MHB Using Snell's Law to determine the fastest escape route

[MarkFL]

On another forum the following question was asked:

If a prisoner can swim at a rate of 6 fps and run at a rate of 12 fps, what angle should the prisoner take to escape as fast as possible? He's going across a river. The distance upstream to the safehouse is 10000 feet and the river is 2000 feet wide. Thanks Everybody! :D

This was my initial reply:

Let:

$L$ = distance upstream to the safehouse
$W$ = width of the river
$v_s$ = swimming speed
$v_r$ = running speed

All variables above are assumed to be positive.

I have set it up as follows:

https://www.physicsforums.com/attachments/2775._xfImport

The prisoner swims from A to B and runs from B to C. The distance from A to B is:

[math]AB=W\csc(\theta)[/math]

The time this takes is:

[math]t_s=\frac{W}{v_s}\csc(\theta)[/math]

The distance from B to C is:

[math]BC=L-W\cot(\theta)[/math]

The time this takes is:

[math]t_r=\frac{1}{v_r}(L-W\cot(\theta))[/math]

The total time then is:

[math]t=t_s+t_r=\frac{W}{v_s}\csc(\theta)+\frac{1}{v_r}(L-W\cot(\theta))[/math]

[math]\frac{dt}{d\theta}=-\frac{W}{v_s}\csc(\theta)\cot(\theta)+\frac{W}{v_r}\csc^2(\theta)=W\csc(\theta)\left(\frac{1}{v_r}\csc(\theta)-\frac{1}{v_s}\cot(\theta) \right)=0[/math]

Observing [math]\csc(\theta)\ne0[/math] for all real $\theta$, we then use:

[math]\frac{1}{v_r}\csc(\theta)-\frac{1}{v_s}\cot(\theta)=0[/math]

[math]\csc(\theta)\left(\frac{1}{v_r}-\frac{1}{v_s}\cos(\theta) \right)=0[/math]

[math]\frac{1}{v_r}-\frac{1}{v_s}\cos(\theta)=0[/math]

[math]\cos(\theta)=\frac{v_s}{v_r}[/math]

[math]\theta=\cos^{-1}\left(\frac{v_s}{v_r} \right)[/math]

This is why I used variables, so we can make a few observations:

(1) the angle is independent of $L$ and $W$. We do however require:

[math]\tan^{-1}\left(\frac{W}{L} \right)\le\theta\le\frac{\pi}{2}[/math]

(2) If [math]v_s=v_r[/math] then:

[math]\theta=\cos^{-1}(1)=0[/math]

However, with $\theta=0$, the prisoner will not cross the river. So, what then is minimum ratio of [math]\frac{v_s}{v_r}=k[/math] which causes the prisoner to swim straight for the safehouse, in other words, to swim the entire distance? We may set the distance BC to zero:

[math]L-W\cot(\theta)=0[/math]

[math]\cot(\theta)=\frac{L}{W}[/math]

[math]\cot\left(\cos^{-1}\left(\frac{v_s}{v_r} \right) \right)=\frac{L}{W}[/math]

[math]\frac{v_s}{\sqrt{v_r^2-v_s^2}}=\frac{L}{W}[/math]

[math]\frac{kv_r}{\sqrt{v_r^2-(kv_r)^2}}=\frac{L}{W}[/math]

[math]\frac{k}{\sqrt{1-k^2}}=\frac{L}{W}[/math]

[math]k=\frac{L}{\sqrt{W^2+L^2}}[/math]

Thus if [math]v_s\ge\frac{L}{\sqrt{W^2+L^2}}v_r[/math] then the prisoner should swim the entire way.

(3) As [math]k\to 0[/math] then [math]\theta\to\frac{\pi}{2}[/math].

Plugging in the data given to finish the problem, we have:

[math]\theta=\cos^{-1}\left(\frac{6\text{ fps}}{12\text{ fps}} \right)=\cos^{-1}\left(\frac{1}{2} \right)=\frac{\pi}{3}=60^{\circ}[/math]

This is a valid angle as it satisfies:

[math]\tan^{-1}\left(\frac{1}{5} \right)\le\theta\le\frac{\pi}{2}[/math]

Then another person made the observation:

The result is equivalent to Snelles-Descartes law of refraction if you consider the index of the two media inversely proportional to the speed of the dude. However does he really deserve we help him to escape? :)

I responded:

Interesting observation! Using this law we could state:

[math]\frac{\sin(\theta_2)}{\sin(\theta_1)}=\frac{v_2}{v_1}[/math]

In this problem (the way I have it set up above), we have:

[math]\theta_1=\frac{\pi}{2}[/math]
[math]\theta_2=\frac{\pi}{2}-\theta[/math]
[math]v_1=v_r[/math]
[math]v_2=v_s[/math]

giving immediately:

[math]\frac{\sin\left(\frac{\pi}{2}-\theta \right)}{\sin\left(\frac{\pi}{2} \right)}=\frac{v_s}{v_r}[/math]

[math]\cos(\theta)=\frac{v_s}{v_r}[/math]

[math]\theta=\cos^{-1}\left(\frac{v_s}{v_r} \right)[/math]

Let's generalize a bit to say we want to find the quickest path across two media:

https://www.physicsforums.com/attachments/2776._xfImport

We are traveling from point A to point B then to point C by the quickest path possible. The question is, where should point B be placed?

I have set up $xy$ coordinate axes such that:

[math]O=(0,0)[/math]

[math]A=(0,-W_1)[/math]

[math]B=(B,0)[/math]

[math]C=(L,W_2)[/math]

The distance from A to B is:

[math]AB=\sqrt{(B-0)^2+(0-(-W_1))^2}=\sqrt{B^2+W_1^2}[/math]

The distance from B to C is:

[math]BC=\sqrt{(L-B)^2+(W_2-0)^2}=\sqrt{(L-B)^2+W_2^2}[/math]

The speed through medium 1 is $v_1$ and the speed through medium 2 is $v_2$. We assume both velocities are positive values. Thus, using the relationship between distance, constant velocity and time, we find the total time as a function of B is:

[math]t(B)=\frac{\sqrt{B^2+W_1^2}}{v_1}+\frac{\sqrt{(L-B)^2+W_2^2}}{v_2}[/math]

Differentiating with respect to B, we find:

[math]t'(B)=\frac{B}{v_1\sqrt{B^2+W_1^2}}+\frac{(B-L)}{v_2\sqrt{(L-B)^2+W_2^2}}=\frac{Bv_2\sqrt{(L-B)^2+W_2^2}+(B-L)v_1\sqrt{B^2+W_1^2}}{v_1v_2\sqrt{B^2+W_1^2}\sqrt{(L-B)^2+W_2^2}}[/math]

The denominator will always be positive, so we need only consider:

[math]Bv_2\sqrt{(L-B)^2+W_2^2}+(B-L)v_1\sqrt{B^2+W_1^2}=0[/math]

[math]Bv_2\sqrt{(L-B)^2+W_2^2}=(L-B)v_1\sqrt{B^2+W_1^2}[/math]

Let's take a moment to rewrite this:

[math]\frac{B}{\sqrt{B^2+W_1^2}}\cdot\frac{\sqrt{(L-B)^2+W_2^2}}{L-B}=\frac{v_1}{v_2}[/math]

[math]\frac{\frac{B}{\sqrt{B^2+W_1^2}}}{\frac{L-B}{\sqrt{(L-B)^2+W_2^2}}}=\frac{v_1}{v_2}[/math]

[math]\frac{\sin(\theta_1)}{\sin(\theta_2)}=\frac{v_1}{v_2}[/math]

Thus, we have shown that Snell's law (or Descartes' law) is satisfied when [math]t'(B)=0[/math].

Comments and questions should be posted here:

http://mathhelpboards.com/commentary-threads-53/commentary-using-snells-law-determine-fastest-escape-route-4204.html

 

[agentredlum]

Moderator edit: This commentary topic pertains to the following tutorial:

http://mathhelpboards.com/math-notes-49/using-snells-law-determine-fastest-escape-route-4178.html

I remember seeing this as Fermat's Principle in an optics book by Pedrotti and Pedrotti but your derivation is much nicer with all the details so someone like me can understand.

Fermat's principle - Wikipedia, the free encyclopedia

:)

 

[MarkFL]

Re: Using Snell's Law to determine the fastest escape route

I remember seeing this as Fermat's Principle in an optics book by Pedrotti and Pedrotti but your derivation is much nicer with all the details so someone like me can understand.

Fermat's principle - Wikipedia, the free encyclopedia

:)

Thank you...I find unless I work out the details myself, I don't really understand either. (Giggle)

I think this is testament to the goals of our site, that we allow the students to solve their problems with only suggestions and hints, and get them to come to a better understanding this way. (Cool)

 

[agentredlum]

Re: Using Snell's Law to determine the fastest escape route

Thank you...I find unless I work out the details myself, I don't really understand either. (Giggle)

I think this is testament to the goals of our site, that we allow the students to solve their problems with only suggestions and hints, and get them to come to a better understanding this way. (Cool)

Yep, the goal of a great educator is to help create students that can think on their own. I usually try to do the same unless i get a feeling the student won't be able to get the result on their own. It took the genius of a fermat to properly explain the principle of least time, not all students are geniuses but it's nice when a teacher get's one that is. (Of course i kept that opinion to myself when I was tutoring)

:)

 

[Nono713]

Re: Using Snell's Law to determine the fastest escape route

It took the genius of a fermat to properly explain the principle of least time

Where can I acquire one of those "fermats"? I could use one :p

 

[agentredlum]

Re: Using Snell's Law to determine the fastest escape route

Where can I acquire one of those "fermats"? I could use one :p

There are probably a few on math forums like this, you can usually tell by quality of postings and the help is free so you don't have to be a marquis to benefit from the knowledge!

:)