Intuition behind variational principle
Let there be given a curve $\vec{p}(t)$ and a real valued function $L$ with the following arguments:
this curve, the time derivative of the curve and the time $t$ itself. Now consider the following integral:
$$
W\left(\vec{p},\dot{\vec{p}}\right) = \int_{t_1}^{t_2} L\left(\vec{p},\dot{\vec{p}},t\right) dt
$$
The integral $W$ is dependent upon the chosen curve $\vec{p}(t)$ ; it is a so-called functional of $\vec{p}$ and $\dot{\vec{p}}$.
We wish to determine for which curves $\vec{p}(t)$ the functional $W(\vec{p},\dot{\vec{p}})$ has an extreme value.
But it is not possible to simply take a derivative and put it to zero. So we must do something else.
Assume that the extreme value is obtained for the curve $\vec{q}(t)$ and let
$\vec{p}(t) = \vec{q}(t) + \epsilon \vec{f}(t)$ , with the same end-points
$\vec{p}(t_1) = \vec{q}(t_1)$ and $\vec{p}(t_2) = \vec{q}(t_2)$, hence
$\vec{f}(t_1) = \vec{f}(t_2) = \vec{0}$. For the rest, the curve $\vec{f}(t)$ is arbitrary
and the variable $\epsilon$ does not depend on $t$. In this way our functional $W$ has become a common function of $\epsilon$:
$$
W\left(\vec{p},\dot{\vec{p}}\right) =
\int_{t_1}^{t_2} L\left(\vec{q} + \epsilon \vec{f},\dot{\vec{q}} + \epsilon \dot{\vec{f}},t\right) dt
$$
A common real-valued function, which can be differentiated in the common way to find extremes:
$$
\frac{dW}{d\epsilon} = 0 \qquad \Longleftrightarrow \qquad
\int_{t_1}^{t_2} \frac{dL\left(\vec{q} + \epsilon \vec{f},\dot{\vec{q}} + \epsilon \dot{\vec{f}},t\right)}
{d\epsilon} dt \quad \mbox{for} \; \epsilon = 0
$$
Let the (vector) components of $\,\vec{p}\,$ be $\,p_k$ . Apply the chain rule:
$$
\frac{dL}{d\epsilon} = \sum_k \frac{\partial L}{\partial p_k} \frac{d p_k}{d\epsilon}
+ \sum_k \frac{\partial L}{\partial \dot{p}_k} \frac{d \dot{p}_k}{d\epsilon}
+ \frac{\partial L}{\partial t} \frac{d t}{d\epsilon} =
\sum_k \frac{\partial L}{\partial p_k} f_k + \sum_k \frac{\partial L}{\partial \dot{p}_k} \dot{f}_k + 0
$$
The last term can be rewritten with help of:
$$
\frac{d}{dt} \left\{ \frac{\partial L}{\partial \dot{p}_k} f_k\right\} =
\frac{\partial L}{\partial \dot{p}_k} \dot{f}_k +
\frac{d}{dt} \left(\frac{\partial L}{\partial \dot{p}_k}\right) f_k
$$
It follows that:
$$
\frac{dW}{d\epsilon} = \sum_k \int_{t_1}^{t_2} \left[\frac{\partial L}{\partial p_k} f_k
- \frac{d}{dt} \left(\frac{\partial L}{\partial \dot{p}_k}\right) f_k
+ \frac{d}{dt} \left\{ \frac{\partial L}{\partial \dot{p}_k} f_k\right\} \right] dt = \\
\sum_k \int_{t_1}^{t_2} \left[\frac{\partial L}{\partial p_k}
- \frac{d}{dt} \left(\frac{\partial L}{\partial \dot{p}_k}\right) \right] f_k\, dt +
\sum_k \left[ \frac{\partial L}{\partial \dot{p}_k} f_k \right]_{t_1}^{t_2}
$$
But we know that $f_k(t_1) = f_k(t_2) = 0$ , so the last term is zero. Furthermore we know that,
for $\epsilon = 0$ : $p_k = q_k$ , and the functions $f_k$ are completely arbitrary. Therefore
it is concluded that the integral is zero if the folowing necessary conditions (not sufficient
eventually) are fulfilled for all $k$ :
$$
\frac{\partial L}{\partial q_k} - \frac{d}{dt} \left(\frac{\partial L}{\partial \dot{q}_k}\right) = 0
$$
These are the well known Euler-Lagrange equations. It is noted that special notations like $\delta W$
for the "variation" of the integral are not needed. The variation is reduced to common differentiation,
which makes the (IMHO confusing) $\delta$ practice completely redundant.