Answer
$${a_N} = \sqrt 2 {e^t}{\text{ and }}{a_T} = \sqrt 3 {e^t}$$
Work Step by Step
$$\eqalign{
& {\bf{r}}\left( t \right) = \left\langle {{e^t}\cos t,{e^t}\sin t,{e^t}} \right\rangle \cr
& {\text{Calculate }}{\bf{v}}\left( t \right){\text{, }}\left| {{\bf{v}}\left( t \right)} \right|{\text{ and }}{\bf{a}}\left( t \right) \cr
& {\bf{v}}\left( t \right) = {\bf{r}}'\left( t \right) \cr
& {\bf{v}}\left( t \right) = \left\langle {{e^t}\cos t - {e^t}\sin t,{e^t}\sin t + {e^t}\cos t,{e^t}} \right\rangle \cr
& \left| {{\bf{v}}\left( t \right)} \right| = \sqrt {{e^{2t}}\left( {1 - 2\sin t\cos t} \right) + {e^{2t}}\left( {1 - 2\sin t\cos t} \right) + {e^t}} \cr
& \left| {{\bf{v}}\left( t \right)} \right| = \sqrt {{e^{2t}} + {e^{2t}} + {e^{2t}}} \cr
& \left| {{\bf{v}}\left( t \right)} \right| = \sqrt 3 {e^t} \cr
& \cr
& {\bf{a}}\left( t \right) = {\bf{v}}'\left( t \right) \cr
& {\bf{a}}\left( t \right) = \frac{d}{{dt}}\left\langle {{e^t}\cos t - {e^t}\sin t,{e^t}\sin t + {e^t}\cos t,{e^t}} \right\rangle \cr
& {\bf{a}}\left( t \right) = \left\langle { - 2{e^t}\sin t,2{e^t}\cos t,{e^t}} \right\rangle \cr
& \cr
& {\text{Find the components of acceleration: }}{a_N}{\bf{N}} + {a_T}{\bf{T}}, \cr
& {\text{Where }}{a_N} = \kappa {\left| {\bf{v}} \right|^2} = \frac{{\left| {{\bf{v}} \times {\bf{a}}} \right|}}{{\left| {\bf{v}} \right|}}{\text{ and }}{a_T} = \frac{{{d^2}s}}{{d{t^2}}} = \frac{{{\bf{v}} \cdot {\bf{a}}}}{{\left| {\bf{v}} \right|}} \cr
& {\text{Then}}{\text{,}} \cr} $$
\[{\mathbf{v}} \times {\mathbf{a}} = \left| {\begin{array}{*{20}{c}}
{\mathbf{i}}&{\mathbf{j}}&{\mathbf{k}} \\
{{e^t}\cos t - {e^t}\sin t}&{{e^t}\sin t + {e^t}\cos t}&{{e^t}} \\
{ - 2{e^t}\sin t}&{2{e^t}\cos t}&{{e^t}}
\end{array}} \right|\]
$$\eqalign{
& {\bf{v}} \times {\bf{a}} = \left( {{e^{2t}}\sin t - {e^{2t}}\cos t} \right){\bf{i}} - \left( {{e^{2t}}\sin t + {e^{2t}}\cos t} \right){\bf{j}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\, + \left( {2{e^{2t}}{{\cos }^2}t + 2{e^{2t}}{{\sin }^2}t} \right){\bf{k}} \cr
& {\bf{v}} \times {\bf{a}} = {e^{2t}}\left( {\sin t - \cos t} \right){\bf{i}} - {e^{2t}}\left( {\sin t + \cos t} \right){\bf{j}} + 2{e^{2t}}{\bf{k}} \cr
& \left| {{\bf{v}} \times {\bf{a}}} \right| = \sqrt {{e^{4t}}\left( {1 - 2\sin t\cos t} \right) + {e^{4t}}\left( {1 + 2\sin t\cos t} \right) + 4{e^{4t}}} \cr
& \left| {{\bf{v}} \times {\bf{a}}} \right| = \sqrt {6{e^{4t}}} = \sqrt 6 {e^{2t}} \cr
& {a_N} = \frac{{\left| {{\bf{v}} \times {\bf{a}}} \right|}}{{\left| {\bf{v}} \right|}} = \frac{{\sqrt 6 {e^{2t}}}}{{\sqrt 3 {e^t}}} \cr
& {a_N} = \sqrt 2 {e^t} \cr
& \cr
& and \cr
& \cr
& {\bf{v}} \cdot {\bf{a}} = \left\langle {{e^t}\cos t - {e^t}\sin t,{e^t}\sin t + {e^t}\cos t,{e^t}} \right\rangle \cdot \left\langle { - 2{e^t}\sin t,2{e^t}\cos t,{e^t}} \right\rangle \cr
& {\bf{v}} \cdot {\bf{a}} = 2{e^{2t}} + {e^{2t}} \cr
& {\bf{v}} \cdot {\bf{a}} = 3{e^{2t}} \cr
& {a_T} = \frac{{{\bf{v}} \cdot {\bf{a}}}}{{\left| {\bf{v}} \right|}} = \frac{{3{e^{2t}}}}{{\sqrt 3 {e^t}}} \cr
& {a_T} = \sqrt 3 {e^t} \cr
& \cr
& {a_N} = \sqrt 2 {e^t}{\text{ and }}{a_T} = \sqrt 3 {e^t} \cr} $$
