Answer
(a)
The principal unit normal vector $N(t)$ is given by:
$$
\begin{aligned}
N(t)&=\frac{1}{\sqrt{5+25 t^{2}}}(-5 t, 1,2)
\end{aligned}
$$
(b)
The curvature is given by:
$$
\begin{aligned}
k(t)&=\frac{\sqrt{5}}{\left(1+5 t^{2}\right)^{3 / 2}}.
\end{aligned}
$$
Work Step by Step
(a)
We have a parametrization
$$
r(t)=\left\langle t, \frac{1}{2} t^{2}, t^{2}\right\rangle
$$
$\Rightarrow $
$$
r^{\prime}(t) =\left\langle 1, t, 2 t \right\rangle
$$
$$
\begin{aligned}
\left|r^{\prime}(t)\right|&=\sqrt{1+t^{2}+4 t^{2}}\\
&=\sqrt{1+5 t^{2}}
\end{aligned}
$$
Then, the unit tangent vector $T(t)$ is given by:
$$
\begin{aligned}
T(t) &=\frac{r^{\prime}(t)}{\left|r^{\prime}(t)\right|}\\
&=\frac{1}{\sqrt{1+5 t^{2}}}\langle 1, t, 2 t\rangle
\end{aligned}
$$
Then, we can find:
$$
\begin{aligned}
T^{\prime}(t)&=\frac{-5 t}{\left(1+5 t^{2}\right)^{3 / 2}}\langle 1, t, 2 t)+\frac{1}{\sqrt{1+5 t^{2}}}\langle 0,1,2\rangle\\
&=\frac{1}{\left(1+5 t^{2}\right)^{3 / 2}}\left(\left\langle-5 t,-5 t^{2},-10 t^{2}\right\rangle+\left\langle 0,1+5 t^{2}, 2+10 t^{2}\right\rangle\right) \\
&=\frac{1}{\left(1+5 t^{2}\right)^{3 / 2}}\langle-5 t, 1,2\rangle
\end{aligned}
$$
$\Rightarrow$
$$
\begin{aligned}
\left|\mathrm{T}^{\prime}(t)\right|& =\frac{1}{\left(1+5 t^{2}\right)^{3 / 2}} \sqrt{25 t^{2}+1+4} \\
&=\frac{1}{\left(1+5 t^{2}\right)^{3 / 2}} \sqrt{25 t^{2}+5}\\
&=\frac{\sqrt{5} \sqrt{5 t^{2}+1}}{\left(1+5 t^{2}\right)^{3 / 2}} \\
&=\frac{\sqrt{5}}{1+5 t^{2}}
\end{aligned}
$$
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,
$$
\begin{aligned}
N(t)&=\frac{\mathbf{T}^{\prime}(t)}{\left|\mathbf{T}^{\prime}(t)\right|} \\
&=\frac{1+5 t^{2}}{\sqrt{5}} \cdot \frac{1}{\left(1+5 t^{2}\right)^{3 / 2}}\langle-5 t, 1,2\rangle\\
&=\frac{1}{\sqrt{5+25 t^{2}}}(-5 t, 1,2)
\end{aligned}
$$
(b)
The curvature is given by:
$$
\begin{aligned}
k(t)&=\frac{\left|\mathbf{T}^{\prime}(t)\right|}{\left|\mathbf{r}^{\prime}(t)\right|} \ \ \text { by using the Formula 9} \\
& =\frac{\sqrt{5} /\left(1+5 t^{2}\right)}{\sqrt{1+5 t^{2}}} \\
&=\frac{\sqrt{5}}{\left(1+5 t^{2}\right)^{3 / 2}}.
\end{aligned}
$$
