I got stuck on a problem. Let $\alpha(s)$ be smooth simple closed curve with curvature $k(s)$ and $0 <k(s) \leq c $. Then show that the length of the curve is limited from below by $2\pi \frac{1}{c}$. I can't formalize the solution very well: in a neighborhood of each point $\alpha(s)$ the curve $\alpha$ has length very near to the length of the osculating circle tangent at $\alpha(s),$ and the osculating circle has radius at least $\frac{1}{c},$ then we should be able to conclude by summing up all the local contributions.
Estimating the length of a curve from below by a bound on its curvature.
curvaturecurvesdifferential-geometrymultivariable-calculus
Related Solutions
(1) If you're driving and you turn the wheel, the direction that you're facing will change. Indeed, the harder you turn, the faster the direction that you're facing changes. Equally true, your resulting trajectory will be bent in a tighter curve. Thus the rate of change of direction-facing measures curvature of your path.
(2) The curvature of a curve should be independent of one's parametrization of it. The most natural way to parametrize a curve is with the natural parametrization - that is why $\kappa:=\|{\bf T}_s\|$. But how do you compute curvature if you're working with a different parametrization? Why, with the chain rule of course:
$$\frac{{\rm d}{\bf T}}{{\rm d}s}=\frac{{\rm d}{\bf T}/{\rm d}t}{{\rm d}s/{\rm d}t} $$
$\large($Technically the chain rule is $\displaystyle\frac{{\rm d}{\bf T}}{{\rm d}t}=\frac{{\rm d}{\bf T}}{{\rm d}s}\frac{{\rm d}s}{{\rm d}t}$, but just divide that by $\displaystyle\frac{{\rm d}s}{{\rm d}t}$ to isolate the $\displaystyle\frac{{\rm d}{\bf T}}{{\rm d}s}$ we want.$\large)$
Note that $\displaystyle\frac{{\rm d}s}{{\rm d}t}=\left\|\frac{{\rm d}{\bf r}}{{\rm d}t}\right\|$.
(3) At every point on a curve there is a unique tangent line. This line is the best local approximation of the curve to the first order. But a line can be thought of as a generalized circle, one of infinite radius and whose center exists "at infinity," so perhaps consider all circles that are tangent to the curve at a point - which is the best fit? Which approximates it best to the second order?
Without loss of generality, we can apply an affine transformation (translation and rotation of the plane) to move the point on the curve to the origin so that the tangent vector there is on the positive $x$-axis; this greatly simplifies the upcoming algebra. The tangent line, i.e. the $x$-axis itself, interpreted as the graph of a (constant) function, already shares the same first derivative as the curve at the origin. We need to find a circular arc (we only consider the arc of a whole circle so that technically we are talking about a function, at least near the origin) that shares the same second derivative as the curve.
Without loss of generality, suppose the curve, interpreted locally as a graph of $y=f(x)$ near $(0,0)$, has a positive second derivative $f''(0)>0$. Now place a circle on the plane of radius $R$ tangent to the $x$-axis at the origin whose center is above the $x$-axis. The center must be at $(0,R)$. Any small enough arc of this circle through the origin is the graph of the function $g(x)=R-\sqrt{R^2-x^2}$. Compute $g''(0)=R^{-1}$. In order for $f$ and $g$ to have the same second derivative at $0$, we must have $f''(0)=R^{-1}$.
Lo and behold... Parametrize ${\bf r}(t)=(t,f(t))$ so that $\left\|\frac{{\rm d}{\bf r}}{{\rm d}t}\right\|=\sqrt{1+f'(t)^2}$. A tangent vector is given by $(1,f'(t))$, which we can normalize to ${\bf T}(t)=(1+f'(t)^2)^{-1/2}(1,f'(t))$ . Compute
$$\frac{{\rm d}{\bf T}}{{\rm d}t}=-\frac{1}{2}(1+f'(t)^2)^{-3/2}(2f'(t)f''(t))~(1,f'(t))+(1+f'(t)^2)^{-1/2}~(0,f''(t))$$
by the product rule (which works on a scalar function times a vector function) and therefore we have $\|{\bf T}_t(0)\|=f''(0)$, but since $\|{\bf r}_t(0)\|=1$ also we obtain $\|{\bf T}_s\|=\|{\bf T}_t\|/\|{\bf r}_t\|=f''(0)$ at the origin.
Thus we have a verification that $\|{\bf T}_s\|=R^{-1}$.
(4) The question and answer you're linking to is about the curvature of a surface, not of a curve. Granted, the answer begins by explaining the curvature of plane curves, but quickly segues into talking about the Gaussian curvature of surfaces that exist in three-dimensional space. You need to get acquainted with the curvature of a plane curve before you venture into visualizing the curvature of a surface. In fact the quote you give has a typo: it should say you need to think in three dimensions.
Here's another perspective. Consider a circle of radius $R$. First off, if you're driving in a circle at a constant speed, the radius is inversely proportional to how tightly you're turning. The prototypical function of $R$ which is inversely proportional to $R$ is the reciprocal function $R^{-1}$, so it makes sense to make this a definition of the circle's curvature $\kappa$.
Now say we wanted to compute $R^{-1}$ using only local data of the circle near a point. For instance, if we're in a vehicle and we can measure small changes in speed and direction but this is the only information we have at our disposal. Without loss of generality suppose the center of the circle is the origin, so that it has natural parametrization ${\bf r}(s)=(R\cos\frac{s}{R},R\sin\frac{s}{R})$. Then ${\bf T}(s)=(-\sin\frac{s}{R},\cos\frac{s}{R})$ and $\|{\bf T}_s\|=R^{-1}$.
Heuristically, then, in order to find $R^{-1}$ where $R$ is the radius of the circle that "best-fits" a curve at a point, simply pretend the curve is a circle at that point and compute $\|{\bf T}_s\|$!
thanks to the coments of @willjagy i could conclude my demonstrations, first i find by derivating $|\alpha(s)|^2$ that $<T,\alpha>=0$, derivating again and using Frenet Formulas i get
$k<N,\alpha>+1=0$ and one more time:
$k^{'}<N,\alpha>-k\tau<B,\alpha>=0$, where $\tau$ is the torsion of the curve, so, as $k$ is contant and positive:
$\tau<B,\alpha>=0$ $ \forall s\in I$,
and then at any cases $\tau=0$ or $<B,\alpha>=0$ i can conclude the $\alpha$ is a circle, for the first i get $\alpha$ in a great circle of the sphere and for the second an arbitrary cirlce with radius less or equal than $r$.
Best Answer
That follows from a simple interpretation of $k$ that you may or may not know.
Let $ \alpha: [0,L] \to \mathbb R^2$ be the arc-length parametrization of the simple closed curve. Then
$$L = \operatorname{length of }\alpha= \int_0^L dt \ge \frac{1}{c} \int_0^L k(t) dt. $$
Since $\alpha$ has arc length parametrization, $\alpha'(t) = (\cos \theta(t), \sin\theta(t))$, where $\theta$ is the angle $c$ make with the $x$-axis. Thus
$$ k(t) = \alpha''(t) \cdot n(t) = \theta'(t). $$
Hence
$$\int_0^L k(t)dt = \theta(L) - \theta(0).$$
Since $\alpha$ is a simple closed curve, the winding number is 1. Thus $\theta(L) - \theta(0) = 2\pi$, which implies that
$$ L \ge \frac{2\pi}{c}.$$