Proof Thuat Curl of the Product of a Function With its Gradient is Zero

Theorem
For any scalar function  
\[f(x,y,x)\]
  with continuous partial derivatives  
\[\mathbf{\nabla} \times (f \nabla f) =0\]

Proof
\[\begin{equation} \begin{aligned} \mathbf{\nabla} \times (f \nabla f) &= (\frac{\partial }{\partial x} \mathbf{i} + \frac{\partial }{\partial y} \mathbf{j} + \frac{\partial }{\partial z} \mathbf{k}) \times (f \frac{\partial f}{\partial x} \mathbf{i} + f \frac{\partial f}{\partial y} \mathbf{j} + f \frac{\partial f}{\partial z} \mathbf{k}) \\ &= (\frac{\partial}{\partial y}(f \frac{\partial f}{\partial z}) - \frac{\partial}{\partial z}(f \frac{\partial f}{\partial y})\mathbf{i} + (\frac{\partial}{\partial z}(f \frac{\partial f}{\partial x}) - \frac{\partial}{\partial x}(f \frac{\partial f}{\partial z})\mathbf{j} + (\frac{\partial}{\partial x}(f \frac{\partial f}{\partial y}) - \frac{\partial}{\partial y}(f \frac{\partial f}{\partial x})\mathbf{k} \\ &= (\frac{\partial f}{\partial y} \frac{\partial f}{\partial z} +f \frac{\partial^2 f}{\partial y \partial z} - \frac{\partial f}{\partial z} \frac{\partial f}{\partial y} - f \frac{\partial^2 f}{\partial z\partial y})\mathbf{i} + (\frac{\partial f}{\partial z} \frac{\partial }{\partial x}+ f \frac{\partial^2 f}{\partial z \partial x} - \frac{\partial f}{\partial x}\frac{\partial f}{\partial z} - f \frac{\partial^2 f}{\partial x \partial z}) \mathbf{j} \\ & + (\frac{\partial f}{\partial x} \frac{\partial f}{\partial y} +f \frac{\partial^2 f}{\partial x \partial y}- \frac{\partial f}{\partial y} \frac{\partial^2 f}{\partial y \partial x})\mathbf{k} \\ &= 0 \end{aligned} \end{equation} \]

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