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Gradient and cross product

  1. Jul 28, 2008 #1
    This is a general question. If we have a parametric equation r(u,v) and we take r_u and r_v, then take their cross product, does it give us the gradient vector? Or just a vector parallel to the gradient vector?
     
    Last edited: Jul 28, 2008
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  3. Jul 28, 2008 #2

    Dick

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    If r(u,v) has a gradient vector, then it's a scalar function. If r_x and r_y are vectors then r is a vector function. Which is it? I don't get you.
     
  4. Jul 28, 2008 #3
    I just edited my original post; I meant r_u X r_v would give a vector orthogonal to the tangent plane. And since the gradient is also (r_u, r_v), is this vector the same vector?
     
  5. Jul 28, 2008 #4

    Defennder

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    You mean to say the gradient of the surface described by a scalar function. Well there are two possible vectors for r_u X r_v, you might end up getting the inward pointing normal vector instead.
     
  6. Jul 28, 2008 #5

    Dick

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    If r is a scalar function, r_u and r_v are scalars. There is no cross product. If r(u,v) is a vector function then r_u x r_v is a normal to the surface defined by r(u,v), but then there is no gradient. Someone is confused here, and I don't think it's me.
     
  7. Jul 29, 2008 #6

    HallsofIvy

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    You are using the terminology incorrectly. A function has a gradient, not a surface.

    If a surface is given by f(x,y,z)= constant, the grad f is perpendicular to the surface.
    If that surface is written as a position vector [itex]\vec{r}(u,v)[/itex], then [itex]\vec{r}_u\times\vec{r}_v[/itex] is also perpendicular to the surface. Your question is whether those two vector must also have the same length.

    The answer is "no" because the gradient of f is a specific vector while different parameterizations of the surface f(x,y,z)= constant will give vectors of different length.
     
  8. Jul 29, 2008 #7

    Why would a surface not have a gradient?

    And yes I am confused, which is why I'm asking questions, but I appreciate your input.
     
  9. Jul 29, 2008 #8

    Dick

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    The big confusion here is that you have two different representations of the surface, as a position vector [itex]\vec{r}(u,v)[/itex] and as a level surface by R(x,y,z)=C. Those two r's are completely different objects. You are confusing me (at least) by labeling them the same and using them interchangeably. As Halls said, if r and R do happen to represent the same surface, then, yes, the cross product and the gradient are parallel, not necessarily equal.
     
  10. Jul 29, 2008 #9

    D H

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    Because a surface isn't a function. Think of it this way: The unit 2-sphere is the locus of points in R3 that satisfy [itex]x^2+y^2+z^2-1=0[/itex]. Note well: [itex]2x^2+2y^2+2z^2-2=0[/itex] also describes the unit 2-sphere. The gradients of the functions [itex]x^2+y^2+z^2-1[/itex] and [itex]2x^2+2y^2+2z^2-2[/itex] are obviously different.
     
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