Answer
(a) $\frac{d}{{dt}}\left( {{{\bf{r}}_1}\left( t \right)\cdot{{\bf{r}}_2}\left( t \right)} \right){|_{t = 1}} = 4{\rm{e}} + 2$
(b) $\frac{d}{{dt}}\left( {{{\bf{r}}_1}\left( t \right)\cdot{{\bf{r}}_2}\left( t \right)} \right){|_{t = 1}} = 4{\rm{e}} + 2$
Part (a) and part (b) give the same answer.
Work Step by Step
(a) We have ${{\bf{r}}_1}\left( t \right) = \left( {{t^2},1,2t} \right)$ and ${{\bf{r}}_2}\left( t \right) = \left( {1,2,{{\rm{e}}^t}} \right)$.
${{\bf{r}}_1}\left( t \right)\cdot{{\bf{r}}_2}\left( t \right) = \left( {{t^2},1,2t} \right)\cdot\left( {1,2,{{\rm{e}}^t}} \right) = {t^2} + 2 + 2t{{\rm{e}}^t}$
$\frac{d}{{dt}}\left( {{{\bf{r}}_1}\left( t \right)\cdot{{\bf{r}}_2}\left( t \right)} \right){|_{t = 1}} = 2t + 2{{\rm{e}}^t} + 2t{{\rm{e}}^t}{|_{t = 1}} = 2 + 2{\rm{e}} + 2{\rm{e}}$
$\frac{d}{{dt}}\left( {{{\bf{r}}_1}\left( t \right)\cdot{{\bf{r}}_2}\left( t \right)} \right){|_{t = 1}} = 4{\rm{e}} + 2$
(b) By Eq. (4) of Theorem 3,
$\frac{d}{{dt}}\left( {{{\bf{r}}_1}\left( t \right)\cdot{{\bf{r}}_2}\left( t \right)} \right) = \left[ {{{\bf{r}}_1}'\left( t \right)\cdot{{\bf{r}}_2}\left( t \right)} \right] + \left[ {{{\bf{r}}_1}\left( t \right)\cdot{{\bf{r}}_2}'\left( t \right)} \right]$
$\frac{d}{{dt}}\left( {{{\bf{r}}_1}\left( t \right)\cdot{{\bf{r}}_2}\left( t \right)} \right) = \left( {2t,0,2} \right)\cdot\left( {1,2,{{\rm{e}}^t}} \right) + \left( {{t^2},1,2t} \right)\cdot\left( {0,0,{{\rm{e}}^t}} \right)$
$\frac{d}{{dt}}\left( {{{\bf{r}}_1}\left( t \right)\cdot{{\bf{r}}_2}\left( t \right)} \right) = 2t + 2{{\rm{e}}^t} + 2t{{\rm{e}}^t}$
$\frac{d}{{dt}}\left( {{{\bf{r}}_1}\left( t \right)\cdot{{\bf{r}}_2}\left( t \right)} \right){|_{t = 1}} = 2 + 2{\rm{e}} + 2{\rm{e}} = 4{\rm{e}} + 2$
The two answers agree.
