Matematické Fórum

kexixex · 20. 01. 2013 20:42

ahoj,
resim ulohu: do rovnice $x^{2}\frac{\partial ^{2}z}{\partial x^{2}}-y^{2}\frac{\partial ^{2}z}{\partial y^{2}}=0$ zavedte souradnice u=xy a v=x/y.

Chtel bych se zeptat, jestli plati vztahy
i) $\frac{\partial z}{\partial x}=\frac{\partial z}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial z}{\partial v}\frac{\partial v}{\partial x}$

ii) $\frac{\partial ^{2}z}{\partial x^{2}}=\frac{\partial }{\partial u}(\frac{\partial z}{\partial x}) + \frac{\partial }{\partial v}(\frac{\partial z}{\partial x})$

Teorii, z niz reseni vychazi, nerozumim natolik, abych si to vymyslel sam. Konkretni vysledek uz si dopocitam.
Diky za rady

Brano · 20. 01. 2013 22:49

↑ kexixex:
Je to asi ia preklep ale v 2) si zabudol nejake cleny
$\frac{\partial ^{2}z}{\partial x^{2}}=\frac{\partial }{\partial u}\left(\frac{\partial z}{\partial x}\right)\cdot\frac{\partial u}{\partial x} + \frac{\partial }{\partial v}\left(\frac{\partial z}{\partial x}\right)\cdot\frac{\partial v}{\partial x}$

samozrejme analogicky pre $y$ .

kexixex · 20. 01. 2013 23:34

↑ Brano:
Jasne, diky.. Chtel jsem napsat toto.

Matematické Fórum

#1 20. 01. 2013 20:42

zavedeni novych souradnic

#2 20. 01. 2013 22:49

Re: zavedeni novych souradnic

#3 20. 01. 2013 23:34

Re: zavedeni novych souradnic

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