What does it mean that the curvature is zero in a single point?

Question

Consider a unit-speed curve $c:I \to \mathbb{R}^2$ where $I$ is an interval in $\mathbb{R}$. Let $\kappa:I \to \mathbb{R}$ be the curvature. Assume there is a point $t_0 \in I$ with $\kappa(t_0)=0$. Does this mean that $c$ is a straight line locally around $t_0$? That is, there is some $\varepsilon >0$ and two elements $a,b \in \mathbb{R}^2$ such that $$c(t)=a+bt\phantom{aaa}\forall t \in (t_0-\varepsilon,t_0+\varepsilon).$$ I am asking this because I try to understand the relationship between curvature, second derivatives and osculating circles in a geometrically accessible way. Intuitively, I guess the answer should be yes because otherwise, we can construct the osculating circle at $c(t_0)$. But if $r$ is the radius, we would have $r=1/\kappa(t_0)$ which is impossible for $\kappa(t_0)=0$. If that is true, is there an easy way to make this argument more formal?

Answer 1

When you throw a ball into the air, the $y$ velocity becomes zero at one moment of time. That does not mean that the $y$ velocity remains $0$ over any interval; that would mean that there was no acceleration, which would correspond to gravity being turned off for a short while.

For the curve $y = \sin x$, the curvature is zero at every multiple of $\pi$. Those are the points where the curve switches from concave up to concave down or vice versa. That does not mean that the sine curve becomes a straight line over an interval.