Answer
(a)
The unit normal vector $N(t)$ is given by:
$$
\begin{aligned}
N(t)&=\frac{1}{e^{2 t}+1}\left\langle 1-e^{2 t}, \sqrt{2} e^{t}, \sqrt{2} e^{t}\right\rangle
\end{aligned}
$$
(b)
The curvature is given by:
$$
\begin{aligned}
k(t)&=\frac{\sqrt{2} e^{2 t}}{\left(e^{2 t}+1\right)^{2}}
\end{aligned}
$$
Work Step by Step
(a)
We have a parametrization
$$
r(t)= \sqrt{2} t \mathbf{i}+e^{t} \mathbf{j}+e^{-t} \mathbf{k} \\
=\left\langle\sqrt{2} t, e^{t}, e^{-t}\right\rangle
$$
$\Rightarrow $
$$
\begin{aligned}
r^{\prime}(t) &= \left\langle\sqrt{2}, e^{t},-e^{-t}\right\rangle
\end{aligned}
$$
$$
\begin{aligned}
\left|r^{\prime}(t)\right|&=\sqrt{2+e^{2 t}+e^{-2 t}}\\
&=\sqrt{\left(e^{t}+e^{-t}\right)^{2}}\\
&=e^{t}+e^{-t}
\end{aligned}
$$
Then, the unit tangent vector $T(t)$ is given by:
$$
\begin{aligned}
T(t) &=\frac{r^{\prime}(t)}{\left|r^{\prime}(t)\right|}\\
&=\frac{1}{e^{t}+e^{-t}}\left\langle\sqrt{2}, e^{t},-e^{-t}\right\rangle , \ \ \ \text{multiplying by} \ \ \frac{e^{t}}{e^{t}},\\
&=\frac{1}{e^{2 t}+1}\left\langle\sqrt{2} e^{t}, e^{2 t},-1\right\rangle
\end{aligned}
$$
Then, we can find:
$$
\begin{aligned}
T^{\prime}(t)&=\frac{1}{e^{2 t}+1}\left\langle\sqrt{2} e^{t}, 2 e^{2 t}, 0\right\rangle -\frac{2 e^{2 t}}{\left(e^{2 t}+1\right)^{2}}\left\langle\sqrt{2} e^{t}, e^{2 t},-1\right\rangle \\
&=\frac{1}{\left(e^{2 t}+1\right)^{2}}\left[\left(e^{2 t}+1\right)\left\langle\sqrt{2} e^{t}, 2 e^{2 t}, 0\right\rangle-2 e^{2 t}\left\langle\sqrt{2} e^{t}, e^{2 t},-1\right\rangle\right] \\
&=\frac{1}{\left(e^{2 t}+1\right)^{2}}\left\langle\sqrt{2} e^{t}\left(1-e^{2 t}\right), 2 e^{2 t}, 2 e^{2 t}\right\rangle
\end{aligned}
$$
$\Rightarrow$
$$
\begin{aligned}
\left|\mathrm{T}^{\prime}(t)\right|& =\frac{1}{\left(e^{2 t}+1\right)^{2}} \sqrt{2 e^{2 t}\left(1-2 e^{2 t}+e^{4 t}\right)+4 e^{4 t}+4 e^{4 t}} \\
&=\frac{1}{\left(e^{2 t}+1\right)^{2}} \sqrt{2 e^{2 t}\left(1+2 e^{2 t}+e^{4 t}\right)}\\
&=\frac{1}{\left(c^{2 t}+1\right)^{2}} \sqrt{2 e^{2 t}\left(1+e^{2 t}\right)^{2}}\\
&=\frac{\sqrt{2} e^{t}\left(1+e^{2 t}\right)}{\left(e^{2 t}+1\right)^{2}} \\
&=\frac{\sqrt{2} e^{t}}{e^{2 t}+1}
\end{aligned}
$$
Since the principal unit normal vector $N(t)$ is given by:
$$
N(t)=\frac{\mathrm{T}^{\prime}(t)}{\left|\mathbf{T}^{\prime}(t)\right|}
$$
Therefore,
$$
\begin{aligned}
N(t)&=\frac{\mathbf{T}^{\prime}(t)}{\left|\mathbf{T}^{\prime}(t)\right|}=\frac{e^{2 t}+1}{\sqrt{2} e^{t}} \frac{1}{\left(e^{2 t}+1\right)^{2}}\left\langle\sqrt{2} e^{t}\left(1-e^{2 t}\right), 2 e^{2 t}, 2 e^{2 t}\right\rangle \\
&=\frac{1}{\sqrt{2} e^{t}\left(e^{2 t}+1\right)}\left\langle\sqrt{2} e^{t}\left(1-e^{2 t}\right), 2 e^{2 t}, 2 e^{2 t}\right\rangle\\
&=\frac{1}{e^{2 t}+1}\left\langle 1-e^{2 t}, \sqrt{2} e^{t}, \sqrt{2} e^{t}\right\rangle
\end{aligned}
$$
(b)
The curvature is given by:
$$
\begin{aligned}
k(t)&=\frac{\left|\mathbf{T}^{\prime}(t)\right|}{\left|\mathbf{r}^{\prime}(t)\right|} \ \ \text { by using the Formula 9} \\
& =\frac{\sqrt{2} e^{t}}{e^{2 t}+1} \cdot \frac{1}{e^{t}+e^{-t}}\\
&=\frac{\sqrt{2} e^{t}}{e^{3 t}+2 e^{t}+e^{-t}} \\
&=\frac{\sqrt{2} e^{2 t}}{e^{4 t}+2 e^{2 t}+1}\\
&=\frac{\sqrt{2} e^{2 t}}{\left(e^{2 t}+1\right)^{2}}.
\end{aligned}
$$
