Let $\displaystyle {x}{\left({t}\right)}={t}^{{3}}-{4}{t}^{{2}}+{1}$

and $\displaystyle {y}{\left({t}\right)}={t}^{{3}}-{5}{t}^{{2}}+{2}{t}+{8}$

At $\displaystyle {t}={2}$,

$\displaystyle {x}{\left({2}\right)}=$
$\displaystyle {y}{\left({2}\right)}=$

$\displaystyle \frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}{\mid}_{{{t}={2}}}=$

$\displaystyle \frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}{\mid}_{{{t}={2}}}=$

$\displaystyle \frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}{\mid}_{{{t}={2}}}$ = tangent slope =

$\displaystyle \text{speed(2) =}$