Answer
The unit tangent vector $t=1$ is
${\bf{T}}\left( 1 \right) = \left( {\frac{4}{{\sqrt {21} }},\frac{1}{{\sqrt {21} }},\frac{2}{{\sqrt {21} }}} \right)$
Work Step by Step
We have ${\bf{r}}\left( t \right) = \left( {{t^2},{{\tan }^{ - 1}}t,t} \right)$.
Since the derivative $\frac{d}{{dt}}{\tan ^{ - 1}}t = \frac{1}{{1 + {t^2}}}$, the tangent vector is
${\bf{r}}'\left( t \right) = \left( {2t,\frac{1}{{1 + {t^2}}},1} \right)$
At $t=1$, we get ${\bf{r}}'\left( 1 \right) = \left( {2,\frac{1}{2},1} \right)$.
The unit tangent vector $t=1$ is
${\bf{T}}\left( 1 \right) = \frac{{{\bf{r}}'\left( 1 \right)}}{{||{\bf{r}}'\left( t \right)||}} = \frac{{\left( {2,1/2,1} \right)}}{{\sqrt {\left( {2,1/2,1} \right)\cdot\left( {2,1/2,1} \right)} }}$
${\bf{T}}\left( 1 \right) = \frac{{\left( {2,1/2,1} \right)}}{{\sqrt {4 + 1/4 + 1} }} = \frac{1}{{\sqrt {21/4} }}\left( {2,\frac{1}{2},1} \right) = \left( {\frac{4}{{\sqrt {21} }},\frac{1}{{\sqrt {21} }},\frac{2}{{\sqrt {21} }}} \right)$
