Diferencia entre revisiones de «Partido 2»
De MateWiki
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<math>\ \bigtriangledown \times \vec u = \frac{1}{\rho} \cdot | <math>\ \bigtriangledown \times \vec u = \frac{1}{\rho} \cdot | ||
| − | \left|\begin{matrix} \vec {e_\rho} & \rho\cdot \vec {e_\theta } & \vec {e_z} \\ \frac{\partial}{\partial \rho } & \frac{\partial}{\partial \theta} & \frac{\partial}{\partial z} \\ 0 & \rho\cdot f(\rho ) & 0 \end{matrix}\right| = \frac{1}{\rho} \cdot [\vec {e_z} \cdot (\frac{\partial}{\partial \rho } \cdot (\rho \cdot f(\rho)))] = \frac{1}{\rho} \cdot | + | \left|\begin{matrix} \vec {e_\rho} & \rho\cdot \vec {e_\theta } & \vec {e_z} \\ \frac{\partial}{\partial \rho } & \frac{\partial}{\partial \theta} & \frac{\partial}{\partial z} \\ 0 & \rho\cdot f(\rho ) & 0 \end{matrix}\right| = \frac{1}{\rho} \cdot [\vec {e_z} \cdot (\frac{\partial}{\partial \rho } \cdot (\rho \cdot f(\rho)))] = \frac{1}{\rho} \cdot ( f(\rho) + \rho \cdot \frac{\partial f(\rho)}{\partial\rho} ) \vec {e_z} |
Revisión del 00:02 7 dic 2022
[math]\ \bigtriangledown \times \vec u = \frac{1}{\rho} \cdot \left|\begin{matrix} \vec {e_\rho} & \rho\cdot \vec {e_\theta } & \vec {e_z} \\ \frac{\partial}{\partial \rho } & \frac{\partial}{\partial \theta} & \frac{\partial}{\partial z} \\ 0 & \rho\cdot f(\rho ) & 0 \end{matrix}\right| = \frac{1}{\rho} \cdot [\vec {e_z} \cdot (\frac{\partial}{\partial \rho } \cdot (\rho \cdot f(\rho)))] = \frac{1}{\rho} \cdot ( f(\rho) + \rho \cdot \frac{\partial f(\rho)}{\partial\rho} ) \vec {e_z}[/math]
