The cockroach needs to run from start (from different positions of the start line) to finish-line in minimum time.
On the field of running there is a section in the form of a circle (see Fig.), where the running speed
differs by V1> V2 or V1 <V2.
Which route (trace) from the start line to the finish line should the cockroach use from different positions?
Better to get numerical, symbolical and plot solution (with animation!).
The case of V2<V1 doesn't make much sense. In that case it's clear that the cockroach must avoid the circle. Then there are an infinite number of equivalent solutions where the cockroach starts at some point on the start line and runs perpendicular to that to the finish line.
If V2>V1 the solution is obvious. Start above the center of the circle, run perendicular to the start line to the finish line. That's the shortest possible path, with as much of it as possible in the circle.
Edit:. You added a new post while I was typing, and now I see that the start point is not a variable, but fixed. That makes the solutions less trvial 
While it's not my problem, I'd suggest the following constraints and clarifications
a.) travel in straight lines only
b.) If the circle is entered, 2 direction changes total are allowed (when you enter circle and when you leave)
c.) If the circle is not entered, 1 direction change is allowed (allowing you to avoid the circle)
d.) the roach is infinitely thin (allowing it to pass tangent to circle)
Just as I finished typing I see your later constraints. But I figured I'd send this anyway.
Sorry, Ai nou Inglish veri plokho!
The problem one more!
I would run from one point on the start line to the finish line with minimal time.
My speed on the field is a function v(x, y).
The 1-st case - in a circle (square, triangle etc) v1>v2 or v1<v2.
What form must have my way (trace) from any point on the start line to finish line?
One solution for case not with 2 lines but with 2 points:
The rules need to be established. Straight lines only, Circle is equidistant from each line.
(i) fastest direct route any condition at V1 is straight across field of play. red, cyan, red
(ii) fastest route if you need to end at same point is directly to and from centre of circle path red, yellow, red. Difference is cyan at V1 or yellow at v2
(iii) fastest route if allowed to arrive at different point may be red,yellow, blue, or red magenta, darker cyan compared to red,cyan, red direct route.
Does this fit the rules.
@terryhendicott wrote:
The rules need to be established. Straight lines only, Circle is equidistant from each line.
(i) fastest direct route any condition at V1 is straight across field of play. red, cyan, red
(ii) fastest route if you need to end at same point is directly to and from centre of circle path red, yellow, red. Difference is cyan at V1 or yellow at v2
(iii) fastest route if allowed to arrive at different point may be red,yellow, blue, or red magenta, darker cyan compared to red,cyan, red direct route.
Does this fit the rules.
1. Straight start and lines
2. Circle is equidistant from each line
3. Allowed to arrive at different point on the finish line
It took a little longer. Unlike (possibly) Valery, I don't have cockroaches in the house. I do have silverfish (Lepisma saccharina) in my house, but they are much harder to train....
Anyway, here's the input:
And the output is shown below, for s=1, r=1 v1=1 and v2=2 and b positions ranging from -2 to 2:
Luc
Luc! I showed your animation to my cockroaches. They liked it. But one (in the middle - the egghead - as you and me) asked to start the animation from the point (case) when they run faster, bypassing the circle - from x =-3 to 3 for example. And for thr case when v.circle<v.field!
My silverfish have come across an interesting, but not completely unexpected phenomenon.
Observe the running field as shown below:
Notice that I've indicated s, the shortest distance between either side of the circle and the start and finish lines, and r is the radius of the circle. Then we have points b, the x-position on the start line, f, the x position on the finish line and i and o the entering and exiting points on the circle. v1 is still the speed of running outside, and v2 inside the circle. For each of these items you'll find values at the top of the picture.
Then there is angles alpha.i and beta.i, which are the angles between the normal (brown line from the circle centre through point i) on the circle edge and the entering path outside and inside the circle corresponding to point i. There's also the normal for point o, and I can define corresponding angles alpha.o and beta.o.
Now watch the animation below and pay attention to alpha.i and beta.i, and the ratio of their sines (shown at the top of the picture).
As you can see, the ratio of the sines of those two angles remains a constant (there's a division by zero problem in the middle of course), and that constant is the ratio of run speeds: v2/v1. This is Snell's law of optics.
Incidentally, I've calculated the run time for each path, and show the value as t. The small blue cross indicates that speed along the y-axis for each of the x-values of b.
Have fun!
Luc
P.S My silverfish are starting to learn how to deal with v1>v2:
My cockroaches long applauded all with their six palms and shouted "Bravo!", when they saw this animation and the remarks!
But the egghead one has some wises - a little and a big.
At first the big. To create not one running trace but a set of traces at v2=0 v2=.. (v2<v1) v2=... v2=v1 v2>v1 etc v1 is fixed!
The little wishes on the picture
My egghead cockroach (with name Pythagoras) long waited for a complete solution. He did not wait, got out of his hole, climbed on my computer and found a simple solution. He likes units and the minimize function - see please the attach.
Now he is thinking over the case when v2<v1 (v2=0 at first). And about optic, angles and the Snell law too!
Two frames from the movie The Run based on the M.Bukgakov book.
https://www.youtube.com/watch?v=MbruWic6JzA
Names of cockroach are Janissar, Button, Pearl of the East, Baba-Yaga, Hooligan and Do not cry, child
