Answer
$\mathrm{T} (\displaystyle \frac{\pi}{4})=\frac{1}{3\sqrt{2}}\mathrm{i}-\frac{1}{3\sqrt{2}}\mathrm{j}+\frac{4}{3\sqrt{2}}\mathrm{k}$.
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: $\mathrm{r}(t):\quad\left\{\begin{array}{ll}
x=\sin^{2}t & \\
y=\cos^{2}t & \\
z=\tan^{2}t &
\end{array}\right.$
Differentiate $(\displaystyle \frac{d}{dt})$ each component function.
x, y, z: chain rule
$\mathrm{r}^{\prime}(t):\quad\left\{\begin{array}{ll}
x=2\sin t\cdot\cos t & \\ \\
y=2\cos t\cdot\sin t\\ & \\
z=2\tan t\cdot\sec^{2}t &
\end{array}\right. $
$\displaystyle \mathrm{r}^{\prime}(\frac{\pi}{4})=(2\cdot\frac{\sqrt{2}}{2}\cdot\frac{\sqrt{2}}{2})\mathrm{i}=(-2\cdot\frac{\sqrt{2}}{2}\cdot\frac{\sqrt{2}}{2})\mathrm{j}+(2\cdot 1\cdot(\sqrt{2})^{2})\mathrm{k}$
$\displaystyle \mathrm{r}^{\prime}(\frac{\pi}{4})==\mathrm{i}-\mathrm{j}+4\mathrm{k}$
$|\displaystyle \mathrm{r}^{\prime}(\frac{\pi}{4})|=\sqrt{1+1+16}=\sqrt{18}=3\sqrt{2}$
$\mathrm{T} (\displaystyle \frac{\pi}{4})=\frac{\mathrm{r}^{\prime}(\frac{\pi}{4})}{|\mathrm{r}^{\prime}(\frac{\pi}{4})|}=\frac{1}{3\sqrt{2}}(\mathrm{i}-\mathrm{j}+4\mathrm{k})=\frac{1}{3\sqrt{2}}\mathrm{i}-\frac{1}{3\sqrt{2}}\mathrm{j}+\frac{4}{3\sqrt{2}}\mathrm{k}$.
