Answer
$\frac{d}{{dt}}\left( {{{\bf{r}}_1}\left( t \right) \times {{\bf{r}}_2}\left( t \right)} \right) = \left( {3{t^2}{{\rm{e}}^t} - {{\rm{e}}^{2t}} + {t^3}{{\rm{e}}^t} - 2t{{\rm{e}}^{2t}}} \right){\bf{i}}$
${\ \ \ \ }$ $ - \left( {2t{{\rm{e}}^t} - {{\rm{e}}^{3t}} + {t^2}{{\rm{e}}^t} - 3t{{\rm{e}}^{3t}}} \right){\bf{j}}$
${\ \ \ \ }$ $ + \left( {2t{{\rm{e}}^{2t}} - 3{t^2}{{\rm{e}}^{3t}} + 2{t^2}{{\rm{e}}^{2t}} - 3{t^3}{{\rm{e}}^{3t}}} \right){\bf{k}}$
Work Step by Step
We have ${{\bf{r}}_1}\left( t \right) = \left( {{t^2},{t^3},t} \right)$ and ${{\bf{r}}_2}\left( t \right) = \left( {{{\rm{e}}^{3t}},{{\rm{e}}^{2t}},{{\rm{e}}^t}} \right)$.
By Eq. (5) of Theorem 3,
$\frac{d}{{dt}}\left( {{{\bf{r}}_1}\left( t \right) \times {{\bf{r}}_2}\left( t \right)} \right) = \left[ {{{\bf{r}}_1}'\left( t \right) \times {{\bf{r}}_2}\left( t \right)} \right] + \left[ {{{\bf{r}}_1}\left( t \right) \times {{\bf{r}}_2}'\left( t \right)} \right]$
So,
(1) ${\ \ \ }$ $\frac{d}{{dt}}\left( {{{\bf{r}}_1}\left( t \right) \times {{\bf{r}}_2}\left( t \right)} \right) = \left( {2t,3{t^2},1} \right) \times \left( {{{\rm{e}}^{3t}},{{\rm{e}}^{2t}},{{\rm{e}}^t}} \right)$
${\ \ \ \ }$ $ + \left( {{t^2},{t^3},t} \right) \times \left( {3{{\rm{e}}^{3t}},2{{\rm{e}}^{2t}},{{\rm{e}}^t}} \right)$
Evaluate the first cross product on the right-hand side of equation (1):
$\left( {2t,3{t^2},1} \right) \times \left( {{{\rm{e}}^{3t}},{{\rm{e}}^{2t}},{{\rm{e}}^t}} \right)$
$\left( {2t,3{t^2},1} \right) \times \left( {{{\rm{e}}^{3t}},{{\rm{e}}^{2t}},{{\rm{e}}^t}} \right) = \left| {\begin{array}{*{20}{c}}
{\bf{i}}&{\bf{j}}&{\bf{k}}\\
{2t}&{3{t^2}}&1\\
{{{\rm{e}}^{3t}}}&{{{\rm{e}}^{2t}}}&{{{\rm{e}}^t}}
\end{array}} \right|$
$ = \left| {\begin{array}{*{20}{c}}
{3{t^2}}&1\\
{{{\rm{e}}^{2t}}}&{{{\rm{e}}^t}}
\end{array}} \right|{\bf{i}} - \left| {\begin{array}{*{20}{c}}
{2t}&1\\
{{{\rm{e}}^{3t}}}&{{{\rm{e}}^t}}
\end{array}} \right|{\bf{j}} + \left| {\begin{array}{*{20}{c}}
{2t}&{3{t^2}}\\
{{{\rm{e}}^{3t}}}&{{{\rm{e}}^{2t}}}
\end{array}} \right|{\bf{k}}$
$ = \left( {3{t^2}{{\rm{e}}^t} - {{\rm{e}}^{2t}}} \right){\bf{i}} - \left( {2t{{\rm{e}}^t} - {{\rm{e}}^{3t}}} \right){\bf{j}} + \left( {2t{{\rm{e}}^{2t}} - 3{t^2}{{\rm{e}}^{3t}}} \right){\bf{k}}$
Evaluate the second cross product on the right-hand side of equation (1):
$\left( {{t^2},{t^3},t} \right) \times \left( {3{{\rm{e}}^{3t}},2{{\rm{e}}^{2t}},{{\rm{e}}^t}} \right)$
$\left( {{t^2},{t^3},t} \right) \times \left( {3{{\rm{e}}^{3t}},2{{\rm{e}}^{2t}},{{\rm{e}}^t}} \right) = \left| {\begin{array}{*{20}{c}}
{\bf{i}}&{\bf{j}}&{\bf{k}}\\
{{t^2}}&{{t^3}}&t\\
{3{{\rm{e}}^{3t}}}&{2{{\rm{e}}^{2t}}}&{{{\rm{e}}^t}}
\end{array}} \right|$
$ = \left| {\begin{array}{*{20}{c}}
{{t^3}}&t\\
{2{{\rm{e}}^{2t}}}&{{{\rm{e}}^t}}
\end{array}} \right|{\bf{i}} - \left| {\begin{array}{*{20}{c}}
{{t^2}}&t\\
{3{{\rm{e}}^{3t}}}&{{{\rm{e}}^t}}
\end{array}} \right|{\bf{j}} + \left| {\begin{array}{*{20}{c}}
{{t^2}}&{{t^3}}\\
{3{{\rm{e}}^{3t}}}&{2{{\rm{e}}^{2t}}}
\end{array}} \right|{\bf{k}}$
$ = \left( {{t^3}{{\rm{e}}^t} - 2t{{\rm{e}}^{2t}}} \right){\bf{i}} - \left( {{t^2}{{\rm{e}}^t} - 3t{{\rm{e}}^{3t}}} \right){\bf{j}} + \left( {2{t^2}{{\rm{e}}^{2t}} - 3{t^3}{{\rm{e}}^{3t}}} \right){\bf{k}}$
From the results above, it follows that
$\frac{d}{{dt}}\left( {{{\bf{r}}_1}\left( t \right) \times {{\bf{r}}_2}\left( t \right)} \right) = \left( {3{t^2}{{\rm{e}}^t} - {{\rm{e}}^{2t}} + {t^3}{{\rm{e}}^t} - 2t{{\rm{e}}^{2t}}} \right){\bf{i}}$
${\ \ \ \ }$ $ - \left( {2t{{\rm{e}}^t} - {{\rm{e}}^{3t}} + {t^2}{{\rm{e}}^t} - 3t{{\rm{e}}^{3t}}} \right){\bf{j}}$
${\ \ \ \ }$ $ + \left( {2t{{\rm{e}}^{2t}} - 3{t^2}{{\rm{e}}^{3t}} + 2{t^2}{{\rm{e}}^{2t}} - 3{t^3}{{\rm{e}}^{3t}}} \right){\bf{k}}$
