Answer
Please see the figure attached.
For $t=1$, we choose the (+) sign to get ${\bf{N}}{\left( 1 \right)_ + } = \left( { - \frac{3}{{\sqrt {10} }},\frac{1}{{\sqrt {10} }}} \right)$.
For $t=-1$, we choose the (-) sign to get ${\bf{N}}{\left( { - 1} \right)_ - } = \left( {\frac{3}{{\sqrt {10} }}, - \frac{1}{{\sqrt {10} }}} \right)$.
Work Step by Step
We have ${\bf{r}}\left( t \right) = \left( {t,{t^3}} \right)$ and ${\bf{r}}'\left( t \right) = \left( {1,3{t^2}} \right)$.
We find the unit tangent vector ${\bf{T}}\left( t \right)$:
${\bf{T}}\left( t \right) = \frac{{{\bf{r}}'\left( t \right)}}{{||{\bf{r}}'\left( t \right)||}} = \frac{{\left( {1,3{t^2}} \right)}}{{\sqrt {\left( {1,3{t^2}} \right)\cdot\left( {1,3{t^2}} \right)} }}$
${\bf{T}}\left( t \right) = \frac{{\left( {1,3{t^2}} \right)}}{{\sqrt {1 + 9{t^4}} }} = \left( {\frac{1}{{\sqrt {1 + 9{t^4}} }},\frac{{3{t^2}}}{{\sqrt {1 + 9{t^4}} }}} \right)$
Taking the derivative, we obtain
${\bf{T}}'\left( t \right) = \left( { - \frac{{18{t^3}}}{{{{\left( {1 + 9{t^4}} \right)}^{3/2}}}},\frac{{6t}}{{{{\left( {1 + 9{t^4}} \right)}^{3/2}}}}} \right)$
By Eq. (6), the normal vector ${\bf{N}}\left( t \right)$ is given by
${\bf{N}}\left( t \right) = \frac{{{\bf{T}}'\left( t \right)}}{{||{\bf{T}}'\left( t \right)||}}$
${\bf{N}}\left( t \right) = \frac{{\left( { - \frac{{18{t^3}}}{{{{\left( {1 + 9{t^4}} \right)}^{3/2}}}},\frac{{6t}}{{{{\left( {1 + 9{t^4}} \right)}^{3/2}}}}} \right)}}{{\sqrt {\left( { - \frac{{18{t^3}}}{{{{\left( {1 + 9{t^4}} \right)}^{3/2}}}},\frac{{6t}}{{{{\left( {1 + 9{t^4}} \right)}^{3/2}}}}} \right)\cdot\left( { - \frac{{18{t^3}}}{{{{\left( {1 + 9{t^4}} \right)}^{3/2}}}},\frac{{6t}}{{{{\left( {1 + 9{t^4}} \right)}^{3/2}}}}} \right)} }}$
${\bf{N}}\left( t \right) = \frac{{\left( { - \frac{{18{t^3}}}{{{{\left( {1 + 9{t^4}} \right)}^{3/2}}}},\frac{{6t}}{{{{\left( {1 + 9{t^4}} \right)}^{3/2}}}}} \right)}}{{\sqrt {\frac{{{{18}^2}{t^6}}}{{{{\left( {1 + 9{t^4}} \right)}^3}}} + \frac{{{6^2}{t^2}}}{{{{\left( {1 + 9{t^4}} \right)}^3}}}} }}$
${\bf{N}}\left( t \right) = \frac{{\left( { - \frac{{18{t^3}}}{{{{\left( {1 + 9{t^4}} \right)}^{3/2}}}},\frac{{6t}}{{{{\left( {1 + 9{t^4}} \right)}^{3/2}}}}} \right)}}{{\sqrt {\frac{{{6^2}{t^2}\left( {1 + 9{t^4}} \right)}}{{{{\left( {1 + 9{t^4}} \right)}^3}}}} }}$
${\bf{N}}\left( t \right) = \pm \left( {\frac{{1 + 9{t^4}}}{{6t}}} \right)\left( { - \frac{{18{t^3}}}{{{{\left( {1 + 9{t^4}} \right)}^{3/2}}}},\frac{{6t}}{{{{\left( {1 + 9{t^4}} \right)}^{3/2}}}}} \right)$
${\bf{N}}\left( t \right) = \pm \left( { - \frac{{3{t^2}}}{{\sqrt {1 + 9{t^4}} }},\frac{1}{{\sqrt {1 + 9{t^4}} }}} \right)$
Notice that the unit normal ${\bf{N}}\left( t \right)$ points in one of the two directions $ \pm \left( { - \frac{{3{t^2}}}{{\sqrt {1 + 9{t^4}} }},\frac{1}{{\sqrt {1 + 9{t^4}} }}} \right)$.
For $t=1$, we get ${\bf{N}}\left( 1 \right) = \pm \left( { - \frac{3}{{\sqrt {10} }},\frac{1}{{\sqrt {10} }}} \right)$. There are two normal vectors, that is,
${\bf{N}}{\left( 1 \right)_ + } = \left( { - \frac{3}{{\sqrt {10} }},\frac{1}{{\sqrt {10} }}} \right)$ ${\ \ }$ and ${\ \ }$ ${\bf{N}}{\left( 1 \right)_ - } = \left( {\frac{3}{{\sqrt {10} }}, - \frac{1}{{\sqrt {10} }}} \right)$
Since the normal vector always point to the inside of the curve, we choose the (+) sign to get ${\bf{N}}{\left( 1 \right)_ + } = \left( { - \frac{3}{{\sqrt {10} }},\frac{1}{{\sqrt {10} }}} \right)$ (see the figure).
For $t=-1$, we get ${\bf{N}}\left( { - 1} \right) = \pm \left( { - \frac{3}{{\sqrt {10} }},\frac{1}{{\sqrt {10} }}} \right)$. There are two normal vectors, that is,
${\bf{N}}{\left( { - 1} \right)_ + } = \left( { - \frac{3}{{\sqrt {10} }},\frac{1}{{\sqrt {10} }}} \right)$ ${\ \ }$ and ${\ \ }$ ${\bf{N}}{\left( { - 1} \right)_ - } = \left( {\frac{3}{{\sqrt {10} }}, - \frac{1}{{\sqrt {10} }}} \right)$
Since the normal vector always point to the inside of the curve, we choose the (-) sign to get ${\bf{N}}{\left( { - 1} \right)_ - } = \left( {\frac{3}{{\sqrt {10} }}, - \frac{1}{{\sqrt {10} }}} \right)$ (see the figure).
