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Is the gradient of the self-intersections of a curve zero?

3

$begingroup$

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

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edited Apr 9 at 17:04

Ernie060

2,940719

asked Apr 9 at 15:50

winstonwinston

544418

$endgroup$

add a comment | 

3

$begingroup$

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

share|cite|improve this question

edited Apr 9 at 17:04

Ernie060

2,940719

asked Apr 9 at 15:50

winstonwinston

544418

$endgroup$

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

share|cite|improve this question

edited Apr 9 at 17:04

Ernie060

2,940719

asked Apr 9 at 15:50

winstonwinston

544418

$endgroup$

share|cite|improve this question

edited Apr 9 at 17:04

Ernie060

2,940719

asked Apr 9 at 15:50

winstonwinston

544418

share|cite|improve this question

edited Apr 9 at 17:04

Ernie060

2,940719

asked Apr 9 at 15:50

winstonwinston

544418

edited Apr 9 at 17:04

Ernie060

2,940719

edited Apr 9 at 17:04

Ernie060

2,940719

asked Apr 9 at 15:50

winstonwinston

544418

asked Apr 9 at 15:50

winstonwinston

544418

2 Answers
2

active

oldest

votes

2

$begingroup$

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 $x$.

share|cite|improve this answer

answered Apr 9 at 16:27

StrantsStrants

5,89921736

$endgroup$

add a comment | 

6

$begingroup$

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 $phi(x,y) = 0$.

share|cite|improve this answer

answered Apr 9 at 16:27

Robert IsraelRobert Israel

332k23222479

$endgroup$

add a comment | 

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2

$begingroup$

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 $x$.

share|cite|improve this answer

answered Apr 9 at 16:27

StrantsStrants

5,89921736

$endgroup$

add a comment | 

2

$begingroup$

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 $x$.

share|cite|improve this answer

answered Apr 9 at 16:27

StrantsStrants

5,89921736

$endgroup$

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 $x$.

share|cite|improve this answer

answered Apr 9 at 16:27

StrantsStrants

5,89921736

$endgroup$

share|cite|improve this answer

answered Apr 9 at 16:27

StrantsStrants

5,89921736

answered Apr 9 at 16:27

StrantsStrants

5,89921736

answered Apr 9 at 16:27

StrantsStrants

5,89921736

6

$begingroup$

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 $phi(x,y) = 0$.

share|cite|improve this answer

answered Apr 9 at 16:27

Robert IsraelRobert Israel

332k23222479

$endgroup$

add a comment | 

6

$begingroup$

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 $phi(x,y) = 0$.

share|cite|improve this answer

answered Apr 9 at 16:27

Robert IsraelRobert Israel

332k23222479

$endgroup$

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 $phi(x,y) = 0$.

share|cite|improve this answer

answered Apr 9 at 16:27

Robert IsraelRobert Israel

332k23222479

$endgroup$

share|cite|improve this answer

answered Apr 9 at 16:27

Robert IsraelRobert Israel

332k23222479

answered Apr 9 at 16:27

Robert IsraelRobert Israel

332k23222479

answered Apr 9 at 16:27

Robert IsraelRobert Israel

332k23222479