Diferenciación ejercicio 8 pagina 153

Ayudame con el ejercicio 8 de la siguiente imagen:

1 respuesta

Respuesta
1

Sea f:R3--->R diferenciable

Tras hacer el cambio

x = ro·cos(teta)·sen(phi)

y = ro·sen(teta)·sen(phi)

z= ro·cos(phi)

Calcular las derivadas parciales respecto de ro, teta, phi en función de las parciales respecto x, y, z

$$\begin{pmatrix}
\frac{\partial f}{\partial \rho}&\frac{\partial f}{\partial \theta} & \frac{\partial f}{\partial \varPhi}
\end{pmatrix}=
\begin{pmatrix}
\frac{\partial f}{\partial x}&\frac{\partial f}{\partial y} & \frac{\partial f}{\partial z}
\end{pmatrix}
\begin{pmatrix}
\frac{\partial x}{\partial \rho}&\frac{\partial x}{\partial \theta}&\frac{\partial x}{\partial \varPhi}\\
\frac{\partial y}{\partial \rho}&\frac{\partial y}{\partial \theta}&\frac{\partial y}{\partial \varPhi}\\
\frac{\partial z}{\partial \rho}&\frac{\partial z}{\partial \theta}&\frac{\partial z}{\partial \varPhi}
\end{pmatrix}
\\
·
\\
\frac{\partial f}{\partial \rho}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial \rho}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial \rho}+\frac{\partial f}{\partial z}\frac{\partial z}{\partial \rho}
\\
·
\\
\frac{\partial f}{\partial \theta}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial \theta}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial \theta}+\frac{\partial f}{\partial z}\frac{\partial z}{\partial \theta}
\\
·
\\
\frac{\partial f}{\partial \varPhi}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial \varPhi}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial \varPhi}+\frac{\partial f}{\partial z}\frac{\partial z}{\partial \varPhi}
\\
Luego
\\
\frac{\partial f}{\partial \rho}=\frac{\partial f}{\partial x}\cos \theta sen \varPhi +\frac{\partial f}{\partial y}sen \theta sen \varPhi+\frac{\partial f}{\partial z}\cos \varphi
\\
.
\\
\frac{\partial f}{\partial \theta}=\frac{\partial f}{\partial x}(-\rho\, sen\theta sen\varPhi)+\frac{\partial f}{\partial y}\rho \cos\theta sen \varPhi+\frac{\partial f}{\partial z}·0
\\
.
\\
\frac{\partial f}{\partial \varPhi}=\frac{\partial f}{\partial x}\rho\, \cos\theta \cos\varPhi+\frac{\partial f}{\partial y}\rho\, sen\theta \cos \varPhi+\frac{\partial f}{\partial z}(-\rho\, sen \varphi)$$

Y eso es todo.

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