Answer
\begin{align*}
\mathbf{T} &=\frac{1}{\sqrt{1+t}} \mathbf{i}+\frac{\sqrt{t}}{\sqrt{1+t}} \mathbf{k}\\
\text{Length }& =\frac{52}{3}
\end{align*}
Work Step by Step
Since
$\mathbf{r}=t \mathbf{i}+\frac{2}{3} t^{3 / 2} \mathbf{k}$
Then $\mathbf{v}=\mathbf{i}+t^{1 / 2} \mathbf{k} \Rightarrow|\mathbf{v}|=\sqrt{1^{2}+\left(t^{1 / 2}\right)^{2}}=\sqrt{1+t}$
Hence
\begin{align*}
\mathbf{T}&=\frac{\mathbf{v}}{|\mathbf{v}|}\\
&=\frac{1}{\sqrt{1+t}} \mathbf{i}+\frac{\sqrt{t}}{\sqrt{1+t}} \mathbf{k}\\
\text{Length }&=\int_{0}^{8} \sqrt{1+t} d t\\
&=\left[\frac{2}{3}(1+t)^{3 / 2}\right]_{0}^{8}\\
&=\frac{52}{3}
\end{align*}
