I have these question but first of all a bit of context.
I have a Lagrangian $L(q,\dot q,t)\in C^\infty(\textbf{T}M \times \mathbb{R})$. We know that the taulogical 1-form in $\textbf{T}M$ is $\theta=\frac{\partial L}{\partial \dot q^{\alpha}}dq^{\alpha}$, so if we apply the Lie derivative then $\mathcal{L}_{\Delta L}\theta=dL$, where $\Delta_{L}$ is the vector field of $L$. So now to add the time dependency the vector field changes to $\Delta_{L}=\dot q^{\alpha}\frac{\partial }{\partial q^{\alpha}}+\ddot q^{\alpha}\frac{\partial}{\partial \dot q^{\alpha}}+\frac{\partial}{\partial t}$. To have the same derivative Lie I suppose that $\theta=\frac{\partial L}{\partial \dot q^{\alpha}}dq^{\alpha}+Ldt$. Is this assumption correct?
On the other hand, I want to work in $\textbf{T}^{*}M\times \mathbb{R}$. We know that the fundamental 1-form on $\textbf{T}^{*}M$ is $\theta=p_{\alpha}dq^{\alpha}$, so if i want to work in $\textbf{T}^{*}M\times \mathbb{R}$, I use the Legendre transform, so now I have $\theta=p_{\alpha}dq^{\alpha}+(p_{\alpha}\dot q^{\alpha}-H)dt$. Is this correct?
Finally, I want to apply the exterior derivative to get the fundamental 2-form on $\textbf{T}^{*}M \times \mathbb{R}$ so I have the following: $$\omega=-d\theta=dq^{\alpha}\wedge dp_{\alpha}+dH \wedge dt-d(p_{\alpha}\dot q^{\alpha})\wedge dt$$ In literature, I found that the fundamental 2-form in $\textbf{T}^{*}M\times \mathbb{R}$ is $\omega=-d\theta=dq^{\alpha}\wedge dp_{\alpha}+dH \wedge dt$. But I don't know how to prove that $d(p_{\alpha}\dot q^{\alpha})\wedge dt=0$, can you give some advice?
Thanks a lot!!!!