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headers already sent by (output started at [ROOT]/includes/functions.php:3887) [phpBB Debug] PHP Warning: in file [ROOT]/includes/functions.php on line 4789: Cannot modify header information - headers already sent by (output started at [ROOT]/includes/functions.php:3887) [phpBB Debug] PHP Warning: in file [ROOT]/includes/functions.php on line 4790: Cannot modify header information - headers already sent by (output started at [ROOT]/includes/functions.php:3887) [phpBB Debug] PHP Warning: in file [ROOT]/includes/functions.php on line 4791: Cannot modify header information - headers already sent by (output started at [ROOT]/includes/functions.php:3887) UBA - CienciaS • Ver Tema - Duda integral de longitud de Arco
Registrado: 25 Ago 2009, 12:04 Mensajes: 371 Ubicación: R^4
Por ahora vengo entendiendo bien todo lo que es curvas. Pero tengo una duda (en realidad una incerteza) con la cuestión de las integrales.
Por ejemplo, si dada una curva C tengo una parametrización: y tengo que calcular la longitud del aro comprendido entre y . Entonces, y
Y la integral a calcular es medio complicada o más bien bastante molesta. Tengo entendido que uno se las puede arreglar encontrando una reparametrización de tal manera que quede una integral más amiga. Hay alguna forma más o menos directa de relacionar una buena reparametrización con la integral? O simplemente me armo una a lo granjero y fue?
En el ejemplo en particular, que parametrización me conviene?
_________________ "What we observe is not nature itself, but nature exposed to our method of questioning..." Werner Heisenberg
mrbean
Asunto: Re: Duda integral de longitud de Arco
Publicado: 01 Mar 2011, 17:18
Vago
Registrado: 21 Dic 2009, 16:32 Mensajes: 13
Son horribles esas integrales, alguien tiene alguna sugerencia?
ivoo
Asunto: Re: Duda integral de longitud de Arco
Publicado: 01 Mar 2011, 21:03
Profesor
Registrado: 13 Abr 2010, 23:16 Mensajes: 290
creo... creo que esa integral salia con una sustitucion por el coseno o el seno hiperbolico
ivoo
Asunto: Re: Duda integral de longitud de Arco
Publicado: 01 Mar 2011, 21:07
Profesor
Registrado: 13 Abr 2010, 23:16 Mensajes: 290
sisis sale con 2t = sh (u) entonces 2 dt = ch (u) du y en consecuencia nos queda |ch (u)| 1/2 ch (u)
KillSchrodingerCat
Asunto: Re: Duda integral de longitud de Arco
Publicado: 06 Ago 2011, 12:27
Doctor
Registrado: 25 Ago 2009, 12:04 Mensajes: 371 Ubicación: R^4
Que bueno que aprobe Mate 3 XD
Esta integral era una mierda directamente entonces. Al final saque esa sustitucion pero no es muy amigable. Bah, que se yo...
Gracias por la respuesta
_________________ "What we observe is not nature itself, but nature exposed to our method of questioning..." Werner Heisenberg
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