Consider a particle moving in some curve $\gamma(t)$ , p.t at points of zero curvature $\dot{ \gamma} \cdot \ddot{\gamma (t) }=0$, assume $\frac{d}{dt} | \dot{\gamma} | \neq 0$ for all $t$
I know curvature is defined as:
$$ \kappa = \frac{|a \times v|}{|v|^3}$$
setting $\kappa=0$, I get $|a \times v | =0$, expanding the vector into curvilinear basis:
$$| \left[ a_{\tau} \tau + a_n n \right] \times \left[ v_{\tau} \tau \right]| = 0$$
This means:
$$ (a_n) ( v_{\tau}) n \times \tau = 0$$
Since $v_{\tau}>0$, this means $a_n=0$ suggesting there is no normal acceleration.. this seems to contradict my intuition.
What I really wanted to prove is the following:
Here note that near the 'flat point' of the curve, we see that the acceleration and velocity vector must be perpendicular, so I wanted to show it generally.. but the above result contradicts it. I say flat point, because I think the describing feature of the 'valley' point here is that the curvature vanishes (?).
If I have made some conceptual mistake, please comment so I can edit the question.
Best Answer
Unfortunately, this is a case where you need to "bite the bullet": your intuition is wrong.
If $\kappa=0$, then $\dot{\gamma}\times\ddot{\gamma}=0$. In 2D, if \begin{gather*} \dot{\gamma}\times\ddot{\gamma}=0 \\ \dot{\gamma}\cdot\ddot{\gamma}=0 \end{gather*} then one of $\dot{\gamma}$ and $\ddot{\gamma}$ must be zero; it is certainly not the former.
In the picture you draw, the labeled point is not "flat"; a tangent circle through that point of sufficiently large radius intersects the curve elsewhere. In fact, your drawn curve looks a lot like a plot of $t\mapsto(t,\frac{1}{2}t^2)$ around $t=0$; at that point, the curvature is precisely $1$.
For an example of a flat curve, consider $t\mapsto(e^t-1,(e^t-1)^4)$ around $t=0$. (A quick plot) There the velocity is $(1,0)$, the acceleration is also $(1,0)$, so that $\dot{\gamma}\cdot\ddot{\gamma}=1$.
On the other hand, if a curve always has constant velocity in a particular direction (say, along the $x$-axis…), then any acceleration must point in a perpendicular direction. So flat points on the curve are places where the acceleration vanishes, and then one can achieve $\dot{\gamma}\cdot\ddot{\gamma}=0$, since the latter factor is just $0$.