Answer
${a_{\bf{T}}} = \frac{7}{{\sqrt 6 }}$, ${\ \ \ }$ ${a_{\bf{N}}} = \sqrt {\frac{{53}}{6}} $
Work Step by Step
We have ${\bf{r}}\left( t \right) = \left( {{{\rm{e}}^{2t}},t,{{\rm{e}}^{ - t}}} \right)$. The velocity and acceleration vectors are ${\bf{v}}\left( t \right) = {\bf{r}}'\left( t \right) = \left( {2{{\rm{e}}^{2t}},1, - {{\rm{e}}^{ - t}}} \right)$ and ${\bf{a}}\left( t \right) = {\bf{r}}{\rm{''}}\left( t \right) = \left( {4{{\rm{e}}^{2t}},0,{{\rm{e}}^{ - t}}} \right)$, respectively.
At $t=0$ we get
${\bf{v}}\left( 0 \right) = \left( {2,1, - 1} \right)$, ${\ \ \ }$ ${\bf{a}}\left( 0 \right) = \left( {4,0,1} \right)$
By Eq. (2) of Theorem 1 we have at $t=0$:
${a_{\bf{T}}} = \frac{{{\bf{a}}\cdot{\bf{v}}}}{{||{\bf{v}}||}} = \frac{{\left( {4,0,1} \right)\cdot\left( {2,1, - 1} \right)}}{{\sqrt {\left( {2,1, - 1} \right)\cdot\left( {2,1, - 1} \right)} }}$
${a_{\bf{T}}} = \frac{7}{{\sqrt 6 }}$
By Eq. (2) of Theorem 1 we have
${a_{\bf{N}}} = \sqrt {||{\bf{a}}|{|^2} - |{a_{\bf{T}}}{|^2}} $
${a_{\bf{N}}} = \sqrt {\left( {4,0,1} \right)\cdot\left( {4,0,1} \right) - \frac{{49}}{6}} $
${a_{\bf{N}}} = \sqrt {17 - \frac{{49}}{6}} = \sqrt {\frac{{53}}{6}} $
