Answer
$\frac{d}{{dt}}\left( {{\bf{r}} \times {\bf{r}}'} \right) = \left( {{t^2}{{\rm{e}}^t} - 2{{\rm{e}}^t}} \right){\bf{i}} - t{{\rm{e}}^t}{\bf{j}} + 2t{\bf{k}}$
Work Step by Step
We have ${\bf{r}}\left( t \right) = \left( {t,{t^2},{{\rm{e}}^t}} \right)$. So,
${\bf{r}}'\left( t \right) = \left( {1,2t,{{\rm{e}}^t}} \right)$
${\bf{r}}{\rm{''}}\left( t \right) = \left( {0,2,{{\rm{e}}^t}} \right)$
From the result in Example 4, we obtain the formula $\frac{d}{{dt}}\left( {{\bf{r}} \times {\bf{r}}'} \right) = {\bf{r}} \times {\bf{r}}{\rm{''}}$. Thus,
$\frac{d}{{dt}}\left( {{\bf{r}} \times {\bf{r}}'} \right) = \left| {\begin{array}{*{20}{c}}
{\bf{i}}&{\bf{j}}&{\bf{k}}\\
t&{{t^2}}&{{{\rm{e}}^t}}\\
0&2&{{{\rm{e}}^t}}
\end{array}} \right|$
$\frac{d}{{dt}}\left( {{\bf{r}} \times {\bf{r}}'} \right) = \left| {\begin{array}{*{20}{c}}
{{t^2}}&{{{\rm{e}}^t}}\\
2&{{{\rm{e}}^t}}
\end{array}} \right|{\bf{i}} - \left| {\begin{array}{*{20}{c}}
t&{{{\rm{e}}^t}}\\
0&{{{\rm{e}}^t}}
\end{array}} \right|{\bf{j}} + \left| {\begin{array}{*{20}{c}}
t&{{t^2}}\\
0&2
\end{array}} \right|{\bf{k}}$
$\frac{d}{{dt}}\left( {{\bf{r}} \times {\bf{r}}'} \right) = \left( {{t^2}{{\rm{e}}^t} - 2{{\rm{e}}^t}} \right){\bf{i}} - t{{\rm{e}}^t}{\bf{j}} + 2t{\bf{k}}$
