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

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

At $\displaystyle {t}={3}$,

$\displaystyle {x}{\left({3}\right)}=$
$\displaystyle {y}{\left({3}\right)}=$

$\displaystyle \frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}{\mid}_{{{t}={3}}}=$

$\displaystyle \frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}{\mid}_{{{t}={3}}}=$

$\displaystyle \frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}{\mid}_{{{t}={3}}}$ = tangent slope = Preview Question 6 Part 5 of 6  

$\displaystyle \text{speed(3) =}$ Preview Question 6 Part 6 of 6