Question

Given that U is a function of x, y, and z

and A a vector field, prove that:

∇×(UA)=(∇U)×A+U(∇×A).

Answer 1

Given:- U is a scalar function of x, y, z and A is a vector field

Let $A_{1}, A_{2}, A_{3}$ are components of A

 $\therefore \nabla \cdot(U A)=\frac{\partial}{\partial x}\left(U A_{1}\right)+\frac{\partial}{\partial y}\left(U A_{2}\right)+\frac{\partial}{\partial z}\left(U A_{3}\right)$  

using product rule

$\begin{aligned} \nabla \cdot(U A) &=\left[\left(\frac{\partial U}{\partial x}\right) A_{1}+U \frac{\partial A_{1}}{\partial x}\right]+\left[\left(\frac{\partial U}{\partial y}\right) A_{2}+U\left(\frac{\partial A_{2}}{\partial y}\right)\right] \\ &+\left[\left(\frac{\partial U}{\partial z}\right) A_{3}+U\left(\frac{\partial A_{3}}{\partial z}\right)\right] \\ &=\frac{\partial U}{\partial x} A_{1}+\frac{\partial U}{\partial y} A_{2}+\frac{\partial U}{\partial z} A_{3}+U\left(\frac{\partial A_{1}}{\partial x}+\frac{\partial A_{2}}{\partial y}+\frac{\partial A_{3}}{\partial z}\right) \\ &=\left(\frac{\partial U}{\partial x} \hat{\imath}+\frac{\partial U}{\partial y} \hat{j}+\frac{\partial U}{\partial z} \hat{k}\right) \cdot\left(A_{1} \hat{i}+A_{2} \hat{\jmath}+A_{3} \hat{k}\right)+U(\nabla \cdot A) \end{aligned}$

we recognize

$\nabla \cdot(\cup A)=(\nabla \cup) \cdot A+U(\nabla \cdot A) \\ \text { Hence proved }$