Answer
The decomposition of ${\bf{a}}$ at $t=1$:
${\bf{a}} = {a_{\bf{T}}}{\bf{T}} + {a_{\bf{N}}}{\bf{N}}$
$\left( {0,1,1} \right) = {\bf{T}} + {\bf{N}}$,
where ${\bf{T}} = \left( {\frac{2}{3},\frac{2}{3},\frac{1}{3}} \right)$ and ${\bf{N}} = \left( { - \frac{2}{3},\frac{1}{3},\frac{2}{3}} \right)$.
Work Step by Step
We have ${\bf{r}}\left( t \right) = \left( {t,\frac{1}{2}{t^2},\frac{1}{6}{t^3}} \right)$. The velocity and acceleration vectors are ${\bf{v}}\left( t \right) = {\bf{r}}'\left( t \right) = \left( {1,t,\frac{1}{2}{t^2}} \right)$ and ${\bf{a}}\left( t \right) = {\bf{r}}{\rm{''}}\left( t \right) = \left( {0,1,t} \right)$, respectively.
At $t=1$, we get ${\bf{v}}\left( 1 \right) = \left( {1,1,\frac{1}{2}} \right)$ and ${\bf{a}}\left( 1 \right) = \left( {0,1,1} \right)$. Thus, the unit tangent vector is
${\bf{T}} = \frac{{\bf{v}}}{{||{\bf{v}}||}} = \frac{{\left( {1,1,\frac{1}{2}} \right)}}{{\sqrt {\left( {1,1,\frac{1}{2}} \right)\cdot\left( {1,1,\frac{1}{2}} \right)} }} = \left( {\frac{2}{3},\frac{2}{3},\frac{1}{3}} \right)$
By Eq. (2) of Theorem 1 we have
${a_{\bf{T}}} = {\bf{a}}\cdot{\bf{T}} = \left( {0,1,1} \right)\cdot\left( {\frac{2}{3},\frac{2}{3},\frac{1}{3}} \right)$
${a_{\bf{T}}} = 1$
Next, we use Eq. (3) of Theorem 1 to find
${a_{\bf{N}}}{\bf{N}} = {\bf{a}} - {a_{\bf{T}}}{\bf{T}}$
${a_{\bf{N}}}{\bf{N}} = \left( {0,1,1} \right) - \left( {\frac{2}{3},\frac{2}{3},\frac{1}{3}} \right)$
${a_{\bf{N}}}{\bf{N}} = \left( { - \frac{2}{3},\frac{1}{3},\frac{2}{3}} \right)$
Since ${\bf{N}}$ is an unit vector, so
${a_{\bf{N}}} = ||{a_{\bf{N}}}{\bf{N}}|| = \sqrt {{{\left( { - \frac{2}{3}} \right)}^2} + {{\left( {\frac{1}{3}} \right)}^2} + {{\left( {\frac{2}{3}} \right)}^2}} $
${a_{\bf{N}}} = 1$
To find ${\bf{N}}$ we use the equation
${\bf{N}} = \frac{{{a_{\bf{N}}}{\bf{N}}}}{{{a_{\bf{N}}}}} = \left( { - \frac{2}{3},\frac{1}{3},\frac{2}{3}} \right)$
Thus, we obtain the decomposition of ${\bf{a}}$ at $t=1$:
${\bf{a}} = {a_{\bf{T}}}{\bf{T}} + {a_{\bf{N}}}{\bf{N}}$
$\left( {0,1,1} \right) = {\bf{T}} + {\bf{N}}$,
where ${\bf{T}} = \left( {\frac{2}{3},\frac{2}{3},\frac{1}{3}} \right)$ and ${\bf{N}} = \left( { - \frac{2}{3},\frac{1}{3},\frac{2}{3}} \right)$.
