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choli [55]
2 years ago
10

If f is a scalar field and F, G are vector fields, then f F, F · G, and F × G are defined by the following. (f F)(x, y, z) = f(x

, y, z) F(x, y, z) (F · G)(x, y, z) = F(x, y, z) · G(x, y, z) (F × G)(x, y, z) = F(x, y, z) × G(x, y, z) Find an identical expression, assuming that the appropriate partial derivatives exist and are continuous.
Mathematics
1 answer:
valentina_108 [34]2 years ago
3 0

Answer:

Step-by-step explanation:

Consider curl (fF) where f is a scalar function and F is a vector function

                  F=F_{1}i +F_{2}j +F_{3}k

                     i         j          k                

curl(fF) = \frac{\partial}{\partial x}      \frac{\partial}{\partial y}      \frac{\partial}{\partial z}        

                  fF_{1}    fF_{2}    fF_{3}  

curl(fF)=i(\frac{\partial}{\partial y}(fF_{3})-\frac{\partial}{\partial z}(fF_{2}))-j(\frac{\partial}{\partial x}(fF_{3})-\frac{\partial}{\partial z}(fF_{1}))+k(\frac{\partial}{\partial x}(fF_{2})-\frac{\partial}{\partial y}(fF_{1}))

=i(\frac{\partial}{\partial y}(F_{3})+\frac{\partial}{\partial y}(f)-\frac{\partial}{\partial z}(F_{2})-\frac{\partial}{\partial z}(f))-j(\frac{\partial}{\partial x}(F_{3})+\frac{\partial}{\partial x}(f)-\frac{\partial}{\partial z}(F_{1})-\frac{\partial}{\partial z}(f))+k(\frac{\partial}{\partial x}(F_{2})+\frac{\partial}{\partial x}(f)-\frac{\partial}{\partial y}(F_{1})-\frac{\partial}{\partial y}(f))

curl(fF)=f(\Delta\times F)+(\Delta f)\times F

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