Tag Archives: vector calculus

puntos críticos de una función suave en el círculo


En esta breve nota demostraremos que cada función f:S^1\to{\mathbb{R}}^1 que tenga un punto crítico aislado debe de tener otro.

Entonces supongamos que existe un punto p en S^1 talque {\rm grad}f(p)=\left[\frac{\partial f}{\partial x}|_p,\frac{\partial f}{\partial y}|_p\right]=\vec{0}, pero si elegimos la parametrización \phi:\ ]0,2\pi[\longrightarrow S^1 dada por t\longmapsto\left(\begin{array}{c}\cos(t)\\ \\ \sin(t)\end{array}\right), entonces tenemos una función g=f\circ\phi para la cual, la regla de la cadena implica que g'=f'(\phi)\phi' satisface

\frac{d g}{dt}|_{t_0}=\left[\frac{\partial f}{\partial x}|_p,\frac{\partial f}{\partial y}|_p\right]\left(\begin{array}{c}-\sin\\ \\ \cos\end{array}\right)_{|_{t_0}}

i.e.

\frac{d g(t_0)}{dt}=-\frac{\partial f(p)}{\partial x}\sin(t_0)+\frac{\partial f(p)}{\partial y}\cos(t_0)

entonces si {\rm grad}f(p)=\vec{0} tendremos \frac{d g(t_0)}{dt}=0, en otras palabras g tiene puntos críticos en t_0 y en t_0+2\pi.

Pero además g(t_0)=f\circ\phi(t_0)=f(p) tanto como

g(t_0+2\pi)=f\circ\phi(t_0+2\pi)=f\circ\phi(t_0)=f(p)

es decir g(t_0)=g(t_0+2\pi) y entonces –por el teorema de Rolle– existe t_1 en el intervalo abierto ]t_0,t_0+2\pi[ talque \frac{d g(t_1)}{dt}=0.

Pero si nos restringimos a S^1\setminus\{p\} entonces f=g\circ\phi^{-1},
y así (también por la regla de la cadena) tenemos {\rm grad}f=\frac{dg}{dt}\ {\rm grad}\ \phi^{-1} i.e.

\left[\frac{\partial f}{\partial x},\frac{\partial f}{\partial y}\right]=\frac{dg}{dt}\left[\frac{\partial\phi^{-1}}{\partial x},\frac{\partial\phi^{-1}}{\partial y}\right]

que evaluando en t_1 implica

\frac{\partial f(q)}{\partial x}=\frac{dg(t_1)}{dt}\frac{\partial\phi^{-1}(q)}{\partial x}=0

tanto como

\frac{\partial f(q)}{\partial y}=\frac{dg(t_1)}{dt}\frac{\partial\phi^{-1}(q)}{\partial y}=0

por lo tanto {\rm grad}f(q)=\vec{0}, donde q\neq p \Box

critical points of functions on the circle

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need more?


¿necesitas o requieres un tema en particular? si es alrededor de álgebra multilineal, anímate a interaccionar. También tenemos topología de dimensiones bajas y más…

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solution to a visual quizz


the correction is:

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vector calculus examples


the following hieroglyph

\left(\begin{array}{c}V\\ W\end{array}\right)\stackrel{\phi}\to\left(\begin{array}{c} \cos V\cos W\\ \cos V\sin W\\ \sin V\end{array}\right)

represent a mapping \mathbb{R}^2\to\mathbb{R}^3. It is enough to take

-\frac{\pi}{2}< V<\frac{\pi}{2}\quad,\quad 0< W<2\pi

to G.P.S -ing (almost) every point in \mathbb{R}^3 which are at distance one from the origin. That is the sphere.

Now the partial derivatives \partial_1=\frac{\partial\phi}{\partial V} and \partial_2=\frac{\partial\phi}{\partial W} are pictorially as:

These derivative are:

\partial_1=\left(\begin{array}{c}-\sin V\cos W\\ -\sin V\sin W\\ \cos V\end{array}\right) and \partial_2=\left(\begin{array}{c}-\cos V\sin W\\ \cos V\cos W\\ 0\end{array}\right)

An easy calculation give that for the inner products \langle\phi,\partial_1\rangle=\langle\phi,\partial_2\rangle=0, so the \partial_i are orthogonal to the position \phi. Then the product \partial_1\times\partial_2, which is orthogonal to the plane determined by the couple of vectors \{\partial_1,\partial_2\}, hence colinear to the position \phi. In fact, after normalization N=\frac{\partial_1\times\partial_2}{||\partial_1\times\partial_2||}.

A good exercise is to unfold the same program for the torus by employing the parameterization given by:

\left(\begin{array}{c}V\\ W\end{array}\right)\stackrel{\phi}\to\left(\begin{array}{c} (2+\cos V)\cos W\\ (2+\cos V)\sin W\\ \sin V\end{array}\right)

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change of basis and change of components


always it amuses me the incredulity of people when they receive an explanation of the covariant and contravariant behavior of linear algebra matters.

If one has a change of basis \mathbb{R}^2\to\mathbb{R}^2, you gotta to specify the basis that is assigned.

If you begin by chosing the canonical basis e_1,e_2, where

e_1=\left(\!\!\begin{array}{c}1\\ 0\end{array}\!\!\right) , e_2=\left(\!\!\begin{array}{c}0\\ 1\end{array}\!\!\right),

and another basis, say

b_1=\left(\!\!\begin{array}{c}1\\ 0\end{array}\!\!\right) , b_2=\left(\!\!\begin{array}{c}1\\ 1\end{array}\!\!\right),

then we have b_1=e_1 and b_2=e_1+e_2. From here one can resolve: e_1=b_1 and e_2=-b_1+b_2.

So if a vector v=Ae_1+Be_2 is an arbitrary (think… A,B\in\mathbb{R}) element in \mathbb{R}^2 then its expression in the new basis is:

v=Ab_1+B(-b_1+b_2)

=(A-B)b_1+Bb_2.

Remember A,B\in\mathbb{R}.

So, if v=\left(\!\!\begin{array}{c}A\\ B\end{array}\!\!\right)_e are the component in the old basis, then v=\left(\!\!\begin{array}{c}A-B\\ B\end{array}\!\!\right)_b are the components in the new one.

Matricially what we see is this:

\left(\!\!\begin{array}{cc}1&-1\\ 0&1\end{array}\!\!\right)\left(\!\!\begin{array}{c}A\\ B\end{array}\!\!\right)=\left(\!\!\begin{array}{c}A-B\\ B\end{array}\!\!\right).

This corresponds to the relation

M^{-1}v_e=v_b, \qquad (*)

which -contrasted against Me_i=b_i in the change between those basis- causes some mind twist  :D…  :p

One says then that the base vectors co-vary, but components of vectors contra-vary.

For more: Covariance and contravariance of vectors.

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re-engineering some maths


digo re-ingeniería del álgebra lineal al querer enfatizar el carácter categórico monoidal de la categoría de los espacios vectoriales sobre los números \mathbb{R}, y re-ingeniería del cálculo vectorial para remediar el remedio de los ingenieros Gibbs-Heavyside-(Adhémar Jean Claude Barré de) Saint-Venant, y substituir  grad, div, rot y Stokes por el cálculo à-la-Cartan, es decir, usando formas diferenciales.

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acerca de un problema della examinazione


estabamos calculando \nabla_{C'}F, que corresponde a

  • una curva \alpha:I\to\Omega
  • una superficie \Phi:\Omega\to\Sigma
  • la curva C=\Phi\!\circ\!\alpha
  • y un campo vectorial en el ambiente  F:{\mathbb{R}}^3\to{\mathbb{R}}^3

Preguntar por \nabla_{C'}F es determinar como varía F intrínseco a la curva en la superficie.

Para calcularle  necesitamos D_{C'}F=\nabla_{C'}F+\langle D_{C'}F, N\rangle N

Una aproximación es considerar la composición F\!\!\circ\!\Phi\!\circ\!\alpha, pues

F\!\!\circ\!\Phi\!\circ\!\alpha=F\!\!\circ\!C

y entonces (F\!\!\circ\!\Phi\!\circ\!\alpha)'=(F\!\!\circ\!C)'=JF\cdot C'=D_{C'}F, que es como varía F con respecto a la curva C considerando el ambiente.

 

Veamos incluido el gráfico que marca la dirección normal:

Finalmente la base tangente \partial_1,\quad\partial_2

La respuesta genéricamente hablando será:

sabemos que queremos encontrar \nabla_{C'}F=D_{C'}F-\langle D_{C'}F,N\rangle N, pero usando C'=v'\partial_1+w'\partial_2 tenemos:

\nabla_{C'}F=\nabla_{v'\partial_1+w'\partial_2}F

=v'\nabla_{\partial_1}F+w'\nabla_{\partial_2}F

donde

\nabla_{\partial_1}F=D_{\partial_1}F-\langle D_{\partial_1}F ,N\rangle N

\nabla_{\partial_2}F=D_{\partial_2}F-\langle D_{\partial_2}F ,N\rangle N

i.e.

\nabla_{\partial_1}F=F_{,1}-\langle F_{,1} ,N\rangle N

\nabla_{\partial_2}F=F_{,2}-\langle F_{,2} ,N\rangle N

teniendo en cuenta F=F^se_s implica F_{,k}=\frac{\partial}{\partial v^k}(F^s)e_s,  en el {\mathbb{R}}^3 con respecto a la superficie en cuestión.

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.. C^{o}

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