Usando a regra da cadeia para z=f(x, y) e w=f(x, y, z), calcule:
partial zpartial t, partial zpartial s e partial wpartial t, partial wpartial s e partial wpartial r
w=raizx^2+y^2+z^2, x=tg(t), y=cos (r), z=sen(s)

Passo 1

O enunciado pede para usar a regra da cadeia em duas etapas. Primeiro, para
z=f(s) com z=sen(s), calculamos partial z/partial t e partial z/partial s.
Depois, para
w=f(x,y,z)=raizx^2+y^2+z^2
com
x=tg(t), y=cos(r), z=sen(s),
calculamos partial w/partial t, partial w/partial s e partial w/partial r usando
partial wpartial u =partial wpartial x,partial xpartial u +partial wpartial y,partial ypartial u +partial wpartial z,partial zpartial u,
para u=t,s,r.

Passo 2

Como z=sen(s) não depende de t, temos
partial zpartial t=0
E como z=sen(s),
partial zpartial s=cos(s)

Passo 3

Para w=raizx^2+y^2+z^2, calculemos as derivadas parciais elementares:
partial wpartial x =xraizx^2+y^2+z^2 =xw, partial wpartial y =yw, partial wpartial z =zw

Passo 4

Agora as derivadas das variáveis em função dos parâmetros:
dxdt=sec^2(t), dydr=-sen(r), dzds=cos(s)
Observe que partial x/partial s=0, partial x/partial r=0, etc.

Passo 5

Aplicando a regra da cadeia para partial w/partial t, como y e z não dependem de t:
partial wpartial t =partial wpartial x,dxdt =xw,sec^2(t) =tg(t),sec^2(t)raiztg^2(t)+cos^2(r)+sen^2(s)

Passo 6

Para partial w/partial s, só z depende de s:
partial wpartial s =partial wpartial z,dzds =zw,cos(s) =sen(s),cos(s)raiztg^2(t)+cos^2(r)+sen^2(s)

Passo 7

Para partial w/partial r, só y depende de r:
partial wpartial r =partial wpartial y,dydr =yw,(-sen(r)) =-,cos(r),sen(r)raiztg^2(t)+cos^2(r)+sen^2(s)

Resposta

partial zpartial t=0, partial zpartial s=cos(s),
partial wpartial t =tg(t),sec^2(t)raiztg^2(t)+cos^2(r)+sen^2(s),
partial wpartial s =sen(s),cos(s)raiztg^2(t)+cos^2(r)+sen^2(s),
partial wpartial r =-,cos(r),sen(r)raiztg^2(t)+cos^2(r)+sen^2(s)

Exercícios relacionados