\section{Introduction}
In introductory physics courses, the student is presented, in general, to the newtonian formalism for classical mechanics, which is based in three simple laws to the bodies motion. This fact require students to think in force before progressing to concepts such as energy, momentum and least action principle. The concept of force has received several criticisms, e.g., Wilczec \cite{wilckzec-essay} has argumented that the force, given by the second's Newton law, has no independent meaning and Jammer \cite{jammer-preface} has suggested that the concept of force is in the end its life cycle. Besides, for introducing the newtonian formalism, the student must deal with the vectorial character of force, which must bring difficulties for problems where vector decomposition is demanded.

On the other hand, there are other ways to describe the motion of a rigid body without resort to the concept of force. Amongst them, we can cite the lagrangian formulation, which describe the system evolution from the concept of action and obtain the motion equations, which agree with the second Newton's law (without the necessity of the concept of force), from the least action principle \cite{greiner}. The lagrangian formalism is important not only in mechanics, but also in several branches of physics, such as quantum field theory \cite{ryder}; for giving theoretical models on condensed matter systems \cite{Atland-book}, e.g., nanomagnetism \cite{Vagson-JAP}-\cite{Vagson-PLA} and nematic liquid crystals \cite{Napoli}; and so forth. Besides, the lagrangian formulation of field theory is behind the Noether's theorem, which connects symmetries and conservational laws \cite{ryder}. Thus, the sooner the student make contact with this formalism, the sooner he will be able to study and discuss about several relevant and current topics in physics.

Despite its importance, in general, in the secondary level or in the beginning of undergraduate courses of engineering, mathematics or physics, no attention is devoted to the lagrangian formalism for classical mechanics. This fact must be associated with the fact that it is not simple to introduce this theme without speaking in the concepts of partial derivative and least action principle, being the last one a difficult theme to be mathematically treated in introductory courses. However, Curtis has proposed the qualitative discussion of the least action principle in introductory courses of physics \cite{curtis-EJP}, as well as other modern physics themes, in order to give a presentation of this discipline without the historical character with which it has been worked out since the 50's years. In this way, we propose the development of a simplification of the ELE for the particles dynamics, in order to obtain a simple equation, accessible untill to high school students. This proposition is in consonance with some recent discussions in the attempt to provide the student with information on relevant topics and relatively advanced in Physics \cite{curtis-EJP}. For example, Organtini \cite{organtini} and Cid \cite{cid} discuss the possibility to introduce the Higgs mechanism to undergraduate students. Besides, the study about surface plasmon resonance, which is often confusing for undergraduate students, was proposed from the link between classical concepts of resonance and the solution of the problem \cite{domini}. Finally,  Bezerra \textit{et al} has proposed the introduction of magnetic dipoles, by considering a localized steady current distribution, where one could do a Taylor expansion directly in the Biot–Savart law to obtain, explicitly, the dominant contribution of the magnetic field at distant points, identifying the magnetic dipole moment of the distribution \cite{rio}.

In this work, we show that it is possible to introduce the lagrangian formulation to describe mechanic systems in introductory physics courses, by simplifing the ELE for one-dimensional motions in a mathematical formulation accessible to high school students. Despite the criticisms in the 'overmathematicalization' in physics education and the proposition that math must be the last thing to be taught, Taber arguments that the mathematical concepts are indispensable in physics courses \cite{taber-PEd}. Thus, we believe that this work must bring new contributions and ways to introduce modern topis on physics to introductory courses.

This work was devided as follows: in section \ref{LagrangeSup} we did a brief review about the Lagrange formalism for classical mechanics; section \ref{LagrangeMed} presents the simplification of the ELE and three simple examples are developed in section \ref{Examples}. Finally, the section \ref{Conc} brings the conclusions and prospects.

\section{The Euler-Lagrange equation}\label{LagrangeSup}
In the lagrangian formalism, the mechanical system is described by $N$ generalized coordinates and $N$ generalized velocities. The system evolves from a configuration in the time $t_1$ to another in the time $t_2$. The ask to be answered is: How does the system evolve from the configuration 1 for the configuration 2? In the newtonian formalism, this evolution is given by the Newton's second law. In the lagrangian one, the Hamilton's principle (also called least action principle) describes the dynamical evolution of the system. The least action principle states that the evolution of the system from configuration 1 to configuration 2 is such that the action is a minimum.

The concept of action is associated with the lagrangian $L$ of the system, which is a function of the generalized coordinates ($q$), generalized velocities ($\dot{q}$) and time ($t$), that is, $L\equiv L(q,\dot{q},t)$. Thus, the action is defined as follows:
\begin{equation}
S=\int_{t_1}^{t_2}L(q,\dot{q},t)dt.
\end{equation}

From the Hamilton's principle, we have that $\delta S=0$, from where one can conclude that:
\begin{equation}
\delta S=\int_{t_1}^{t_2}\sum_{i=1}^{N}\left(\frac{\partial L}{\partial q_i}-\frac{d}{dt}\frac{\partial L}{\partial \dot{q}_i}\right)\delta q_i(t)dt=0.
\end{equation}
Once the $\delta q_i$ are arbitrary functions of $t$, if there is no links between the $q_i$'s, they will be independent and:
\begin{equation}\label{EQL}
\frac{\partial L}{\partial q}-\frac{\partial}{\partial t}\left(\frac{\partial L}{\partial\dot{q}}\right)=0,
\end{equation}
where we have ommited the $i$ subindex. This is the ELE, which gives us the time evolution of the system. This equation is as important to the lagrangian formalism as the Newton's second law is to the newtonian one.

The lagrangian is defined to be $L=T-V$, where $T$ is the kinetic energy and $V$ is the potential energy. The lagrangian written in this way makes the ELE equivalent to the Newton's second law. For proofing it in the case of conservative forces, considere a particle in a region where there is an interaction potential:
\begin{equation}
L=\frac{1}{2}mv^2-V(\mathbf{r}).
\end{equation}
By taking the generalized coordinates as the cartesian coordinates, we have:
$$\frac{\partial L}{\partial x_i}=\frac{\partial V}{\partial x_i},$$
\begin{equation}
\frac{d}{dt}\frac{\partial L}{\partial\dot{x}_i}=m\ddot{x}_i.
\end{equation}
Thus, the ELE gives:
\begin{equation}
m\ddot{x}_i=-\frac{\partial V}{\partial x_i},
\end{equation}
which, as expected, is the Newton's second law for conservative forces.

Now, we are in conditions to simplify the ELE for one-dimensional motions and study it for some particular problems.

\section{Simplification of the ELE equation}\label{LagrangeMed}
From Eq. (\ref{EQL}), one can see that, to solve the ELE equation for a particular mechanical system, one must calculate two partial derivatives, which is not a simple task for several students, in special, that ones which are in the high school level or in the beginning of undergraduate courses. In order to simplify the ELE for one-dimensional problems, without lost of generality, we will adopt the $x$ axis as the direction of the motion and will start form the concept of partial derivative.
 
Given an arbitrary function $f(x,y)$, the partial derivatives of $f$ in relation to $x$ and $y$ at the point $(x_0,y_0)$ are defined, respectively, by:
\begin{equation}
\frac{\partial f}{\partial x}(x,y)=\lim_{\Delta x\rightarrow 0}\frac{f(x+\Delta x,y)-f(x,y)}{\Delta x}
\end{equation}
and
\begin{equation}
\frac{\partial f}{\partial y}(x,y)=\lim_{\Delta y\rightarrow 0}\frac{f(x,y+\Delta y)-f(x,y)}{\Delta y},
\end{equation}
where $\Delta x=x-x_0$ and $\Delta y=y-y_0$.

In this way, we can take:
\begin{equation}
\frac{\partial L}{\partial v}=\lim_{v\rightarrow v_0}\frac{L(x_0,v)-L(x_0,v_0)}{v-v_0}
\end{equation}
and
\begin{equation}
\frac{\partial L}{\partial x}=\lim_{x\rightarrow x_0}\frac{L(x,v_0)-L(x_0,v_0)}{x-x_0},
\end{equation}
which are the partial derivative definitions for the two terms of the ELE.

Now, for one-dimensional motions in a constant force, we will define two operators, $\Delta_x$ and $\Delta_v$, here called \textit{mean delta operators}, which act at the lagrangian in the below form:
\begin{itemize}
\item the $\Delta_x$ operator acts in the lagrangian terms that depend only on the $x$ coordinates. That is: 
\begin{equation}
\Delta_x L=L(x)-L(x_0);
\end{equation} 
\item the $\Delta_v$ operator, which acts in the lagrangian terms that depends only on $v$. That is:
\begin{equation}
\Delta_v L=L(v)-L(v_0),
\end{equation} 
\end{itemize}
where $L(x)$ and $L(v)$ are the terms in the lagrangian depending only on $x$ and $v$, respectively. Thus, we have that:
\begin{equation}
\frac{\Delta_{v}L}{\Delta v}=\frac{L(v)-L(v_0)}{v-v_0}\label{EQ3}
\end{equation} and
\begin{equation}
\frac{\Delta_{x}L}{\Delta x}=\frac{L(x)-L(x_0)}{x-x_0}\label{EQ4}
\end{equation}

Now, as well as we call the mean velocity of a particle as $v_m=\Delta x/\Delta t$, we will call the Eqs. (\ref{EQ3}) and (\ref{EQ4}) as the \textit{mean lagrangian} in relation to $v$ and $x$, respectively. With that definitions, we will rewrite the Eq. (\ref{EQL}), for one-dimensional motions subject to a constant force, as:
\begin{equation}
\frac{\Delta}{\Delta t}\left(\frac{\Delta_{v}L}{\Delta v}\right)-\frac{\Delta_{x}L}{\Delta x }=0,\label{zen0}
\end{equation} 
where $\Delta x=x-x_{0}$, $\Delta v=v-v_{0}$ and $\Delta t=t-t_0$. From Eq. (\ref{zen0}) we can conclude that the temporal variation of the \textit{mean lagrangian} with relation to $v$ is equal to the \textit{mean lagrangian} in relation to $x$. Here, the Eq. (\ref{zen0}) is the simplification of the ELE for one-dimensional motions subject to a constant force, without the partial derivatives use.

After the presentation of the simplification of the ELE equation, we can study three simple examples for testing the vality of this approximation: a particle subject to the gravitational field (free fall body), the Atwood's machine and the inclinated plan.

\section{Three simple examples}\label{Examples}
\subsection{Free fall body}
In the case of a free fall body, the lagrangian is given by: 
\begin{equation}
L=\frac{1}{2}mv^{2}-mgy,
\end{equation}
where $y$ is the height, measured from the soil. In this case, the lagrangian components depending on the position and on the velocity are given by $L(v)=\frac{1}{2}mv^{2}$ and $L(y)=-mgy$. Obviously, $L(v_{0})=\frac{1}{2}mv_{0}^{2}$ and $L(y_{0})=-mgy_{0},$ where $y_0$ and $v_0$ are the initial height and velocity of the particle. Thus:
\begin{equation}\label{anterior}
\frac{\Delta_vL}{\Delta v}=\frac{\frac{1}{2}mv^{2}-\frac{1}{2}mv_0^{2}}{v-v_{0}}=\frac{1}{2}m(v+v_0)
\end{equation}
and
\begin{equation}
\frac{\Delta_yL}{\Delta y}=\frac{-mgy+mgy_0}{y-y_0}=-mg,
\end{equation}
where we have used the factorization property $(v^{2}-v_{0}^2)=(v+v_{0})(v-v_{0})$ in Eq. (\ref{anterior}). 

For continuing our analisys, we will define ${(v+v_0)}/{2}\equiv v_{_M}$, 
where $v_{_M}$ is the average of the velocities between two arbitrary points in the trajectory of the free fall body, which, for uniformly varied rectilinear motions is equal to the mean velocity of the particle. For proofing it, we have that $v_m=\Delta x/\Delta t$, however, from the Torricelli's equation:
\begin{equation}
\Delta x=\frac{(v+v_0)(v-v_0)}{2a}=\frac{(v+v_0)\Delta t}{2}\Rightarrow v_m=\frac{v+v_0}{2}=v_{_M}.
\end{equation} 
Thus:
\begin{equation}
\frac{\Delta}{\Delta t}\left(\frac{\Delta_{v}L}{\Delta v}\right)=m\frac{\Delta v_{_M}}{\Delta t}.
\end{equation}

Note that we are not talking about variation of the mean velocity of a particle, but the variation of the average of velocities between two points of the trajectory, which changes if we consider two different excerpts in the body motion. For example, when we compare the average of velocities during the first and second halves of the motion of a free fall body, we note that it has different values​​. In fact, if some body is dropped from a height $h$, in a place where the acceleration of gravity is $g$, when it reaches the height ${h}/{2}$, its speed is $ v_1=\sqrt{gh} $. Once it was abandoned, we have $v_0=0$ and in the first half of the fall, this average is:
\begin{equation}
v_{_{M1}}=\frac{v_1+v_0}{2}=\frac{\sqrt{gh}}{2}.
\end{equation}
However, upon reaching the soil, the body velocity is $v_2=\sqrt{2gh}$, in such way that during the second half of the trajectory, the average between the velocities of the free fall body is: \begin{equation}
v_{_{M2}}=\frac{v_2+v_1}{2}=\frac{\sqrt{3gh}}{2},
\end{equation}
which proofs the fact that there exists a variation of the average of velocities of the particle along the trajectory. Finally, from the definition of mean acceleration, we will take the body acceleration as $a\equiv({\Delta v_{_M}}/{\Delta t})$, and from Eq. (\ref{zen0}), we have $a=-g$.

Since the gravitational acceleration is constant next to the Earth surface, the particle realizes a uniformly accelerated rectilinear motion, with the velocity and position being given by:
\begin{equation}\label{Eq-mot}
v=v_0-gt,\hspace{1cm}y=y_0+v_{0}t-\frac{1}{2}gt^2,
\end{equation}
which are the motion equations to a particle moving under the Newton's laws, with constant force $\mathbf{P}=m\mathbf{g}$.

\subsection{The Atwood's machine}
The Atwood's machine system can be used for introducing students to more complex problems involving ropes and chains. Despite it is an ancient problem, it will be considered here by having pedagogical advantages for secondary and high school students, and still, it is useful to study mass variable systems \cite{sousa-PEd}.

The Atwood's machine consists in two blocks of masses $m_1$ and $m_2$, which are connected by a massless string with lenght $\ell$ passing over a frictionless pulley of negligible mass and radius $R$. By taking $m_1>m_2$ and $\ell=\pi R+x_1+x_2$, with (see Fig. \ref{atwoodFig}) we have:
\begin{equation}
L(v_{1})=\frac{1}{2}m_1v_1^{2}\,\,\,\,\,\,\,\, e\,\,\,\,\,\,\,\, L(x_1)=-m_1gx
\end{equation}
and
\begin{equation}
L(v_{2})=\frac{1}{2}m_{2}v_2^{2}\,\,\,\,\,\,\,\, e\,\,\,\,\,\,\,\, L(x_{2})=-m_{2}g(\ell-x),
\end{equation}
where we have used the fact that when the block 1 falls to the height $x$, the block 2 rises to $\ell-x$. Since the blocks are linked by a string with constant lenght $\ell$, its velocities $v_1$ and $v_2$ are equal, that is, $v_1=v_2\equiv v$. Thus, the total lagrangian of the system is: 
\begin{equation}
L=\frac{1}{2}(m_1+m_2)v^2-m_1gx-m_2g(\ell-x).
\end{equation}
At the time $t_0$, we have:
 \begin{equation}
 L_0=\frac{1}{2}(m_1+m_2)v_0^{2}-m_1gx_0-m_2g(\ell-x_0).
 \end{equation}
\begin{center}
\begin{figure}
\includegraphics[scale=0.5]{atwood.eps}\caption{[Color online] Atwood's machine scheme. Two blocks with mass $m_1$ and $m_2$ connected by a string with lenght $\ell=x_1+x_2+\pi R$.}\label{atwoodFig}
\end{figure}
\end{center}

We can now use the Eq. (\ref{zen0}) to obtain the acceleration of the system. In this case, we will have:
  $$\frac{\Delta_{v}L}{\Delta v}=\frac{\frac{1}{2}(m_1+m_2)v^2 -\frac{1}{2}(m_1+m_2)v_0^2}{v-v_0}$$
\begin{equation}=\frac{\frac{1}{2}(m_1+m_2)(v^2-v_0^2)}{v-v_0}=\frac{1}{2}(m_1+m_2)(v+v_0)\label{zen1}
\end{equation}
and
\begin{equation}\frac{\Delta_{x}L}{\Delta x}=\frac{(-m_1gx-m_2g(\ell-x))+(m_1gx_0+m_2g(\ell-x_0))}{x-x_0}=g(m_2-m_1).\label{zen2}
  \end{equation}

By taking (\ref{zen1}) and (\ref{zen2}) in Eq. (\ref{zen0}), we get:
\begin{equation}
(m_1+m_2)\frac{\Delta v_{_M}}{\Delta t}-g(m_1-m_2)=0\Rightarrow a=\left(\frac{m_2-m_1}{m_2+m_1}\right)g.
  \end{equation}
Then, as expected, we have found the same acceleration predicted by the Newton's law \cite{sousa-PEd}. Since the acceleration is a constant, the motion equations are given by the Eq. (\ref{Eq-mot}) if we replace $g$ by $g({m_2-m_1})/({m_2+m_1})$.

\subsection{The inclinated plan}
The last example to be treated in this paper is the inclinated plan, which consists in a block with mass $m$ on a plan inclinated by an angle $\theta$, in a gravitational field. The gravitational acceleration is $g$ (See Fig. \ref{incplan}).

Supposing that the block is abandoned from the rest, we have that:
\begin{equation}
L=\frac{1}{2}mv^2-mg{x}{\sin\theta},
\end{equation}
where $x$ is the distance measured on the surface plan ($x=0$ in the bottom of the plan). Now, we have that:
\begin{equation}\frac{\Delta_{v}L}{\Delta v}=\frac{\frac{1}{2}mv^2}{v}=mv_{_M}\end{equation}
and,
\begin{equation}
\frac{\Delta_{x}L}{\Delta x}=\frac{-mgx\sin\theta}{x}=-mg\sin\theta.
\end{equation}
Finally, from Eq. (\ref{zen0}), we have that:
\begin{equation}
a=-g\sin\theta,
\end{equation}
which is the acceleration predicted by the Newton's law, as it should be. Since the angle is not variable, the acceleration of the block is constant, and it is slipping on the plan performing an uniformly variable rectilinear motion.
\begin{figure}
\includegraphics[scale=0.8]{incplane.eps}
\caption{Inclinated plan scheme. The body falls from a height $y$ on an inclinated plan by an angle $\theta$.}\label{incplan}
\end{figure}
It is important to note that we have gotten this result without the use forces and without the need of vectors decomposition.

\section{Conclusions and prospects}\label{Conc}
In this work, we have discussed the ELE for one-dimensional motions in a constant force. By introducing two operators here called \textit{mean delta operators}, we have simplified the ELE in such way that it may be solved by students in introductory physics courses. The simplification was obtained from the definition of partial derivative and the extrapolation of the limits for greater intervals of distances and velocities of rectilinear motions.

Three examples were explicitly solved: the free fall body, the Atwood's machine and the inclinated plan. In the three cases, the obtained acceleration agrees with that predicted by the Newton's law, as it should be.

The simplification proposed in this work has the limitation to be applied only to one-dimensional problems in a constant force. Furthermore the equality between the acceleration obtained by the Newton's law and the acceleration defined from the temporal variation of average of velocities occurs only for uniformly varied rectilinear motions. However, this work opens new possibilities in the discussions about the removal of the force concept from mechanical courses, once this one is in the end of its life cycle \cite{jammer-preface}. Among the perspectives that can be opened up with this work, there is the possibility to extend this formulation for more complex problems.\\

\begin{center}
\textbf{Acknowledgements}
\end{center}
The authors would thanks to CNPq (Grant number 562867/2010-4) by finantial support. We are also greatful to Ivan S Costa and Maria Aparecida (Cidinha) by encouraging the development of this work. V L Carvalho-Santos thanks the hospitality of the Universidade do Estado da Bahia - Campus VII, where part of this work was developed.\\

