calculo integral 3
Derivada da função
Anexos:
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Soluções para a tarefa
Respondido por
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Correção ao enunciado da questão: A função
não representa um campo vetorial, é apenas uma função real de três variáveis.
Esta função dada é

não é campo vetorial!!!
Para encontrar o campo vetorial gradiente de
precisamos das derivadas parciais de
em relação a cada uma das variáveis.
Derivada parcial de
em relação a 
( considera as outras variáveis
e
como constantes, e deriva normalmente em relação a
)

_____________
Derivada parcial de
em relação a 
( considera as outras variáveis
e
como constantes, e deriva normalmente em relação a
)

_____________
Derivada parcial de
em relação a 
( considera as outras variáveis
e
como constantes, e deriva normalmente em relação a
)

_____________________
O campo vetorial gradiente de
é

( este sim é um campo vetorial. O gradiente... )
Resposta: alternativa
Esta função dada é
Para encontrar o campo vetorial gradiente de
( considera as outras variáveis
_____________
( considera as outras variáveis
_____________
( considera as outras variáveis
_____________________
O campo vetorial gradiente de
( este sim é um campo vetorial. O gradiente... )
Resposta: alternativa
Lukyo:
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