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Suppose we have a smooth differentiable planar curve given as $F(x,y)=0$. Suppose that the centroid of the curve $F=0$ is $F(x_0,y_0)$.

My question is:- $$\nabla F(x_0,y_0)=0$$ That is, does the gradient of $F$ becomes zero at $(x_0,y_0)$?

Now some examples supporting this statement are circles and ellipses.

I would be grateful if anyone proves the statement or gives a counter example. Also I welcome anyone who would like to generalise this statement in any direction.

Thanks!

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  • $\begingroup$ Can you recall what is the centroid of the curve? $\endgroup$ Jul 16, 2021 at 17:19
  • $\begingroup$ For parametric smooth curve it probably is $\int_0 ^1 \vec{a}(s)ds$ where $s$ is arc length parameterizing $\vec{a}$, if the total length of curve is 1. $\endgroup$ Jul 16, 2021 at 17:49

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Let $$F(x,y) = x^4 - 4 x^3 + 2 x^2 y^2 - 4 x y^2 + 16 x + y^4 - 16.$$

The graph of $F(x,y) = 0$ is a circle of radius $2$ centered at the origin, so its centroid is $(x_0,y_0) = (0,0).$

But

$$ \nabla F(x,y) = (4 (x^3 - 3 x^2 + x y^2 - y^2 + 4), 4 y (x^2 - 2 x + y^2)),$$

which is not zero at $(x,y)=(0,0).$

(To obtain this function, I treated the coordinates $x,y$ as the real and imaginary parts of a complex number $z = x+iy,$ applied the Möbius transformation $z \mapsto (z+1)/(z-1),$ separated the result into real and imaginary parts $x'$ and $y'$, and applied the circle formula $x'^2 + y'^2 = 4.$ This produced a circle with center at $(-1,0)$, so I changed $x \to x-1$ to translate the circle's center to the origin, then expanded the polynomial. I was hoping that the Möbius transformation, while it mapped a circle to a circle, would distort things enough to put the zero gradient off-center, and this turned out to be a good guess.)

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  • $\begingroup$ Thanks for such an excellent answer and also sharing the root of the intuition how you came up with the solution. I plugged in the equation in Wolfram and then gave graph as the circle you mentioned and also other forms of y in terms of X two of which were "y is complex for all x in R" now my question in continuation to this is this why we are getting the non zero directional derivatives and if for all x in some open set all roots of y is real them can we guarantee my statement in question is true. $\endgroup$ Jul 17, 2021 at 14:50
  • $\begingroup$ And as another continuation to the last comment, is this weaker version of my given statement in the question true, i.e. there exist a vector, directional derivative along which is zero? In your example along y axis direction the directional derivative is clearly zero. $\endgroup$ Jul 17, 2021 at 14:51
  • $\begingroup$ The second question is trivially true, because no matter what direction the gradient is, you can just take the direction perpendicular to that and then the directional derivative in that direction will be zero. If you are only concerned about the coordinate axis directions, we could transform the equation by a $45$-degree rotation of the $x$ and $y$ coordinates in order to get a gradient whose $x$ and $y$ components were both non-zero. $\endgroup$
    – David K
    Jul 17, 2021 at 15:08
  • $\begingroup$ From what you say, Wolfram shows you something a little different than what I see when I give the equation to Wolfram Alpha. I see four solutions for $y$ in terms of $x,$ of which two are zero at $x = 2$ and non-zero imaginary for all other real $x$ and two are real for $-2\leq x\leq 2$ and non-zero imaginary for all other real $x$. I didn't see anything "complex for all $x$ in R" if "complex" signifies "not real". $\endgroup$
    – David K
    Jul 17, 2021 at 15:17
  • $\begingroup$ wolframalpha.com/input/…*y%5E2%2B16x%2By%5E4%3D16+plot+%28x%2Cy%29 $\endgroup$ Jul 17, 2021 at 15:21

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