Answer
${a_{\bf{T}}} = 0$, ${\ \ \ }$ ${a_{\bf{N}}} = 1$
Work Step by Step
We have ${\bf{r}}\left( t \right) = \left( {t,\cos t,\sin t} \right)$. The velocity and acceleration vectors are ${\bf{v}}\left( t \right) = {\bf{r}}'\left( t \right) = \left( {1, - \sin t,\cos t} \right)$ and ${\bf{a}}\left( t \right) = {\bf{r}}{\rm{''}}\left( t \right) = \left( {0, - \cos t, - \sin t} \right)$, respectively.
By Eq. (2) of Theorem 1 we have
${a_{\bf{T}}} = \frac{{{\bf{a}}\cdot{\bf{v}}}}{{||{\bf{v}}||}} = \frac{{\left( {0, - \cos t, - \sin t} \right)\cdot\left( {1, - \sin t,\cos t} \right)}}{{\sqrt {\left( {1, - \sin t,\cos t} \right)\cdot\left( {1, - \sin t,\cos t} \right)} }}$
${a_{\bf{T}}} = 0$
By Eq. (2) of Theorem 1 we have
${a_{\bf{N}}} = \sqrt {||{\bf{a}}|{|^2} - |{a_{\bf{T}}}{|^2}} $
${a_{\bf{N}}} = \sqrt {\left( {0, - \cos t, - \sin t} \right)\cdot\left( {0, - \cos t, - \sin t} \right)} $
${a_{\bf{N}}} = \sqrt 1 = 1$
