Answer
At $t=2$:
${\bf{a}}\left( 2 \right) = \left( {2, - 2,0} \right) = \frac{{12}}{{\sqrt {21} }}{\bf{T}}\left( 2 \right) + \sqrt {\frac{8}{7}} {\bf{N}}\left( 2 \right)$,
where ${\bf{T}}\left( 2 \right) = \frac{1}{{\sqrt {21} }}\left( {4, - 2,1} \right)$ and ${\bf{N}}\left( 2 \right) = \left( { - \frac{1}{{\sqrt {14} }}, - \frac{3}{{\sqrt {14} }}, - \frac{2}{{\sqrt {14} }}} \right)$.
Work Step by Step
We have ${\bf{r}}\left( t \right) = \left( {{t^2},2t - {t^2},t} \right)$. So, the tangential and the acceleration vectors are
${\bf{r}}'\left( t \right) = {\bf{v}}\left( t \right) = \left( {2t,2 - 2t,1} \right)$
and
${\bf{r}}{\rm{''}}\left( t \right) = {\bf{a}}\left( t \right) = \left( {2, - 2,0} \right)$,
respectively.
At $t=2$, we get
${\bf{v}}\left( 2 \right) = \left( {4, - 2,1} \right)$
${\bf{a}}\left( 2 \right) = \left( {2, - 2,0} \right)$
The unit tangent vector at $t=2$ is
${\bf{T}}\left( 2 \right) = \frac{{{\bf{r}}'\left( 2 \right)}}{{||{\bf{r}}'\left( 2 \right)||}} = \frac{{\left( {4, - 2,1} \right)}}{{\sqrt {\left( {4, - 2,1} \right)\cdot\left( {4, - 2,1} \right)} }}$
${\bf{T}}\left( 2 \right) = \frac{1}{{\sqrt {21} }}\left( {4, - 2,1} \right)$
By Eq. (2) of Theorem 1 (Section 14.5), the tangential component of acceleration is
${a_{\bf{T}}} = {\bf{a}}\left( 2 \right)\cdot{\bf{T}}\left( 2 \right) = \frac{1}{{\sqrt {21} }}\left( {2, - 2,0} \right)\cdot\left( {4, - 2,1} \right)$
${a_{\bf{T}}} = \frac{{12}}{{\sqrt {21} }}$
By Eq. (3) of Theorem 1 (Section 14.5),
${a_{\bf{N}}}{\bf{N}} = {\bf{a}} - {a_{\bf{T}}}{\bf{T}}$
$ = \left( {2, - 2,0} \right) - \frac{{12}}{{\sqrt {21} }}\frac{1}{{\sqrt {21} }}\left( {4, - 2,1} \right)$
$ = \left( {2, - 2,0} \right) - \frac{{12}}{{21}}\left( {4, - 2,1} \right)$
$ = \left( { - \frac{2}{7}, - \frac{6}{7}, - \frac{4}{7}} \right)$
Since ${\bf{N}}$ is unit vector, so
$||{a_{\bf{N}}}{\bf{N}}|| = {a_{\bf{N}}} = \sqrt {\left( { - \frac{2}{7}, - \frac{6}{7}, - \frac{4}{7}} \right)\cdot\left( { - \frac{2}{7}, - \frac{6}{7}, - \frac{4}{7}} \right)} $
${a_{\bf{N}}} = \sqrt {\frac{8}{7}} $
So,
${\bf{N}} = \frac{{{a_{\bf{N}}}{\bf{N}}}}{{{a_{\bf{N}}}}} = \sqrt {\frac{7}{8}} \left( { - \frac{2}{7}, - \frac{6}{7}, - \frac{4}{7}} \right)$
${\bf{N}} = \left( { - \frac{1}{{\sqrt {14} }}, - \frac{3}{{\sqrt {14} }}, - \frac{2}{{\sqrt {14} }}} \right)$
Finally, we obtain the decomposition of ${\bf{a}}$:
${\bf{a}}\left( t \right) = {a_{\bf{T}}}{\bf{T}} + {a_{\bf{N}}}{\bf{N}}$
at $t=2$:
${\bf{a}}\left( 2 \right) = \left( {2, - 2,0} \right) = \frac{{12}}{{\sqrt {21} }}{\bf{T}}\left( 2 \right) + \sqrt {\frac{8}{7}} {\bf{N}}\left( 2 \right)$,
where ${\bf{T}}\left( 2 \right) = \frac{1}{{\sqrt {21} }}\left( {4, - 2,1} \right)$ and ${\bf{N}}\left( 2 \right) = \left( { - \frac{1}{{\sqrt {14} }}, - \frac{3}{{\sqrt {14} }}, - \frac{2}{{\sqrt {14} }}} \right)$.
