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 7 hours ago
Ernie060
2,940719
asked 8 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 7 hours ago
Ernie060
2,940719
asked 8 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 7 hours ago
Ernie060
2,940719
asked 8 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 7 hours ago
Ernie060
2,940719
asked 8 hours ago
winstonwinston
537418
edited 7 hours ago
Ernie060
2,940719
asked 8 hours ago
winstonwinston
537418
edited 7 hours ago
Ernie060
2,940719
edited 7 hours ago
Ernie060
2,940719
edited 7 hours ago
Ernie060
2,940719
2,940719
asked 8 hours ago
winstonwinston
537418
asked 8 hours ago
winstonwinston
537418
asked 8 hours ago
winstonwinston
537418
537418
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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 7 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 7 hours ago
StrantsStrants
5,84921736
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
active
oldest
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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 7 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 7 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 7 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 7 hours ago
Robert IsraelRobert Israel
331k23220475
answered 7 hours ago
Robert IsraelRobert Israel
331k23220475
answered 7 hours ago
Robert IsraelRobert Israel
331k23220475
answered 7 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 7 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 7 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 7 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 7 hours ago
StrantsStrants
5,84921736
answered 7 hours ago
StrantsStrants
5,84921736
answered 7 hours ago
StrantsStrants
5,84921736
answered 7 hours ago
StrantsStrants
5,84921736
5,84921736
add a comment |
add a comment |
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