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\begin{document}

\noindent Solution for assignment 11a: \bigskip

\noindent For a bilinear kinetic Energy we find by differentiation and 
carrying the sum over $i$ out
\begin{eqnarray} \nonumber
   \sum_i \dot{q}_i\,\frac{\partial T}{\partial \dot{q}_i} &=&
   \sum_i \dot{q}_i\,\frac{\partial\ }{\partial \dot{q}_i}  
   \sum_{j,k} a_{jk}\,\dot{q}_j\dot{q}_k\ =\
   \sum_i \dot{q}_i\,\sum_{j,k}a_{jk}\,\left(\delta_{ij}\,\dot{q}_k 
   + \dot{q}_j\,\delta_{ik}\right) \\ \nonumber &=&
   \sum_{j,k}a_{jk}\,\left(\dot{q}_j\,\dot{q}_k + 
   \dot{q}_j\,\dot{q}_k\right)\ =\ 2\,T\ .
\end{eqnarray}

\noindent Solution for assignment 11b: Motivation of generalized 
momentum definition. \bigskip

$$ \frac{\partial\ }{\partial\dot{x}_i}\left(\frac{1}{2}\,m\,
   \vec{v}^2 - V(\vec{x})\right)\ =\ m\,\dot{x}_i\ =\ p_i\ . $$
\vfil 

\end{document}
