Answer
The normal vector ${\bf{N}}\left( t \right)$ at $t=-1$ is
${\bf{N}}\left( { - 1} \right) = \left( { - \frac{7}{{\sqrt {66} }},\frac{1}{{\sqrt {66} }},2\sqrt {\frac{2}{{33}}} } \right)$
Work Step by Step
We have ${\bf{r}}\left( t \right) = \left( {{t^{ - 1}},t,{t^2}} \right)$. The derivatives are ${\bf{r}}'\left( t \right) = \left( { - {t^{ - 2}},1,2t} \right)$ and ${\bf{r}}{\rm{''}}\left( t \right) = \left( { - 2{t^{ - 3}},0,2} \right)$.
Let ${\rm{v}}\left( t \right) = ||{\bf{r}}'\left( t \right)||$. So,
${\rm{v}}\left( t \right) = \sqrt {{t^{ - 4}} + 1 + 4{t^2}} $
$v'\left( t \right) = \frac{{ - 4{t^{ - 5}} + 8t}}{{2\sqrt {{t^{ - 4}} + 1 + 4{t^2}} }} = \frac{{ - 2{t^{ - 5}} + 4t}}{{\sqrt {{t^{ - 4}} + 1 + 4{t^2}} }}$
Evaluate $v\left( t \right){\bf{r}}{\rm{''}}\left( t \right) - v'\left( t \right){\bf{r}}'\left( t \right)$ at $t=-1$:
$v\left( { - 1} \right){\bf{r}}{\rm{''}}\left( { - 1} \right) - v'\left( { - 1} \right){\bf{r}}'\left( { - 1} \right) = \sqrt 6 \left( { - 2,0,2} \right) - \left( {\frac{{ - 2}}{{\sqrt 6 }}} \right)\left( { - 1,1, - 2} \right)$
$v\left( { - 1} \right){\bf{r}}{\rm{''}}\left( { - 1} \right) - v'\left( { - 1} \right){\bf{r}}'\left( { - 1} \right) = \left( { - 2\sqrt 6 - \sqrt {\frac{2}{3}} ,\sqrt {\frac{2}{3}} ,2\sqrt 6 - 2\sqrt {\frac{2}{3}} } \right)$
By Eq. (12), the normal vector ${\bf{N}}\left( t \right)$ at $t=-1$ is given by
${\bf{N}}\left( { - 1} \right) = \frac{{v\left( { - 1} \right){\bf{r}}{\rm{''}}\left( { - 1} \right) - v'\left( { - 1} \right){\bf{r}}'\left( { - 1} \right)}}{{||v\left( { - 1} \right){\bf{r}}{\rm{''}}\left( { - 1} \right) - v'\left( { - 1} \right){\bf{r}}'\left( { - 1} \right)||}}$
${\bf{N}}\left( { - 1} \right) = \frac{{\left( { - 2\sqrt 6 - \sqrt {\frac{2}{3}} ,\sqrt {\frac{2}{3}} ,2\sqrt 6 - 2\sqrt {\frac{2}{3}} } \right)}}{{||\left( { - 2\sqrt 6 - \sqrt {\frac{2}{3}} ,\sqrt {\frac{2}{3}} ,2\sqrt 6 - 2\sqrt {\frac{2}{3}} } \right)||}}$
${\bf{N}}\left( { - 1} \right) = \frac{{\left( { - 2\sqrt 6 - \sqrt {\frac{2}{3}} ,\sqrt {\frac{2}{3}} ,2\sqrt 6 - 2\sqrt {\frac{2}{3}} } \right)}}{{\sqrt {\left( { - 2\sqrt 6 - \sqrt {\frac{2}{3}} ,\sqrt {\frac{2}{3}} ,2\sqrt 6 - 2\sqrt {\frac{2}{3}} } \right)\cdot\left( { - 2\sqrt 6 - \sqrt {\frac{2}{3}} ,\sqrt {\frac{2}{3}} ,2\sqrt 6 - 2\sqrt {\frac{2}{3}} } \right)} }}$
${\bf{N}}\left( { - 1} \right) = \left( { - \frac{7}{{\sqrt {66} }},\frac{1}{{\sqrt {66} }},2\sqrt {\frac{2}{{33}}} } \right)$
