Solving the Boat Crossing a River Problem using the Calculus of Variations

[erobz]

I was hoping to explore the Calculus of Variations.

How do we prove by Calculus of Variations that the minimum time for boat crossing a river (perpendicular to the current for starters) with current $v_r$, and boat velocity in still water $v_b$ that the path will be a straight line?

I tried, but I got lost.

Let $x$ be the distance from launch point A to point B (directly opposite A) on the other bank.

$\rightarrow^{x^+} \downarrow^{y^+}$

I get the following equations:

$$ \frac{dx}{dt} = v_b \cos \theta \tag{1}$$

$$ \frac{dy}{dt} = v_b \sin \theta - v_r \tag{2}$$

I believe we are supposed to minimize the following:

$$ T = \frac{1}{v_b} \int \frac{dx}{ \cos \theta} $$

I notice that using the Chain Rule on $(2)$:

$$ \frac{dy}{dx} = \tan \theta - \frac{v_r}{v_b} \sec \theta $$

Thats implies that:

$$ 0 = \int dy = \int \left( \tan \theta - \frac{v_r}{v_b} \sec \theta \right) dx $$

Is it then not only the case that the integral on the RHS is zero if:

$$ \tan \theta - \frac{v_r}{v_b} \sec \theta = 0 \implies \sin \theta = \frac{v_r}{v_b} $$

Which means that $\theta$ is a constant... i.e. the path is a straight line. I realize this doesn't show the time is minimized though.

What is the proper way to go about proving $(1)$ is minimized using the Calculus of Variations?

 

[kuruman]

I was hoping to explore the Calculus of Variations.

How do we prove by Calculus of Variations that the minimum time for boat crossing a river (perpendicular to the current for starters) with current $v_r$, and boat velocity in still water $v_b$ that the path will be a straight line?

The path is always a straight line because the acceleration is zero. If you are thinking about the angle at which the boat must be aimed, clearly the time is $T=\frac{L}{v_0\cos\theta}$. Using calculus of variations to find the minimum of that is like using a sledgehammer to kill a flea.

Perhaps you might wish to explore the following situation. A lifeguard can run on sand at top speed $v_s$ and swim in water at top speed $v_w$ ($v_w<v_s.$) He sees a swimmer in distress at coordinates (X,Y) relative to him. Assuming that he runs and swims in straight lines, at what angle must he aim himself while running in order to reach the swimmer in the shortest possible time? Hint: You might recognize this as an analog of Snell's law.

 

[erobz]

Perhaps you might wish to explore the following situation. A lifeguard can run on sand at top speed $v_s$ and swim in water at top speed $v_w$ ($v_w<v_s.$) He sees a swimmer in distress at coordinates (X,Y) relative to him. Assuming that he runs and swims in straight lines, at what angle must he aim himself while running in order to reach the swimmer in the shortest possible time? Hint: You might recognize this as an analog of Snell's law.

That is just a standard optimization problem though, no Calculus of Variations required. Unless you are asking me to solve it with the Calculus of Varations?

I don't know how it works, which is why I stated:

I was hoping to explore the Calculus of Variations.

 

[erobz]

Using calculus of variations to find the minimum of that is like using a sledgehammer to kill a flea.

I don't think that's totally fair. One of the most common first examples in C.o.V. is using it to determine the straight line is the shortest path between two points. Using it on simple examples demonstrates the machinery without carrying along the added complexity of the "problem"?

 

[kuruman]

One of the most common first examples in C.o.V. is using it to determine the straight line is the shortest path between two points.

Which is exactly what you have here. If that's what you want to show, then do it. I thought you wanted to look into something more interesting.

 

[erobz]

Which is exactly what you have here. If that's what you want to show, then do it. I thought you wanted to look into something more interesting.

Sure, I'll examine the problem using C.o.V. with you (or anyone), I though I was helping by making it as simple as possible. Like I've said, apart from knowing basically (and that is a stretch) how it is supposed to function, I've never applied it to solve any problem.

Does it start in the same way as the standard optimization?

Let:
$T$ be the time it takes the lifeguard to get to the swimmer
$v_r$ be the velocity of the lifeguard running on the beach.
$v_s$ be the velocity of the lifeguard swimming
$w$ is the horizontal distance between the lifeguard and the point of entry into the water
$\theta_r$ is the angle ( relative to vertical ) which the lifeguard runs
$\theta_s$ is the angle ( relative to vertical) which the lifeguard swims

Then:

$$ T = \frac{w}{v_r \sin \theta_r} + \frac{x-w}{v_s \sin \theta_s} $$

 

[kuruman]

Good start, but now what? What are you varying to get an extremum?

 

[erobz]

From there the standard optimization techniques proceed in two ways:

$$ \frac{\partial T}{ \partial \theta_r } = 0 = \frac{\partial T}{ \partial \theta_s } $$

or write $\sin \theta_s$ as a function of $\theta_r$

I suspect the first method to be the cleanest approach.

 

[kuruman]

Try both and see what happens. You said you wanted to explore, so explore.

 

[erobz]

Try both and see what happens. You said you wanted to explore, so explore.

The second approach looks nasty ( smart people often have clever ways to simplify expressions like this that I rarely see):

$$ \sin \theta_s = \sqrt{ \frac{\tan^2 \theta_s}{ 1 + \tan^2 \theta_s } } $$

Where;

$$ \tan \theta_s = \frac{ \left( x-w \right) \tan \theta_r}{ y \tan \theta_r - w}$$

Based on that result I'm not going to waste my time on that one( I'de expect to be differintating for about a week or so).

 

[erobz]

The other way is also nasty, but I think I can get a result?

$$ \frac{\partial T}{ \partial \theta_r } = -\frac{w}{v_r} \frac{ \cos \theta_r}{1 - \cos^2 \theta_r} \tag{1}$$$$ \frac{\partial T}{ \partial \theta_s } = -\frac{x-w}{v_s} \frac{ \cos \theta_s}{1 - \cos^2 \theta_s} \tag{2}$$

Let:

$$ \lambda = \cos \theta_s $$

$$ \beta = \cos \theta_r $$

By equating $(1)$ and $(2)$


$$ \frac{x-w}{v_s} \lambda \beta^2 + \frac{w}{v_r} \left( 1 - \lambda^2 \right) \beta - \frac{x-w}{v_s} \lambda = 0 $$

So I would solve that for $\beta$ with the quadratic formula which would give me $\cos \theta_r$ as a function of $\cos \theta_s$ i.e. $\beta( \lambda)$

Then I should be able to use this:

$$ \tan \theta_s = \frac{ \left( x-w \right) \tan \theta_r}{ y \tan \theta_r - w}$$

along with:

$$ 1 + \tan^2\theta_s = \sec^2 \theta_s $$

and at least in principal, ( graphically or numerically...maybe analytically ) solve for $\theta_r$

I hope the point in all this was just to show me how complicated (ugly) the standard optimization techniques are (or I've completely fumbled this approach)?

Reducing this to Snells Law would appear to be quite the transformation, so I don’t know.

 

[kuruman]

No, the point is to show you how to apply the basic idea of the calculus of variations to this simple problem. You equated that idea to taking derivatives and, as you found out, it's not the best way to proceed. The goal is to find point $x$ so that the time when the lifeguard crosses at that point is at a minimum. To achieve this goal,

1. Assume that you have found such a point $x$.
2. Vary the path by moving away from it by a small amount $\delta<<x$.
3. Find what condition must be satisfied for the difference between times along each path to be zero in the limit $\delta\rightarrow 0$.

I know this looks suspiciously the same as taking derivatives, but you have already seen where that took you, so bear with me. I will get you started and you will have to wrap it up if you are so inclined.
The two times to reach the swimmer are $$\begin{align} & T=\frac{\sqrt{h_s^2+x ^2}}{v_s}+\frac{\sqrt{h_w^2+(w-x) ^2}}{v_w} \nonumber \\ & T'=\frac{\sqrt{h_s^2+(x+\delta) ^2}}{v_s}+\frac{\sqrt{h_w^2+[w-(x+\delta) ]^2}}{v_w} \nonumber \end{align}$$We want$$\delta T=T'-T=(T'_s+T'_w)-(T_s+T_w)=\delta T_s+\delta T_w=0.$$ I will take the first difference and rely on you to find the second, add them and set the sum equal to zero.$$\delta T_s=\frac{1}{v_s}\left[{\sqrt{h_s^2+(x+\delta) ^2}}-{\sqrt{h_s^2+x ^2}}\right]
=\frac{{h_s^2+(x+\delta) ^2}- ( {h_s^2+x ^2}) }{v_s[\sqrt{h_s^2+(x+\delta) ^2}+\sqrt{h_s^2+x ^2}]}.$$ Upon expansion of the terms in the numerator, we get

$$\delta T_s=\frac{2x\delta+\delta^2 }{v_s[\sqrt{h_s^2+(x+\delta) ^2}+\sqrt{h_s^2+x ^2}]}.$$ Up to this point the expression is exact. In the limit $\delta <<x$, $$\delta T_s \approx \frac{2x\delta}{2v_s\sqrt{h_s^2+x ^2}}=\frac{\delta}{v_s}\frac{x}{\sqrt{h_s^2+x ^2}}=\frac{\delta}{v_s}\sin\theta_{\!s}.$$See how it works? It would be nice if you completed the derivation and posted it here for everyone's benefit. I used this to derive Snell's law using Fermat's principle when I taught introductory algebra-based physics.

On edit : I used velocity subscripts "$s$" for "sand" and "$w$" for "water" to denote the media in which the velocity has the given values. Same thing for the distances.

 

[erobz]

No, the point is to show you how to apply the basic idea of the calculus of variations to this simple problem.

I used this to derive Snell's law using Fermat's principle when I taught introductory algebra-based physics.

Have I offended you in some way? This language on how simple and basic this problem is sure makes it seem that way. The highest math I left high school with...20 years ago(to no ones fault but my own) was Algebra 1. I started college as an adult learning to do fractions...just to give you some perspective. I don't recall claiming in here that I'm gifted in any way? I'm confused, because if I were helping someone that was having trouble, I wouldn't use that kind of language to indicate my level of boredom.

 

[kuruman]

Have I offended you in some way?

Absolutely not in any way. You misunderstood me. The statement was intended as a "How to" for deriving Snell's law from Fermat's principle without calculus. I came up with this derivation when the algebra-based textbook I used at the time just plopped down Snell's law with a statement like "Derivation of the law requires calculus which is beyond the scope of this book." That is what offended me because there was a derivation of the law of reflection based on Fermat's principle in the previous section.

 

[malawi_glenn]

"Derivation of the law requires calculus which beyond the scope of this book."

I do a similar derivation of snells law in my physics class. We also do it "exploatory" where I assign a path with some angle to a pair of students and they are asked to calculate the time their path took. Then we go over their results on the blackboard. This make them realize that there should be a path which actually minimize the time. And when we get to the part in calculus where they learn how to differentiate those functions, we do the derivation of Snells law again using calculus.

 

[erobz]

Absolutely not in any way. You misunderstood me.

I once asked a question to one of my physics professor in a lab (I became confused about something...I don't even remember what it was), and his reply in front of the whole class was "that I should immediately drop out of engineering and pursue a business degree" completely serious expression - class fell dead silent. I feel like that past experience shades my view of what people are saying.

 

[kuruman]

I once asked a question to one of my physics professors in a lab (I became confused about something...I don't even remember what it was), and his reply in front of the whole class was "that I should immediately drop out of engineering and pursue a business degree". I feel like that past experience shades my view of what people are saying.

I understand why you feel the way you do. Belittling a student, whether in front of the class (or in private), is not conducive to learning, it accomplishes nothing and in my book it is tantamount to bullying.

 

[erobz]

He had achieved some small level of fame by regularly appearing on the Jay Leno Show as "The Mad Scientist" years prior to this experience...he was fighting some internal demons too.

 

[kuruman]

He had achieved some small level of fame by regularly appearing on the Jay Leno Show as "The Mad Scientist" years prior to this experience...he was fighting some internal demons too.

Still, this sort of behavior is not conducive to learning which his job is to facilitate. You must be a charitable person.

 

[erobz]

Anyhow... I'll try to figure it out and post the attempt.

 

[kuruman]

Anyhow... I'll try to figure it out and post the attempt.

Great! Just follow my method for the second term and, if you get stuck, we're here to help.

 

[kuruman]

I do a similar derivation of snells law in my physics class. We also do it "exploatory" where I assign a path with some angle to a pair of students and they are asked to calculate the time their path took. Then we go over their results on the blackboard. This make them realize that there should be a path which actually minimize the time. And when we get to the part in calculus where they learn how to differentiate those functions, we do the derivation of Snells law again using calculus.

In this case, the "variation" and the existence of the minimum need to be established first. In the diagram below the lifeguard starts at point A and wants to end up at point B. He could go along the straight path ACB, but that's not minimum time. If crosses to the left of C, clearly the path is longer both on sand and in water and so is the time. However, if he crosses at a point to the right of C, he can do better because he trades distance in the medium where he moves faster for distance in the medium where he moves slower.

Can he do so indefinitely to the right of C? No. If he goes to the right of D, the distances in both media increase and so will the corresponding times. Therefore, somewhere between points C and D there must be an optimum crossing point. It follows by looking at the drawing that the faster he can move on sand relative to water, the closer the optimum point is to D.

I think this problem encapsulates the essence of the calculus of variations and that is why I proposed it to @erobz. It is simple to understand, simple to explain and simple to deal with.

 

[erobz]

In this case, the "variation" and the existence of the minimum need to be established first. In the diagram below the lifeguard starts at point A and wants to end up at point B. He could go along the straight path ACB, but that's not minimum time. If crosses to the left of C, clearly the path is longer both on sand and in water and so is the time. However, if he crosses at a point to the right of C, he can do better because he trades distance in the medium where he moves faster for distance in the medium where he moves slower.

Can he do so indefinitely to the right of C? No. If he goes to the right of D, the distances in both media increase and so will the corresponding times. Therefore, somewhere between points C and D there must be an optimum crossing point. It follows by looking at the drawing that the faster he can move on sand relative to water, the closer the optimum point is to D.

I think this problem encapsulates the essence of the calculus of variations and that is why I proposed it to @erobz. It is simple to understand, simple to explain and simple to deal with.

View attachment 305382

Yeah, I find that to be very well explained!

 

[malawi_glenn]

In this case, the "variation" and the existence of the minimum need to be established first

Yes, some students get your path ACB and some get ADB. Some get where crossing is to the left of C and some where the crossing is to the right of D. But most get a path between C and D. Now I do not tell them at first we are looking for the path with minimal time but just "calculate the time of travel for your path, the speed in medium 1 is... and the speed in medium 2 is". Then I ask the students who got those four paths about their result and I present it on the blackboard. Then I say to them "we can see something interesting here already, can you discuss with each other what you think that is".

 

[erobz]

No, the point is to show you how to apply the basic idea of the calculus of variations to this simple problem. You equated that idea to taking derivatives and, as you found out, it's not the best way to proceed. The goal is to find point $x$ so that the time when the lifeguard crosses at that point is at a minimum. To achieve this goal,

1. Assume that you have found such a point $x$.
2. Vary the path by moving away from it by a small amount $\delta<<x$.
3. Find what condition must be satisfied for the difference between times along each path to be zero in the limit $\delta\rightarrow 0$.

I know this looks suspiciously the same as taking derivatives, but you have already seen where that took you, so bear with me. I will get you started and you will have to wrap it up if you are so inclined.
The two times to reach the swimmer are $$\begin{align} & T=\frac{\sqrt{h_s^2+x ^2}}{v_s}+\frac{\sqrt{h_w^2+(w-x) ^2}}{v_w} \nonumber \\ & T'=\frac{\sqrt{h_s^2+(x+\delta) ^2}}{v_s}+\frac{\sqrt{h_w^2+[w-(x+\delta) ]^2}}{v_w} \nonumber \end{align}$$We want$$\delta T=T'-T=(T'_s+T'_w)-(T_s+T_w)=\delta T_s+\delta T_w=0.$$ I will take the first difference and rely on you to find the second, add them and set the sum equal to zero.$$\delta T_s=\frac{1}{v_s}\left[{\sqrt{h_s^2+(x+\delta) ^2}}-{\sqrt{h_s^2+x ^2}}\right]
=\frac{{h_s^2+(x+\delta) ^2}- ( {h_s^2+x ^2}) }{v_s[\sqrt{h_s^2+(x+\delta) ^2}+\sqrt{h_s^2+x ^2}]}.$$ Upon expansion of the terms in the numerator, we get

$$\delta T_s=\frac{2x\delta+\delta^2 }{v_s[\sqrt{h_s^2+(x+\delta) ^2}+\sqrt{h_s^2+x ^2}]}.$$ Up to this point the expression is exact. In the limit $\delta <<x$, $$\delta T_s \approx \frac{2x\delta}{2v_s\sqrt{h_s^2+x ^2}}=\frac{\delta}{v_s}\frac{x}{\sqrt{h_s^2+x ^2}}=\frac{\delta}{v_s}\sin\theta_{\!s}.$$See how it works? It would be nice if you completed the derivation and posted it here for everyone's benefit. I used this to derive Snell's law using Fermat's principle when I taught introductory algebra-based physics.

On edit : I used velocity subscripts "$s$" for "sand" and "$w$" for "water" to denote the media in which the velocity has the given values. Same thing for the distances.

Doing the algebra:

$$ \delta T_w = \frac{1}{v_w} \left( \sqrt{ {h_w}^2 + \left( w - \left( x- \delta\right) \right)^2} - \sqrt{ {h_w}^2 + \left( w-x\right)^2 } \right) $$

Multiply the numerator by its conjugate, expand, and cancel opposite terms:

$$ \delta T_w = \frac{1}{v_w} \left( \frac{ -2\left( w - x \right) \delta + \delta^2 }{ \sqrt{ {h_w}^2 + \left( w - \left( x- \delta \right) \right)^2} + \sqrt{ {h_w}^2 + \left( w-x\right)^2 } }\right)$$

Then apply the condition $\delta \ll x$:

$$ \delta T_w \approx - \frac{ \left( w - x \right) \delta }{ \sqrt{ {h_w}^2 + \left( w-x\right)^2 } } \implies - \frac{1}{v_w} \sin \theta_w $$

Putting that all together with $\delta T_s + \delta T_w = 0$ we arrive at ( spoiler alert ) Snells Law:

$$ \frac{\sin \theta_s}{v_s} = \frac{ \sin \theta_w}{ v_w} $$

 

[kuruman]

Doing the algebra:

$$ \delta T_w = \frac{1}{v_w} \left( \sqrt{ {h_w}^2 + \left( w - \left( x- \delta\right) \right)^2} - \sqrt{ {h_w}^2 + \left( w-x\right)^2 } \right) $$

Multiply the numerator by its conjugate, and expand and cancel opposite terms:

$$ \delta T_w = -\frac{1}{v_w} \left( \frac{ 2\left( x-w \right) \delta + \delta^2 }{ \sqrt{ {h_w}^2 + \left( w - \left( x- \delta\right) \right)^2} + \sqrt{ {h_w}^2 + \left( w-x\right)^2 } \right)$$

Then apply the condition $\delta \ll x$:

$$ \delta T_w = - \frac{ \left( x-w \right) \delta }{ \sqrt{ {h_w}^2 + \left( w-x\right)^2 } } \implies - \frac{1}{v_w} \sin \theta_w $$

Putting that all together with $\delta T_s + \delta T_w = 0$ we arrive at ( spoiler alert ) Snells Law:

$$ \frac{\sin \theta_s}{v_s} = \frac{ \sin \theta_w}{ v_w} $$

Please fix the LaTeX to make it legible.

 

[erobz]

Please fix the LaTeX to make it legible.

Yeah, miss a bracket and it all goes to crap. It's hard to track those down.

 

[malawi_glenn]

Yeah, miss a bracket and it all goes to crap. It's hard to track those down.

You can use preview :)

 

[erobz]

You can use preview :)

I was under the impression that function didn't work?

 

[malawi_glenn]

I was under the impression that function didn't work?

Works for me. But one might have to hit refresh on the browser now and then

 

[erobz]

Now can we talk about how this works?

Choose an optimum set of paths ( for this problem ), and deviate from it by some amount $\delta$.

For this problem we demanded the total deviation from optimum be zero.

Firstly, how is that possible? I would think we should not be able to deviate at all from an optimum path with no net change in the optimized parameter.

Secondly, what happens if there is no optimum path. i.e. we try to optimize something that is not optimizable with this technique? It seems like we should have to have prior knowledge about the existence of an optimum before this can be applied?

 

[erobz]

Works for me. But one might have to hit refresh on the browser now and then

Thats right, if there is latex in the thread this will work. I forgot.

 

[kuruman]

These are good questions.

Now can we talk about how this works?

Choose an optimum set of paths ( for this problem ), and deviate from it by some amount $\delta$.

That's not what I wrote in item 1, post #12. There is only one optimum path and you start by assuming that you have found it. You give a placeholder name to a parameter that specifies it and the task is to find the value of this placeholder parameter in terms of the given quantities.

Firstly, how is that possible? I would think we should not be able to deviate at all from an optimum path with no net change in the optimized parameter.

You can calculate times for any path for a choice of $\delta$. If two different values of $\delta$ give the same time, then neither of the two paths can be optimum. However, the optimum should be between the two values of $\delta$, one of which must be positive and the other negative. Do you see why?

Secondly, what happens if there is no optimum path. i.e. we try to optimize something that is not optimizable with this technique? It seems like we should have to have prior knowledge about the existence of an optimum before this can be applied?

You cannot find something that doesn't exist. That is why before starting out to optimize a function, you need to ascertain that it can be optimized. That is what I did in post #22 when I argued that the optimum path must be at a crossing point situated between points C and D.

 

[erobz]

You can calculate times for any path for a choice of $\delta$. If two different values of $\delta$ give the same time, then neither of the two paths can be optimum. However, the optimum should be between the two values of $\delta$, one of which must be positive and the other negative. Do you see why?

Well, I think because we've chosen the coordinate $x$ to be at an extremum, that is either concave up ( local minimum ) or concave down ( local maximum ) in the vicinity of $x$. So I agree that the change in time could be zero with $\pm \delta$.

You emphasized that we deviate a small amount $\delta$. What is the significance of that in this context?

I see that $\delta \ll x$ is significant, because it can't become small arbitrarily small or else, we get $0 + 0 = 0$. This Is like a linearization in that way around $x$.