How will I represent the scalar function?

  • Thread starter Thread starter bugatti79
  • Start date Start date
  • Tags Tags
    Function Scalar
bugatti79
Messages
786
Reaction score
4

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
 
Physics news on Phys.org


bugatti79 said:

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.
 


SammyS said:
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
 


bugatti79 said:
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
az 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)
 


SammyS said:
az 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.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top