dog (dog2) would walk in a circle centered at dog1, and dog 1 moving towards dog 2. the circular motion is not going to affect the distance between the dogs since the radius stays the same. thus it is is equal to dog1 moving to a stationary dog2. Really neat problem, with a nice solution.
4:58 pretty sure it wouldn't keep getting smaller as long as time lag is large enough. At some point, the dogs would just form a square with sidelengths smaller than t, meaning they overshoot the position of the next dog in the next timestep. No?
Very good question. The fact that every square is smaller than the previous one does not necessarily mean that they will, at some point be smaller than t. Infact, the sidelength, in a sense, approaches 't'. This becomes visible when we prove S_n > S_n+1 > t for any n.
@@Mathing Right, guess my intuition was off (unless t>1 but that's stupid). Thanks for clarifying, and thinking back on it, it makes sense that this wouldn't work. ;)
Because of the symmetry of the initial system with n dogs (in the problem statement, n=4) it will continue to be symmetric at all times t. Therefore, at any time t, to get the location of the next dog, we just rotate the location of the current dog by the angle 2pi/n . The location of the center, our dog, and the next dog, produces an isosceles triangle. The central angle is (2pi/n). The other two angles are therefore (pi - (2pi/n))/2. Then, we look at the angle relative to the perpendicular, to see how to split the speed of our dog between the inward direction and the direction perpendicular to that. The angle made with the perpendicular direction is (pi/2) - ((pi - (2pi/n))/2) = pi/n. So, the tangent-to-circle part of the direction is cos(pi/n) And the normal to circle part is sine(pi/n) inwards. So, radius at time t is 1 - t sine(pi/n). And, the part along circle should be like, 2 pi r(t) (d theta / dt) Putting in our expression for r(t) this is 2 pi (1 - t sine(pi/n)) (d theta / dt) So, cos(pi/n) = 2 pi (1 - t sine(pi/n)) (d theta / dt) Solving this for (d theta/dt) we get (d theta/dt) = (cos(pi/n) / ((2 pi) (1 - t sine(pi/n)))) Solving this differential equation we get theta(t) = (-1/(2 pi)) cot(pi/n) ln(1 - t sine(pi/n))
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
Nice problem! I solved it by solving the differential equation y’ = (i-1)y, and then finding the arc length, but your solution is nicer and generalises to other polygons easily
A weakness of mine would be that I still do not know how to solve differential equations- a big tool missing in my set of tools- so I tried explaining it such a way that pre-calc student too have something to take from this video! I'll make sure to look into your method!
So solving the differential equation gives y = exp((i-1)t). For the dogs to travel at unit speed you need to rescale t so that |y'| = 1. exp((i-1)f(t))' = exp((i-1)f(t))(i-1)f'(t) f should be monotonically increasing, so |f'| = f', giving us 1 = exp(-f) sqrt(2)f' Solving this equation, using that f(0)=0, gives f(t) = -ln(1-t/sqrt(2)). From this we can immediately tell that the distance traveled will be sqrt(2) (which is the side length of the square in my setup) because that's when f reaches a singularity. Plugging f into the expression we get that the dog will have a distance of 1 - t/sqrt(2) away from the center, and will make an angle of -ln(1 - t/sqrt(2)) with its original position.
I had a differential equation is the Cartesian plane which I didn't know how to solve. (See my comment if interested) I see you have a differential equation in the complex plane and it is much simpler! Can you explain how you got it?
@@joseville I got the same differential equation by recognizing that at any point the derivative has to be dt * (the dog vector rotated by 45 degrees). thats because the dogs are always on a square's diagonal and theyre moving along the sides - > pi/4 radians BUT THATS NOT RIGHT IF THE SPEED IS CONSTANT
Nice problem! Me as a person familiar with differential equations, made one for this and ignored, that the velocity is constant (because it seems, that velocity doesn't matter, i didn't prove that, and otherwise the system will be too complex). Without normalizing velocity I've got system of linear diff.eq., which gives a path of parameter t (not the actual time). Then I substituted s(t) - parameter to make velocity equal to 1, and got that s=1-exp(-t) approaches to 1. Hence the distance is 1 P.S. path for dog in (0,0) x(t) = 1/2 * [1 - exp(-t)*(cos(t)-sin(t)] y(t) = 1/2 * [1 - exp(-t)*(cos(t)+sin(t)] (for real distanse substitute t = -ln(1-s)) So basically dogs will have to circle around the center infinitely many times, until they meet up :)
The fact thay the dogs circle around the center infinitely many times was really hard for me to swallow when I found it while tinkering with it. Also, now that you pointed out the fact that the velocity doesn't matter as long as they are the same for all the dogs to determine the trajectory- it is interesting, probably this can be shown using my method, since whatever the velocity is, the 't' should converge 0 and should not play any role in changing the slope of the graph I made! Cool!
@@Mathing : That's an interesting way to see that the dogs circle the center infinitely many times. We can start with the fact that the path shape is independent of the speed, and we also know (from your drawing) that the dogs will go at least 45 degrees around the center. So we can see that the problem of finding the shape starting at a point when the dogs have gone 45 degrees around the center is the same as the original problem, just smaller and rotated -- and so the shape starting at 45 degrees is the same as the whole shape, just smaller and rotated. That means that the dogs aren't at the center when they've gone 45+45=90 degrees around the center either, and we can repeat this shrinking-and-rotating the shape ad infinitum. And, for extra fun, it's almost exactly a retelling of Zeno's paradox. Instead of using 45 degrees (which is pretty arbitrary), we can take the point they get to after running for 1/2 time unit (which is also pretty arbitrary). In that time, they've gone some angle around the center -- call it A. So we follow the angle, and the next rotation of A around the center takes them 1/4 time unit, and they travel 1/4 distance unit. And the next one takes them 1/8 time unit and they travel 1/8 distance unit, and so on. In effect, the "steps" in Zeno's paradox are being counted out by the rotation angle around the center, and the resolution is the same as the resolution of the paradox: The sum of this particular infinite series is finite. Also the dogs when they do reach each other will obviously be infinitely dizzy.
Agreed, this seems like the most obvious solution to me. Dog A is always running directly towards Dog B. Dog b's motion is perpendicular to Dog A's. Therefore the only factor which affects the distance between the 2 is Dog A's movement. Therefore it's just time = distance/speed = 1/1 = 1
I am pretty sure the curves are logirmatic spiral, we reparametrize the time so that the speed is proportional to the length of the enclosing square as in the video. the speed is proportional to the length of the side of the enclosing square but the rate of rotation should be constant (noticing in this reparametrization the curve will never reach the origin but the problem under rotation around the origin scaling), now it is too see the curve is of the form ke^at (reversing the time) where a and k are or some complex numbers or the solution curve of the vector field y'=ay, a can be found by noting the velocity is tangent to the side of the square at the corner and has speed one, now plugging in the initial condition we find the solution curve I guess the difficult part is to notice that 5 point together with origin will form a fixed configuration as time passes, let a1, a2, a3 , a4 be 4 corner, one can show that they always form a square with the origin in the center by computing various derivative, for example (a1+a2+a3+a4)/4 to show that the center of the square is unchanged. now with this in mind, the solution is time invariant (if f(t) is a solution so is f(t+s))
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.
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.
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........
When thinking about the problem on my own i felt like a fractal pattern of a spiral would be created, which could then mean that the distance traveled of each dog is (theoretically) infinite. But your video clearly disproves that, good work.
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.
1) At the end of their chase are the dogs spinning infinitely fast? If not, which direction are they facing when they cross? 2) What happens if one of the dogs is blind and always moves in the direction it started? How do the other dogs move? 3) What if one is a plush dog and doesn’t move?
Also you could apply some physics to the problem and, if you take, say, Dog 1 as your reference then Dog 2 by definition must be moving towards it in a straight line hence they must meet at t=1.
This is very good video , amazing method , i have done similar question but it was triangle , but couldn't think of method like this. Also I liked the quote at last, please keep posting amazing videos ,I shared and subbed.
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 paused and solved it like this: 1. Observe that the entire problem retains 90 degree rotation symmetry as the dogs progress. 2. Observe that at any point in time, the current state is a scaled & rotated version of the original state. 3. From here, you can see that we're ultimately just modelling motion where at any point, the velocity is 1 unit/s at a 45 degree angle counterclockwise away from the line to the middle. 4. Time for some actual equations: In polar coordinates (with theta counted clockwise from "north"), dr/dt = -sqrt(2)/2, and dtheta/dt = -sqrt(2)/2/r(t) 5. Integrate with initial conditions (assuming the top-left dog): r(t) = sqrt(2)/2-sqrt(2)/2*t, theta(t) = -pi/4-(-ln(1-t)) 6. Simplify: r(t) = sqrt(2)/2*(1-t), theta(t) = ln(1-t)-pi/4 7. r=0 at t=1, so with a constant speed of 1 unit/s, that means total arc length is 1 unit.
Nice! Atleast you didn't take a week to solve it hehe :) In my case, I wanted to solve it using the least bit of calculus I could use, and that cost me this solution
This was a brilliant video - super engaging! As an educational video creator myself, I understand how much effort must have been put into this. Liked and subscribed, always enjoy supporting fellow small creators :)
fun fact: the trajectory of the dogs is described by a logarithmic spiral. You can see this by constructing the differential equation of one trajectory and solving it with the substitution x = r(theta)cos(theta), y = r(theta)sin(theta), pluging in the parameters of the initial position
You can also skip cartesian coordinates altogether by looking at a small right triangle with sides dr and r*dθ. Through some geometry you can find that the two other angles must be 45 degrees, and so dr/dθ=r.
A logarithmic spiral feels like it makes sense to be the answer, especially since zooming in to the cente of the spiral yields the same spiral, albeit rotated, just in the same way the sequence of squares in the slow dogs approximation are scaled and rotated versions of each other, the next square being scaled from the previous by the same ratio.
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.
Wow - I never would have guessed that the spiral length is one! SO enlightening. Did you try zooming into the center? Do the spirals keep winding around at the limit? This is a great submission for SoME2!
Thank you so much phi! Your encouraging words pushed me a lot to complete this video❤️ I didn't try zooming in the center, but I tried to find out a similar fact mathematically.. The question I asked myself was, it looks as if the intermediate squares are just a rotated and scaled version of the original square- so I tried finding the total amount of rotation that the square connecting the dogs would undergo till they reach the center- and I found that to be infinity- Which I think should mean that the spiral will go on forever towards the center, getting closer and closer each time. Although the idea of zooming into the center is interesting, I might try it out!
In polar coordinates, I found the parameterized path (for the dog at the top left) is r(t) = sqrt(2)/2*(1-t), theta(t) = ln(1-t)-pi/4, from which you can see that theta does increase (negatively) without bound. Thus, it completes an unlimited number of windings, despite it not looking like it when the path is graphed. Zooming in should look identical, no matter how far you zoom, since the state of the dogs at any point is just a scaled and rotated version of the original state.
Let s = sqrt(1/2). Note the initial distance from the center is s. Split velocities into radial part vr (towards the center) and tangential part vt (perpendicular to the radial line) Notice that the dog always runs along the side of the square, which will always be the bisector of the radial velocity and tangential velocity. So vr = vt = -s, constant. For convenience I'll have the dogs run in whichever direction (clockwise vs anticlockwise) lets this be true. r = r0 + vr * t = s - s t = s * (1-t) dtheta/dr = dtheta/dt = vt / r = -s / r so theta - theta0 = -s ln r + s ln r0 r = r0 e^-(theta - theta0)/s Each dog reaches r=0 at t=1 after moving 1 unit along a logarithmic spiral.
ooo i figured out an easy way, bc of symetry the dogs are always in a square formation that is rotating, if you counter rotate the square at the same rate and keep dog one stationary visually (like the camera is in their pov) it would look like dog two has to run in a straight line to dog one wich is one unit so in one second
I feel like one approach is to take the approximation you used further: It should be possible to prove that this one step approximation is converging to the real solution using methods of numerical differential equations. Then, we can add up the lengths we are interested in, take the limit, and the answer should be the same. I don't know how hard it would be to go through all the necessary hoops, but it seems like ot should be possible
Tried this problem and correctly guessed the polar form was of the type r=root(2) - c*theta (with origin in center of square), but I got caught up in the equations of the curves and didn't realize you could just solve for their length without actually having the equations. My second attempt tried to parametrize the motion and equate two curves using a rotation matrix but that also came up short. Maybe I could've gotten it but it would have definitely been a much longer solution
It is much simpler. Dog 1 is initially separated from Dog 2 by distance s, where s is the side of the square. Dog 2 goes in a direction perpendicular to the one in which Dog 1 goes. In an infinitesimal time dt, you can assume that Dog 2 does not move towards or away from Dog 1. Dog 1 moves towards Dog 2 by some distance. This distance is the same as the reduction in the side of the next square formed after dt time has elapsed. So the dogs must travel a total distance of s before they meet at center of the original squares or any of the intermediate squares.
@8:24 wait. They're travelling at 1/s on a 1x1 square towards the center. A straight line is (1/2)x✓2 or 0.7(something). The fact that it takes 1s is a little mindblowing. Considering that ✓2 is irrational
just the last semester I had to solve that very same problem for a diferential geomtry homework... I remember the proffesor giving us a hint about squares that I didn't understand, it was probably something like this... I used complex numbers to do it , but this way is also quite nice
@@kamakshigarg1289 I put the four dogs in the 4 fourth roots of 1 (ie +/-1 and +/-i) as a srart point. Then the derivative over time of a dog would be its position minus the position of the following dog (meaning that they always run toward the next one at a constant rate). Finally, because of symmetry, you have that the position of any dog is the position of the next one times the fourth root of 1 and there you have a differential equation that results in a complex exponential and therefore a exponential spiral... also if you change the 4th roots of 1 with nth roots you'll have the same problem but with n dogs 🤭
I paused and my hypothesis is that the dog they follow always move perpendicularly: If the distance didn't change they would move in circles, i.e. they would always remain at the same distance. Therefore they don't alter the amount of distance they have to travel.
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
Have the dogs each move N units forwards, and turn to face the dog they are to go towards, then repeat until they are within n units of the dog they are supposed to go forwards. Next, take the limit of that as N approaches 0 from the positive side. You can construct a nice drawing of this by drawing concentric squares, where each new square has it's vertices n units along the edge of the outer square. When looking at it this way, it become obvious that perhaps instead of n units, it should be 1/n th of the way along the edge. This makes the next square have the same relationship with the current square as the current square has with the previous square. This symmetry is likely useful for simplifying the math. Again, we would take the limit to get the exact solution, but this time it would be as n approaches infinity (cause we want 1/n to get approach 0) I'm not good enough to come up with an exact function for these curves (perhaps using polar coordinates would be helpful?). Actually, doing some back of the hand math, that would have to be the polar function equivalent of a 45 degree (pi/4) line, would it not? The derivitive should be exactly 1 at every point along the line? So like, radius=angle?
Great, you can actually use this same technique to solve it for any polygon, including your equilateral triangle! In your case the graph will still be a straight line, but the slope will not be -1, it will be a different value. Also for finding the slope, you'll need to derive a relation between any two consecutive triangles- here, you might find the extension of the Pythagoras Theorem for Acute angles useful (Or even the cosine rule of Trigonometry)
Awesome video! My approach, which didn't pan out for me, but might for someone else. Let dog C be in the origin and mark it's position as (x, f(x)) Then dog D's position is (1-f(x), x) Since dog C is always directed towards dog D, we have f'(x) = rise/run = (D.y - C.y) / (D.x - C.y) = (x - f(x)) / (1 - f(x) - x) Unfortunately I wasn't able to get further, but solving this differential equation would get f(x), a closed form for the trajectory of dog C. Then one could do a line integral to get the length of the trajectory. I'm speculating, but since the trajectory turns out to a spiral which fails the vertical line test, the trajectory isn't actually the function f(x). f(x) is just a the first portion of the trajectory. Dealing with this could get messy - I think the full trajectory would be a union of piecewise functions each of which is a solution to a slightly different differential equation or maybe each is a solution to the same differential equation, but with different boundary conditions...?
Interesting, the way you positioned D with respect to C using symmetry is cool! (EDIT: Skip the next passage where I make a fool out of myself hehe:) I'm not (yet) familiar with solving differential equations, so pardon any mis-observations: I think in case of f'(x) = (Dy - Cy) / (Dx - Cy) Here f'(x) is calculating the slope of the line connecting C and D together right? This should not be tangent to their trajectory, so I think this might not give the desired results? Also, the trajectory, amazingly swirls inside infinitely- so if you keep zooming more and more, you'll find the same curve rotating (Pardon me but I am out of formal language to describe what I'm trying to say:))
@@Mathing yes, f'(x) is notation for the derivative of f(x). The way it's set up guarantees that C's velocity vector points to D always. Interesting, the curve has finite length, but makes infinitely many turns around the center point. So that means the dog make an infinite number of turns in finite time. As you may know acceleration not measures only change in speed, but also change in direction - velocity is a vector having magnitude and direction, after all. Here the acceleration is purely due to change in direction as the speed remains 1 unit/sec at all times. Acceleration due to change in direction is proportional to the curvature of the trajectory. At the center, I think acceleration might go to infinity since (I think) the curvature also goes to infinity at the center.
you can make this idea work. as you have observed, the dogs make infinitely many turns around the center of the square, so the trajectory of a dog cannot be given by a function y=f(x) - what you instead need to do is make both y _and x_ functions, so the trajectory is P(t) = (x(t), y(t)). This will lead to a system of differential equations which you can solve to determine x(t) and y(t), and then you can calculate the arclength of the trajectory using an integral. (If you decide to try this out then you should put the vertices of the square at (0, ±1) and (±1, 0) instead of (±1, ±1) as this will make the calculations a lot nicer)
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
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.
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?
hello I really enjoyed your video it was beautiful I just wanted to ask you a question, how did you get the limit to be equal to -1? I would like to see your calculations 😁
Thanks! I'll provide it here: Lim(t→0) (sqrt( (S_n - t)^2 + t^2 ) - S_n) /t Here, as t approaches 0, the sum (S_n - t)^2 + t^2 can be considered as (S_n - t)^2 since t^2 quickly converges to 0. So this makes the limit simpler: Lim(t→0) (sqrt( (S_n - t)^2 ) - S_n) /t = Lim (t→0) (S_n - t - S_n)/t = Lim (t→0) (-t)/t = -1
@@Mathing to show the limit rigorously (saying that one part of an expression converges "more quickly" than another is good heuristically but it is not a proof) you can just use L'Hopital's rule (which says that if f(t)/g(t) is an indeterminate form when you plug in t=a then lim(t→a) f(t)/g(t) = lim(t→a) f'(t)/g'(t). in this case the indeterminate form is of type 0/0)
@@EebstertheGreat Indeed each S_n is a function of t, but L'Hopital's rule still works. (since S_0(t) = 1 and S_{n+1}(t) = sqrt((S_n(t) - t)^2 + t^2) we have S_n(0) = 1 for all n. You get terms involving S_n'(t) when you calculate d/dt[sqrt((S_n(t) - t)^2 + t^2) - S_n(t)], but when you evaluate at t=0 you don't actually need to know S_n'(0) because the terms cancel in the end.
@@schweinmachtbree1013 I don't get it. That holds for S(0), but we are interested in showing that the slope is constant, so we can't just look at one point. We need it to hold for every point.
Masha Allah congratulations Saad. You did it! This is speach lessly amazing! You are very gifted bhaiya. Allah has blessed you with a very unique talent! ❤️ Never give up no matter what the situation is because it doesn't ever take away the awesomeness you have in you. You always have you and your gift, no matter where you are. I believe in you that you will create magic from being anywhere in the world! You will make it a better place Inshallah ❤️ Sawda apu loves you a lot! May Allah bless you with the best always 🤗❤️
I made a video that generalizes this problem and solve in detail the system of differential equations tha arises to derive the exact equation of the trajectory. I also show where those spirals appear. Look it up on my channel, it is called : Zeno's Mice (Ants) Problem and the Logarithmic Spirals.
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.
This is a wonderful submission! It really reflects how I think most mathematicians might solve a problem - yeah, you can exploit symmetry in *this* case, but if you were to make slight tweaks to the problem, your solution generalizes, while other solutions do not. Kudos!
So I'm thinking that it'll be a square inscribed in a circle, and then it's just figuring out the length of a spiral like that. Not that I know how to calculate that. Edit: after giving it a bit more thought, since it'll always be a square with a shrinking side length, it should take one second, and the distance traveled should be 1 unit, even though the math to figure it out is probably hard. Basically we can imagine the dogs on the corners of a circumscribed square where the r is slowly getting smaller and the theta is increasing(for polar coordinates)
Posting this before watching the rest of the video: I'm considering putting together a system of differential parametric equations, and using Eulers method to solve it numerically. From there finding the distance of each path should be simple. But it's late, so I'll watch the video now.
Drawing the shape of the spiral was a challenge, but right away I noticed that as the dogs move, they're always on the verteces of a square, thus they're moving perpendicular to their neighbors. Therefore they neither increase or decrease the distance from their chaser, who always closes in on them in 1u/s. That's how I figured they'll all meet after 1s or 1u
The answer is not paradoxical. The dogs move toward each other so that they are always moving at right angles to the target dog. Therefore the motion of the target dog does not have any influence on the distance the pursuing dog most travel. This is intuitively obvious.
Here you go: Lim(t→0) (sqrt( (S_n - t)^2 + t^2 ) - S_n) /t Here, as t approaches 0, the sum (S_n - t)^2 + t^2 can be considered as (S_n - t)^2 since t^2 quickly converges to 0. So this makes the limit simpler: Lim(t→0) (sqrt( (S_n - t)^2 ) - S_n) /t = Lim (t→0) (S_n - t - S_n)/t = Lim (t→0) (-t)/t = -1
You can actually use this same technique to solve it for any polygon, including an equilateral triangle! In your case the graph will still be a straight line, but the slope will not be -1, it will be a different value. Also for finding the slope, you'll need to derive a relation between any two consecutive triangles- here, you might find the extension of the Pythagoras Theorem for Acute angles useful (Or even the cosine rule of Trigonometry)
Hi, do you know what happens when t is negative, could you possibly test it out with your code :) Another question i had was the nature of the tilings. are there smaller sub square(s) that is a complete 360 degree rotation from the big square we start with?
Interesting point! Here's what it looks like when t = - 0.2 ...Amazing! drive.google.com/file/d/1GsIMW-I-PUjiBRjOXvcfzrWSuGZSBSp-/view?usp=sharing About the nature of the sub-squares, when t approaches 0, the curve as it tends towards the center, rotates for an infinite amount- so you can in theory keep on zooming towards the center for infinity and you'll find the same curve just rotating around! So this means yes, there are infinitely many sub-squares formed by the dogs that are a complete 360 degree rotation from the original one!
Nice video, though I do feel like you are missing some kind of argument about the uniform convergence of slopes to make your reasoning truly rigorous. Anyway, great video! Btw, I'm not sure I fully agree with your last statement. Usually mathematicians are way more interested in developing powerful methods that can solve multiple problems than to know solutions of a particuliar problem. Nevertheless, it's true that solving one problem multiple ways can be pedagogically very enlightening
Thank you so much! This submission has been really helpful for me in ways that it has given me the opportunity to learn more and go deeper than the shallow depth I'm in, and I am really grateful for this. I will make sure to check out 'uniform convergence of slopes' to get better at rigorous terms, thank you for this! And yes, the ending statement is only partially true, that for inspiring people to take different journeys that might bring up newer approaches and 'more powerful tools'. But I agree with the case of powerful methods- they do feel magical in their own way. Thanks for your insight!
I'm really glad to see this "classical" problem getting solved once again =) The last time I saw it was in my 9th grade at a certain physics olympiad school, when the solution didn't invoke any limits or line integrals, since we didn't even know what a derivative was (and weren't supposed to).
The proposed solution was based upon noticing the fact that at any moment any two adjacent dogs travel perpendicular to one another, hence the distance between them shrinks at a speed equal to one of the dogs' velocity, which is 1 unit/s and is constant. Therefore the time it takes before all the dogs meet is equal to 1 unit (the initial distance between 2 adjacent dogs) divided by this speed, which amounts to 1 second. And since the dog's velocity is constant, the required distance is just time times velocity, in this case the answer is 1 unit. I think this might be one of the solutions you found online =)
There was also a slightly more complicated version of this problem, with the room being in the shape of an equilateral triangle (or in general an n-sided regular polygon) and with 3 (or n) dogs in the corners respectively. Maybe you'd want to cover this one in one of your next videos =)
Yep, this was the solution I referred to as being too clever & short hehe.
About the triangular case, you can actually solve this puzzle for any regular n-gon using the very method in this video! This will include the same steps for example creating a graph of sidelength vs distance, proving that the graph is a straight line, finding relation between two consecutive sidelengths to get the slope.. But in this particular case, the slope won't be equal to -1, it'll just turn out to be a different value instead.
@@Mathing Yeah, I thought so as well. Then here's a couple of different pursuit-type problems that might catch your interest. They're fascinating in their own way, but the soution isn't as straightforward.
1. A rabbit is running along a straight road with a constant speed of 10 m/s. A fox is trying to catch the rabbit by running with a constant speed of 20 m/s, but its velocity vector is always pointing at the rabbit. Initially the fox is standing 300 m from the road and then starts running when the rabbit is 300 m away from the fox. Find the time it takes for the fox to catch the rabbit.
2. Two ships - A and B - are departing simultaneously from two ports located on a straight coastline at a distance L from one another. Ship A is moving perpendicular to the coast with a constant speed, whereas ship B has the same constant speed as ship A, but its direction is always towards ship A. Find the distance between the ships after a long time.
As a bonus you can try to derive the trajectories in both problems, but the solutions don't require that. In fact, the answers actually look pretty nice =)
Also in the n-gon problem there can be an additional question about the initial accelerations of the dogs and the initial curvatures of their trajectories, so that's that for a change =P
About the triangular case: That is the form of the problem I have first seen also in preparation for the Physics Olympiad.
The way I solved it:
All velocity vectors summed up at any instance will be zero (for all constellations not only triangular obviously) hence the "center of mass" of the system will not move. Therefore we know that our dogs or whatever will meet in the center (if they ever do).
Then we notice that the situation is the same as perceived from any of the dogs. That is to say all connecting lines between the dogs (how many there might be) must remain in the starting configuration bc we can just copy what one dog sees to the other rotating according to the direction it's heading(they might be scaled as the problem does not have any dependency on the dogs distances, but only directions). Now it is rather straightforward: We may take the rotating frame of reference, then all dogs are moving straight to the center with constant velocity (one decomposes the initial velocity into radial and tangential components) and finds the distance to the center.
It is nice to see such a seemingly complex problem to have such an easy (to calculate) solution. No saying that your way is not cool as well especially the bit finding the geometric manifestation of the derivative ; ) just probably unlikely to pull off in an exam.
@@invictus1132 Oh, that solution I've never seen actually =)
I made a video that generalizes this problem and solve in detail the system of differential equations tha arises to derive the exact equation of the trajectory.
I also show where those spirals appear. Look it up on my channel, it is called : Zeno's Mice (Ants) Problem and the Logarithmic Spirals.
I think there is a more intuitive solution.
Observe that:
1. Each dog is moving towards another dog, which is moving perpendicular to it.
(for example, at the start it is easy to see the velocity vectors are perpendicular and stay this way due to symmetry)
2. Perpendicular motion is well-approximated by a circular motion.
Thus, dog A is moving towards dog B, while dog B is moving along a circular arc around it (for small time intervals). Thus the distance between A and B gets shorter by 1 unit/sec as A comes closer to B, while B is trying to stay at the same distance.
Since they approach each other at 1 unit/sec, they will meet in 1 second.
(the approximation works well on a thin segment of a disk)
Congratulations that you got it! In fact, this was the very solution I referred to as 'short' and 'elegant'. A more general version of the same solution can be found here: www.theguardian.com/science/2019/jun/03/did-you-solve-it-dogs-in-pursuit
now I feel clever, this is what I came up as well! Though I will say, the revelation that the distance between the dogs closes at a constant rate only gave me the distance traveled and the general sense that the radius of the curve must linearly decrease, not an actual shape. I had no idea how to construct that curve, which the video's method gets for free...though as I type this I realize I'm dumb and you just plot the curve in the polar plane
I believe this needs a proof that at any point in time they are indeed moving perpendicular to each other. That is not given.
@@grantofat6438 from symmetry we know that the dogs form a square. From the statement of the problem we know that their velocities are pointed toward each other. So we have 4 velocities pointed along the sides of the square.
This implies the velocities are perpendicular.
this is not an approximation, 1 is an exact answer, and to simplify your explaination, i can provide a more intuitive explaination
imagine you are one of the dogs, and you think that you are stationary (even when the dog is moving), another dog is approaching you directly in a straight path, meaning that the reletive velocitty of the dog wrt to you is always v towards you, and distance between you and the dog is d, t = d/v
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.
We can simply use velocity of approach(taking one of the dog as frame of reference)
This was the exact question we were given to solve during our first year of Highschool 🙌🙌
I'm glad that you showed a different way around
What the dog doing?
Two additional thoughts:
1. If t is negative, we'll get the curve in the opposite direction, i.e. a spiral outwards, continuing the logarithmic spiral pattern outwards, with the center missing.
2. Wikipedia's article (en.wikipedia.org/wiki/Mice_problem ) has mistakenly stated that it will take "infinite time" to approach the center. This is impossible because the distance is finite, and we're moving at constant speed.
1) The value of 't' being negative never occured to me, it would surely look cool. Thanks for pointing it out!
2) I found the same thing, but yes they were
wrong.
It's interesting to think about t, we know that for t=1 the dogs will just run around the square, and we know that for all 0
That mistake had been introduced by someone on August 2 and, fortunately, been corrected again by August 28. en.wikipedia.org/w/index.php?title=Mice_problem&action=history
Hi, just tried out what the graph looks like when t = - 0.2 and yes, it is as you said, a spiral outwards for infinity: drive.google.com/file/d/1GsIMW-I-PUjiBRjOXvcfzrWSuGZSBSp-/view?usp=sharing
The correct thing here is that the spirals will wind an infinite number of times around the center. This confuses some people, since it does so in finite time and distance.
Complex numbers make this easier. By symmetry, each dog's motion is one 90 degree rotation with respect to the previous one. If we place the center of the square at the origin of the complex plane, then b = ia, c = ib, d = ic, a = id, where a, b, c, d are complex-valued functions.
Hence we have
a' = b - a = ia - a = (-1+i)a
so a = e^((-1+i)t), i.e. an exponential spiral
Note that I'm removing the "constant speed" constraint here, in order to find the paths more easily.
I've seen this puzzle before. The result is not surprising at all when you consider this: Let's say dog A is walking towards dog B. Since B is always moving at a right angle to A's direction of travel, it is neither getting closer nor farther from A. So, for A, it's just as if B doesn't move; and so A walks 1 unit to get to B. That's not a proof but it's sure compelling, I think.
Such 'mathing' is definitely my obsession! Thanks for the effort!
Very nice video! I'll note that the limit at 10:20 can found in a few ways; it's a nice exercise for L'Hopital's rule or, you can do it with elementary methods with the difference of two squares.
I feel like there should be some nice argument to prove the linearity of the distance vs time using the temporal symmetry of the problem (the system looks very alike at any point in time up to rescalling)
Could be! This problem becomes the a rotated version of itself after infinitesimally small time, so this is interesting!
@@Mathing The problem is the constant speed, which gets ever bigger relative to the square as the square shrinks. You can drive this by considering smaller time steps.
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.
There's a very nice physics-y way to get the answer. We already know by symmetry the answer must be in the center of all 4 of them (otherwise rotating 90° would change the answer but otherwise leave the problem unchanged) so we can just think about how long it takes to reach the center. To do that we find the radial velocity of any dog by drawing a radial line and projecting the velocity on to it (fortunately since the shape stays the same, the radial velocity does too).
The radial velocity is cos(45) = sqrt(2)/2 and the radial distance is sqrt(2(l/2)^2) = lsqrt(2)/2. In terms of theta that's l/(2cos(theta/2). Dividing the gives the time required to get to the center
Neat!
l is the side length of the current square described by the positions of the 4 dogs?
What is theta?
Yes, I also have used this method
If you rotate the figure by 45 deg witt origin at the center of the square, a very straight forward differential equation is easy to derive and the result is a well known curve.
if you compute it with radial coordinates it's even easier because you prove that n^r=θ and that the derivative at the point (0;1) is equal to -1 without using differential equations.
@@lorenzodiambra5210 Nice. But do you get the length of the curve? There is also a simple orthogonality argument that says the length of the path = one side of the square. No equations or arithmetic.
My solution/proof for this would be that every dog is moving towards a dog that is moving perpendicular to it at all times, therefore the distance only changes by the amount the first one is moving. The target dog doesn't ever change it's distance from the follower, only the follower changes distance, therefore the distance would be exactly the same as at start, 1 unit.
Edit to add, I hadn't read any other comments or watched the whole video when I wrote this. I see others came up with essentially the same solution.
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.
Solving a system of differential equations (made fairly simple in polar coordinates) allows one to calculate the equations for the dog's path as
r(t) = (1-t)/sqrt(2)
θ(t) = π/4 - log(1-t).
I made a video with this solution exactly, look it up on my channel, it is called : Zeno's Mice (Ants) Problem and the Logarithmic Spirals.
Beautiful proof! I solved this many year ago, when there was no Internet... I used polar coordinates and was quite simple when you realize the 45° constant!
you can write a differential equation In the coordinates of a particular dog using the fact that the next dog is always ninety degrees rotated from the current dog because this whole system has to keep fourfold. then take derivative of the radius using the chain role and observed that is constant. unlike the solution using perpendicularity this will easily generalize to higher polygon.
I solved this problem too this year and generalized with n dogs !
After 3:04 I just saw there triangles and pythagorean theorem, so my solution was just
Lim [t->0] { √( t² + (1-t)² ) } (what would happen if dog's brain lag was getting shorter and shorter, so it creates new triangles and calculates new distance faster and faster). And it just goes to 1 straight.
upd: probably i messed something up, but can't understand that
when you think about it, it is actually quite obvious why the length has to be 1 unit. Since every dog is moving straight towards another dog and the other dog is only moving perpendicular to that direction so the distance traveled by the dog has to be equal to 1 minus the distance between the dogs (only true in the limit because then for every single instance in time your dog is moving on a circle (or rather on the tangent of a circle at the point where tangent and circle meet) around another dog while moving towards a third dog. so if every dog is moving with a slope equal to the tangent of a circle around another dog you can center your point of view on one of the dogs (dog1) and it would look like another dog (dog2) would walk in a circle centered at dog1, and dog 1 moving towards dog 2. the circular motion is not going to affect the distance between the dogs since the radius stays the same. thus it is is equal to dog1 moving to a stationary dog2.
Really neat problem, with a nice solution.
4:58 pretty sure it wouldn't keep getting smaller as long as time lag is large enough. At some point, the dogs would just form a square with sidelengths smaller than t, meaning they overshoot the position of the next dog in the next timestep. No?
Very good question. The fact that every square is smaller than the previous one does not necessarily mean that they will, at some point be smaller than t. Infact, the sidelength, in a sense, approaches 't'. This becomes visible when we prove S_n > S_n+1 > t for any n.
@@Mathing Right, guess my intuition was off (unless t>1 but that's stupid). Thanks for clarifying, and thinking back on it, it makes sense that this wouldn't work. ;)
this channel is insanely underrated!!!
Because of the symmetry of the initial system with n dogs (in the problem statement, n=4) it will continue to be symmetric at all times t.
Therefore, at any time t, to get the location of the next dog, we just rotate the location of the current dog by the angle 2pi/n .
The location of the center, our dog, and the next dog, produces an isosceles triangle. The central angle is (2pi/n). The other two angles are therefore (pi - (2pi/n))/2.
Then, we look at the angle relative to the perpendicular, to see how to split the speed of our dog between the inward direction and the direction perpendicular to that.
The angle made with the perpendicular direction is
(pi/2) - ((pi - (2pi/n))/2)
= pi/n.
So, the tangent-to-circle part of the direction is cos(pi/n)
And the normal to circle part is sine(pi/n) inwards.
So, radius at time t is 1 - t sine(pi/n).
And, the part along circle should be like, 2 pi r(t) (d theta / dt)
Putting in our expression for r(t) this is
2 pi (1 - t sine(pi/n)) (d theta / dt)
So, cos(pi/n) = 2 pi (1 - t sine(pi/n)) (d theta / dt)
Solving this for (d theta/dt) we get
(d theta/dt) = (cos(pi/n) / ((2 pi) (1 - t sine(pi/n))))
Solving this differential equation we get
theta(t) = (-1/(2 pi)) cot(pi/n) ln(1 - t sine(pi/n))
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
Nice problem! I solved it by solving the differential equation y’ = (i-1)y, and then finding the arc length, but your solution is nicer and generalises to other polygons easily
A weakness of mine would be that I still do not know how to solve differential equations- a big tool missing in my set of tools- so I tried explaining it such a way that pre-calc student too have something to take from this video! I'll make sure to look into your method!
So solving the differential equation gives y = exp((i-1)t). For the dogs to travel at unit speed you need to rescale t so that |y'| = 1.
exp((i-1)f(t))' = exp((i-1)f(t))(i-1)f'(t)
f should be monotonically increasing, so |f'| = f', giving us
1 = exp(-f) sqrt(2)f'
Solving this equation, using that f(0)=0, gives f(t) = -ln(1-t/sqrt(2)).
From this we can immediately tell that the distance traveled will be sqrt(2) (which is the side length of the square in my setup) because that's when f reaches a singularity.
Plugging f into the expression we get that the dog will have a distance of 1 - t/sqrt(2) away from the center, and will make an angle of -ln(1 - t/sqrt(2)) with its original position.
this solution forgets about the constant speed, y' has to have the same length for every y but for very small y, y' is very small
I had a differential equation is the Cartesian plane which I didn't know how to solve. (See my comment if interested)
I see you have a differential equation in the complex plane and it is much simpler! Can you explain how you got it?
@@joseville I got the same differential equation by recognizing that at any point the derivative has to be dt * (the dog vector rotated by 45 degrees). thats because the dogs are always on a square's diagonal and theyre moving along the sides - > pi/4 radians
BUT THATS NOT RIGHT IF THE SPEED IS CONSTANT
Nice problem!
Me as a person familiar with differential equations, made one for this and ignored, that the velocity is constant (because it seems, that velocity doesn't matter, i didn't prove that, and otherwise the system will be too complex). Without normalizing velocity I've got system of linear diff.eq., which gives a path of parameter t (not the actual time). Then I substituted s(t) - parameter to make velocity equal to 1, and got that s=1-exp(-t) approaches to 1. Hence the distance is 1
P.S. path for dog in (0,0)
x(t) = 1/2 * [1 - exp(-t)*(cos(t)-sin(t)]
y(t) = 1/2 * [1 - exp(-t)*(cos(t)+sin(t)]
(for real distanse substitute t = -ln(1-s))
So basically dogs will have to circle around the center infinitely many times, until they meet up :)
The fact thay the dogs circle around the center infinitely many times was really hard for me to swallow when I found it while tinkering with it.
Also, now that you pointed out the fact that the velocity doesn't matter as long as they are the same for all the dogs to determine the trajectory- it is interesting, probably this can be shown using my method, since whatever the velocity is, the 't' should converge 0 and should not play any role in changing the slope of the graph I made! Cool!
@@Mathing : That's an interesting way to see that the dogs circle the center infinitely many times. We can start with the fact that the path shape is independent of the speed, and we also know (from your drawing) that the dogs will go at least 45 degrees around the center. So we can see that the problem of finding the shape starting at a point when the dogs have gone 45 degrees around the center is the same as the original problem, just smaller and rotated -- and so the shape starting at 45 degrees is the same as the whole shape, just smaller and rotated. That means that the dogs aren't at the center when they've gone 45+45=90 degrees around the center either, and we can repeat this shrinking-and-rotating the shape ad infinitum.
And, for extra fun, it's almost exactly a retelling of Zeno's paradox. Instead of using 45 degrees (which is pretty arbitrary), we can take the point they get to after running for 1/2 time unit (which is also pretty arbitrary). In that time, they've gone some angle around the center -- call it A. So we follow the angle, and the next rotation of A around the center takes them 1/4 time unit, and they travel 1/4 distance unit. And the next one takes them 1/8 time unit and they travel 1/8 distance unit, and so on. In effect, the "steps" in Zeno's paradox are being counted out by the rotation angle around the center, and the resolution is the same as the resolution of the paradox: The sum of this particular infinite series is finite.
Also the dogs when they do reach each other will obviously be infinitely dizzy.
The dogs always move at right angles to each other, so from the point of view of one of the dogs it is just running straight to it's target.
Nice observation!
Agreed, this seems like the most obvious solution to me. Dog A is always running directly towards Dog B. Dog b's motion is perpendicular to Dog A's. Therefore the only factor which affects the distance between the 2 is Dog A's movement. Therefore it's just time = distance/speed = 1/1 = 1
I am pretty sure the curves are logirmatic spiral, we reparametrize the time so that the speed is proportional to the length of the enclosing square as in the video. the speed is proportional to the length of the side of the enclosing square but the rate of rotation should be constant (noticing in this reparametrization the curve will never reach the origin but the problem under rotation around the origin scaling), now it is too see the curve is of the form ke^at (reversing the time) where a and k are or some complex numbers or the solution curve of the vector field y'=ay, a can be found by noting the velocity is tangent to the side of the square at the corner and has speed one, now plugging in the initial condition we find the solution curve
I guess the difficult part is to notice that 5 point together with origin will form a fixed configuration as time passes, let a1, a2, a3 , a4 be 4 corner, one can show that they always form a square with the origin in the center by computing various derivative, for example (a1+a2+a3+a4)/4 to show that the center of the square is unchanged. now with this in mind, the solution is time invariant (if f(t) is a solution so is f(t+s))
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.
I had my moment of surprise. Thank you for sharing.
My pleasure❤️ Really appreciate it:)
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.
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........
When thinking about the problem on my own i felt like a fractal pattern of a spiral would be created, which could then mean that the distance traveled of each dog is (theoretically) infinite. But your video clearly disproves that, good work.
Thanks so much!
Your intuition is almost correct! It is a self-similar spiral; it's just that not all self-similar curves are fractal.
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.
1) At the end of their chase are the dogs spinning infinitely fast? If not, which direction are they facing when they cross?
2) What happens if one of the dogs is blind and always moves in the direction it started? How do the other dogs move?
3) What if one is a plush dog and doesn’t move?
Also you could apply some physics to the problem and, if you take, say, Dog 1 as your reference then Dog 2 by definition must be moving towards it in a straight line hence they must meet at t=1.
For the first one of these, you can use self-similarity of the problem to prove that the dogs are spinning infinitely fast.
This is very good video , amazing method , i have done similar question but it was triangle , but couldn't think of method like this. Also I liked the quote at last, please keep posting amazing videos ,I shared and subbed.
Thank you so much! Glad you liked it:)
Excellent explanation! I really enjoy your pace and process.
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 🧑🎓).
Nice video! I thought you would end up using integrals to solve it but I'm glad you didn't
I paused and solved it like this:
1. Observe that the entire problem retains 90 degree rotation symmetry as the dogs progress.
2. Observe that at any point in time, the current state is a scaled & rotated version of the original state.
3. From here, you can see that we're ultimately just modelling motion where at any point, the velocity is 1 unit/s at a 45 degree angle counterclockwise away from the line to the middle.
4. Time for some actual equations: In polar coordinates (with theta counted clockwise from "north"), dr/dt = -sqrt(2)/2, and dtheta/dt = -sqrt(2)/2/r(t)
5. Integrate with initial conditions (assuming the top-left dog): r(t) = sqrt(2)/2-sqrt(2)/2*t, theta(t) = -pi/4-(-ln(1-t))
6. Simplify: r(t) = sqrt(2)/2*(1-t), theta(t) = ln(1-t)-pi/4
7. r=0 at t=1, so with a constant speed of 1 unit/s, that means total arc length is 1 unit.
Nice! Atleast you didn't take a week to solve it hehe :) In my case, I wanted to solve it using the least bit of calculus I could use, and that cost me this solution
This was a brilliant video - super engaging! As an educational video creator myself, I understand how much effort must have been put into this. Liked and subscribed, always enjoy supporting fellow small creators :)
Much appreciated!
fun fact: the trajectory of the dogs is described by a logarithmic spiral. You can see this by constructing the differential equation of one trajectory and solving it with the substitution x = r(theta)cos(theta), y = r(theta)sin(theta), pluging in the parameters of the initial position
You can also skip cartesian coordinates altogether by looking at a small right triangle with sides dr and r*dθ. Through some geometry you can find that the two other angles must be 45 degrees, and so dr/dθ=r.
A logarithmic spiral feels like it makes sense to be the answer, especially since zooming in to the cente of the spiral yields the same spiral, albeit rotated, just in the same way the sequence of squares in the slow dogs approximation are scaled and rotated versions of each other, the next square being scaled from the previous by the same ratio.
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.
Wow - I never would have guessed that the spiral length is one! SO enlightening.
Did you try zooming into the center? Do the spirals keep winding around at the limit?
This is a great submission for SoME2!
Thank you so much phi! Your encouraging words pushed me a lot to complete this video❤️
I didn't try zooming in the center, but I tried to find out a similar fact mathematically..
The question I asked myself was, it looks as if the intermediate squares are just a rotated and scaled version of the original square- so I tried finding the total amount of rotation that the square connecting the dogs would undergo till they reach the center- and I found that to be infinity- Which I think should mean that the spiral will go on forever towards the center, getting closer and closer each time. Although the idea of zooming into the center is interesting, I might try it out!
@@Mathing "4 Dogs - The Sequel" :)
In polar coordinates, I found the parameterized path (for the dog at the top left) is r(t) = sqrt(2)/2*(1-t), theta(t) = ln(1-t)-pi/4, from which you can see that theta does increase (negatively) without bound. Thus, it completes an unlimited number of windings, despite it not looking like it when the path is graphed.
Zooming in should look identical, no matter how far you zoom, since the state of the dogs at any point is just a scaled and rotated version of the original state.
@@tejing2001 As long as the dogs are really tiny :)
@@phiarchitect What kind of mathematician wouldn't assume the dogs are infinitesimal? That's just crazy! :-P
Let s = sqrt(1/2).
Note the initial distance from the center is s.
Split velocities into radial part vr (towards the center) and tangential part vt (perpendicular to the radial line)
Notice that the dog always runs along the side of the square, which will always be the bisector of the radial velocity and tangential velocity. So vr = vt = -s, constant. For convenience I'll have the dogs run in whichever direction (clockwise vs anticlockwise) lets this be true.
r = r0 + vr * t = s - s t = s * (1-t)
dtheta/dr = dtheta/dt = vt / r = -s / r
so theta - theta0 = -s ln r + s ln r0
r = r0 e^-(theta - theta0)/s
Each dog reaches r=0 at t=1 after moving 1 unit along a logarithmic spiral.
ooo i figured out an easy way, bc of symetry the dogs are always in a square formation that is rotating, if you counter rotate the square at the same rate and keep dog one stationary visually (like the camera is in their pov) it would look like dog two has to run in a straight line to dog one wich is one unit so in one second
The clever solution I referred to- yes, congratulations that you got it!
I feel like one approach is to take the approximation you used further: It should be possible to prove that this one step approximation is converging to the real solution using methods of numerical differential equations. Then, we can add up the lengths we are interested in, take the limit, and the answer should be the same. I don't know how hard it would be to go through all the necessary hoops, but it seems like ot should be possible
Tried this problem and correctly guessed the polar form was of the type r=root(2) - c*theta (with origin in center of square), but I got caught up in the equations of the curves and didn't realize you could just solve for their length without actually having the equations. My second attempt tried to parametrize the motion and equate two curves using a rotation matrix but that also came up short. Maybe I could've gotten it but it would have definitely been a much longer solution
It is much simpler. Dog 1 is initially separated from Dog 2 by distance s, where s is the side of the square.
Dog 2 goes in a direction perpendicular to the one in which Dog 1 goes. In an infinitesimal time dt, you can assume that Dog 2 does not move towards or away from Dog 1. Dog 1 moves towards Dog 2 by some distance. This distance is the same as the reduction in the side of the next square formed after dt time has elapsed. So the dogs must travel a total distance of s before they meet at center of the original squares or any of the intermediate squares.
@8:24 wait. They're travelling at 1/s on a 1x1 square towards the center. A straight line is (1/2)x✓2 or 0.7(something). The fact that it takes 1s is a little mindblowing. Considering that ✓2 is irrational
What if there are 3 dogs in the triangle - will the conclusion stand?
2/3
just the last semester I had to solve that very same problem for a diferential geomtry homework... I remember the proffesor giving us a hint about squares that I didn't understand, it was probably something like this... I used complex numbers to do it , but this way is also quite nice
how did you do it with complex no.s??
@@kamakshigarg1289 I put the four dogs in the 4 fourth roots of 1 (ie +/-1 and +/-i) as a srart point. Then the derivative over time of a dog would be its position minus the position of the following dog (meaning that they always run toward the next one at a constant rate). Finally, because of symmetry, you have that the position of any dog is the position of the next one times the fourth root of 1 and there you have a differential equation that results in a complex exponential and therefore a exponential spiral... also if you change the 4th roots of 1 with nth roots you'll have the same problem but with n dogs 🤭
It might feel paraDOGcical :)
Great video!
10/10 pun
I paused and my hypothesis is that the dog they follow always move perpendicularly: If the distance didn't change they would move in circles, i.e. they would always remain at the same distance. Therefore they don't alter the amount of distance they have to travel.
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
Have the dogs each move N units forwards, and turn to face the dog they are to go towards, then repeat until they are within n units of the dog they are supposed to go forwards.
Next, take the limit of that as N approaches 0 from the positive side.
You can construct a nice drawing of this by drawing concentric squares, where each new square has it's vertices n units along the edge of the outer square. When looking at it this way, it become obvious that perhaps instead of n units, it should be 1/n th of the way along the edge. This makes the next square have the same relationship with the current square as the current square has with the previous square. This symmetry is likely useful for simplifying the math.
Again, we would take the limit to get the exact solution, but this time it would be as n approaches infinity (cause we want 1/n to get approach 0)
I'm not good enough to come up with an exact function for these curves (perhaps using polar coordinates would be helpful?).
Actually, doing some back of the hand math, that would have to be the polar function equivalent of a 45 degree (pi/4) line, would it not? The derivitive should be exactly 1 at every point along the line? So like, radius=angle?
Our physics teacher gave us similar puzzle with 3 flys in equaledral triangle
Great, you can actually use this same technique to solve it for any polygon, including your equilateral triangle!
In your case the graph will still be a straight line, but the slope will not be -1, it will be a different value.
Also for finding the slope, you'll need to derive a relation between any two consecutive triangles- here, you might find the extension of the Pythagoras Theorem for Acute angles useful (Or even the cosine rule of Trigonometry)
Awesome video!
My approach, which didn't pan out for me, but might for someone else.
Let dog C be in the origin and mark it's position as (x, f(x))
Then dog D's position is (1-f(x), x)
Since dog C is always directed towards dog D, we have
f'(x) = rise/run
= (D.y - C.y) / (D.x - C.y)
= (x - f(x)) / (1 - f(x) - x)
Unfortunately I wasn't able to get further, but solving this differential equation would get f(x), a closed form for the trajectory of dog C. Then one could do a line integral to get the length of the trajectory.
I'm speculating, but since the trajectory turns out to a spiral which fails the vertical line test, the trajectory isn't actually the function f(x). f(x) is just a the first portion of the trajectory. Dealing with this could get messy - I think the full trajectory would be a union of piecewise functions each of which is a solution to a slightly different differential equation or maybe each is a solution to the same differential equation, but with different boundary conditions...?
Interesting, the way you positioned D with respect to C using symmetry is cool!
(EDIT: Skip the next passage where I make a fool out of myself hehe:)
I'm not (yet) familiar with solving differential equations, so pardon any mis-observations: I think in case of
f'(x) = (Dy - Cy) / (Dx - Cy)
Here f'(x) is calculating the slope of the line connecting C and D together right? This should not be tangent to their trajectory, so I think this might not give the desired results?
Also, the trajectory, amazingly swirls inside infinitely- so if you keep zooming more and more, you'll find the same curve rotating (Pardon me but I am out of formal language to describe what I'm trying to say:))
CORRECTION: My bad, yes it must be tangent because C is by definition following D :)
@@Mathing yes, f'(x) is notation for the derivative of f(x).
The way it's set up guarantees that C's velocity vector points to D always.
Interesting, the curve has finite length, but makes infinitely many turns around the center point.
So that means the dog make an infinite number of turns in finite time.
As you may know acceleration not measures only change in speed, but also change in direction - velocity is a vector having magnitude and direction, after all. Here the acceleration is purely due to change in direction as the speed remains 1 unit/sec at all times. Acceleration due to change in direction is proportional to the curvature of the trajectory. At the center, I think acceleration might go to infinity since (I think) the curvature also goes to infinity at the center.
you can make this idea work. as you have observed, the dogs make infinitely many turns around the center of the square, so the trajectory of a dog cannot be given by a function y=f(x) - what you instead need to do is make both y _and x_ functions, so the trajectory is P(t) = (x(t), y(t)). This will lead to a system of differential equations which you can solve to determine x(t) and y(t), and then you can calculate the arclength of the trajectory using an integral. (If you decide to try this out then you should put the vertices of the square at (0, ±1) and (±1, 0) instead of (±1, ±1) as this will make the calculations a lot nicer)
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
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.
Interesting approach!❤🧡💛
Thanks! Glad you like it:)
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?
hello I really enjoyed your video it was beautiful I just wanted to ask you a question, how did you get the limit to be equal to -1?
I would like to see your calculations 😁
Thanks! I'll provide it here:
Lim(t→0) (sqrt( (S_n - t)^2 + t^2 ) - S_n) /t
Here, as t approaches 0, the sum
(S_n - t)^2 + t^2 can be considered as
(S_n - t)^2 since t^2 quickly converges to 0.
So this makes the limit simpler:
Lim(t→0) (sqrt( (S_n - t)^2 ) - S_n) /t
= Lim (t→0) (S_n - t - S_n)/t
= Lim (t→0) (-t)/t
= -1
@@Mathing to show the limit rigorously (saying that one part of an expression converges "more quickly" than another is good heuristically but it is not a proof) you can just use L'Hopital's rule (which says that if f(t)/g(t) is an indeterminate form when you plug in t=a then lim(t→a) f(t)/g(t) = lim(t→a) f'(t)/g'(t). in this case the indeterminate form is of type 0/0)
@@schweinmachtbree1013 I think you need more than that. S is a function of t, so you actually have d/dt √(S²−2t) − S = (SS'−1)/√(S²−2t) − S'.
@@EebstertheGreat Indeed each S_n is a function of t, but L'Hopital's rule still works. (since S_0(t) = 1 and S_{n+1}(t) = sqrt((S_n(t) - t)^2 + t^2) we have S_n(0) = 1 for all n. You get terms involving S_n'(t) when you calculate d/dt[sqrt((S_n(t) - t)^2 + t^2) - S_n(t)], but when you evaluate at t=0 you don't actually need to know S_n'(0) because the terms cancel in the end.
@@schweinmachtbree1013 I don't get it. That holds for S(0), but we are interested in showing that the slope is constant, so we can't just look at one point. We need it to hold for every point.
Masha Allah congratulations Saad. You did it! This is speach lessly amazing! You are very gifted bhaiya. Allah has blessed you with a very unique talent! ❤️ Never give up no matter what the situation is because it doesn't ever take away the awesomeness you have in you. You always have you and your gift, no matter where you are. I believe in you that you will create magic from being anywhere in the world! You will make it a better place Inshallah ❤️ Sawda apu loves you a lot! May Allah bless you with the best always 🤗❤️
Thank you so much apu❤️ These words mean a lott
@@Mathing ❤️❤️❤️🤗🤗😅
Wowza, awesome, immediately subscribed.
Thanks so much
I made a video that generalizes this problem and solve in detail the system of differential equations tha arises to derive the exact equation of the trajectory.
I also show where those spirals appear. Look it up on my channel, it is called : Zeno's Mice (Ants) Problem and the Logarithmic Spirals.
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.
This is a wonderful submission! It really reflects how I think most mathematicians might solve a problem - yeah, you can exploit symmetry in *this* case, but if you were to make slight tweaks to the problem, your solution generalizes, while other solutions do not. Kudos!
Thanks so much
Really cool solution!
fantastic video! keep up the good work!
So I'm thinking that it'll be a square inscribed in a circle, and then it's just figuring out the length of a spiral like that. Not that I know how to calculate that.
Edit: after giving it a bit more thought, since it'll always be a square with a shrinking side length, it should take one second, and the distance traveled should be 1 unit, even though the math to figure it out is probably hard.
Basically we can imagine the dogs on the corners of a circumscribed square where the r is slowly getting smaller and the theta is increasing(for polar coordinates)
The dogs might move on a square geometry whose dimention is 1 unit in a non-euclidean reference system at constant speed
Posting this before watching the rest of the video: I'm considering putting together a system of differential parametric equations, and using Eulers method to solve it numerically. From there finding the distance of each path should be simple.
But it's late, so I'll watch the video now.
Subbed after i saw the thumbnail
Drawing the shape of the spiral was a challenge, but right away I noticed that as the dogs move, they're always on the verteces of a square, thus they're moving perpendicular to their neighbors. Therefore they neither increase or decrease the distance from their chaser, who always closes in on them in 1u/s. That's how I figured they'll all meet after 1s or 1u
If they neither increase nor decrease their relative distances, then they never meet
@@ethandavis7310 each dog doesn't get away/toward the one chasing them, but the chaser closes in on them
The answer is not paradoxical. The dogs move toward each other so that they are always moving at right angles to the target dog. Therefore the motion of the target dog does not have any influence on the distance the pursuing dog most travel. This is intuitively obvious.
I would have loved to see the limit calculation
Here you go:
Lim(t→0) (sqrt( (S_n - t)^2 + t^2 ) - S_n) /t
Here, as t approaches 0, the sum
(S_n - t)^2 + t^2 can be considered as
(S_n - t)^2 since t^2 quickly converges to 0.
So this makes the limit simpler:
Lim(t→0) (sqrt( (S_n - t)^2 ) - S_n) /t
= Lim (t→0) (S_n - t - S_n)/t
= Lim (t→0) (-t)/t
= -1
Where can I find similar questions?
What about the 3 dogs in a triangle problem?
You can actually use this same technique to solve it for any polygon, including an equilateral triangle!
In your case the graph will still be a straight line, but the slope will not be -1, it will be a different value.
Also for finding the slope, you'll need to derive a relation between any two consecutive triangles- here, you might find the extension of the Pythagoras Theorem for Acute angles useful (Or even the cosine rule of Trigonometry)
AMAZING!!!
i wonder what the formula for the path is, i would almost guess a 5th degree bezier curve, but those have nonconstant velocities
It's a polar exponential curve; r = e^(k*theta) for some constant k.
Very nice video.
But I think the soundquality still got potential
Thanks for your feedback! I agree, I need to improve my sound equipment
This is a very interesting video
Firstly, science bujina🙂. Secondly, english bujina. Kmne ki krbo bol🥺!!
You are doing absolutely great, bro!!
Your participation is much appreciated brother❤️ Thanks
What is the short way to solve it ?
Wooow, this is fantastic
Hi, do you know what happens when t is negative, could you possibly test it out with your code :)
Another question i had was the nature of the tilings. are there smaller sub square(s) that is a complete 360 degree rotation from the big square we start with?
Interesting point!
Here's what it looks like when t = - 0.2 ...Amazing!
drive.google.com/file/d/1GsIMW-I-PUjiBRjOXvcfzrWSuGZSBSp-/view?usp=sharing
About the nature of the sub-squares, when t approaches 0, the curve as it tends towards the center, rotates for an infinite amount- so you can in theory keep on zooming towards the center for infinity and you'll find the same curve just rotating around! So this means yes, there are infinitely many sub-squares formed by the dogs that are a complete 360 degree rotation from the original one!
This will help me in jee!
Also, does the speed even matter? Looks like the path should be the same as long as all dogs have the same speed.
That was a fun watch!
Thanks so much!
Masha Allah
mathematically bred dogs made my day
Nice video, though I do feel like you are missing some kind of argument about the uniform convergence of slopes to make your reasoning truly rigorous.
Anyway, great video!
Btw, I'm not sure I fully agree with your last statement. Usually mathematicians are way more interested in developing powerful methods that can solve multiple problems than to know solutions of a particuliar problem. Nevertheless, it's true that solving one problem multiple ways can be pedagogically very enlightening
Thank you so much! This submission has been really helpful for me in ways that it has given me the opportunity to learn more and go deeper than the shallow depth I'm in, and I am really grateful for this. I will make sure to check out 'uniform convergence of slopes' to get better at rigorous terms, thank you for this!
And yes, the ending statement is only partially true, that for inspiring people to take different journeys that might bring up newer approaches and 'more powerful tools'. But I agree with the case of powerful methods- they do feel magical in their own way.
Thanks for your insight!
What the name of this shape/image (spiral in a square)?
I believe it’s called a logarithmic spiral
@@Mathing okay thank you sir
perfect
Good solution
Thanks man!