Answer
(a)
The unit normal vector $N(t)$ is given by:
$$
\begin{aligned}
N(t)&=\langle 0,-\cos t,-\sin t\rangle \\
\end{aligned}
$$
(b)
The curvature is given by:
$$
\begin{aligned}
& k(t)=\frac{3}{10}.
\end{aligned}
$$
Work Step by Step
(a)
We have a parametrization
$$
r(t)= t \mathbf{i}+3 \cos t \mathbf{j}+3 \sin t \mathbf{k} \\
= \langle t, 3 \cos t, 3 \sin t\rangle
$$
$\Rightarrow $
$$
r^{\prime}(t)=\langle 1,-3 \sin t, 3 \cos t\rangle
$$
$$
\mathrm\Rightarrow\left|\mathrm{r}^{\prime}(t)\right|=\sqrt{1+9 \sin ^{2} t+9 \cos ^{2} t}=\sqrt{10}
$$
The unit tangent vector $T(t)$ is given by:
$$
\begin{aligned}
T(t) &=\frac{r^{\prime}(t)}{\left|r^{\prime}(t)\right|}\\
&=\frac{1}{\sqrt{10}}\langle 1,-3 \sin t, 3 \cos t\rangle
\end{aligned}
$$
Then
$$
\begin{aligned}
T^{\prime}(t)=\frac{1}{\sqrt{10}}(0,-3 \cos t,-3 \sin t)
\end{aligned}
$$
$\Rightarrow$
$$
\left|\mathrm{T}^{\prime}(t)\right|=\frac{1}{\sqrt{15}} \sqrt{0+9 \cos ^{2} t+9 \sin ^{2} t}=\frac{3}{\sqrt{10}}
$$
The principal unit normal vector $N(t)$ is given by:
$$
N(t)=\frac{\mathrm{T}^{\prime}(t)}{\left|\mathbf{T}^{\prime}(t)\right|}
$$
Therefore,
$$
\begin{aligned}
N(t)&=\frac{\mathrm{T}^{\prime}(t)}{\left|\mathbf{T}^{\prime}(t)\right|} \\
&=\frac{1 / \sqrt{10}}{3 / \sqrt{10}}\langle 0,-3 \cos t,-3 \sin t\rangle \\
& =\langle 0,-\cos t,-\sin t\rangle \\
\end{aligned}
$$
(b)
The curvature is given by:
$$
\begin{aligned}
k(t)&=\frac{\left|\mathbf{T}^{\prime}(t)\right|}{\left|\mathbf{r}^{\prime}(t)\right|} \ \ \text { by using the Formula 9} \\
& =\frac{3 / \sqrt{10}}{\sqrt{10}}\\
&=\frac{3}{10}.
\end{aligned}
$$
