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On another forum the following question was asked:

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:

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

\(\displaystyle AB=W\csc(\theta)\)

The time this takes is:

\(\displaystyle t_s=\frac{W}{v_s}\csc(\theta)\)

The distance from B to C is:

\(\displaystyle BC=L-W\cot(\theta)\)

The time this takes is:

\(\displaystyle t_r=\frac{1}{v_r}(L-W\cot(\theta))\)

The total time then is:

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

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

Observing \(\displaystyle \csc(\theta)\ne0\) for all real $\theta$, we then use:

\(\displaystyle \frac{1}{v_r}\csc(\theta)-\frac{1}{v_s}\cot(\theta)=0\)

\(\displaystyle \csc(\theta)\left(\frac{1}{v_r}-\frac{1}{v_s}\cos(\theta) \right)=0\)

\(\displaystyle \frac{1}{v_r}-\frac{1}{v_s}\cos(\theta)=0\)

\(\displaystyle \cos(\theta)=\frac{v_s}{v_r}\)

\(\displaystyle \theta=\cos^{-1}\left(\frac{v_s}{v_r} \right)\)

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:

\(\displaystyle \tan^{-1}\left(\frac{W}{L} \right)\le\theta\le\frac{\pi}{2}\)

(2) If \(\displaystyle v_s=v_r\) then:

\(\displaystyle \theta=\cos^{-1}(1)=0\)

However, with $\theta=0$, the prisoner will not cross the river. So, what then is minimum ratio of \(\displaystyle \frac{v_s}{v_r}=k\) 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:

\(\displaystyle L-W\cot(\theta)=0\)

\(\displaystyle \cot(\theta)=\frac{L}{W}\)

\(\displaystyle \cot\left(\cos^{-1}\left(\frac{v_s}{v_r} \right) \right)=\frac{L}{W}\)

\(\displaystyle \frac{v_s}{\sqrt{v_r^2-v_s^2}}=\frac{L}{W}\)

\(\displaystyle \frac{kv_r}{\sqrt{v_r^2-(kv_r)^2}}=\frac{L}{W}\)

\(\displaystyle \frac{k}{\sqrt{1-k^2}}=\frac{L}{W}\)

\(\displaystyle k=\frac{L}{\sqrt{W^2+L^2}}\)

Thus if \(\displaystyle v_s\ge\frac{L}{\sqrt{W^2+L^2}}v_r\) then the prisoner should swim the entire way.

(3) As \(\displaystyle k\to 0\) then \(\displaystyle \theta\to\frac{\pi}{2}\).

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

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

This is a valid angle as it satisfies:

\(\displaystyle \tan^{-1}\left(\frac{1}{5} \right)\le\theta\le\frac{\pi}{2}\)

Then another person made the observation:

Interesting observation! Using this law we could state:

\(\displaystyle \frac{\sin(\theta_2)}{\sin(\theta_1)}=\frac{v_2}{v_1}\)

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

\(\displaystyle \theta_1=\frac{\pi}{2}\)

\(\displaystyle \theta_2=\frac{\pi}{2}-\theta\)

\(\displaystyle v_1=v_r\)

\(\displaystyle v_2=v_s\)

giving immediately:

\(\displaystyle \frac{\sin\left(\frac{\pi}{2}-\theta \right)}{\sin\left(\frac{\pi}{2} \right)}=\frac{v_s}{v_r}\)

\(\displaystyle \cos(\theta)=\frac{v_s}{v_r}\)

\(\displaystyle \theta=\cos^{-1}\left(\frac{v_s}{v_r} \right)\)

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

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:

\(\displaystyle O=(0,0)\)

\(\displaystyle A=(0,-W_1)\)

\(\displaystyle B=(B,0)\)

\(\displaystyle C=(L,W_2)\)

The distance from A to B is:

\(\displaystyle AB=\sqrt{(B-0)^2+(0-(-W_1))^2}=\sqrt{B^2+W_1^2}\)

The distance from B to C is:

\(\displaystyle BC=\sqrt{(L-B)^2+(W_2-0)^2}=\sqrt{(L-B)^2+W_2^2}\)

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:

\(\displaystyle t(B)=\frac{\sqrt{B^2+W_1^2}}{v_1}+\frac{\sqrt{(L-B)^2+W_2^2}}{v_2}\)

Differentiating with respect to B, we find:

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

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

\(\displaystyle Bv_2\sqrt{(L-B)^2+W_2^2}+(B-L)v_1\sqrt{B^2+W_1^2}=0\)

\(\displaystyle Bv_2\sqrt{(L-B)^2+W_2^2}=(L-B)v_1\sqrt{B^2+W_1^2}\)

Let's take a moment to rewrite this:

\(\displaystyle \frac{B}{\sqrt{B^2+W_1^2}}\cdot\frac{\sqrt{(L-B)^2+W_2^2}}{L-B}=\frac{v_1}{v_2}\)

\(\displaystyle \frac{\frac{B}{\sqrt{B^2+W_1^2}}}{\frac{L-B}{\sqrt{(L-B)^2+W_2^2}}}=\frac{v_1}{v_2}\)

\(\displaystyle \frac{\sin(\theta_1)}{\sin(\theta_2)}=\frac{v_1}{v_2}\)

Thus, we have shown that Snell's law (or Descartes' law) is satisfied when \(\displaystyle t'(B)=0\).

Comments and questions should be posted here:

http://mathhelpboards.com/commentar...-law-determine-fastest-escape-route-4204.html

This was my initial reply: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!

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:

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

\(\displaystyle AB=W\csc(\theta)\)

The time this takes is:

\(\displaystyle t_s=\frac{W}{v_s}\csc(\theta)\)

The distance from B to C is:

\(\displaystyle BC=L-W\cot(\theta)\)

The time this takes is:

\(\displaystyle t_r=\frac{1}{v_r}(L-W\cot(\theta))\)

The total time then is:

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

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

Observing \(\displaystyle \csc(\theta)\ne0\) for all real $\theta$, we then use:

\(\displaystyle \frac{1}{v_r}\csc(\theta)-\frac{1}{v_s}\cot(\theta)=0\)

\(\displaystyle \csc(\theta)\left(\frac{1}{v_r}-\frac{1}{v_s}\cos(\theta) \right)=0\)

\(\displaystyle \frac{1}{v_r}-\frac{1}{v_s}\cos(\theta)=0\)

\(\displaystyle \cos(\theta)=\frac{v_s}{v_r}\)

\(\displaystyle \theta=\cos^{-1}\left(\frac{v_s}{v_r} \right)\)

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:

\(\displaystyle \tan^{-1}\left(\frac{W}{L} \right)\le\theta\le\frac{\pi}{2}\)

(2) If \(\displaystyle v_s=v_r\) then:

\(\displaystyle \theta=\cos^{-1}(1)=0\)

However, with $\theta=0$, the prisoner will not cross the river. So, what then is minimum ratio of \(\displaystyle \frac{v_s}{v_r}=k\) 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:

\(\displaystyle L-W\cot(\theta)=0\)

\(\displaystyle \cot(\theta)=\frac{L}{W}\)

\(\displaystyle \cot\left(\cos^{-1}\left(\frac{v_s}{v_r} \right) \right)=\frac{L}{W}\)

\(\displaystyle \frac{v_s}{\sqrt{v_r^2-v_s^2}}=\frac{L}{W}\)

\(\displaystyle \frac{kv_r}{\sqrt{v_r^2-(kv_r)^2}}=\frac{L}{W}\)

\(\displaystyle \frac{k}{\sqrt{1-k^2}}=\frac{L}{W}\)

\(\displaystyle k=\frac{L}{\sqrt{W^2+L^2}}\)

Thus if \(\displaystyle v_s\ge\frac{L}{\sqrt{W^2+L^2}}v_r\) then the prisoner should swim the entire way.

(3) As \(\displaystyle k\to 0\) then \(\displaystyle \theta\to\frac{\pi}{2}\).

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

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

This is a valid angle as it satisfies:

\(\displaystyle \tan^{-1}\left(\frac{1}{5} \right)\le\theta\le\frac{\pi}{2}\)

Then another person made the observation:

I responded: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?

Interesting observation! Using this law we could state:

\(\displaystyle \frac{\sin(\theta_2)}{\sin(\theta_1)}=\frac{v_2}{v_1}\)

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

\(\displaystyle \theta_1=\frac{\pi}{2}\)

\(\displaystyle \theta_2=\frac{\pi}{2}-\theta\)

\(\displaystyle v_1=v_r\)

\(\displaystyle v_2=v_s\)

giving immediately:

\(\displaystyle \frac{\sin\left(\frac{\pi}{2}-\theta \right)}{\sin\left(\frac{\pi}{2} \right)}=\frac{v_s}{v_r}\)

\(\displaystyle \cos(\theta)=\frac{v_s}{v_r}\)

\(\displaystyle \theta=\cos^{-1}\left(\frac{v_s}{v_r} \right)\)

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

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:

\(\displaystyle O=(0,0)\)

\(\displaystyle A=(0,-W_1)\)

\(\displaystyle B=(B,0)\)

\(\displaystyle C=(L,W_2)\)

The distance from A to B is:

\(\displaystyle AB=\sqrt{(B-0)^2+(0-(-W_1))^2}=\sqrt{B^2+W_1^2}\)

The distance from B to C is:

\(\displaystyle BC=\sqrt{(L-B)^2+(W_2-0)^2}=\sqrt{(L-B)^2+W_2^2}\)

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:

\(\displaystyle t(B)=\frac{\sqrt{B^2+W_1^2}}{v_1}+\frac{\sqrt{(L-B)^2+W_2^2}}{v_2}\)

Differentiating with respect to B, we find:

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

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

\(\displaystyle Bv_2\sqrt{(L-B)^2+W_2^2}+(B-L)v_1\sqrt{B^2+W_1^2}=0\)

\(\displaystyle Bv_2\sqrt{(L-B)^2+W_2^2}=(L-B)v_1\sqrt{B^2+W_1^2}\)

Let's take a moment to rewrite this:

\(\displaystyle \frac{B}{\sqrt{B^2+W_1^2}}\cdot\frac{\sqrt{(L-B)^2+W_2^2}}{L-B}=\frac{v_1}{v_2}\)

\(\displaystyle \frac{\frac{B}{\sqrt{B^2+W_1^2}}}{\frac{L-B}{\sqrt{(L-B)^2+W_2^2}}}=\frac{v_1}{v_2}\)

\(\displaystyle \frac{\sin(\theta_1)}{\sin(\theta_2)}=\frac{v_1}{v_2}\)

Thus, we have shown that Snell's law (or Descartes' law) is satisfied when \(\displaystyle t'(B)=0\).

Comments and questions should be posted here:

http://mathhelpboards.com/commentar...-law-determine-fastest-escape-route-4204.html

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