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Suppose a curve with self-intersections can be described by $\phi(x,y)=0$. Suppose the intersections are $T_i$, $i=1,2,...$ and the gradient $\nabla \phi$ at those intersections are well defined. Then is it true that $\nabla\phi(T_i)=0$ for all $i$? In other words, are the gradients at those intersections all zero?

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If we agree that $\phi$ is continuously differentiable (so $\nabla \phi(x,y)$ is a continuous function of $x$ and $y$), then yes, this must be true.

The reason is that, if $\nabla \phi(x_0, y_0) \neq 0$ for some $(x_0, y_0)$, then the implicit function theorem guarantees that (locally) we can write $y$ as a function of $x$ or $x$ as a function of $y$. However, at a self-intersection $T_i$, our curve fails the horizontal and vertical line tests, so we cannot express $x$ as a function of $y$ or $y$ as a function of $x$.

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Assuming $\phi(x,y)$ is continuously differentiable in a neighbourhood of $T_i$, yes, because otherwise you could use the Implicit Function Theorem to get a unique curve in a neighourhood of $T_i$ satisfying $\phi(x,y) = 0$.

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A different argument

(For the case where two branches of the curve have distinct tangents at the intersection point)

Since $\phi$'s values do not change along the curve, we know that if $\phi$ is continuously differentiable then any of its directional derivatives along the curve is $0$.

At an intersection point, we would have a null directional derivative along two linearly independent directions. Having projection $0$ along two linearly independent vectors, we know the gradient is $0$.

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  • $\begingroup$ This assumes the two branches of the curve have distinct tangents at the intersection point. But what if they have the same tangent there? $\endgroup$ – Robert Israel Apr 21 at 18:16
  • $\begingroup$ @RobertIsrael Thanks for the observation! I hadn't thought about that, but it seems the argument does not apply in that case. I'll try to think if it can be expanded to cover it. $\endgroup$ – dafinguzman Apr 22 at 20:03

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