Variational Calculus - Minimal Arc lenght for given surface

[Zaknife]

Homework Statement

A curve is enclosing constant area P. By means of variational calculus show, that the curve with minimal arc length is a circle,

Homework Equations

The Attempt at a Solution

$$F(t)= \int_{x_{1}}^{x_{2}}\sqrt{1+(f^{'})^{2}}dt$$
$$G(t)= \int_{x_{1}}^{x_{2}}f(t)=const$$

If i use Lagrange theorem for functionals i will get

$$\int_{x_{1}}^{x_{2}}=\sqrt{1+(f^{'})^{2}} +\lambda f dx$$

Since, upper functional is not a function of T i can use Euler-Lagrange equation, after differentiation solution will be:
$$1+(f^{'})^{2}=\frac{1}{(C-\lambda f)^{2}}$$
What should i do now ?

 

[mark726]

First, let's clarify what the variables in your equations represent. The variable t is typically used to represent time, but in this case it seems to represent the parameterization of the curve. Let's use x instead to avoid confusion. Also, the functions F and G are not defined, so we can't use them in our solution.

Now, let's start by writing out the functional that we want to minimize:

J[f]=\int_{x_{1}}^{x_{2}}\sqrt{1+(f^{'})^{2}}dx

We want to find the function f(x) that minimizes this functional subject to the constraint that the area enclosed by the curve is constant:

A=\int_{x_{1}}^{x_{2}}f(x)dx=P

To do this, we will use the method of Lagrange multipliers. We introduce a new variable, \lambda, and define a new functional:

L[f,\lambda]=J[f]+\lambda(G[f]-P)

Where G[f] is the constraint function. Now, we can apply the Euler-Lagrange equation to this new functional:

\frac{\partial L}{\partial f}-\frac{d}{dx}\frac{\partial L}{\partial f'}=0

Plugging in our definitions for J and G, we get:

\frac{\partial}{\partial f}\left(\int_{x_{1}}^{x_{2}}\sqrt{1+(f^{'})^{2}}dx+\lambda\int_{x_{1}}^{x_{2}}f(x)dx\right)-\frac{d}{dx}\left(\frac{\partial}{\partial f'}\left(\int_{x_{1}}^{x_{2}}\sqrt{1+(f^{'})^{2}}dx+\lambda\int_{x_{1}}^{x_{2}}f(x)dx\right)\right)=0

Simplifying and rearranging, we get:

\frac{f^{'}}{\sqrt{1+(f^{'})^{2}}}-\lambda=0

Now, we can solve for f'(x):

f'(x)=\lambda\sqrt{1+(f^{'})^{2}}

Integrating both sides with respect to x, we get:

f(x)=\lambda x+\beta

Where \beta is a constant of integration. Plugging this back into our constraint equation,