VC3.36

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Let $V=V_1i+V_2j+V_3k,V^2=V_1^{2}+V_2^{2}+V_3^{2}\,$ --(1)

Therefore $\mathrm{curl}V=[\frac{\partial V_3}{\partial y}-\frac{\partial V_2}{\partial z}]i+[\frac{\partial V_1}{\partial z}-\frac{\partial V_3}{\partial x}]j+[\frac{\partial V_2}{\partial x}-\frac{\partial V_1}{\partial y}]k\,$ --(2)

Using (1) and (2)

$V\times\mathrm{curl}V=\begin{vmatrix} i & j & k \\ V_1 & V_2 & V_3 \\ \frac{\partial V_3}{\partial y}-\frac{\partial V_2}{\partial z} & \frac{\partial V_1}{\partial z}-\frac{\partial V_3}{\partial x} & \frac{\partial V_2}{\partial x}-\frac{\partial V_1}{\partial y}\end{vmatrix}\,$

= $\sum i{V_2[\frac{\partial V_2}{\partial x}-\frac{\partial V_1}{\partial y}]-V_3[\frac{\partial V_1}{\partial z}-\frac{\partial V_3}{\partial x}]}\,$

= $\sum i{[V_2\frac{\partial V_2}{\partial x}+V_3\frac{\partial V_3}{\partial x}]-[V_2\frac{\partial V_1}{\partial y}+V_3\frac{\partial V_1}{\partial z}]}\,$

Adding and subtracting $V_1\frac{\partial V_1}{\partial x}\,$ to the above, we get

= $\sum i{[V_1\frac{\partial V_1}{\partial x}+V_2\frac{\partial V_2}{\partial x}+V_3\frac{\partial V_3}{\partial x}]-[V_1\frac{\partial V_1}{\partial x}+V_2\frac{\partial V_1}{\partial y}+V_3\frac{\partial V_1}{\partial z}]}\,$

= $\sum i{\frac{1}{2}\frac{\partial}{\partial x}(V_1^{2}+V_2^{2}+V_3^{2})}-\sum i[V_1\frac{\partial}{\partial x}+V_2\frac{\partial}{\partial y}+V_3\frac{\partial}{\partial z}]V_1\,$

= $\frac{1}{2}\sum i\frac{\partial V^2}{\partial x}-[V_1\frac{\partial}{\partial x}+V_2\frac{\partial}{\partial y}+V_3\frac{\partial}{\partial z}]\sum V_1 i\,$ , by (1)

= $\frac{1}{2}[i\frac{\partial}{\partial x}+j\frac{\partial}{\partial y}+k\frac{\partial}{\partial z}]V^2-(V_1i+V_2j+V_3k)\cdot[i\frac{\partial}{\partial x}+j\frac{\partial}{\partial y}+k\frac{\partial}{\partial z}]V\,$ [Since $V=V_1i+V_2j+V_3k=\sum V_1 i\,$ ]

= $\frac{1}{2}\nabla V^2-(V\cdot\nabla)V\,$

Hence,the required is proved.

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VC3.36

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