At time t=0, point A is at the origin (0,0) and point B is distance 1 appart at (0,1). A moves to the right (on the x-axis) with velocity 1 and B moves with velocity 1 towards A. What will be the distance between A and B when t goes to infinity?
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18 Responses to Test Your Intuition 35: What is the Limiting Distance?
Not sure what you mean by "towards A". what happens if it reaches A as A approaches B. Since A continues to move right I understand that B will "follow" A with the same speed (both 1) and thus distance is 0.
But, can B reach A?
Ah, I misread B as starting at (1,0). This makes more sense.
I couldn't see how one can get intuition about this problem, so I cheated and solved the ODE.
Conclusion: as long as B does not start exactly to the right of A it will never catch up: there will be a positive limiting distance.
Very very nice! I got it wrong, although I realized the flaw in my intuition shortly after guessing.
After some though, I came up with an elementary way to get the right answer without solving the ODE (which I didn't do mostly out of laziness), but I'm not sure if it counts as "intuition".
A follow up: Suppose that the same process actually started before t=0 (and the conditions given are what was observed in medias res). At the negative starting time, B was actually on the line y = 3. What was the starting distance between A and B?
can you give a hint on your method?
I also think I got it with an elegant method. I can give you hint if you send me an e-mail at parity@gmail.com, as there might be people here who don't want to see any hint.
Can anyone provide a hint for this? I believe the limiting distance is smaller than 1/2, but I don't know how to find the precise one.
I'm curious why you think the limiting distance is less than 1/2.
Write $|AB|=:s$. Then $t-x=s \sin\theta= s\dot x$, y=s\cos\theta=-s\dot y$. Differentiate $(t-x)^2+y^2=s^2$ and simplify. You obtain $\dot s=\dot x-1=-(s\dot x)^\dot$. Draw conclusions.
I'm proud of my solution, but perhaps others do not want the spoiler, so here it is in cipher (enter text at https://www.rot13.com/ to decode).
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where W is the LambertW function.