Question

Se w=e^(-c^2 t).sencx, mostre que (∂^2 w)/(∂x^2 )=∂w/∂t . Considere c constante ????

Answer 1

Vamos começar calculando a derivada segunda de w(x) em relação a variável "x":

$\dfrac{\partial^2w}{\partial x^2}~=~\dfrac{\partial}{\partial x}\left(\dfrac{\partial w}{\partial x}\right)\\\\\\\dfrac{\partial^2w}{\partial x^2}~=~\dfrac{\partial}{\partial x}\left(\dfrac{\partial}{\partial x}~\left(e^{-c^2t}\cdot sen(cx)\right)~\right)\\\\\\O~termo~e^{c^2t}~independe~de~x,~logo~pode~ser~considerado~''constante''\\Assim,~utilizando~a~regra~da~cadeia~em~sen(cx)$

$\dfrac{\partial^2w}{\partial x^2}~=~\dfrac{\partial}{\partial x}\left(sen(cx)'\cdot e^{-c^2t}\right)\\\\\\\dfrac{\partial^2w}{\partial x^2}~=~\dfrac{\partial}{\partial x}\left(c\cdot cos(cx)\cdot e^{-c^2t}\right)$

$Os~termos~''c''~e~e^{-c^2t}~sao~independents~de~x,~logo~podem~ser\\~consideradas~''constantes''.\\Aplicando~a~regra~da~cadeia~em~cos(cx)\\\\\\\dfrac{\partial^2w}{\partial x^2}~=~c\cdot e^{-c^2t}\cdot c\cdot -sen(cx)\\\\\\\boxed{\dfrac{\partial^2w}{\partial x^2}~=~-c^2\cdot e^{-c^2t}\cdot sen(cx)}$

Podemos agora calcular a derivada parcial de w em relação a variável t:

$\dfrac{\partial w}{\partial t}~=~\dfrac{\partial }{\partial t}\left(e^{-c^2t}\cdot sen(cx)\right)\\\\\\O~termo~sen(cx)~independe~de~t,~logo~podemos~considerar~''constante''.\\Aplicando~a~regra~da~cadeia~no termo~e^{-c^2t}\\\\\\\boxed{\dfrac{\partial w}{\partial t}~=~-c^2\cdot e^{-c^2t}\cdot sen(cx)}$

Como podemos ver pelos resultados obtidos,  (∂^2 w)/(∂x^2 )=∂w/∂t.