Answer
${\bf{r}}{\rm{''}}\left( { - 3} \right) = \left( {\frac{2}{{27}}, - \frac{1}{4}, - 18} \right)$
Work Step by Step
We have ${\bf{r}}\left( t \right) = \left( {{t^{ - 2}},{{\left( {t + 1} \right)}^{ - 1}},{t^3} - t} \right)$. So,
${\bf{r}}'\left( t \right) = \left( { - 2{t^{ - 3}}, - {{\left( {t + 1} \right)}^{ - 2}},3{t^2} - 1} \right)$
${\bf{r}}{\rm{''}}\left( t \right) = \left( {6{t^{ - 4}},2{{\left( {t + 1} \right)}^{ - 3}},6t} \right)$
${\bf{r}}{\rm{''}}\left( { - 3} \right) = \left( {\frac{2}{{27}}, - \frac{1}{4}, - 18} \right)$
