Unit vector field and gradient

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  • #1
-EquinoX-
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Homework Statement



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).

Homework Equations





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?
 
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Answers and Replies

  • #2
CFDFEAGURU
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The grad (del operator) by itself is simply the derivative taken in 3 directions, x,y,z for your problem.
 
  • #3
dx
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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.
 
  • #4
-EquinoX-
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is the g'(4) there for a reason?
 
  • #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?
 
  • #6
-EquinoX-
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I just corrected the question
 
  • #7
dx
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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.
 
  • #8
-EquinoX-
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I believe I just the derivative of r = √(x² + y² + z²) each with respect to x, y, and z to get the gradf?
 
  • #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)?
 
  • #10
-EquinoX-
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hmm.. it points outwards as you said right?? can I say f(x,y,z) instead of f(rho)
 
  • #11
dx
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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.
 
  • #12
-EquinoX-
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gradf = ∂f/∂x √(x² + y² + z²)i + ∂f/∂y √(x² + y² + z²)j + ∂f/∂z √(x² + y² + z²)k

it's just that right?
 
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  • #13
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No..

What's the definition of grad f?
 
  • #14
-EquinoX-
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as stated above:

The grad (del operator) by itself is simply the derivative taken in 3 directions, x,y,z for your problem.
 
  • #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?
 
  • #16
-EquinoX-
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yes I know it's a vector I just forgot to put my i,j,k above.. right?
 
  • #17
dx
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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.
 
  • #18
-EquinoX-
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I really don't understand how post #3 could help in this case
 
  • #20
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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))
 
  • #21
-EquinoX-
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hmm.. after reading your third post again:

the gradient is just then ∂f/∂r

is it just 1/3rho?
 
  • #22
-EquinoX-
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bump!
 
  • #23
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The gradient cannot be ∂f/∂r, because ∂f/∂r is not a vector!
 
  • #24
-EquinoX-
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yes I know... but it seems like it's like that.. ∂f/∂r at point (1,2,-2) is 1/3.. so is it 1/3r
 
  • #25
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Depeds, what is r in (1/3)r ?
 
  • #26
-EquinoX-
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it's x²i + y²j + z²k, right?
 
  • #27
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Then how is (1/3)r a vector?
 
  • #28
-EquinoX-
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I think it's something like this 1/3x²i + 1/3y²j + 1/3z²k
 
  • #29
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No. The gradient at (1, 2, -2) is (1/3)(1/3i + 2/3j - 2/3k).
 
  • #30
-EquinoX-
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why is r here 1/3i + 2/3j -2/3k?
 
  • #31
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That's the answer you get if you evaluate the gradient using the definition ∇f = (∂f/∂x, ∂f/∂y, ∂f/∂z).
 
  • #32
-EquinoX-
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and so is (1/3)(1/3i + 2/3j - 2/3k) equal to [tex] \vec{F} [/tex]
 
  • #33
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Yes, at the point (1, 2, -2) .
 
  • #34
-EquinoX-
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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 [tex]\vec{F} [/tex] at (0, 3, 0)?
 
  • #35
dx
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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.
 
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