Deduce the behavior of the following $f(s)$ and $\gamma(s)$

Question

Let $\gamma(s)$ be a smooth curve in $\mathbb{R}^3$ parametrized by arclength. Supposed that for some function $f(s)$, $\gamma''(s) = f(s)\gamma(s)$. What can you deduce about $f(s)$ and $\gamma(s)$?

Hint: Consider taking dot product by the vector $\gamma' = T$ and interpret the outcome.

We are having difficulty interpreting the outcome of this exercise. We did the following:

$\gamma'\cdot\gamma''=T.\kappa N=(T.N)\kappa=0$, since we are in the $\{T,N,B\}$ frame. From this we have that $\gamma'\cdot f\gamma=0$. Hence $\gamma'$ is perpendicular to $f\gamma$. Would this be sufficient or is there any other interpretation we are overlooking?

Answer 1

First, let us consider $\gamma''.$ We have $\gamma' = {\bf T}$ and so $\gamma'' = {\bf T}' = \kappa{\bf N}.$ If $\gamma'' = f\gamma$ then it tells us that $\gamma \propto {\bf N}$ and so both the osculating and the normal planes pass through the origin.

Next, let us differentiate the expression $\gamma'' = f\gamma.$ We get $\gamma''' = f'\gamma + f \gamma' = f'\gamma + f{\bf T}$. Consider the left hand side, we know that $\gamma' = {\bf T}$ and so $\gamma'' = {\bf T}' = \kappa{\bf N}$, and finally

$$\gamma''' = (\kappa{\bf N})' = \kappa'{\bf N} + \kappa{\bf N}' = \kappa'{\bf N} + \kappa(\tau{\bf B}-\kappa{\bf T}) = -\kappa^2{\bf T} + \kappa'{\bf N} + \kappa\tau{\bf B} \, . $$

Thus, $\gamma''' = f'\gamma + f{\bf T}$ implies $f'\gamma + f{\bf T} = -\kappa^2{\bf T} + \kappa'{\bf N} + \kappa\tau{\bf B}.$ We have already seen that $\gamma \propto {\bf N}$. Let us write $\gamma = \alpha{\bf N}.$ Fianlly we get:

$$\alpha f'{\bf N} + f{\bf T} = -\kappa^2{\bf T} + \kappa'{\bf N} + \kappa\tau{\bf B} \, . $$

Since ${\bf T},$ ${\bf N}$ and ${\bf B}$ are linearly independent it follows that $f=-\kappa^2$, $\alpha f' = \kappa'$ and $\kappa\tau=0.$

One possibility is that $\kappa \equiv 0$.

Let us assume $\kappa \not\equiv 0$. It follows from $\kappa\tau = 0$ that $\tau \equiv 0,$ and $f = - \kappa^2$.