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UBA - CienciaS Foro de Alumnos 2009-03-04T07:40:44-03:00 http://ubacs.com.ar/ubacs/feed.php?f=37&t=887 2009-03-04T07:40:44-03:00 2009-03-04T07:40:44-03:00 http://ubacs.com.ar/ubacs/viewtopic.php?t=887&p=4849#p4849 <![CDATA[Re: [No resuelto]diciembre 2006 ej. 4)]]> Estadísticas: Publicado por Ger — 04 Mar 2009, 07:40

]]> 2009-02-26T16:37:05-03:00 2009-02-26T16:37:05-03:00 http://ubacs.com.ar/ubacs/viewtopic.php?t=887&p=4761#p4761 <![CDATA[Re: [No resuelto]diciembre 2006 ej. 4)]]>

Estadísticas: Publicado por daniel_htm — 26 Feb 2009, 16:37

]]> 2009-02-26T13:51:00-03:00 2009-02-26T13:51:00-03:00 http://ubacs.com.ar/ubacs/viewtopic.php?t=887&p=4756#p4756 <![CDATA[Re: [No resuelto]diciembre 2006 ej. 4)]]>
Para hallar el máximo y el mínimo se puede hacer cualquiera de las dos siguientes cosas:

1) Utilizar la desigualdad de Cauchy-Schwartz:


Y mostrar que la igualdad se cumple para algún punto.

2) Utilizar la identidad:


Donde es el ángulo que forman los vectores.

Saludos.

Estadísticas: Publicado por Don Equis — 26 Feb 2009, 13:51

]]> 2009-02-26T11:02:29-03:00 2009-02-26T11:02:29-03:00 http://ubacs.com.ar/ubacs/viewtopic.php?t=887&p=4755#p4755 <![CDATA[Re: [No resuelto]diciembre 2006 ej. 4)]]>
Exe, seguí los pasos que indicaste pero no me termina de quedar claro cuando analizo el caso donde t es distinto de {0, pi/2, pi, 3pi/2, 2pi}. La función arctg solo se anula en cero por lo que si t es distinto de cero no habría puntos críticos no?

Después analizo los puntos críticos (0, pi/2, pi, 3pi/2, 2pi) en la composición h(t) y me dan los valores a, b, -a, -b, a, respectivamente.
Ejemplo h(0)= acos(0) + bsen(0) = a

Ahora, ¿cómo discrepo quién es máximo y quién es mínimo, de manera de poder definir la imagen?
Tampoco caso cómo a don equis le da esa imagen...

Estadísticas: Publicado por daniel_htm — 26 Feb 2009, 11:02

]]> 2009-02-26T10:24:52-03:00 2009-02-26T10:24:52-03:00 http://ubacs.com.ar/ubacs/viewtopic.php?t=887&p=4754#p4754 <![CDATA[Re: [No resuelto]diciembre 2006 ej. 4)]]> Estadísticas: Publicado por Juosja — 26 Feb 2009, 10:24

]]> 2009-02-26T05:46:08-03:00 2009-02-26T05:46:08-03:00 http://ubacs.com.ar/ubacs/viewtopic.php?t=887&p=4753#p4753 <![CDATA[Re: [No resuelto]diciembre 2006 ej. 4)]]>
Edito: De hecho, se obtiene que el resultado es que

Estadísticas: Publicado por Don Equis — 26 Feb 2009, 05:46

]]> 2009-02-26T05:33:27-03:00 2009-02-26T05:33:27-03:00 http://ubacs.com.ar/ubacs/viewtopic.php?t=887&p=4752#p4752 <![CDATA[Re: [No resuelto]diciembre 2006 ej. 4)]]>
Perdón que haya sido tan desordenado(la hora no ayudó, con lo que puede haber errores). Con esto debería salir el ejercicio, creo. Si no sale, avisá. Saludos

PD: ya postearon, pero ya lo escribí, así que lo dejo. Perdón

Estadísticas: Publicado por exequiel131719 — 26 Feb 2009, 05:33

]]> 2009-02-26T05:27:53-03:00 2009-02-26T05:27:53-03:00 http://ubacs.com.ar/ubacs/viewtopic.php?t=887&p=4750#p4750 <![CDATA[Re: [No resuelto]diciembre 2006 ej. 4)]]> Estadísticas: Publicado por Don Equis — 26 Feb 2009, 05:27

]]> 2009-02-25T19:21:22-03:00 2009-02-25T19:21:22-03:00 http://ubacs.com.ar/ubacs/viewtopic.php?t=887&p=4745#p4745 <![CDATA[[No resuelto]diciembre 2006 ej. 4)]]>
g(v) = df/dv(0,0)

Hallar Im(g). Justificar.
_____________________

Yo pensé lo siguiente:
Como f es diferenciable en (0,0) entonces vale df/dv(0,0) = grad f(0,0) . v , con v = (v1,v2)
Reemplazo gradiente por (a,b)
Luego, df/dv(0,0) = av1 + bv2

¿Y ahora?

Estadísticas: Publicado por daniel_htm — 25 Feb 2009, 19:21

]]>