Answer
$\textbf{r}'(\pi/2) = \langle 1, 0, 0\rangle$
Work Step by Step
$\textbf{r}(t) = \langle f(t), g(t), h(t)\rangle$
$\textbf{r}'(t) = \langle f'(t), g'(t), h'(t)\rangle$
$\textbf{r}(t) = \langle t,cos(2t),2sin(t)\rangle$
$\textbf{r}'(t) = \langle 1, -2sin(2t), 2cos(t)\rangle$
$\textbf{r}'(\pi/2) = \langle 1, -2sin(2(\pi/2)), 2cos(\pi/2)\rangle = \langle 1, 0, 0\rangle$