Unit vector field and gradient

[dx]

That's the answer you get if you evaluate the gradient using the definition ∇f = (∂f/∂x, ∂f/∂y, ∂f/∂z).

 

[-EquinoX-]

and so is (1/3)(1/3i + 2/3j - 2/3k) equal to \vec{F}

 

[dx]

Yes, at the point (1, 2, -2) .

 

[-EquinoX-]

The points (1, 2, -2) and (0, 3, 0) are both on the sphere x^2 + y^2 + z^2 = 9. How can I estimate \vec{F} at (0, 3, 0)?

 

[dx]

You don't need to estimate it. The exact answer can be calculated from the definition. Please show your work/attempt if you have any more questions.

 

[-EquinoX-]

well from what I know when I put the points at ρ = (x2 + y2 + z2)1/2 it gives me the same result so I am curious that they are the same

 

[-EquinoX-]

hmm... I guess it's not the same... can someone give me some guides...

 

[-EquinoX-]

anyone?

 

[dx]

I've practically already given you the answer -EquinoX-. No on will be able to help you if you don't show a detailed attempt.