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Unit vector field and gradient

  1. May 2, 2009 #1
    1. The problem statement, all variables and given/known data

    http://img245.imageshack.us/img245/2353/87006064.th.jpg [Broken]

    I need to find the unit vector in the direction of [tex] \vec{F} [/tex] at the point (1, 2, -2).

    2. Relevant equations



    3. The attempt at a solution

    well first of all I need to find what F is right, which is gradf.. how can I get that?? what is the level surface of f? is it a sphere?
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. May 2, 2009 #2
    The grad (del operator) by itself is simply the derivative taken in 3 directions, x,y,z for your problem.
     
  4. May 2, 2009 #3

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    grad f for a function which depends only on r = √(x² + y² + z²) always points radially, and has a magnitude ∂f/∂r.

    To see why, use the chain rule. For example ∂f/∂x = (∂f/∂r)(∂r/∂x) = (∂f/∂r)(x/r).

    Also, √(1² + 2² + (-2)²) = 3, and you already know ∂f/∂r at r = 3.
     
  5. May 2, 2009 #4
    is the g'(4) there for a reason?
     
  6. May 2, 2009 #5

    dx

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    What was the question? I think you missed some words in your post. Unit vector in the direction of what?
     
  7. May 2, 2009 #6
    I just corrected the question
     
  8. May 2, 2009 #7

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    Ok, g'(4) is not required.

    Hint: Read my first post. Pay attention to the fact that g'(3) is positive. The magnitude of g'(3) is irrelevant.
     
  9. May 2, 2009 #8
    I believe I just the derivative of r = √(x² + y² + z²) each with respect to x, y, and z to get the gradf?
     
  10. May 2, 2009 #9

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    The definition of grad f is (∂f/∂x, ∂f/∂y, ∂f/∂z). Which direction does this point in if f is a function of only √(x² + y² + z²)?
     
  11. May 2, 2009 #10
    hmm.. it points outwards as you said right?? can I say f(x,y,z) instead of f(rho)
     
  12. May 2, 2009 #11

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    No, I didn't say it points outward. I said it points radially. It could be radially outward, or radially inward. Just do what I suggested: evaluate grad f in detail, from the definition of grad, using my previous hints.
     
  13. May 2, 2009 #12
    gradf = ∂f/∂x √(x² + y² + z²)i + ∂f/∂y √(x² + y² + z²)j + ∂f/∂z √(x² + y² + z²)k

    it's just that right?
     
    Last edited: May 2, 2009
  14. May 2, 2009 #13

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    No..

    What's the definition of grad f?
     
  15. May 2, 2009 #14
    as stated above:

    The grad (del operator) by itself is simply the derivative taken in 3 directions, x,y,z for your problem.
     
  16. May 2, 2009 #15

    dx

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    grad f is a vector field. It as components. The x-component is ∂f/∂x. The y-component is ∂f/∂y. The z-component is ∂f/∂z.

    Evaluate these derivatives using the chain rule, as I have done in post #3. What is the vector you get when you do this?
     
  17. May 2, 2009 #16
    yes I know it's a vector I just forgot to put my i,j,k above.. right?
     
  18. May 2, 2009 #17

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    Nope, still not right. Don't try to guess it. Do it in detail, as I've already shown you how to do #3.
     
  19. May 2, 2009 #18
    I really don't understand how post #3 could help in this case
     
  20. May 2, 2009 #19
    Last edited by a moderator: May 4, 2017
  21. May 2, 2009 #20

    dx

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    No. √(x² + y² + z²) is not the value of the function. Reread the question. It says that f is a function of √(x² + y² + z²), which means that the value of f depends only on √(x² + y² + z²). For example, if x = 1, y = 0, z = 0, then the value of f is g(√(x² + y² + z²)) = g(√(1² + 0² + 0²)) = g(1) which from the graph in the original question is approximately 1.4.


    f(x, y, z) = g(√(x² + y² + z²))
     
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