Answer
The velocity vector at time $t$ is
${\bf{r}}'\left( t \right) = \left( {\frac{{20}}{{\sqrt {17} }}, - \frac{{\sqrt {17} }}{{20}}{t^{ - 2}}} \right)$
The velocity vector at location $\left( {2,\frac{1}{2}} \right)$, that is at $t = \frac{2}{\alpha } = \frac{{\sqrt {17} }}{{10}}$ is
${\bf{r}}'\left( {\frac{{\sqrt {17} }}{{10}}} \right) = \left( {\frac{{20}}{{\sqrt {17} }}, - \frac{5}{{\sqrt {17} }}} \right)$
Work Step by Step
We have $y = {x^{ - 1}}$ to the right of the $y$-axis. Since parametrization is not unique, let $x = \alpha t$, where $\alpha$ is a scalar. So, we may parametrize the curve by ${\bf{r}}\left( t \right) = \left( {\alpha t,\frac{1}{\alpha }{t^{ - 1}}} \right)$. The derivative is ${\bf{r}}'\left( t \right) = \left( {\alpha , - \frac{1}{\alpha }{t^{ - 2}}} \right)$.
When the particle is at the point $\left( {2,\frac{1}{2}} \right)$, it has a speed of $5$ cm/s. In our parametrization, the point $\left( {2,\frac{1}{2}} \right)$ corresponds to $t = \frac{2}{\alpha }$. So, at $t = \frac{2}{\alpha }$, $||{\bf{r}}'\left( {\frac{2}{\alpha }} \right)|| = 5$ cm/s.
Since ${\bf{r}}'\left( t \right) = \left( {\alpha , - \frac{1}{\alpha }{t^{ - 2}}} \right)$, so
$||{\bf{r}}'\left( t \right)|{|^2} = \left( {\alpha , - \frac{1}{\alpha }{t^{ - 2}}} \right)\cdot\left( {\alpha , - \frac{1}{\alpha }{t^{ - 2}}} \right)$
$||{\bf{r}}'\left( t \right)|{|^2} = {\alpha ^2} + \frac{1}{{{\alpha ^2}}}{t^{ - 4}}$
$||{\bf{r}}'\left( t \right)|| = \sqrt {{\alpha ^2} + \frac{1}{{{\alpha ^2}}}{t^{ - 4}}} $
Since $||{\bf{r}}'\left( {\frac{2}{\alpha }} \right)|| = 5$, so
$||{\bf{r}}'\left( {\frac{2}{\alpha }} \right)|| = \sqrt {{\alpha ^2} + \frac{1}{{{\alpha ^2}}}\frac{{{\alpha ^4}}}{{16}}} = 5$
$\frac{{17}}{{16}}{\alpha ^2} = 25$
$\alpha = \sqrt {\frac{{400}}{{17}}} = \frac{{20}}{{\sqrt {17} }}$
Thus, the parametrization is
${\bf{r}}\left( t \right) = \left( {\alpha t,\frac{1}{\alpha }{t^{ - 1}}} \right) = \left( {\frac{{20}}{{\sqrt {17} }}t,\frac{{\sqrt {17} }}{{20}}{t^{ - 1}}} \right)$
The velocity vector at time $t$ is
${\bf{r}}'\left( t \right) = \left( {\frac{{20}}{{\sqrt {17} }}, - \frac{{\sqrt {17} }}{{20}}{t^{ - 2}}} \right)$
The velocity vector at location $\left( {2,\frac{1}{2}} \right)$, that is at $t = \frac{2}{\alpha } = \frac{{\sqrt {17} }}{{10}}$ is
${\bf{r}}'\left( {\frac{{\sqrt {17} }}{{10}}} \right) = \left( {\frac{{20}}{{\sqrt {17} }}, - \frac{5}{{\sqrt {17} }}} \right)$
