Chapter 14 - Calculus of Vector-Valued Functions - 14.2 Calculus of Vector-Valued Functions - Exercises - Page 720: 22

(a) $\frac{d}{{dt}}\left( {{{\bf{r}}_1}\left( t \right) \times {{\bf{r}}_2}\left( t \right)} \right){|_{t = 1}} = \left( {{\rm{e}} - 4} \right){\bf{i}} - \left( {3{\rm{e}} - 2} \right){\bf{j}} + 4{\bf{k}}$ (b) $\frac{d}{{dt}}\left( {{{\bf{r}}_1}\left( t \right) \times {{\bf{r}}_2}\left( t \right)} \right){|_{t = 1}} = \left( {{\rm{e}} - 4} \right){\bf{i}} - \left( {3{\rm{e}} - 2} \right){\bf{j}} + 4{\bf{k}}$ Part (a) and part (b) give the same answer.

(a) We have ${{\bf{r}}_1}\left( t \right) = \left( {{t^2},1,2t} \right)$ and ${{\bf{r}}_2}\left( t \right) = \left( {1,2,{{\rm{e}}^t}} \right)$. ${{\bf{r}}_1}\left( t \right) \times {{\bf{r}}_2}\left( t \right) = \left| {\begin{array}{*{20}{c}} {\bf{i}}&{\bf{j}}&{\bf{k}}\\ {{t^2}}&1&{2t}\\ 1&2&{{{\rm{e}}^t}} \end{array}} \right|$ ${{\bf{r}}_1}\left( t \right) \times {{\bf{r}}_2}\left( t \right) = \left| {\begin{array}{*{20}{c}} 1&{2t}\\ 2&{{{\rm{e}}^t}} \end{array}} \right|{\bf{i}} - \left| {\begin{array}{*{20}{c}} {{t^2}}&{2t}\\ 1&{{{\rm{e}}^t}} \end{array}} \right|{\bf{j}} + \left| {\begin{array}{*{20}{c}} {{t^2}}&1\\ 1&2 \end{array}} \right|{\bf{k}}$ ${{\bf{r}}_1}\left( t \right) \times {{\bf{r}}_2}\left( t \right) = \left( {{{\rm{e}}^t} - 4t} \right){\bf{i}} - \left( {{t^2}{{\rm{e}}^t} - 2t} \right){\bf{j}} + \left( {2{t^2} - 1} \right){\bf{k}}$ $\frac{d}{{dt}}\left( {{{\bf{r}}_1}\left( t \right) \times {{\bf{r}}_2}\left( t \right)} \right) = \left( {{{\rm{e}}^t} - 4} \right){\bf{i}} - \left( {2t{{\rm{e}}^t} + {t^2}{{\rm{e}}^t} - 2} \right){\bf{j}} + 4t{\bf{k}}$ $\frac{d}{{dt}}\left( {{{\bf{r}}_1}\left( t \right) \times {{\bf{r}}_2}\left( t \right)} \right){|_{t = 1}} = \left( {{\rm{e}} - 4} \right){\bf{i}} - \left( {3{\rm{e}} - 2} \right){\bf{j}} + 4{\bf{k}}$ (b) 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) ${\ \ \ }$ $\frac{d}{{dt}}\left( {{{\bf{r}}_1}\left( t \right) \times {{\bf{r}}_2}\left( t \right)} \right) = \left( {2t,0,2} \right) \times \left( {1,2,{{\rm{e}}^t}} \right)$ ${\ \ \ \ }$ $ + \left( {{t^2},1,2t} \right) \times \left( {0,0,{{\rm{e}}^t}} \right)$ Evaluate the first cross product on the right-hand side of equation (1): $\left( {2t,0,2} \right) \times \left( {1,2,{{\rm{e}}^t}} \right)$ $\left( {2t,0,2} \right) \times \left( {1,2,{{\rm{e}}^t}} \right) = \left| {\begin{array}{*{20}{c}} {\bf{i}}&{\bf{j}}&{\bf{k}}\\ {2t}&0&2\\ 1&2&{{{\rm{e}}^t}} \end{array}} \right|$ $ = \left| {\begin{array}{*{20}{c}} 0&2\\ 2&{{{\rm{e}}^t}} \end{array}} \right|{\bf{i}} - \left| {\begin{array}{*{20}{c}} {2t}&2\\ 1&{{{\rm{e}}^t}} \end{array}} \right|{\bf{j}} + \left| {\begin{array}{*{20}{c}} {2t}&0\\ 1&2 \end{array}} \right|{\bf{k}}$ $ = - 4{\bf{i}} - \left( {2t{{\rm{e}}^t} - 2} \right){\bf{j}} + 4t{\bf{k}}$ Evaluate the second cross product on the right-hand side of equation (1): $\left( {{t^2},1,2t} \right) \times \left( {0,0,{{\rm{e}}^t}} \right)$ $\left( {{t^2},1,2t} \right) \times \left( {0,0,{{\rm{e}}^t}} \right) = \left| {\begin{array}{*{20}{c}} {\bf{i}}&{\bf{j}}&{\bf{k}}\\ {{t^2}}&1&{2t}\\ 0&0&{{{\rm{e}}^t}} \end{array}} \right|$ $ = \left| {\begin{array}{*{20}{c}} 1&{2t}\\ 0&{{{\rm{e}}^t}} \end{array}} \right|{\bf{i}} - \left| {\begin{array}{*{20}{c}} {{t^2}}&{2t}\\ 0&{{{\rm{e}}^t}} \end{array}} \right|{\bf{j}} + \left| {\begin{array}{*{20}{c}} {{t^2}}&1\\ 0&0 \end{array}} \right|{\bf{k}}$ $ = {{\rm{e}}^t}{\bf{i}} - {t^2}{{\rm{e}}^t}{\bf{j}}$ From the results above, it follows that $\frac{d}{{dt}}\left( {{{\bf{r}}_1}\left( t \right) \times {{\bf{r}}_2}\left( t \right)} \right) = \left( { - 4 + {{\rm{e}}^t}} \right){\bf{i}} - \left( {2t{{\rm{e}}^t} - 2 + {t^2}{{\rm{e}}^t}} \right){\bf{j}} + 4t{\bf{k}}$ So, $\frac{d}{{dt}}\left( {{{\bf{r}}_1}\left( t \right) \times {{\bf{r}}_2}\left( t \right)} \right){|_{t = 1}} = \left( {{\rm{e}} - 4} \right){\bf{i}} - \left( {3{\rm{e}} - 2} \right){\bf{j}} + 4{\bf{k}}$ The two answers agree.