NY Times math problem

Nope, when the rabbit is close enough to a point on shore that he can reach it before the agent, the rabbit will take a straight line path. It no longer matters if it is 180 degrees from the agent.

- William Hughes

Beacause the sped of agents is four times the rabbits speed, then the rabbit get away.

The rabbit escapes if speed of agents is

;

Please give the greatest agent speed for which the rabbit can escape using this strategy.

Dave

Point. But the path should be a spiral before that.

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How is this figure derived?

t e

Missing a semicolon somewhere?

Michael

Assuming a pond's radius =3D 1 and the rabbit's speed =3D 1: If the agent's speed is v, the rabbit can position himself 1/v from the center on the opposite side of the pond from the agent. He then heads radially toward the shore and watches to see which way the agent runs. When the agent has committed himself to a direction, the rabbit turns

90 degrees in the opposite direction and heads in a straight line toward the shore. If the agent reverses direction, the rabbit will again head out on a radius until the agent, the rabbit, and the center of the pond again are collinear or the agent reverses again. At this point, the rabbit will be closer to the shore before and the agent will find himself in a worse position; thus, once he commits to a direction, the agent cannot afford to reverse it. The rabbit must swim sqrt(1 - 1 / v^2) while the agent runs pi + arccos(1 / v). So solving

sqrt(1 - 1/v^2) =3D ( pi + arccos(1 / v) ) / v

for v gives the number in question.

Dave

That solution presumes the fox/agent doesn't move immediately.

Of course the problem is poorly stated, much like all requests for circuit help here ;-)

...Jim Thompson

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If the pool is a point, the agent wins, trivially. But in the limit as the pool shrinks to radius zero, the rabbit wins.

(and this is why first-year calculus students get headaches, at least until they learn to just push the symbols around and forget about it :-) )

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e

No. It presumes that both the rabbit and the agent react instantly. So the instant the rabbit gets more than 1/v from the center, the agent chooses a direction and starts running. If he doesn't, the rabbit can just swim straight to the shore.

Dave

M Rath wrote:

No the compiler is not accepting "ads:= Pi / 2" as a test value to develop formulas in a loop. Now the compiler might be saying "What's the point of calculating the same result over and over in a loop ?" but it did not accept "ads:= Pi/2" with the loop commented out. But in both cases it accepts the numerical value...

Observation: I once chased a jackrabbit. I was on a motorcycle, he was not. The rabbit got away.

Given his preternatural speed, the agent should fox the rabbit into quitting the safety of the pond, then give chase--he'll have the bunny in a few strides.

Cheers, James Arthur

Okay the circular pond has North and East coordinates at the radius point of

10 , 10 and the radius is 10 units . The agent is at North and East coordinates of 20 , 10 and moves clockwise in 1-degree steps. The rabbit at the center of the pond moves away from the agent at each step on a line between the agent and the rabbit. And the rabbit moves 1/4 the distance that the agent moves.

The rabbit gets out at North and East coordinates of 1.3437 , 4.9694 when the agent is at an azimuth of 232 degrees on the circle edge. But the rabbit is out at an azimuth of 210 degrees or so. In other words the rabbit doubled back while the agent kept going. Before the rabbit doubled back it looks like a slow outward spiral but I will write a ScratchPlot file later.

Of course this is one systematic method with no answer for what the agent does when the rabbit is within one unit of shore and near the agent.

Here is the KBH code:

Var ads, nag, eag, rba, rds, sec, rbn, rbe, nd, ed, dgr, dst: double; flg: integer;

Procedure D; Begin If (flg = 1) Then Begin {Procedure E} rba:= rba + Pi; Exit; End; rba:= rba + (Pi * 2); {The structure of procedures D, E, & F are from the KBH Survey Progam for the HP35S} End;

Procedure F; Begin If (flg = 1) Then Exit; rba:= rba + Pi; End;

begin {KBH Code} flg:= 0; sec:= 1 / 3600; rbn:= 10; rbe:= 10; rds:= 0; dgr:= 1;

While (dgr < 233) Do Begin ads:= (dgr * Pi) / 180; nag:= 10 + Cos(ads); eag:= 10 + Sin(ads); nd:= rbn - nag; If (nd = 0) Then nd:= 0.000000000001; ed:= rbe - eag; If (ed >= 0) Then flg:= 1; rba:= ArcTan(ed / nd); If (rba < 0.000000000001) Then D Else F; rbn:= rbn + (Cos(rba) * 0.043633231); rbe:= rbe + (Sin(rba) * 0.043633231); dst:= Sqrt(Sqr(10 - rbn) + Sqr(10 - rbe)); WriteLn(rbn:15:4, rbe:15:4, dgr:15:4, dst:15:4); dgr:= dgr + 1; End;

ReadLn; end.

Oh...that KBH Code will run in Delphi Console Mode.

The following is a file for KBH ScratchPlot. Copy to a text file and save the file as "Name.plt" with the quotation marks and then the file will run in ScratchPlot. And of course the file should contain text...not the html being picked up here.

I didn't convert the entire coordinate file to a ScratchPlot file...I just entered 15 points for the rabbit as lines. Then I laid out 15 points for the agent from the radius point with the Fan and Azimuth features. Then I added

15 points to the rabbit line so that the points of the agent could be matched up to the points of the rabbit. For instance both the agent and the rabbit were at 220 degrees clockwise azimuth but at different times. I could have added a full circle but that would make the file very large... 663&137@F+bkN0E0 41&759@S+bkN20E20 0 352&448@P+bkN10E10 353&447@P+rdN9.9564E9.9992 385&442@L+rdN8.9295E9.8114 416&429@L+rdN7.9231E9.3937 445&412@L+rdN6.9807E8.8454 473&392@L+rdN6.0868E8.2205 500&372@L+rdN5.2194E7.5591 527&351@L+rdN4.3545E6.8944 555&331@L+rdN3.4713E6.2543 583&312@L+rdN2.5578E5.6584 595&306@L+rdN2.1834E5.4343 600&302@L+rdN1.9944E5.3254 606&299@L+rdN1.8042E5.2184 612&296@L+rdN1.6130E5.1134 618&292@L+rdN1.4208E5.0102 621&291@L+rdN1.3437E4.9694 352&448@P+bkN10E10 352&448@P+bkN10E10 70&579@P+bkN19.0631E14.2262 352&448@+bkN10E10 41&453@P+bkN19.9985E10.1745 352&448@+bkN10E10 152&686@P+bkN16.4279E17.6604 352&448@+bkN10E10 271&748@P+bkN12.5882E19.6593 352&448@+bkN10E10 406&754@P+bkN8.2635E19.8481 352&448@+bkN10E10 530&702@P+bkN4.2642E18.1915 352&448@+bkN10E10 621&603@P+bkN1.3397E15.0000 352&448@+bkN10E10 661&475@P+bkN0.0381E10.8716 352&448@+bkN10E10 644&341@P+bkN0.6031E6.5798 352&448@+bkN10E10 621&292@P+bkN1.3397E5.0000 352&448@+bkN10E10 606&269@P+bkN1.8085E4.2642 352&448@+bkN10E10 590&248@P+bkN2.3396E3.5721 352&448@+bkN10E10 571&228@P+bkN2.9289E2.9289 352&448@+bkN10E10 551&209@P+bkN3.5721E2.3396 352&448@+bkN10E10 543&202@P+bkN3.8434E2.1199 352&448@+bkN10E10 352&448@P+bkN10E10 41&453@P+prN19.9985E10.1745 352&448@+prN10E10 41&453@P+bkN19.9985E10.1745 352&448@+bkN10E10 385&442@P+bkN8.9295E9.8114 416&429@P+bkN7.9231E9.3937 445&412@P+bkN6.9807E8.8454 473&392@P+bkN6.0868E8.2205 500&372@P+bkN5.2194E7.5591 527&351@P+bkN4.3545E6.8944 555&331@P+bkN3.4713E6.2543 583&312@P+bkN2.5578E5.6584 595&306@P+bkN2.1834E5.4343 600&302@P+bkN1.9944E5.3254 606&299@P+bkN1.8042E5.2184 612&296@P+bkN1.6130E5.1134 618&292@P+bkN1.4208E5.0102 621&291@P+bkN1.3437E4.9694

CORRECTED CODE !

No...the plot didn't look right. And so in the previous code the coordinates of the agent were based on a radius of 1 but should have been based on a radius of 10.

And with this corrected code the Rabbit does not get out of the pond with this systematic method ! But I'm sure the plot will make some nice curves...which is what I was looking for.

Here is the corrected KBH code:

Var ads, nag, eag, rba, rds, sec, rbn, rbe, nd, ed, dgr, dst: double; flg: integer;

Procedure D; Begin If (flg = 1) Then Begin {Procedure E} rba:= rba + Pi; Exit; End; rba:= rba + (Pi * 2); {The structure of procedures D, E, & F are from the KBH Survey Progam for the HP35S} End;

Procedure F; Begin If (flg = 1) Then Exit; rba:= rba + Pi; End;

begin {KBH Code} flg:= 0; sec:= 1 / 3600; rbn:= 10; rbe:= 10; rds:= 0; dgr:= 1;

While (dgr < 360) Do Begin ads:= (dgr * Pi) / 180; nag:= 10 + (Cos(ads) * 10); eag:= 10 + (Sin(ads) * 10); nd:= rbn - nag; If (nd = 0) Then nd:= 0.000000000001; ed:= rbe - eag; If (ed >= 0) Then flg:= 1; rba:= ArcTan(ed / nd); If (rba < 0.000000000001) Then D Else F; rbn:= rbn + (Cos(rba) * 0.043633231); rbe:= rbe + (Sin(rba) * 0.043633231); dst:= Sqrt(Sqr(10 - rbn) + Sqr(10 - rbe)); WriteLn(rbn:15:4, rbe:15:4, dgr:15:4, dst:15:4); dgr:= dgr + 1; End;

ReadLn; end.

And that KBH Code will run in Delphi Console mode...

Then plot the coordinates in KBH Scratch Plot...