Tag Archives: calculus

wedge product example


When bivectors are defined by

\beta^i\wedge\beta^j:=\beta^i\otimes\beta^j-\beta^j\otimes\beta^i,

so, for two generic covectors

\theta=a\beta^1+b\beta^2+c\beta^3 and \phi=d\beta^1+e\beta^2+f\beta^3,

we have the bivector

\theta\wedge\phi=(bf-ce)\beta^2\wedge\beta^3+(cd-af)\beta^3\wedge\beta^1+(ad-be)\beta^1\wedge\beta^2.

 

Otherwise,

Cf. this with the data \left(\begin{array}{c}a\\b\\c\end{array}\right) and \left(\begin{array}{c}d\\e\\f\end{array}\right) to construct the famous

\left(\begin{array}{c}a\\b\\c\end{array}\right)\times\left(\begin{array}{c}d\\e\\f\end{array}\right)=\left(\begin{array}{c}bf-ce\\cd-af\\ad-be\end{array}\right)

So, nobody should be confused about the uses of the symbol \wedge dans le calcul vectoriel XD

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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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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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gradient of a scalar on a surface


This image represent how to calculate the gradient of a scalar function f:\Sigma\to\mathbb{R}, a measurement on a surface (surface1 , surface2 or surface3)

On the final two lines you gotta remember that the notation Xf means \langle X,{\rm grad}f\rangle which is also equals to \langle {\rm grad}f,X\rangle, this, codes how f varies relatively to X vector or vector field, see

Remember also that \Omega is an open set of \mathbb{R}^2 and  \Phi is injective with jacobian J\Phi having rank two “along” \Omega

With this device you can transport the Euclidean calculus in \mathbb{R}^2 to calculus in the surface \Sigma\subset\mathbb{R}^3

See how the chain rule is adapted: J(f\circ\Phi)=Jf\cdot J\Phi, from where we can evaluate: J(f\circ\Phi)(a)=Jf(\Phi(a))\cdot J\Phi(a)… What are \partial_1,\partial_2 here?

As a nontrivial example consider the notion of round functions

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four geodesics in the band


on the right of this post you can see 4 (approximately) geodesics in the mobius strip, one is transversal to the core curve and the others (three)  begin at that transversal… One almost can see that all geodesics in the surface have curvature but without 3d torsion… would it be a theorem?  

Rendered with Mathematica-6.

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