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







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    $begingroup$


    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











    $endgroup$















      2












      2








      2





      $begingroup$


      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











      $endgroup$




      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






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      share|cite|improve this question













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      edited 8 hours ago









      Ernie060

      2,940719




      2,940719










      asked 9 hours ago









      winstonwinston

      537418




      537418






















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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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            $begingroup$

            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 $x$.






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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 $phi(x,y) = 0$.






              share|cite|improve this answer









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                5












                $begingroup$

                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









                $endgroup$
















                  5












                  5








                  5





                  $begingroup$

                  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









                  $endgroup$



                  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 $phi(x,y) = 0$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 9 hours ago









                  Robert IsraelRobert Israel

                  331k23220475




                  331k23220475























                      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









                      $endgroup$


















                        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









                        $endgroup$
















                          2












                          2








                          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









                          $endgroup$



                          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 $x$.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered 9 hours ago









                          StrantsStrants

                          5,84921736




                          5,84921736






























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