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{\bf Question}

Use the results  that $\dot {\bf e}_r  = \dot \phi {\bf e}_\phi $
and $\dot {\bf e}_\phi = -\dot \phi {\bf e}_r$ to show that in
polar coordinates the time derivative of the position vector of a
particle, ${\bf r} = r {\bf e}_r, $ is $\dot  {\bf r} = \dot r
{\bf e}_r + r \dot \phi {\bf e}_\phi$



\vspace{.25in}

{\bf Answer}

$\ds{\bf v} = \frac{d}{dt}(r {\bf e}_r) = \dot r {\bf e}_r  + r
\dot {\bf e}_r = \dot r {\bf e}_r + r \dot \phi {\bf e}_\phi$



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