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Analysis of a vector field

  1. Apr 24, 2015 #1
    Oops. I just realized that this is the physics homework forum... This is actually calculus homework...

    1. The problem statement, all variables and given/known data


    b.png a.png

    2. Relevant equations
    n/a

    3. The attempt at a solution
    I read, analyzed and reread the text, but I am still confused.

    1) How was the position vector ##<x,y>## determined?

    2) How is it related to the circle with centered at the origin?

    3) We are trying to prove that ##F(x,y)## is tangent to a circle with center at the origin. Why then, are we taking a dot product? The dot product shows that ##F(x.y)## is perpendicular to the position vector ##<x,y>##. But isn't being perpendicular the opposite of being tangent?

    4) The magnitude of the vector ##F(x,y)## is equal to the radius of the circle. If the circle wasn't centered at the origin, I believe that it will still be true. What is the significance of the magnitude in other vector fields, ones that don"t involve circles?

    Thank you all so much!
     
    Last edited: Apr 24, 2015
  2. jcsd
  3. Apr 24, 2015 #2

    Orodruin

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    The position vector is simply the vector from the origin to a given point. A circle centered on the origin fulfills ##F(\vec x) = \vec x^2 = R^2## and is therefore a level surface of ##F##. The normal of a level surface is the gradient of the function and in this case it is proportional to ##\vec x##.

    A tangent to a surface is orthogonal to its normal, so a tangent to the circle is necessarily also orthogonal to ##\vec x##.
     
  4. Apr 24, 2015 #3
    While your answer has helped me understand the answer to my first question, it has also raised new several new questions. For the circle centered in the origin, how was ##F(\vec x) = \vec x^2 = R^2## determined? Does ##R^2## mean two dimensional? A level surface is a projection (or shadow) of a 3D object onto the xy plane. The circle is not 3D. I agree that the normal to a level surface is the gradient, but I don't see how it's proportional to ##\vec x ##

    Thank you Orodriun.

     
    Last edited: Apr 24, 2015
  5. Apr 24, 2015 #4

    Orodruin

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    No, ##R## is the radius of the circle.

    No, a level surface is the set of points where a function has a given value. It has a dimension of one less than the full space. For this reason, a level surface in two dimensions is a one dimensional curve.

    Did you try taking the gradient of the function ##F(\vec x) = \vec x^2##?
     
  6. Apr 24, 2015 #5
    Is ##F(\vec x) = \vec x^2 \hat i = R^2 ## ?
     
  7. Apr 24, 2015 #6

    Orodruin

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    No, ##\vec x^2 = \vec x \cdot \vec x = x^2 + y^2##, which is a scalar quantity. A vector can never be equal to a scalar. This is just the equation for a circle, ##x^2 + y^2 = R^2##.
     
  8. Apr 24, 2015 #7
    Is the ##\vec x## the same as the ##x## two original equations? The one found in ##F(x,y) = -y \hat + x \hat j ## and/or ##x = x \hat i + y \hat j ##
     
  9. Apr 24, 2015 #8

    Orodruin

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    ##\vec x## is the position vector ##x\hat i + y \hat j##. The vector arrow is important.
     
  10. Apr 24, 2015 #9
    Dear @Calpalned,
    I would recommend you re-read everything you have on vectors to clear up some basic concepts before proceeding in your studies in Calculus III.
    Otherwise I fear you will face a lot of confusion.
     
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