Is the Variational Principle an introductory example for Calculus of Variations?

  • MHB
  • Thread starter mathmari
  • Start date
  • Tags
    Minimum
In summary, the conversation discusses the best way to travel from point A to point B without crossing a lake. Three different options are presented, each with a different distance calculation. The speaker also mentions using the Pythagorean theorem and finding the lengths of different parts of the path. Ultimately, it is determined that the third option is the most optimal, as any small changes to the path would not significantly affect the distance traveled. This is confirmed by the Variational Principle.
  • #1
mathmari
Gold Member
MHB
5,049
7
Hey! :eek:

We want to get from the poit $A$ to the point $B$ avoiding the lake. How should we move to minimize the distance?

View attachment 4795

There are the following possibilities:
1.
View attachment 4794

(We go along the red lines)

In this case the distance is the following:

Let $y_1$ be the first red line and $y_2$ the second one. Then from the Pythagorean theorem we have the following:
$$y_1^2=a^2+p^2 \Rightarrow y_1=\sqrt{a^2+p^2} \\ y_2^2=(b-a)^2+p^2 \Rightarrow y_2=\sqrt{(b-a)^2+p^2}$$

So the length of the way is equal to $y_1+y_2$. Is this correct? 2.
View attachment 4796

To find the the lengh ofthe first red line we apply the pythagorean theorem, right? So let $x$ be this part, then $a^2=x^2+p^2 \Rightarrow x=\sqrt{a^2-p^2}$. The last part of the red line is equal to $b-a$. How can we find the length of the red part of the circle? 3.
View attachment 4797

How can we find in this case the length of the red line?
 

Attachments

  • circle.png
    circle.png
    4 KB · Views: 62
  • lake.png
    lake.png
    2 KB · Views: 67
  • circle2.png
    circle2.png
    2.5 KB · Views: 53
  • circle3.png
    circle3.png
    2.2 KB · Views: 57
Physics news on Phys.org
  • #2
mathmari said:
1.
In this case the distance is the following:
Let $y_1$ be the first red line and $y_2$ the second one. Then from the Pythagorean theorem we have the following:
$$y_1^2=a^2+p^2 \Rightarrow y_1=\sqrt{a^2+p^2} \\ y_2^2=(b-a)^2+p^2 \Rightarrow y_2=\sqrt{(b-a)^2+p^2}$$

So the length of the way is equal to $y_1+y_2$. Is this correct?

Hi mathmari! (Smile)

I don't think so...

Isn't the distance $y_1$ to the top of the lake instead of slightly higher as it should be? (Wondering)

2.
To find the the lengh ofthe first red line we apply the pythagorean theorem, right? So let $x$ be this part, then $a^2=x^2+p^2 \Rightarrow x=\sqrt{a^2-p^2}$. The last part of the red line is equal to $b-a$. How can we find the length of the red part of the circle?

The length along a circle with radius $p$ is $p\Delta\phi$ where $\phi$ is the angle that is traversed. (Nerd)

3.
How can we find in this case the length of the red line?

It's a combination of (1) and (2).
First we travel in a straight line to the point that touches the lake.
Then we travel along the circumference of the lake.
After which we continue in a straight line to the destination. (Thinking)
 
  • #3
I like Serena said:
I don't think so...

Isn't the distance $y_1$ to the top of the lake instead of slightly higher as it should be? (Wondering)

What do you mean? (Thinking)
I like Serena said:
The length along a circle with radius $p$ is $p\Delta\phi$ where $\phi$ is the angle that is traversed. (Nerd)

Is it as follows?

View attachment 4798

$$\cos \theta=\frac{p}{a} \Rightarrow \cos (\pi -\phi )=\frac{p}{a} \Rightarrow \cos (\phi )=-\frac{p}{a} \Rightarrow \phi =\arccos \left (-\frac{p}{a}\right )$$
I like Serena said:
It's a combination of (1) and (2).
First we travel in a straight line to the point that touches the lake.
Then we travel along the circumference of the lake.
After which we continue in a straight line to the destination. (Thinking)

Is it as follows?

$$\sqrt{a^2-p^2}+p\Delta \phi +\sqrt{(b-a)^2-p^2}$$
 

Attachments

  • circle2.png
    circle2.png
    3.1 KB · Views: 57
  • #4
mathmari said:
What do you mean? (Thinking)

Which are the sides of the Pythagorean triangle exactly?
Does the hypotenuse correspond to the full length of the red line? (Wondering)

Is it as follows?

$$\cos \theta=\frac{p}{a} \Rightarrow \cos (\pi -\phi )=\frac{p}{a} \Rightarrow \cos (\phi )=-\frac{p}{a} \Rightarrow \phi =\arccos \left (-\frac{p}{a}\right )$$

Yup. (Nod)

Is it as follows?

$$\sqrt{a^2-p^2}+p\Delta \phi +\sqrt{(b-a)^2-p^2}$$

Yep. (Smile)
 
  • #5
I like Serena said:
Which are the sides of the Pythagorean triangle exactly?
Does the hypotenuse correspond to the full length of the red line? (Wondering)

View attachment 4801

$$(AK)=\sqrt{a^2-p^2} \ \ , \ \ (LB)=\sqrt{(b-a)^2-p^2}$$

Can we find the length of the parts $KC$ and $CL$ ?
 

Attachments

  • circle.png
    circle.png
    4.5 KB · Views: 55
  • #6
mathmari said:
$$(AK)=\sqrt{a^2-p^2} \ \ , \ \ (LB)=\sqrt{(b-a)^2-p^2}$$

Can we find the length of the parts $KC$ and $CL$ ?

Sure!
$$\tan\theta=\frac P{KC}$$
 
  • #7
I like Serena said:
Sure!
$$\tan\theta=\frac P{KC}$$

Is it as follows?

View attachment 4802

$$\cos \theta =\frac{p}{a} \Rightarrow \theta=\arccos \left (\frac{p}{a}\right ) \\ \tan \theta=\frac{(KC)}{p} \Rightarrow (KC)=p \tan \theta , \text{ with } \theta=\arccos \left (\frac{p}{a}\right )$$

and

$$\cos \phi =\frac{p}{b-a} \Rightarrow \phi=\arccos \left (\frac{p}{b-a}\right ) \\ \tan \phi=\frac{(CL)}{p} \Rightarrow (CL)=p \tan \phi , \text{ with } \phi=\arccos \left (\frac{p}{b-a}\right )$$
 

Attachments

  • circle.png
    circle.png
    4.7 KB · Views: 53
  • #8
Are the total lengths at each case the following?

  1. $$\sqrt{a^2-p^2}+p\tan \theta+p\tan \phi +\sqrt{(b-a)^2-p^2}$$
  2. $$\sqrt{a^2-p^2}+p\arccos \left (-\frac{p}{a}\right )+b-a-p$$
  3. $$\sqrt{a^2-p^2}+p\left (\pi -\arccos \left (\frac{p}{a}\right )-\arccos \left (\frac{p}{b-a}\right )\right )+\sqrt{(b-a)^2-p^2}$$

Having this information, how do we know which of them is the minimum?
 
  • #9
mathmari said:
Is it as follows?



$$\cos \theta =\frac{p}{a} \Rightarrow \theta=\arccos \left (\frac{p}{a}\right ) \\ \tan \theta=\frac{(KC)}{p} \Rightarrow (KC)=p \tan \theta , \text{ with } \theta=\arccos \left (\frac{p}{a}\right )$$

and

$$\cos \phi =\frac{p}{b-a} \Rightarrow \phi=\arccos \left (\frac{p}{b-a}\right ) \\ \tan \phi=\frac{(CL)}{p} \Rightarrow (CL)=p \tan \phi , \text{ with } \phi=\arccos \left (\frac{p}{b-a}\right )$$

With the angle for $\theta$ that you picked, I think it should be: $\sin \theta =\frac{p}{a}$.
mathmari said:
Are the total lengths at each case the following?

  1. $$\sqrt{a^2-p^2}+p\tan \theta+p\tan \phi +\sqrt{(b-a)^2-p^2}$$
  2. $$\sqrt{a^2-p^2}+p\arccos \left (-\frac{p}{a}\right )+b-a-p$$
  3. $$\sqrt{a^2-p^2}+p\left (\pi -\arccos \left (\frac{p}{a}\right )-\arccos \left (\frac{p}{b-a}\right )\right )+\sqrt{(b-a)^2-p^2}$$

Having this information, how do we know which of them is the minimum?

That looks correct.

Well, in (1) we can shorten the length by pulling the intersection at the top down to the lake.
In (2) we can shorten the length by pulling the angle on the right hand side of the lake outward.

It's only in (3) that there is nothing that we can do to shorten the length.
Moreover, wherever we pull a bit on the path, in the "first order" the length won't even change.
This confirms that according to the Variational Principle, it's an optimal path.
 
  • #10
I like Serena said:
With the angle for $\theta$ that you picked, I think it should be: $\sin \theta =\frac{p}{a}$.

Why is it $\sin$ and not $\cos$ ?

I like Serena said:
Well, in (1) we can shorten the length by pulling the intersection at the top down to the lake.
Do you mean that we can shorten the length by the following green line?

View attachment 4808

I like Serena said:
In (2) we can shorten the length by pulling the angle on the right hand side of the lake outward.
What exactly do you mean?
I like Serena said:
It's only in (3) that there is nothing that we can do to shorten the length.
Moreover, wherever we pull a bit on the path, in the "first order" the length won't even change.
This confirms that according to the Variational Principle, it's an optimal path.
What exactly is the Variantonal Principle?
 

Attachments

  • circle.png
    circle.png
    4.6 KB · Views: 52
  • #11
mathmari said:
Why is it $\sin$ and not $\cos$ ?

Because the angle $\theta$ that you picked, is the same as the angle at $A$.
And the sine of the angle at $A$ is $\frac p a$.

Do you mean that we can shorten the length by the following green line?

Yes.

What exactly do you mean?

We can shorten the path by making it more like (3), removing the angle at the right hand side of the lake.

What exactly is the Variantonal Principle?

The Variational Principle lies at the root of Calculus of Variations.

In essence it says that if some path solution doesn't become better in the first order if you change it anywhere, it's an optimal solution.

We can compare it to thinking of an elastic string.
If we put an elastic string on the red path and tighten it, both solutions (1) and (2) will change until they reach solution (3), which is the tightest possible.

Since you have already asked a question about Calculus of Variations in the chat room, can it be that this is an introductory example? (Wondering)
 

FAQ: Is the Variational Principle an introductory example for Calculus of Variations?

What is the purpose of "Get the minimum distance" in scientific research?

The purpose of "Get the minimum distance" in scientific research is to determine the shortest distance between two points or objects. This is important in many fields of science, such as physics, chemistry, and biology, as it allows researchers to calculate relationships and interactions between objects in a precise manner.

How is the minimum distance calculated?

The minimum distance is calculated using various mathematical formulas and techniques, depending on the specific scenario and variables involved. In general, it involves finding the distance between two points by using their coordinates or determining the closest approach between two objects based on their sizes and positions.

What factors can affect the minimum distance between two objects?

The minimum distance between two objects can be affected by factors such as the size and shape of the objects, their velocities, and any external forces acting on them. Other factors may include the medium in which the objects are moving through, such as air or water, and any obstacles or barriers present in their path.

Why is it important to accurately measure the minimum distance in scientific experiments?

The accurate measurement of the minimum distance is crucial in scientific experiments as it can greatly impact the results and conclusions drawn from the data. Inaccurate measurements can lead to incorrect calculations and interpretations, which can ultimately affect the validity and reliability of the research.

Can technology be used to assist in obtaining the minimum distance?

Yes, technology plays a significant role in helping scientists obtain the minimum distance between objects. Tools such as rulers, measuring tapes, and calipers are commonly used in experiments to measure distances. In more advanced research, technologies such as lasers, sonar, and GPS can be utilized to accurately measure distances in complex scenarios.

Similar threads

Replies
12
Views
2K
Replies
3
Views
2K
Replies
15
Views
1K
Replies
3
Views
2K
Replies
2
Views
2K
Replies
8
Views
1K
Replies
4
Views
1K
Back
Top