Is the gradient of the self-intersections of a curve zero?
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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 $nablaphi(T_i)=0$ for all $i$? In other words, are the gradients at those intersections all zero?
real-analysis calculus geometry differential-geometry
edited 9 hours ago
Ernie060
2,940719
asked 11 hours ago
winstonwinston
537418
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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 9 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 9 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 9 hours ago
Ernie060
2,940719
asked 11 hours ago
winstonwinston
537418
edited 9 hours ago
Ernie060
2,940719
asked 11 hours ago
winstonwinston
537418
edited 9 hours ago
Ernie060
2,940719
edited 9 hours ago
Ernie060
2,940719
edited 9 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
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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 10 hours ago
Robert IsraelRobert Israel
331k23220475
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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 10 hours ago
StrantsStrants
5,84921736
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add a comment |
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2 Answers
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2 Answers
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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 10 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 10 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 10 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 10 hours ago
Robert IsraelRobert Israel
331k23220475
answered 10 hours ago
Robert IsraelRobert Israel
331k23220475
answered 10 hours ago
Robert IsraelRobert Israel
331k23220475
answered 10 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 10 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 10 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 10 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 10 hours ago
StrantsStrants
5,84921736
answered 10 hours ago
StrantsStrants
5,84921736
answered 10 hours ago
StrantsStrants
5,84921736
answered 10 hours ago
StrantsStrants
5,84921736
5,84921736
add a comment |
add a comment |
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