** — 1. Variational Calculus — **

Definition 1 A functional is a mapping from vector spaces to into real numbers. |

Let . Suppose that and are constants, the functional form of is known.

According to definition 1 is a functional and the goal of the Calculus of Variations is to determine such that the value of is an extremum.

Let be a parametric representation of such that is the function that makes an extremum.

We can write , where is a function of of the class (that means that is a continuous function whose derivative is also continuous) with .

Now is of the form

Therefore the condition for to be an extremum is

Exercise 1 Given the points and , calculate the equation of the curve that minimizes the distance between the points.
Now . It is , And it is with . The rest is left as an exercise for the reader. |

** — 2. Euler Equations — **

In the following section we’ll analyze the condition for to be an extremum:

Since it is and it follows

Now .

For the first term it is since by hypothesis.

Hence

Remembering that and taking into account the fact that is an arbitrary function one can conclude that

The previous equation is known as the Euler’s Equation

To close our thoughts on the Euler equation let us say that there also is a second form for the Euler equation. The second form is

and is used in the cases where doesn’t depend explicitly on .

** — 3. Euler Equation for variables — **

Let be of the form .

Now we have and for each of the values of . Since are independent functions it follows that for

That is to say we have independent Euler equations.

[…] function can be identified with the function that we saw on Newtonian Mechanics 04 if one makes the obvious […]

“Hence {\displaystyle J(\alpha)= \int_0^{2\pi}(1+\alpha\cos x +\alpha^2\cos ^2x)dx=2\pi+\alpha^2\pi}.”

Should the second term be ?

“Hence .”

Should the second term be ?

Yes it should. Sorry for the typo and thanks for the help.

In Example 2 where you derive the equations of motion for the cycloid… is because is not dependent on , or is there some other reason?

Edit:

Attempt 3:

That’s exactly the reason.

And I assume Exercise 1 ( without actually completing the entire integral) that the solution comes from trigonometric substitution which its second to last step is

After evaluation with the limits in the theta domain the right side becomes 0, hence y = 0 is the equation of the line?

You have to make a trigonometric change of variable to calculate this integral and then you need to make an approximation but it is slightly more complicated than what you present.