Answer
The derivative at $t=3$:
$\frac{d}{{dt}}\left( {{{\bf{r}}_1}\left( t \right)\cdot{{\bf{r}}_2}\left( t \right)} \right){|_{t = 3}} = 2$
Work Step by Step
$\frac{d}{{dt}}\left( {{{\bf{r}}_1}\left( t \right)\cdot{{\bf{r}}_2}\left( t \right)} \right) = \left( {\frac{d}{{dt}}{{\bf{r}}_1}\left( t \right)} \right)\cdot{{\bf{r}}_2}\left( t \right) + {{\bf{r}}_1}\left( t \right)\cdot\left( {\frac{d}{{dt}}{{\bf{r}}_2}\left( t \right)} \right)$
$ = {{\bf{r}}_1}'\left( t \right)\cdot{{\bf{r}}_2}\left( t \right) + {{\bf{r}}_1}\left( t \right)\cdot{{\bf{r}}_2}'\left( t \right)$
The derivative at $t=3$:
$\frac{d}{{dt}}\left( {{{\bf{r}}_1}\left( t \right)\cdot{{\bf{r}}_2}\left( t \right)} \right){|_{t = 3}} = {{\bf{r}}_1}'\left( 3 \right)\cdot{{\bf{r}}_2}\left( 3 \right) + {{\bf{r}}_1}\left( 3 \right)\cdot{{\bf{r}}_2}'\left( 3 \right)$
$ = \left( {0,0,1} \right)\cdot\left( {1,1,0} \right) + \left( {1,1,0} \right)\cdot\left( {0,2,4} \right)$
$\frac{d}{{dt}}\left( {{{\bf{r}}_1}\left( t \right)\cdot{{\bf{r}}_2}\left( t \right)} \right){|_{t = 3}} = 2$
