Is the gradient of the self-intersections of a curve zero? The 2019 Stack Overflow Developer Survey Results Are InMonotonic curvature and self intersections.Parallel translation along a self intersecting curveSelf adjoint total covariant derivativeStokes Theorem for Manifolds with Self-IntersectionsIntersections of two curves in $mathbbR^n$Self intersections of a smooth closed curve being deformedProving that strictly monotonic curvature implies no self intersections (more specifically, using the following inequalities)Does an immersed curve in general position has finite self-intersections?Can we describe Injective and non-Injective functions through intersections?Problem understanding the gradient of a field.
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Is the gradient of the self-intersections of a curve zero?
The 2019 Stack Overflow Developer Survey Results Are InMonotonic curvature and self intersections.Parallel translation along a self intersecting curveSelf adjoint total covariant derivativeStokes Theorem for Manifolds with Self-IntersectionsIntersections of two curves in $mathbbR^n$Self intersections of a smooth closed curve being deformedProving that strictly monotonic curvature implies no self intersections (more specifically, using the following inequalities)Does an immersed curve in general position has finite self-intersections?Can we describe Injective and non-Injective functions through intersections?Problem understanding the gradient of a field.
$begingroup$
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 $nablaphi(T_i)=0$ for all $i$? In other words, are the gradients at those intersections all zero?
real-analysis calculus geometry differential-geometry
edited 10 hours ago
Ernie060
2,940719
asked 11 hours ago
winstonwinston
537418
$endgroup$
add a comment |
$begingroup$
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 $nablaphi(T_i)=0$ for all $i$? In other words, are the gradients at those intersections all zero?
real-analysis calculus geometry differential-geometry
edited 10 hours ago
Ernie060
2,940719
asked 11 hours ago
winstonwinston
537418
$endgroup$
add a comment |
$begingroup$
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 $nablaphi(T_i)=0$ for all $i$? In other words, are the gradients at those intersections all zero?
real-analysis calculus geometry differential-geometry
edited 10 hours ago
Ernie060
2,940719
asked 11 hours ago
winstonwinston
537418
$endgroup$
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 $nablaphi(T_i)=0$ for all $i$? In other words, are the gradients at those intersections all zero?
real-analysis calculus geometry differential-geometry
real-analysis calculus geometry differential-geometry
edited 10 hours ago
Ernie060
2,940719
asked 11 hours ago
winstonwinston
537418
edited 10 hours ago
Ernie060
2,940719
asked 11 hours ago
winstonwinston
537418
edited 10 hours ago
Ernie060
2,940719
edited 10 hours ago
Ernie060
2,940719
edited 10 hours ago
Ernie060
2,940719
2,940719
asked 11 hours ago
winstonwinston
537418
asked 11 hours ago
winstonwinston
537418
asked 11 hours ago
winstonwinston
537418
537418
add a comment |
add a comment |
2 Answers
2
active
oldest
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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 $phi(x,y) = 0$.
answered 11 hours ago
Robert IsraelRobert Israel
331k23220475
$endgroup$
add a comment |
$begingroup$
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 $x$.
answered 11 hours ago
StrantsStrants
5,84921736
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
active
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$begingroup$
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 $phi(x,y) = 0$.
answered 11 hours ago
Robert IsraelRobert Israel
331k23220475
$endgroup$
add a comment |
$begingroup$
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 $phi(x,y) = 0$.
answered 11 hours ago
Robert IsraelRobert Israel
331k23220475
$endgroup$
add a comment |
$begingroup$
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 $phi(x,y) = 0$.
answered 11 hours ago
Robert IsraelRobert Israel
331k23220475
$endgroup$
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 $phi(x,y) = 0$.
answered 11 hours ago
Robert IsraelRobert Israel
331k23220475
answered 11 hours ago
Robert IsraelRobert Israel
331k23220475
answered 11 hours ago
Robert IsraelRobert Israel
331k23220475
answered 11 hours ago
Robert IsraelRobert Israel
331k23220475
331k23220475
add a comment |
add a comment |
$begingroup$
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 $x$.
answered 11 hours ago
StrantsStrants
5,84921736
$endgroup$
add a comment |
$begingroup$
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 $x$.
answered 11 hours ago
StrantsStrants
5,84921736
$endgroup$
add a comment |
$begingroup$
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 $x$.
answered 11 hours ago
StrantsStrants
5,84921736
$endgroup$
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 $x$.
answered 11 hours ago
StrantsStrants
5,84921736
answered 11 hours ago
StrantsStrants
5,84921736
answered 11 hours ago
StrantsStrants
5,84921736
answered 11 hours ago
StrantsStrants
5,84921736
5,84921736
add a comment |
add a comment |
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