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Question:

I was reading this analogy of Fermat's Principle of Least Time:

In Figure, our problem is to go from A to B in the shortest time. To illustrate that the best thing to do is not just to go in a straight line, let us imagine that a beautiful girl has fallen out of a boat, and she is screaming for help in the water at point B. The line marked x is the shoreline. We are at point A on land, and we see the accident, and we can run and can also swim. But we can run faster than we can swim. What do we do? Do we go in a straight line? (Yes, no doubt!) However, by using a little more intelligence we would realize that it would be advantageous to travel a little greater distance on land to decrease the distance in the water because we go so much slower in the water.

I am thinking that our speed on land is faster than in water so to reach in the least time, we must minimize our distance in the water. So we would take the path $AMB$ ($MB\perp x$).

Why would we take path $ACB$?

Is it because $AM$ is increased by a large factor?

Diagram

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  • $\begingroup$ Hi @Mayank. Welcome to PSE! I hace put a brief answer to your question below, and I noticed yo said you were in year 10, and since I omitted a trivial proof below, let me know if you would like to see it explicitly. $\endgroup$ – joseph h Sep 6 '20 at 6:14
  • $\begingroup$ I would like to give it a try. Maybe I can learn something new from it. $\endgroup$ – Wolgwang Sep 6 '20 at 6:28
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For comparable distances, $AM \approx AC$ and $CB \approx MB$ you would think that both ACB and AMB will take similar times since you are running and swimming for roughly equal distances. But you have said that "AM is increased by a large factor" meaning AM >> AC (note that this condition would imply that the angle x will not necessarily be 90 degrees) and therefore MB > CB. It is then trivial to show that since $t_{AC} < t_{AM}$ and $t_{CB} < t_{MB}$ and that since

$ t_{AMB} = t_{AM} + t_{MB}$

and

$ t_{ACB} = t_{AC} + t_{CB}$

then $t_{ACB}$ will be smaller than $t_{AMB}$.

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  • $\begingroup$ I haven't said“$AM$ is increased by large factor”. I was thinking that it is the reason for taking path$ACB$. $\endgroup$ – Wolgwang Sep 6 '20 at 6:35
  • $\begingroup$ @Mayank so you would like to compute this problem ignoring the "red line" path? Do you want to do it considering the path light would take? $\endgroup$ – joseph h Sep 6 '20 at 7:52
  • $\begingroup$ @Mayank. OK. What you would need to do is open another question asking about light and Fermatt's principle. All you would need to is using the same diagram and ask why would light take that particular path. Thanks. $\endgroup$ – joseph h Sep 6 '20 at 8:15
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Yes, you're right. If you go to $M$ instead of $C$, the increased time on land is so much more than the decreased time in the water that the overall time is longer. You can show this mathematically by setting variables for the relevant lengths and speeds and then minimizing the total time required; you'll find that the location of $C$ is always between the foot of the perpendicular from $A$ to $x$ and the foot of the perpendicular from $B$ to $x$.

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Which path you must take to save the girl before drowning (so in the shortest time) depends on the difference between your running velocity and your swimming velocity. If this difference is zero then you have to go to her in a straight line AB. When the difference is tiny, this straight path will be a bit different from the straight path. The bigger the difference the more the path will become one with one of both paths (swimming or running) perpendicular to the shore. There is no path that includes a point to the right of M (or to the left of the point beneath S).

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