• CALCULE partial w / partial u E partial w / partial v PARA CADA UMA DAS SEGUINTES FUNÇÕES E EXPRESSE O RESULTADO EM FUNÇÃO SOMENTE DAS VARÁVEIS u E v
& w=f(x, y, z)=3 x^2-2 x y+4 z^2 & x=x(u, v)=e^2 u cos 3 v & y=y(u, v)=e^2 u sin 3 v & z=z(u, v)=e^2 u

Passo 1

Resumindo, temos
w=f(x,y,z)=3x^2-2xy+4z^2,
com
x=e^2ucos3v, y=e^2usin3v, z=e^2u
Para achar partial w/partial u e partial w/partial v, aplicamos a regra da cadeia multivariada, que diz
partial wpartial u =fx,xu+fy,yu+fz,zu, partial wpartial v =fx,xv+fy,yv+fz,zv

Passo 2

Calculamos as derivadas parciais de f em relação a x,y,z:
fx=partialpartial x(3x^2-2xy+4z^2)=6x-2y,
fy=partialpartial y(3x^2-2xy+4z^2)=-2x,
fz=partialpartial z(3x^2-2xy+4z^2)=8z

Passo 3

Agora as derivadas de x,y,z em u,v:
xu=partialpartial u(e^2ucos3v)=2e^2ucos3v=2x, xv=partialpartial v(e^2ucos3v)=-3e^2usin3v=-3y,
yu=2e^2usin3v=2y, yv=3e^2ucos3v=3x,
zu=partialpartial u(e^2u)=2e^2u=2z, zv=0

Passo 4

Aplicando a regra da cadeia para partial w/partial u:
partial wpartial u =(6x-2y),2x+(-2x),2y+(8z),2z =12x^2-4xy-4xy+16z^2 =12x^2-8xy+16z^2
Substituindo x=e^2ucos3v, y=e^2usin3v, z=e^2u:
partial wpartial u =12e^4ucos^23v-8e^4ucos3vsin3v+16e^4u =e^4ul(12cos^23v-8cos3vsin3v+16r)

Passo 5

Para partial w/partial v:
partial wpartial v =(6x-2y)(-3y)+(-2x)(3x)+(8z)cdot0 =-18xy+6y^2-6x^2 =-6x^2-18xy+6y^2
Substituindo x,y:
partial wpartial v =-6e^4ucos^23v-18e^4ucos3vsin3v+6e^4usin^23v =6e^4ul(-cos^23v-3cos3vsin3v+sin^23vr)

Resposta

partial wpartial u =e^4ul(12cos^23v-8cos3vsin3v+16r)
partial wpartial v =6e^4ul(-cos^23v-3cos3vsin3v+sin^23vr)

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