Answer
\begin{align*}
\mathbf{T}&= (-\cos t) \mathbf{j}+(\sin t) \mathbf{k}\\
\text{Length }&= \frac{3}{2}
\end{align*}
Work Step by Step
Since
$\mathbf{r}=\left(\cos ^{3} t\right) \mathbf{j}+\left(\sin ^{3} t\right) \mathbf{k} $
Then
$ \mathbf{v}=\left(-3 \cos ^{2} t \sin t\right) \mathbf{j}+\left(3 \sin ^{2} t \cos t\right) \mathbf{k}$
and
\begin{align*}
|v|&=\sqrt{\left(-3 \cos ^{2} t \sin t\right)^{2}+\left(3 \sin ^{2} t \cos t\right)^{2}}\\
&=\sqrt{\left(9 \cos ^{2} t \sin ^{2} t\right)\left(\cos ^{2} t+\sin ^{2} t\right)}\\
&=3|\cos t \sin t|
\end{align*}
Hence
\begin{align*}
\mathbf{T}&=\frac{\mathrm{v}}{|\mathrm{v}|}\\
&=\frac{-3 \cos ^{2} t \sin t}{3|\cos t \sin t|} \mathbf{j}+\frac{3 \sin ^{2} t \cos t}{3|\cos t \sin t|} \mathbf{k}\\
&=(-\cos t) \mathbf{j}+(\sin t) \mathbf{k}\\
\text{Length }&=\int_{0}^{\pi / 2} 3|\cos t \sin t| d t\\
&=\int_{0}^{\pi / 2} 3 \cos t \sin t d t\\
&=\int_{0}^{\pi / 2} \frac{3}{2} \sin 2 t d t\\
&=\left[-\frac{3}{4} \cos 2 t\right]_{0}^{\pi / 2}\\
&=\frac{3}{2}
\end{align*}
