Unit vector field and gradient

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Homework Help Overview

The discussion revolves around finding the unit vector in the direction of a vector field \(\vec{F}\) at a specific point (1, 2, -2). The context involves concepts from vector calculus, specifically the gradient of a scalar function and its relationship to unit vectors.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the definition of the gradient and its components, questioning how to derive \(\vec{F}\) from the function \(f\). There are discussions about the implications of using the chain rule and the nature of the function \(f\) in relation to spherical coordinates.

Discussion Status

The discussion is ongoing, with participants providing hints and corrections regarding the evaluation of the gradient. Some participants express confusion over the definitions and calculations, while others emphasize the need for detailed attempts to clarify misunderstandings.

Contextual Notes

There are indications of missing information and assumptions about the function \(f\) and its dependence on the radial distance \(r\). Participants also note the relevance of specific points in relation to a sphere defined by \(x^2 + y^2 + z^2 = 9\).

  • #31
That's the answer you get if you evaluate the gradient using the definition ∇f = (∂f/∂x, ∂f/∂y, ∂f/∂z).
 
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  • #32
and so is (1/3)(1/3i + 2/3j - 2/3k) equal to \vec{F}
 
  • #33
Yes, at the point (1, 2, -2) .
 
  • #34
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)?
 
  • #35
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.
 
Last edited:
  • #36
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
 
  • #37
hmm... I guess it's not the same... can someone give me some guides...
 
  • #38
anyone?
 
  • #39
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.
 

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