Is a closed simple curve of the plane entirely determined by the points of extremal or stationary curvature? The title is rather explicit: consider a $1$-periodic smooth map $f:[0,1)\to \mathbb{R}^{2}$ injective on $[0,1)$ and let $C_f$ be the image of $[0,1]$ by $f$. Let $s$ be the abscissae on the curve and consider the points thereon where the derivative of the curvature with respect to $s$ vanishes. Do these points entirely determine $C_f$? Is the center of gravity of the considered curve given by the barycenter of these points with the value of the curvature at them as coefficients?
Thanks in advance.
 A: No. Consider 
$$
\alpha(t) = (4 \cos 2 \pi t,  \sin 2 \pi t)
$$
which is a nice ellipse. Now slightly adjust the parameterization, by letting
$$
s(t) = t + c * \sin(4 \pi t)
$$
where $c = 0.03$, and define
$$
\beta(t) = (4 \cos 2 \pi s(t),  \sin 2 \pi s(t))
$$
Clearly $\alpha$ and $\beta$ have the same image, but they differ as parameterized curves. Now let 
$$
\gamma(t) = \frac{1}{2}(\alpha(t) + \beta(t))
$$
The curvature extrema of $\gamma$ are once again at $t = 0, 1/4, 1/2, 3/4$, and for these values of $t$, we have $\alpha(t) = \beta(t) = \gamma(t)$. But elsewhere, $\alpha$ and $\gamma$ differ. 
Have I actually done the derivative and curvature calculations? No. But I wrote a matlab program that pretty strongly convinced me. Here it is, if it'll be of any help to you. 
function ell()
a = 4;
b = 1;
n = 200;
t = linspace(0, 1, 2000)
s = 0.03;
tp = t + s * sin(4 * pi * t)
dt = 1/n;
x = a * cos(2*pi*t);
y = b * sin(2*pi*t);

k = curvature(x, y, dt)
clf;
subplot(2, 2, 1);
plot(x, y, 'r');
subplot(2, 2, 3);
plot(k);

x1 = a* cos(2*pi*tp);
y1 = b * sin(2*pi*tp);
x2 = (x + x1)/2;
y2 = (y + y1)/2;
k2 = curvature(x2, y2, dt)
subplot(2, 2, 2);
plot(x2, y2, 'r');
subplot(2, 2, 4);
plot(k2);
figure(gcf)
end

function k = curvature(x, y, dt)
xp = x(2:end) - x(1:end-1);
yp = y(2:end) - y(1:end-1);
xpc = 0.5 * (x(3:end) - x(1:end-2));
ypc = 0.5 * (y(3:end) - y(1:end-2));
xpp = xp(2:end) - xp(1:end-1);
ypp = yp(2:end) - yp(1:end-1);
n = xpp .* ypc - ypp .* xpc;
d = (xpc.^2 + ypc .^2) .^ 1.5;
k = n ./ d;
end

