Please Help: Utter confusion - First Variation of a function - Calculus of variations So i've been at this for most of the night. i was originally asked to find the first and second variation of the problem $$\int_{0}^{1} \sqrt{\dot{x}^{2}+\dot{y}^2}~dt$$ 
but at this point i'll settle with just understanding what the problem entails
so. Given a functional $J(x,u,u') = \int_{a}^{b} \Lambda(x,u,u') dx$ to find the first variation, is all we have to do simply to find the euler-legrange equation?
so would $\delta J = \int_{a}^{b} \left(\frac{\partial \Lambda}{\partial u} - \frac{d}{dx} \frac{\partial \Lambda}{\partial u'} \right) \delta u~dx$
would this be accepted as "the answer"?
then if we wish to find the second variation, would we simply have to set $$\delta J = \int_{a}^{b}\Phi(x,u,u') \delta u~dx$$ and find
$$\delta^2 J = \int_{a}^{b} \left[\frac{\partial \Phi}{\partial u} - \frac{d}{dt}\frac{\partial \Phi}{\partial u'} + \frac{d^2}{dt^2}\frac{\partial \Phi}{\partial u''} \right]\delta^{2}u~dt$$
where $\delta u$  and $\delta^2 u$ are arbitary functions (and thus do not have to be explicitly stated as $\delta u =~....$
is this all there is to it? i know there are a variety of different ways to do this, including the directional derivative $$\delta J = \left. \Lambda(u+\epsilon h) \right|_{\epsilon = 0}$$ and ive seen also $$\delta J = \Lambda(u+\epsilon h) - \Lambda(u)$$ etc...
finally if all of the above is correct... how does this work with multiple variables like the above parametric form? i would know how to derive the Euler-Legrange equations for both $\dot{x}$ and $\dot{y}$ would i just sum them in the integrand? or is there a specific format for the answer?
Thanks for the help...im kind of going nuts here.
 A: Let me give you a few pointers. Let's define $u=(x,y)$. You start with
$$
S(u,u') = \int_0^1 dt L(u,u')
$$
where $u'=du/dt$. To compute the first order variation we can discard quadratic terms in $\delta u $. Hence we get the following equation true up to first order
$$
\delta S = S(u+\delta u ) - S(u) = \int_0^1 dt \frac{\partial L}{\partial u} \delta u + \frac{\partial L}{\partial u'} \delta u'
$$
Now note: $\delta u' = d/dt \delta u $. So we can integrate by part the second term:
$$
\int_0^1 dt \frac{\partial L}{\partial u'} \frac{d}{dt} \delta u  = \left .   \frac{\partial L}{\partial u'}  \delta u \right |_0^1 - \int_0^1 dt \left ( \frac{d}{dt} \frac{\partial L}{\partial u'} \right ) \delta u
$$
If the path $u$ is fixed at the border (as is your case) we must have $\delta u(0) = \delta u(1)=0$ and the border term vanishes. Hence we get
$$ 
\delta S = \int_0^1 dt \left [ \frac{\partial L}{\partial u} - \left ( \frac{d}{dt} \frac{\partial L}{\partial u'} \right )  \right ] \delta u . 
$$
This is the formula you quote (except that you called suddenly $t=x$). If you require $\delta S =0$ for all allowed variation $\delta u$ it can be shown that this implies
$$
\left [ \frac{\partial L}{\partial u} - \left ( \frac{d}{dt} \frac{\partial L}{\partial u'} \right )  \right ] =0
$$
that is, the Euler-Lagrange equations. This is sometimes called the fundamental lemma of calculus of variations. 
Of course in your case $u=(x,y)$ is a vector and so you get
\begin{align}
\delta S &= \int_0^1 dt \left [ \frac{\partial L}{\partial x} - \left ( \frac{d}{dt} \frac{\partial L}{\partial x'} \right )  \right ] \delta x  \\
&+ \left [ \frac{\partial L}{\partial y} - \left ( \frac{d}{dt} \frac{\partial L}{\partial y'} \right )  \right ] \delta y
\end{align}
Again, specializing to your case 
$$
\frac{\partial L}{\partial x}= \frac{\partial L}{\partial y}=0
$$
and so you obtain
$$
\delta S = - \int_0^1 dt \left [  \left ( \frac{d}{dt} \frac{\partial L}{\partial x'} \right )   \delta x + \left ( \frac{d}{dt} \frac{\partial L}{\partial y'} \right )   \delta y \right ] 
$$
Since your action measures the length of the path $u$, it can be checked that the Euler-Lagrange equation gives back the equations for a geodesic in 2D, that is a straight line. 
A: From my understanding of the problem now; and if i'm incorrect by all means someone feel free to correct me. 
the parameterisation above of the arc length can be solved using the standard Euler-Legrange equation format (Summing the EL equations WRT 
the explicit variables) the confusion stems over what we do with the "end points", Since in the current terms of the course the end points vanish, this essentially means that we're left with a general function in the direction of u, $$\delta u$$ before hand...now given variation at the end points the increment makes it so that we need to consider this and essentially shift the first variation resulting in (apologies for notation change)
$$\int_{x_0}^{x_1}\left[\Lambda_{u}-\frac{d}{dx} \Lambda_{u'} \right] h(x) dx +  \Lambda_{u'} \delta u \bigg\rvert^{x=x_1}_{x=x_0}+\left(\Lambda - \Lambda_u u'\right)\bigg\rvert^{x = x_1}_{x=x_0}$$
but this is apparently (as described by Gelfand and Fomin) to be the basic formula for the general variation of the functional if the end points are contrained to line on the straight line $x=x_0, x=x_1$ so this would further assume there are other formulations of the variation leading to the idea that the only sure fire way depending on the classification of problem is to do it the incremental way. ie
calculating $$\Delta \Lambda = \Lambda[u+h] - \Lambda[u]$$
hope this has helped anyone in a similair situation as i was. and thank you all for the comments
