Usando a regra da cadeia para z=f(x, y) e w=f(x, y, z), calcule \frac{d z}{d t} e \frac{d w}{d t} :
(l) w=\left(x^{2}-y^{2}\right) \ln \left(\sqrt{\frac{x^{3}}{x^{2}-y^{2}}}\right), x=\operatorname{cosh} (t), y=\operatorname{senh}(t), z=i

Passo 1

Resumindo: temos
w=(x^2 - y^2)\,\ln\!\Bigl(\sqrt{\frac{x^3}{x^2-y^2}}\Bigr), \quad x=\operatorname{cosh} t,\quad y=\operatorname{sinh} t,\quad z=t.
Precisamos de \dfrac{dz}{dt} e \dfrac{dw}{dt} usando a regra da cadeia
\frac{dw}{dt} =\frac{\partial w}{\partial x}\,\frac{dx}{dt} +\frac{\partial w}{\partial y}\,\frac{dy}{dt} +\frac{\partial w}{\partial z}\,\frac{dz}{dt}.

Passo 2

Primeiro simplificamos w. Note que
\sqrt{\frac{x^3}{x^2-y^2}} =\bigl(x^3/(x^2-y^2)\bigr)^{1/2} = x^{3/2}(x^2-y^2)^{-1/2},
logo
\ln\!\Bigl(\sqrt{\tfrac{x^3}{x^2-y^2}}\Bigr) =\tfrac32\ln x \;-\;\tfrac12\ln(x^2-y^2).
Definindo A=x^2-y^2 e
B=\tfrac32\ln x - \tfrac12\ln A,
temos
w=A\,B.

Passo 3

Como z=t,
\frac{dz}{dt}=1.
Além disso \partial w/\partial z=0 (pois w não depende de z).

Passo 4

Calculamos \partial w/\partial x e \partial w/\partial y. Usando w=AB:
\frac{\partial w}{\partial x} =A\frac{\partial B}{\partial x}+B\frac{\partial A}{\partial x} =A\Bigl(\tfrac32\frac1x-\tfrac12\frac{2x}{A}\Bigr)+B\,(2x) =\frac{3A}{2x}-x+2xB,
\frac{\partial w}{\partial y} =A\frac{\partial B}{\partial y}+B\frac{\partial A}{\partial y} =A\Bigl(-\tfrac12\frac{2(-y)}{A}\Bigr)+B\,(-2y) =y-2yB.
E também
\frac{dx}{dt}=\operatorname{sinh} t, \quad \frac{dy}{dt}=\operatorname{cosh} t.

Passo 5

Em (x,y)=(\operatorname{cosh} t,\operatorname{sinh} t) vale A=x^2-y^2=1 e
B=\tfrac32\ln(\operatorname{cosh} t).
Logo
\frac{dw}{dt} =\Bigl(\frac{3}{2x}-x+2xB\Bigr)\operatorname{sinh} t +\bigl(y-2yB\bigr)\operatorname{cosh} t.
Substituindo x=\operatorname{cosh} t, y=\operatorname{sinh} t e B=\tfrac32\ln(\operatorname{cosh} t), as parcelas em \ln(\operatorname{cosh} t) se cancelam e sobra
\frac{dw}{dt}=\frac32\operatorname{tanh} t.

Resposta

\displaystyle \frac{dz}{dt}=1,\qquad \frac{dw}{dt}=\frac32\,\operatorname{tanh} t.

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