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{\bf Question}

Express the basis vectors ${\bf e}_r$ and ${\bf e}_\phi$ of polar
coordinates the time derivative of the basis vectors of Cartesian
coordinates ${\bf i}$ and ${\bf j}$.  Suppose that ${\bf e}_r$ and
${\bf e}_\phi$ depend on time.  Show that
\begin{description}
\item[(a)] $\dot {\bf e}_r  = \dot \phi {\bf e}_\phi $
\item[(b)] $\dot {\bf e}_\phi = -\dot \phi {\bf e}_r$
\end{description}

\vspace{.25in}

{\bf Answer}


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\begin{picture}(3.5,3.5)

\put(0,0){\vector(1,0){3}}

\put(0,0){\vector(0,1){3}}

\put(3.2,0){\makebox(0,0){{\bf i}}}

\put(0,3.5){\makebox(0,0){{\bf j}}}

\put(1.4,1.4){$\bullet$}

\put(0,0){\line(1,1){0.3}}

\put(0.6,0.6){\line(1,1){0.3}}

\put(1.2,1.2){\line(1,1){0.3}}

\put(1.5,1.5){\vector(1,1){1}}

\put(1.5,1.5){\line(0,1){0.3}}

\put(1.5,2.1){\line(0,1){0.3}}

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\put(2.7,1.5){\line(1,0){0.3}}

\put(1.5,1.5){\vector(-1,1){1}}

\put(1.9,1.7){\makebox(0,0){$\phi$}}

\put(1.4,1.9){\makebox(0,0){$\phi$}}

\put(2.7,2.7){\makebox(0,0){${\bf e}_r$}}

\put(0.3,2.7){\makebox(0,0){${\bf e}_\phi$}}

\end{picture}
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\begin{eqnarray*} {\bf e}_r & = & {\bf i} \cos \phi + {\bf j} \sin \phi \\
{\bf e}_\phi & = & -{\bf i} \cos\left( \frac{\pi}{2} - \phi\right)
+ {\bf j} \cos \phi \\ & = & -{\bf i} \sin \phi + {\bf j} \cos
\phi
\end{eqnarray*}

$\ds \dot {\bf e}_r = \frac{d}{dt}( {\bf i} \cos \phi + {\bf j}
\sin \phi) = \dot \phi ( -{\bf i} \sin \phi + {\bf j} \cos \phi) =
\dot \phi {\bf e}_\phi$

$\ds \dot {\bf e}_\phi = \frac{d}{dt} (-{\bf i} \sin \phi + {\bf
j} \cos \phi) = \dot \phi (-{\bf i} \cos \phi - {\bf j} \sin \phi)
= - \dot \phi {\bf e} _r$



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