Applications of the Euler-Lagrange Equation

[aaj92]

Homework Statement

Find and describe the path y = y(x) for which the integral $\int$$\sqrt{x}$$\sqrt{1+y^{' 2}}$dx (the integral goes from x1 to x2... wasn't sure how to put that in. sorry) is stationary.

Homework Equations

$\frac{\partial f}{\partial y}$ - $\frac{d}{dx}$$\frac{\partial f}{\partial y^{'}}$=0

The Attempt at a Solution

$\frac{\partial f}{\partial y}$ = 0 so that means $\frac{d}{dx}$$\frac{\partial f}{\partial y^{'}}$ also has to equal zero. this means that $\frac{\partial f}{\partial y^{'}}$ is a constant. So I set it equal to a random constant c and solved for x. I ended up getting x = 2C$^{2}$ which is really wrong :p. The answer is supposed to be x = C + $\frac{(y-D)^{2}}{4C}$
where C and D are both constants... I don't understand how they ended up with two constants. I'm not really sure how the constants work. In class it seems like he just sort of pulls them out of the air and so now I'm confused

 

[anathema]

.

Thank you for your question. The path y = y(x) for which the integral is stationary can be found by using the Euler-Lagrange equation, which you have correctly identified. However, there seems to be an error in your attempt at a solution.

Instead of setting \frac{\partial f}{\partial y} = 0, you should set \frac{\partial f}{\partial y^{'}} = c, where c is a constant. This is because the Euler-Lagrange equation states that \frac{\partial f}{\partial y} - \frac{d}{dx}\frac{\partial f}{\partial y^{'}} = 0, and since we want the integral to be stationary, \frac{\partial f}{\partial y} should also be equal to 0.

Now, using the given equation \int\sqrt{x}\sqrt{1+y^{' 2}}dx, we can find the partial derivative with respect to y': \frac{\partial f}{\partial y^{'}} = \frac{1}{2}\frac{\sqrt{x}}{\sqrt{1+y^{' 2}}}.

Setting this equal to c and solving for y', we get y' = \pm\sqrt{\frac{x}{c^2-x}}. Integrating this with respect to x, we get y = \pm\frac{1}{2}(\sqrt{c^2-x}-c\ln(\sqrt{c^2-x}+c)) + D, where D is another constant of integration.

Now, we can simplify this expression by setting c = \frac{1}{2} and D = 0. This gives us the desired path y = \frac{(y-D)^2}{4c} = x, where c and D are both constants. This means that for any value of c and D, the integral will be stationary for the path y = \frac{(y-D)^2}{4c}.

I hope this explanation helps you understand the concept of constants in the Euler-Lagrange equation. Please let me know if you have any further questions.