$\alpha(t)$ is a regular, closed, and simple curve in $\mathbb{R}^2$. Assuming $\alpha(t):[0,1]\to\mathbb{R}^2$ regular, simple and closed curve. For what $c\in\mathbb{R}$, can the function $\kappa(t)=c\sin{(2\pi t)}$ be the curvature of $\alpha(t)$?
 A: Strictly speaking, curvature is defined to be positive, or at least non-negative; we may see this from the Frenet-Serret formulas for a plane curve, which are may be expressed in terms of the arc-length parametrization $s$ as
$T(s) = \dot \alpha(s), \tag 1$
$\dot T(s) = \kappa(s) N(s), \tag 2$
$\dot N(s) = -\kappa(s) T(s); \tag 3$
here $T(s)$, $N(s)$, and $\kappa(s)$ are repsectively the unit tangent and normal fields and curvature of $\alpha(s)$; we restrict
$\kappa(s) > 0 \tag 4$
in order to ensure $N(s)$ may be unambiguously defined via (2), which then yields
$\kappa(s) = \Vert \dot T(s) \Vert > 0; \tag 5$
of course, this formulation also requires $\dot T(s) \ne 0$, so technically it applies to those regions of a curve where $\dot T(s) \ne 0$ binds.
It follows that, according to these definitions, there is no allowable value of $c$.
The above gives what I believe to be the main-line, most widely accepted approach to this issue.  Of course, there exist formulations in which $\kappa(s) = 0$ is allowed (and $N(s)$ is not defined for such $s$), and some authors allow for a signed curvature $k(s)$ which can become negative, but I think the stipulation $\kappa(s) > 0$ is the most widely adopted.
