How will I represent the scalar function?

[bugatti79]

Homework Statement

show that $\nabla \times (f F)= f \nabla \times F+ (\nabla f) \times F$

The Attempt at a Solution

How will I represent the scalar function? Do I write $f=\psi(x,y,z)$ or

$f=A_x+A_y+A_z$

I chose $F=a_x \vec i +a_y \vec j +a_z \vec k$

Using $f=\psi(x,y,z)$ I work out the LHS of question as

$(\psi a_z)_y-(\psi a_y)_z-(\psi a_z)_x+(\psi a_x)_z+(\psi a_y)_x-(\psi a_x)_y$....How to go further?

Thanks

 

[SammyS]

Homework Statement

show that $\nabla \times (f F)= f \nabla \times F+ (\nabla f) \times F$

The Attempt at a Solution

How will I represent the scalar function? Do I write $f=\psi(x,y,z)$ or

$f=A_x+A_y+A_z$

I chose $F=a_x \vec i +a_y \vec j +a_z \vec k$

Using $f=\psi(x,y,z)$ I work out the LHS of question as

$(\psi a_z)_y-(\psi a_y)_z-(\psi a_z)_x+(\psi a_x)_z+(\psi a_y)_x-(\psi a_x)_y$....How to go further?

Thanks

Just saying that f is a function of x, y, and z would have been sufficient, but $f=\psi(x,y,z)$ is perfectly fine.

Use the product rule (of differentiation) for each of $(\psi a_z)_y\,,\ (\psi a_y)_z\,,\ (\psi a_z)_x\,,\ (\psi a_x)_z\,,\ (\psi a_y)_x\,,\ (\psi a_x)_y\,.$

Using subscripts for partial derivatives may not be such a good idea here. There may be some confusion as to the meaning of such quantities as $(a_x)_y$ for example.

 

[bugatti79]

Just saying that f is a function of x, y, and z would have been sufficient, but $f=\psi(x,y,z)$ is perfectly fine.

Use the product rule (of differentiation) for each of $(\psi a_z)_y\,,\ (\psi a_y)_z\,,\ (\psi a_z)_x\,,\ (\psi a_x)_z\,,\ (\psi a_y)_x\,,\ (\psi a_x)_y\,.$

Using subscripts for partial derivatives may not be such a good idea here. There may be some confusion as to the meaning of such quantities as $(a_x)_y$ for example.

Ok but why would we use the product rule though, looking at the first term for example

if psi is a function of x,y and z but since we are partially differentiating wrt y doesn't it just become $a_z \psi_y$. Ie, if the function was something like $\psi=x^2y^2z^2$ then $\psi_y= 2yx^2z^2$ etc

 

[SammyS]

Ok but why would we use the product rule though, looking at the first term for example

if psi is a function of x,y and z but since we are partially differentiating wrt y doesn't it just become $a_z \psi_y$. Ie, if the function was something like $\psi=x^2y^2z^2$ then $\psi_y= 2yx^2z^2$ etc

a_z is also a function of x, y, and z.

$\displaystyle\frac{\partial}{\partial y}\left(\psi a_y\right)=\left(\frac{\partial\psi}{\partial y}\right)a_y+\psi\left(\frac{\partial a_y}{\partial y}\right)$

 

[bugatti79]

a_z is also a function of x, y, and z.

$\displaystyle\frac{\partial}{\partial y}\left(\psi a_y\right)=\left(\frac{\partial\psi}{\partial y}\right)a_y+\psi\left(\frac{\partial a_y}{\partial y}\right)$

I got this, thank you.