How Can You Optimize Travel Time with Varying Speeds in Calculus of Variations?

In summary, the conversation discusses a difficult question regarding calculus of variations, specifically the time it takes for a particle to travel from one point to another in the (x,y) plane. The particle's speed depends on its distance from the x-axis and the direction of travel can be controlled to minimize the travel time. The equation for the time is given, and the final direction needed to be proven is found to be equal to tan^{-1}(e\sqrt{2}-1). The conversation then delves into the use of Euler's equation and the Beltrami identity to calculate partial derivatives and solve the problem. The value of h is also discussed, with the conclusion that the results hold for h=1.
  • #1
touqra
287
0
I have another difficult question regarding calculus of variations.
A particle travels in the (x,y) plane has a speed u(y) that depends on the distance of the particle from the x-axis. The direction of travel subtends an angle [tex] \theta [/tex] with the x-axis that can be controlled to give the minimum time to move between two points.
Let [tex] u(y) = Ue^{-y/h} [/tex] whereby, U and h are constants.
If the particles starts at (0,0) and has to get to the point [tex](\pi{h}/4,h)[/tex] in the shortest time, show that, the final direction is [tex]\theta = tan^{-1}(e\sqrt{2}-1)[/tex]
 
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  • #2
touqra said:
I have another difficult question regarding calculus of variations.
A particle travels in the (x,y) plane has a speed u(y) that depends on the distance of the particle from the x-axis. The direction of travel subtends an angle [tex] \theta [/tex] with the x-axis that can be controlled to give the minimum time to move between two points.
Let [tex] u(y) = Ue^{-y/h} [/tex] whereby, U and h are constants.
If the particles starts at (0,0) and has to get to the point [tex](\pi{h}/4,h)[/tex] in the shortest time, show that, the final direction is [tex]\theta = tan^{-1}(e\sqrt{2}-1)[/tex]

The time it takes to travel from point a to b is given by:

[tex]t_{a,b}=\int_{p_1}^{p_2}\frac{ds}{v}[/tex]

Where s is the arclength and v is the speed. Note that:

[tex]ds=\sqrt{dx^2+dy^2}=\sqrt{1+(y^{'})^2}dx[/tex]

The final direction is just the slope of the extremal function at the end point right?
 
  • #3
saltydog said:
The time it takes to travel from point a to b is given by:
[tex]t_{a,b}=\int_{p_1}^{p_2}\frac{ds}{v}[/tex]
Where s is the arclength and v is the speed. Note that:
[tex]ds=\sqrt{dx^2+dy^2}=\sqrt{1+(y^{'})^2}dx[/tex]
The final direction is just the slope of the extremal function at the end point right?
yup, that's what I did initially. But, when I used the Euler equation to get me an equation, I have an equation with some unknown constants, ie, U and h. But the final direction which was needed to be proven does not contain any constants.
What I get was:
[tex]F=\sqrt{1+y'^2}{e^{y/h}}/U[/tex]
Using Euler equation and the Beltrami identity (since F is independent of x),
[tex]F-{y'}\frac{\partialF}{\partialy'}=constant[/tex]
[tex]\sqrt{1+y'^2}{e^{y/h}}-{y'^2}{(1+y'^2)}^{-1/2}{e^{y/h}}/2= constant [/tex]
 
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  • #4
touqra said:
yup, that's what I did initially. But, when I used the Euler equation to get me an equation, I have an equation with some unknown constants, ie, U and h. But the final direction which was needed to be proven does not contain any constants.
What I get was:
[tex]F=\sqrt{1+y'^2}{e^{y/h}}/U[/tex]
Using Euler equation and the Beltrami identity (since F is independent of x),
[tex]F-{y'}\frac{\partialF}{\partialy'}=constant[/tex]
[tex]\sqrt{1+y'^2}{e^{y/h}}-{y'^2}{(1+y'^2)}^{-1/2}{e^{y/h}}/2= constant [/tex]

For the moment, let h=1, leave U as just some constant: This is sufficient to lead to the angle you posted above.

So we start here:

[tex]t_{a,b}=\int_{p_1}^{p_2}\frac{ds}{v}[/tex]

and extremize:

[tex]\mathbf{T}[y(x)]=\int_{p_1}^{p_2}\frac{\sqrt{1+(y^{'})^2}}{Ue^{-y/h}}dx[/tex]

(v above in my post is just u(y(x)), the speed of travel)

with the boundary conditions you posted.

so:

[tex]F(x,y,y^{'})=\frac{\sqrt{1+(y^{'})^2}}{Ue^{-y/h}}[/tex]

So we employ Euler's equation and calculate 3 derivatives. Can you post the first two?

[tex]\frac{\partial F}{\partial y}=[/tex]

[tex]\frac{\partial F}{\partial y^{'}}=[/tex]
 
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  • #5
saltydog said:
For the moment, let h=1, leave U as just some constant: This is sufficient to lead to the angle you posted above.
So we start here:
[tex]t_{a,b}=\int_{p_1}^{p_2}\frac{ds}{v}[/tex]
and extremize:
[tex]\mathbf{T}[y(x)]=\int_{p_1}^{p_2}\frac{\sqrt{1+(y^{'})^2}}{Ue^{-y/h}}dx[/tex]
(v above in my post is just u(y(x)), the speed of travel)
with the boundary conditions you posted.
so:
[tex]F(x,y,y^{'})=\frac{\sqrt{1+(y^{'})^2}}{Ue^{-y/h}}[/tex]
So we employ Euler's equation and calculate 3 derivatives. Can you post the first two?
[tex]\frac{\partial F}{\partial y}=[/tex]
[tex]\frac{\partial F}{\partial y^{'}}=[/tex]
I obtain
[tex]\frac{\partial F}{\partial y'} = Uy'e^{y/h}/\sqrt{1+y'^2} = J[/tex]
[tex]\frac{\partial F}{\partial y} = U\sqrt{(1+y'^2)}e^{y/h}/h[/tex]
[tex]\frac{dJ}{dx}=\frac{K-L}{M}[/tex]
where, [tex]K = e^{y/h}(1+y'^2)(y''+\frac{y'^2}{h}) [/tex]
[tex]L = y''y'e^{y/h}[/tex]
[tex]M = (1+y'^2)\sqrt{(1+y'^2)} [/tex]
How do you proceed from here?
Why do you say that putting h=1 will give the final angle?
 
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  • #6
Touqra, it looks like you're having problems calculating partials. Looks so, maybe not.

First, let's clean up the functional: Really, if we want to minimize that integral, we can just move U across the integral sign right, and let's put the exponential in the numerator:

[tex]\mathbf{T}[y(x)]=\frac{1}{U}\int_{p_1}^{p_2} e^{y/h}\sqrt{1+(y^{'})^2}dx[/tex]


So:

[tex]F(x,y,y^{'})=e^{y/h}\sqrt{1+(y^{'})^2}[/tex]

and therefore:

[tex]\frac{\partial F}{\partial y}=\frac{e^{y/h}\sqrt{1+(y^{'})^2}}{h}[/tex]

and:

[tex]\frac{\partial F}{\partial y^{'}}=\frac{e^{y/h}y^{'}}{\sqrt{1+(y^{'})^2}}[/tex]

and so:

[tex]
\begin{align*}
\frac{d}{dx}\left(\frac{\partial F}{\partial y^{'}}\right)&=\frac{d}{dx}\left[\frac{e^{y/h}y^{'}}{\sqrt{1+(y^{'})^2}}\right] \\

&=\frac{d}{dx}\left[e^{y/h}y^{'} \cdot \frac{1}{\sqrt{1+(y^{'})^2}}\right] \\

&=e^{y/h}y^{'}\cdot\frac{-1/2}{(1+(y^{'})^2)^{3/2}}\cdot 2 y^{'}y^{''} \\

&+\frac{1}{\sqrt{1+(y^{'})^2}}\left(e^{y/h}y^{''}+y^{'}\frac{1}{h}y^{'}e^{y/h}\right) \\

&=\frac{e^{y/h}y^{''}}{\sqrt{1+(y^{'})^2}}-\frac{e^{y/h}(y^{'})^2y^{''}}{(1+(y^{'})^2)^{3/2}}+
\frac{e^{y/h}(y^{'})^2}{h\sqrt{1+(y^{'})^2}}

\end{align}
[/tex]

So, once we obtain the partials, then we substitute them into the Euler equation and equate the expression to zero. Now, can you substitute these expressions into:

[tex]\frac{\partial F}{\partial y}-\frac{d}{dx}\left(\frac{\partial F}{\partial y^{'}}\right)=0[/tex]

simplify, and then post the results?

Also, with regards to h=1: I just worked the problem with that value and obtained the results you indicated. Perhaps it works for any value of h. Not sure.
 
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  • #7
saltydog said:
Touqra, it looks like you're having problems calculating partials. Looks so, maybe not.
First, let's clean up the functional: Really, if we want to minimize that integral, we can just move U across the integral sign right, and let's put the exponential in the numerator:
[tex]\mathbf{T}[y(x)]=\frac{1}{U}\int_{p_1}^{p_2} e^{y/h}\sqrt{1+(y^{'})^2}dx[/tex]
So:
[tex]F(x,y,y^{'})=e^{y/h}\sqrt{1+(y^{'})^2}[/tex]
and therefore:
[tex]\frac{\partial F}{\partial y}=\frac{e^{y/h}\sqrt{1+(y^{'})^2}}{h}[/tex]
and:
[tex]\frac{\partial F}{\partial y^{'}}=\frac{e^{y/h}y^{'}}{\sqrt{1+(y^{'})^2}}[/tex]
and so:
[tex]
\begin{align*}
\frac{d}{dx}\left(\frac{\partial F}{\partial y^{'}}\right)&=\frac{d}{dx}\left[\frac{e^{y/h}y^{'}}{\sqrt{1+(y^{'})^2}}\right] \\
&=\frac{d}{dx}\left[e^{y/h}y^{'} \cdot \frac{1}{\sqrt{1+(y^{'})^2}}\right] \\
&=e^{y/h}y^{'}\cdot\frac{-1/2}{(1+(y^{'})^2)^{3/2}}\cdot 2 y^{'}y^{''} \\
&+\frac{1}{\sqrt{1+(y^{'})^2}}\left(e^{y/h}y^{''}+y^{'}\frac{1}{h}y^{'}e^{y/h}\right) \\
&=\frac{e^{y/h}y^{''}}{\sqrt{1+(y^{'})^2}}-\frac{e^{y/h}(y^{'})^2y^{''}}{(1+(y^{'})^2)^{3/2}}+
\frac{e^{y/h}(y^{'})^2}{h\sqrt{1+(y^{'})^2}}
\end{align}
[/tex]
So, once we obtain the partials, then we substitute them into the Euler equation and equate the expression to zero. Now, can you substitute these expressions into:
[tex]\frac{\partial F}{\partial y}-\frac{d}{dx}\left(\frac{\partial F}{\partial y^{'}}\right)=0[/tex]
simplify, and then post the results?
Also, with regards to h=1: I just worked the problem with that value and obtained the results you indicated. Perhaps it works for any value of h. Not sure.
I simplified the result and obtained
[tex]hy'' = 1 + y'^2 [/tex]
solving the above equation, we have
[tex]y' = tan (\frac{(x+c)}{h}) [/tex]
since, [tex] y' = tan (angle) [/tex]
this means, angle of direction [tex] = \frac{(x+c)}{h} [/tex]
Then, how do I proceed?
 
  • #8
touqra said:
I simplified the result and obtained
[tex]hy'' = 1 + y'^2 [/tex]
solving the above equation, we have
[tex]y' = tan (\frac{(x+c)}{h}) [/tex]
since, [tex] y' = tan (angle) [/tex]
this means, angle of direction [tex] = \frac{(x+c)}{h} [/tex]
Then, how do I proceed?

Well, would be nice to integrate it one more time to get the actual function y(x). That of course brings in another constant of integration. But you have two boundary condtions right? Two equation, two unknowns, little algebra, should be able to figure out what the constants are.

Edit: Oh yea, just use h=1.
 
  • #9
saltydog said:
Well, would be nice to integrate it one more time to get the actual function y(x). That of course brings in another constant of integration. But you have two boundary condtions right? Two equation, two unknowns, little algebra, should be able to figure out what the constants are.
Edit: Oh yea, just use h=1.

Thanks. I got it.
But I still have to figure out why h=1 works. When I didn't make that assumption, it turns out that I have [tex]e^h[/tex]. h = 1 works because [tex]e^h = e [/tex]
Hmmm...
 

What is the purpose of the calculus of variations?

The calculus of variations is a branch of mathematics that deals with finding the optimal solution to a problem where there are many possible solutions. It is used to find the path, function, or shape that minimizes or maximizes a given quantity.

What is the difference between the calculus of variations and traditional calculus?

The main difference between the calculus of variations and traditional calculus is that the calculus of variations deals with optimizing a function over a range of values, while traditional calculus deals with optimizing a function at a specific point.

What are some real-world applications of the calculus of variations?

The calculus of variations has many real-world applications, including in physics, engineering, economics, and biology. It can be used to optimize the shape of structures, such as bridges, to minimize energy consumption in a system, or to model the behavior of particles in motion.

What are some techniques used in the calculus of variations?

Some common techniques used in the calculus of variations include the Euler-Lagrange equation, the direct method, and the principle of least action. These techniques allow for the formulation and solution of variational problems.

What are some challenges in solving problems using the calculus of variations?

One of the main challenges in solving problems using the calculus of variations is that the solution often involves solving a differential equation, which can be complex and require advanced mathematical techniques. Additionally, finding the optimal solution may not always be possible or may require a lot of computational power.

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