Answer
$\boldsymbol{T}(0)=\langle\frac{1}{9},\frac{4}{9},\frac{8}{9}\rangle$.
Work Step by Step
$\boldsymbol{r}(t)=\langle \arctan{t},2e^{2t},8te^t\rangle$
$\boldsymbol{T}(t)=\frac{\boldsymbol{r'}(t)}{|\boldsymbol{r'}(t)|}=\frac{1}{\sqrt{(\frac{1}{1+t^2})^2+(4e^{2t})^2+(8e^t+8te^t)^2}}\langle\frac{1}{1+t^2},4e^{2t},8e^t+8te^t\rangle \Rightarrow \boldsymbol{T}(0)=\frac{1}{\sqrt{(\frac{1}{1+(0)^2})^2+(4e^{2(0)})^2+(8e^{(0)}+8(0)e^{(0)})^2}}\langle\frac{1}{1+(0)^2},4e^{2(0)},8e^{(0)}+8(0)e^{(0)}\rangle=\frac{1}{9}\langle1,4,8\rangle=\langle\frac{1}{9},\frac{4}{9},\frac{8}{9}\rangle$
