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UBA - CienciaS Foro de Alumnos 2013-08-12T22:43:01-03:00 http://ubacs.com.ar/ubacs/feed.php?f=37&t=2954 2013-08-12T22:43:01-03:00 2013-08-12T22:43:01-03:00 http://ubacs.com.ar/ubacs/viewtopic.php?t=2954&p=15904#p15904 <![CDATA[Re: Pedido Final 23/07/2013]]>
modulo de f(x) - f (p) - Tp (x-p) sobre norma de (x-p) tiende a cero, cuando X tiende a P (espero que se entienda, sino lo escribo en formula la proxima)

Como el limite del enunciado existe, en particular existe componiendo con cualquier curva que tenga limite P . Tomamos X= P + tV (con V de norma 1)

y tenemos que, por un lado X-P= tV y por otro lado norma de (X-P)= t , reemplazando estas cosa en la expresion de arriba queda;

Limite (cuando t tiende a 0) de f (P+tV) - f(P) -Tp (tV) / t = 0 (todo en modulo claro)

ahora distribuyendo la t: lim [ f (P+tV) - f(P) ] / t -Tp (V) = 0 (recordar que vale sacar el escalar t afuera en la TL )

en definitiva lim [f (P+tV) - f(P)]/t = Tp (V) ya que lo de la derecha dejo de depender de t

El limite de la izquierda es la derivada direccional de f en el punto P, entonces como Tp existe, tambien existe cualquier derivada direccional, en particular existen todas las derivadas parciales...

Estadísticas: Publicado por placopas — 12 Ago 2013, 22:43


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2013-08-09T12:40:46-03:00 2013-08-09T12:40:46-03:00 http://ubacs.com.ar/ubacs/viewtopic.php?t=2954&p=15898#p15898 <![CDATA[Re: Pedido Final 23/07/2013]]> Estadísticas: Publicado por dxdt — 09 Ago 2013, 12:40


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2013-08-08T12:41:33-03:00 2013-08-08T12:41:33-03:00 http://ubacs.com.ar/ubacs/viewtopic.php?t=2954&p=15895#p15895 <![CDATA[Re: Pedido Final 23/07/2013]]> Estadísticas: Publicado por inkosoft — 08 Ago 2013, 12:41


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2013-08-08T02:09:30-03:00 2013-08-08T02:09:30-03:00 http://ubacs.com.ar/ubacs/viewtopic.php?t=2954&p=15893#p15893 <![CDATA[Re: Pedido Final 23/07/2013]]>












Pero f no es diferenciable en (0,0):







Si me acerco por x=y (recta que pasa por (0,0):







Mientras que






Espero haber ayudado

Saludos!

Estadísticas: Publicado por -Teano — 08 Ago 2013, 02:09


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2013-08-08T02:13:57-03:00 2013-08-08T01:48:09-03:00 http://ubacs.com.ar/ubacs/viewtopic.php?t=2954&p=15892#p15892 <![CDATA[Re: Pedido Final 23/07/2013]]> Estadísticas: Publicado por Teano — 08 Ago 2013, 01:48


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2013-08-03T12:48:04-03:00 2013-08-03T12:48:04-03:00 http://ubacs.com.ar/ubacs/viewtopic.php?t=2954&p=15852#p15852 <![CDATA[Re: Pedido Final 23/07/2013]]> Estadísticas: Publicado por marsden — 03 Ago 2013, 12:48


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2013-08-03T12:28:39-03:00 2013-08-03T12:28:39-03:00 http://ubacs.com.ar/ubacs/viewtopic.php?t=2954&p=15848#p15848 <![CDATA[Re: Pedido Final 23/07/2013]]> Estadísticas: Publicado por inkosoft — 03 Ago 2013, 12:28


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2013-08-03T12:01:39-03:00 2013-08-03T12:01:39-03:00 http://ubacs.com.ar/ubacs/viewtopic.php?t=2954&p=15847#p15847 <![CDATA[Re: Pedido Final 23/07/2013]]> Estadísticas: Publicado por marsden — 03 Ago 2013, 12:01


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2013-08-02T16:57:05-03:00 2013-08-02T16:57:05-03:00 http://ubacs.com.ar/ubacs/viewtopic.php?t=2954&p=15840#p15840 <![CDATA[Re: Pedido Final 23/07/2013]]> Estadísticas: Publicado por inkosoft — 02 Ago 2013, 16:57


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2013-08-02T16:07:39-03:00 2013-08-02T16:07:39-03:00 http://ubacs.com.ar/ubacs/viewtopic.php?t=2954&p=15838#p15838 <![CDATA[Re: Pedido Final 23/07/2013]]>
Ya lo resolví, bastante fácil :/

Lo de que el gradiente es la dirección de máximo crecimiento si g es diferenciable supongo que si, habrá que demostrarlo. Sino el ejercicio sería nada más que acordarse la ec. del plano y hacer regla de la cadena...

La dirección me quedó (0,4).
Saludos

Estadísticas: Publicado por dxdt — 02 Ago 2013, 16:07


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