Answer
$\displaystyle \langle\frac{1}{3},\ \displaystyle \frac{2}{3},\ \displaystyle \frac{2}{3}\rangle$
Work Step by Step
Unit tangent vector at t:$\quad \mathrm{T}(t)= \displaystyle \frac{\mathrm{r}^{\prime}(t)}{|\mathrm{r}^{\prime}(t)|}$
Parametric equations: $\mathrm{r}(t):\quad\left\{\begin{array}{ll}
x=te^{-t} & \\
y=2\arctan t & \\
z=2e^{t} &
\end{array}\right.$
Differentiate $(\displaystyle \frac{d}{dt})$ each component function.
x: product and chain rule
y, z: tabular derivative
$\mathrm{r}^{\prime}(t):\quad\left\{\begin{array}{ll}
x=-te^{-t}+e^{-t} & \\ \\
y=\dfrac{2}{1+t^{2}}\\ & \\
z=2e^{t} &
\end{array}\right. $
$\mathrm{r}^{\prime}(0)=\langle-(0)e^{-0}+e^{-0},\ 2/(1+0),\ 2e^{0}\rangle=\langle 1,2,2\rangle$
$|\mathrm{r}^{\prime}(0)|=\sqrt{1^{2}+2^{2}+2^{2}}=\sqrt{9}=3$
$\displaystyle \mathrm{T}(0)=\frac{1}{|\mathrm{r^{\prime}}(0)|}\mathrm{r}^{\prime}(0)=\frac{1}{3}\langle 1,2,2\rangle=\langle\frac{1}{3},\ \displaystyle \frac{2}{3},\ \displaystyle \frac{2}{3}\rangle$.
