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ChesterDog
ChesterDog
Joined: Jul 26, 2010
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April 8th, 2020 at 10:03:50 PM permalink
Can anyone solve this set of differential equations:

dy/dt = 1 - (y + t -12)2/18
(dx/dt)2 + (dy/dt)2 = 1
with dx/dt>0 before x reaches 6, and dx/dt < 0 afterwards.
x(6) = 3*sqrt(3)
y(6)=3
?

(Edit: I corrected the initial conditions' time.)

Even a numerical solution would be nice.

I believe the solution is the curved part of the path to avoid the selfish walker.
Last edited by: ChesterDog on Apr 9, 2020
ChesterDog
ChesterDog
Joined: Jul 26, 2010
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Thanks for this post from:
tringlomane
April 9th, 2020 at 11:12:09 AM permalink

This solution assumes the two walkers must maintain the same constant speed. I set my unit of time to make the speed 1 foot / unit.

I couldn't solve the above differential equation, but typing y' = 1 - (y+t-12)^2/18; y(6)=3 into WolframAlpha gives a nice solution for y.

Then x = sqrt[36 - (y + t - 12)^2 ] to keep the distance to the selfish walker 6 feet.

x' = - sqrt( 1 - (y')^2 ) for after x becomes 6. Next, dy/dx = y' / x' which we set equal to the slope of the line to the destination, (12 - y) / (0 - x).

Trial and error gives t = 8.723448, x = 5.634036, and y = 5.339957.

The length of the straight line between (x, y) and (0, 12) is sqrt( (12-y)^2 + (0-x)^2 ) = 8.723448, which happens to equal the time for the sum of the initial straight path and the curved path!

So, the total distance is 17.446895 feet.


rsactuary
rsactuary
Joined: Sep 6, 2014
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Thanks for this post from:
ChesterDog
April 11th, 2020 at 5:45:47 PM permalink
https://fivethirtyeight.com/features/can-you-catch-the-free-t-shirt/

Answers to this puzzle are now posted. Along with new puzzles. ChesterDog appears to have the right solution (at least numerically, didn't check the formulae).
Wizard
Administrator
Wizard 
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April 11th, 2020 at 6:59:16 PM permalink
There is a nice animated solution at that link. I would hate to say how many pieces of paper are in my trash right now about that problem. Not to mention a spreadsheet that attempted to do it in micro-steps. Big fail on my part with that one. I'm going to think twice about earning extra credit next time.
It's not whether you win or lose; it's whether or not you had a good bet.
rsactuary
rsactuary
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April 11th, 2020 at 7:59:53 PM permalink
It's surprising to me that the path isn't symmetric, along the lines of what was proposed in this thread. I never would have thought outside of that box.
ssho88
ssho88
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April 11th, 2020 at 8:35:14 PM permalink
Quote: rsactuary

It's surprising to me that the path isn't symmetric, along the lines of what was proposed in this thread. I never would have thought outside of that box.



May I know what is the exact angle between the INITIAL RED STRAIGHT PATH and the horizontal BLUE line ? Not exact 60 dgeree ?
ChesterDog
ChesterDog
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April 11th, 2020 at 9:00:51 PM permalink
Here's the derivation of the differential equation, but not its solution:

Starting at coordinates (0,12), the selfish walker's path is described by x=0 and y=12 - t, assuming his speed is 1 foot/unit time.

A 6-foot circle around him would be x2+ (y - 12 + t)2 = 36 (Eqn. 1).

The other walker's trajectory is counterclockwise about this circle, with a speed of 1. His speed squared can be expressed as: xt2 + yt2 = 1 (Eqn. 2).

Differentiating the path with respect to t gives: 2xxt + 2(y-12+t)(yt+1) = 0. And solving for xt gives: xt = (12-y-t)(yt+1)/x (Eqn. 3).

Equations 1, 2, and 3 can be combined to eliminate x and xt to give:
18 yt2 + (y-12+t)2yt -18 + (y-12+t)2 = 0

This can be solved for yt using the quadratic formula (using the + instead of the -) to give:
yt = 1 - (y+t-12)2/18.

And the initial condition is y(6) = 3.
ssho88
ssho88
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April 11th, 2020 at 9:19:27 PM permalink
ChesterDog, is the angle between initial red straight path and horizontal blue line EXACT 60 degree ?
ChesterDog
ChesterDog
Joined: Jul 26, 2010
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April 11th, 2020 at 10:08:59 PM permalink
Quote: ssho88

ChesterDog, is the angle between initial red straight path and horizontal blue line EXACT 60 degree ?



Yes--it's exactly 60 degrees.

A good way to prove that is this: If the selfish walker starts at B (0,12), then if he would walk for 6 feet, he would be at O (0,6). Imagine a 6-foot-radius circle centered at O (0,6).

If the other person started at A (0,0) walking 60 degrees from straight ahead, he would touch that circle at C, 6 feet from his starting point. And he would arrive at the circle at the same time the selfish walker reached (0,6).

In other words, both walkers went 6 feet to be 6 feet away from each other. Think of the circle O with an equilateral triangle AOC having the circle's radius AO as its base.

Edit: Now, I see that the above doesn't prove that the minimum total distance is achieved by starting at 60 degrees.
Last edited by: ChesterDog on Apr 11, 2020

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