In other words, consider $\gamma: I \subset \mathbb{R} \mapsto \mathbb{R}^3$ a normal curve with positive curvature. We then assume that the normal unit vector $N$ of the curve is everywhere parallel with the position vector $\gamma(t)$, in other words
\begin{align}
N(t) = c \gamma(t).
\end{align}
Now we do not know if the curve is of arc length or not, so we are obliged to use the generalized Serret-Frenet formulas. It's known that the first three give
\begin{align}
\begin{cases}
T(t) &= \dfrac{\gamma'(t)}{||\gamma'(t)||}\\
N(t) &= T(t) \times B(t) = \dfrac{\gamma'(t) \times\left\{ \gamma'(t)\times\gamma''(t) \right\}}{||\gamma'(t)||\cdot||\gamma'(t) \times \gamma''(t)||}\\
B(t) &= \dfrac{\gamma'(t) \times \gamma''(t)}{||\gamma'(t) \times \gamma''(t)||}
\end{cases}
\end{align}
which, tested by my calculations does not lead to anything that is of interest. If we take the first derivatives, it's again known that
\begin{align}
N'(t) = -u(t) k(t)T(t) + u(t) \tau(t)B(t)
\end{align}
where $u$ is a function of the Euclideian norm of the length, again useless since we do not know the form of the curve.
Any ideas would be welcome.
Best Answer
Vector $N(t)$ is perpendicular to $\gamma'(t)$, hence: $$ \gamma'(t)\cdot\gamma(t)=0\ \implies {d\over dt}\big(\gamma(t)\cdot\gamma(t)\big)=0 \ \implies \|\gamma(t)\|=\text{constant}. $$