How will I represent the scalar function?

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Homework Statement



show that [itex]\nabla \times (f F)= f \nabla \times F+ (\nabla f) \times F[/itex]


The Attempt at a Solution



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

[itex]f=A_x+A_y+A_z[/itex]

I chose [itex]F=a_x \vec i +a_y \vec j +a_z \vec k[/itex]

Using [itex]f=\psi(x,y,z)[/itex] I work out the LHS of question as

[itex](\psi a_z)_y-(\psi a_y)_z-(\psi a_z)_x+(\psi a_x)_z+(\psi a_y)_x-(\psi a_x)_y[/itex].............How to go further?

Thanks
 

Answers and Replies

  • #2
SammyS
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Homework Statement



show that [itex]\nabla \times (f F)= f \nabla \times F+ (\nabla f) \times F[/itex]


The Attempt at a Solution



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

[itex]f=A_x+A_y+A_z[/itex]

I chose [itex]F=a_x \vec i +a_y \vec j +a_z \vec k[/itex]

Using [itex]f=\psi(x,y,z)[/itex] I work out the LHS of question as

[itex](\psi a_z)_y-(\psi a_y)_z-(\psi a_z)_x+(\psi a_x)_z+(\psi a_y)_x-(\psi a_x)_y[/itex].............How to go further?

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

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

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 [itex](a_x)_y[/itex] for example.
 
  • #3
719
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Just saying that f is a function of x, y, and z would have been sufficient, but [itex]f=\psi(x,y,z)[/itex] is perfectly fine.

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

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 [itex](a_x)_y[/itex] 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 [itex]a_z \psi_y[/itex]. Ie, if the function was something like [itex]\psi=x^2y^2z^2[/itex] then [itex]\psi_y= 2yx^2z^2[/itex] etc
 
  • #4
SammyS
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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 [itex]a_z \psi_y[/itex]. Ie, if the function was something like [itex]\psi=x^2y^2z^2[/itex] then [itex]\psi_y= 2yx^2z^2[/itex] etc
az is also a function of x, y, and z.

[itex]\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)[/itex]
 
  • #5
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az is also a function of x, y, and z.

[itex]\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)[/itex]


I got this, thank you.
 

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