Answer
$\displaystyle \mathrm{T}(0)= \frac{3}{5}\mathrm{j}+\frac{4}{5}\mathrm{k}$.
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: $\mathrm{r}(t):\quad\left\{\begin{array}{ll}
x=\cos t & \\
y=3t & \\
z=2\sin 2t &
\end{array}\right.$
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=-\sin t & \\ \\
y=3\\ & \\
z=2(\cos 2t\cdot 2) & =4\cos 2t
\end{array}\right. $
$\mathrm{r}^{\prime}(0)=\langle-\sin 0,\ 3,\ 4\cos 0\rangle=\langle 0,3,4\rangle$
$|\mathrm{r}^{\prime}(0)|=\sqrt{0+9+16}=5$
$\displaystyle \mathrm{T}(0)=\frac{1}{|\mathrm{r}^{\prime}(0)|}\mathrm{r}^{\prime}(0)=\frac{1}{5}\langle 0,3,4\rangle=\langle 0,\ \displaystyle \frac{3}{5},\ \displaystyle \frac{4}{5}\rangle$.
$\displaystyle \mathrm{T}(0)= \frac{3}{5}\mathrm{j}+\frac{4}{5}\mathrm{k}$.
