A question about the derivation of the Euler-Lagrange equation

[planck42]

In the book Mathematical Methods for Engineers and Scientists 3, the derivation of the Euler-Lagrange equation starts roughly along the lines of this:

In order to minimize the functional $$I=\int_{x_1}^{x_2}{f(x,y,y')dx}$$, one should define two families of functions Y(x) and Y'(x), where Y(x) is $$y(x)+{\alpha}{\eta}(x)$$ and Y'(x) is the x-derivative of y(x), $$\alpha$$ is an arbitrary constant coefficient of $${\eta}(x)$$, which is an arbitrary function that must be zero at $$x_1$$ and $$x_2$$.

These steps aren't difficult to understand; it allows for any possible function that could render the functional stationary while maintaining the boundary conditions.

The minimum value of I should occur when $$\frac{dI}{d\alpha}$$ is zero at $${\alpha}=0$$

Whoa! Why are we dealing in $$\alpha$$'s all of a sudden? How does this minimize I and not the derivative with respect to x? Does working it out with x always produce a trivial solution? I need help with this one step!

 

[Office_Shredder]

I isn't a function of x (x is a dummy variable inside the integral, not a variable that you put into I).

On the other hand I DOES change as alpha changes (and as eta changes, but we're not going to worry about that)

 

[planck42]

Makes sense. However, I still have one question that I'll attempt to answer: why does $${\alpha}=0$$ produce an extreme, and not, for instance, $${\alpha}=1$$? My guess is that the value of $$\alpha$$ is irrelevant; any value of $$\alpha$$ can make the first derivative of I with respect to $$\alpha$$ vanish. Hence the simplest choice, $${\alpha}=0$$.

 

[Galileo]

The point is that $y(x)$ is a function that minimizes (or extremizes) the integral.
So in [tex]y(x)+{\alpha}{\eta}(x)[/itex], the $$y(x)$$ is the sought after solution and $${\alpha}{\eta}(x)$$ is an arbitrary deviation from that solution.
Therefore, the minimum occurs at $${\alpha}=0$$