Answer
$$T(1)= \langle \frac{2}{3}, \frac{2}{3}, \frac{1}{3} \rangle$$
$$\mathrm{N}(1) =\left\langle \frac{-1}{3}, \frac{2}{3} ,\frac{-2}{3}\right\rangle$$
$$
\begin{aligned}
\mathbf{B}(1)=\left\langle-\frac{2}{3}, \frac{1}{3}, \frac{2}{3}\right\rangle
\end{aligned}
$$
Work Step by Step
We have the curve:
$$
\mathbf{r}(t)=\left\langle t^{2}, \frac{2}{3} t^{3}, t\right\rangle
$$
At $t=1$, we find that:
$$
\mathbf{r}(1)=\left\langle (1)^{2}, \frac{2}{3} (1)^{3}, (1)\right\rangle= \left(1, \frac{2}{3}, 1 \right)
$$
We first compute the ingredients needed for the unit normal vector:
$$
r^{\prime}(t) =\left\langle 2t, 2 t^{2}, 1\right\rangle
$$
$\Rightarrow$
$$
\begin{aligned}
\left|r^{\prime}(t)\right|&=\sqrt{ 4t^{2}+4t^{4}+1}\\
&=\sqrt{(1+2 t^{2})^{2}}\\
&=1+2 t^{2}
\end{aligned}
$$
Then, the unit tangent vector $T(t)$ is given by:
$$
\begin{aligned}
T(t) &=\frac{r^{\prime}(t)}{\left|r^{\prime}(t)\right|}\\
&=\frac{1}{1+2 t^{2}}\left\langle 2t, 2 t^{2}, 1\right\rangle\\
&=\frac{\left\langle 2t, 2 t^{2}, 1\right\rangle}{1+2 t^{2}}
\end{aligned}
$$
at the given point, corresponding to $t=1$, we find that:
$$
\begin{aligned}
T(1) &=\frac{\left\langle 2(1), 2 (1)^{2}, 1\right\rangle}{1+2 (1)^{2}}\\
&=\frac{\left\langle 2, 2, 1 \right\rangle}{1+2 }\\
&=\left\langle \frac{2}{3}, \frac{2}{3}, \frac{1}{3} \right\rangle
\end{aligned}
$$
Then, we can find:
$$
\begin{aligned}
\mathbf{T}^{\prime}(t) &=-4 t\left(2 t^{2}+1\right)^{-2}\left\langle 2 t, 2 t^{2}, 1\right\rangle+\left(2 t^{2}+1\right)^{-1}\langle 2,4 t, 0\rangle\\
&=\left(2 t^{2}+1\right)^{-2}\left\langle-8 t^{2}+4 t^{2}+2,-8 t^{3}+8 t^{3}+4 t,-4 t\right\rangle \\
& =2\left(2 t^{2}+1\right)^{-2}\left\langle 1-2 t^{2}, 2 t,-2 t\right\rangle
\end{aligned}
$$
$\Rightarrow$
$$
\begin{aligned}
\left|\mathrm{T}^{\prime}(t)\right|& =2\left(2 t^{2}+1\right)^{-2} \sqrt{\left(1-2 t^{2}\right)^{2}+(2 t)^{2}+(-2 t)^{2}}
\end{aligned}
$$
Then, the principal unit normal vector $N(t)$ is given by:
$$
\begin{aligned}
\mathrm{N}(t)&=\frac{\mathbf{T}^{\prime}(t)}{\left|\mathbf{T}^{\prime}(t)\right|} \\
&=\frac{2\left(2 t^{2}+1\right)^{-2}\left\langle 1-2 t^{2}, 2 t,-2 t\right\rangle}{2\left(2 t^{2}+1\right)^{-2} \sqrt{\left(1-2 t^{2}\right)^{2}+(2 t)^{2}+(-2 t)^{2}}} \\
&=\frac{\left\langle 1-2 t^{2}, 2 t,-2 t\right\rangle}{\sqrt{1-4 t^{2}+4 t^{4}+8 t^{2}}} \\
&=\frac{\left\langle 1-2 t^{2}, 2 t,-2 t\right\rangle}{1+2 t^{2}}
\end{aligned}
$$
at the given point, corresponding to $t=1$, we find that:
$$
\begin{aligned}
\mathrm{N}(1)&=\frac{\left\langle 1-2 (1)^{2}, 2 (1),-2(1)\right\rangle}{1+2 (1)^{2}}\\
&=\frac{\left\langle -1, 2 ,-2\right\rangle}{3 }\\
&=\left\langle \frac{-1}{3}, \frac{2}{3} ,\frac{-2}{3}\right\rangle
\end{aligned}
$$
The binormal vector is given by:
$$
\begin{aligned}
\mathbf{B}(1)=\mathbf{T}(1) \times \mathbf{N}(1)&=\left[\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\frac{2}{3} & \frac{2}{3} & \frac{1}{3} \\
\frac{-1}{3}, & \frac{2}{3} & \frac{-2}{3}\\
\end{array}\right]\\
&=\left\langle-\frac{4}{9}-\frac{2}{9},-\left(-\frac{4}{9}+\frac{1}{9}\right), \frac{4}{9}+\frac{2}{9}\right\rangle \\
&=\left\langle-\frac{2}{3}, \frac{1}{3}, \frac{2}{3}\right\rangle.
\end{aligned}
$$
