[example: Particle In Euclidean Space]
For a particle moving in $\mathbb R^n$, the configuration space is $Q=\mathbb R^n$, and each tangent space is canonically $T_q\mathbb R^n\cong \mathbb R^n$. Hence
\begin{align*}
TQ\cong \mathbb R^n\times \mathbb R^n,
\end{align*}
where a point of $TQ$ is written $(q,v)$ with $q=(q_1,\dots,q_n)$ and $v=(v_1,\dots,v_n)$. For a path $q:I\to\mathbb R^n$, the velocity is
\begin{align*}
\dot q(t)=(\dot q_1(t),\dots,\dot q_n(t)).
\end{align*}
The Lagrangian
\begin{align*}
L(q,v)=\frac{1}{2}m|v|^2-V(q)
\end{align*}
means explicitly
\begin{align*}
L(q,v)=\frac{1}{2}m\sum_{j=1}^n v_j^2-V(q_1,\dots,q_n).
\end{align*}
Applying the *Euler-Lagrange equations from stationary action*, we compute the two derivatives in the $i$th coordinate. Since $q_i$ occurs only in the potential term,
\begin{align*}
\frac{\partial L}{\partial q_i}(q,v)=-\frac{\partial V}{\partial q_i}(q).
\end{align*}
Since $v_i$ occurs in the kinetic term as $\frac{1}{2}m v_i^2$,
\begin{align*}
\frac{\partial L}{\partial v_i}(q,v)=m v_i.
\end{align*}
Along the path, $v_i=\dot q_i(t)$, so
\begin{align*}
\frac{d}{dt}\left(\frac{\partial L}{\partial \dot q_i}(q(t),\dot q(t))\right)=\frac{d}{dt}(m\dot q_i(t))=m\ddot q_i(t).
\end{align*}
The Euler-Lagrange equation therefore becomes
\begin{align*}
m\ddot q_i(t)-\left(-\frac{\partial V}{\partial q_i}(q(t))\right)=0.
\end{align*}
Equivalently,
\begin{align*}
m\ddot q_i(t)=-\frac{\partial V}{\partial q_i}(q(t)).
\end{align*}
Putting the $n$ scalar equations together gives
\begin{align*}
m\ddot q(t)=-\nabla V(q(t)).
\end{align*}
Thus the variational equation for kinetic minus potential energy recovers Newton's force law with conservative force $F(q)=-\nabla V(q)$.
[/example]