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Gradient vector as Normal vector

  1. May 21, 2008 #1
    I'm trying to understand why the gradient vector is always normal to a surface in space. My textbook describes r(t) as a curve along the surface in space. Subsequently, r'(t) is tanget to this curve and perpendicular to the gradient vector at some point P, which implies the gradient vector to be a normal vector.

    My question is, how can r(t) be a curve? I thought the position vector was a straight vector that stems from the origin of the coordinate system. My textbook shows r(t) as a curved double arrow that lies on the surface in space.

    Any help is appreciated.
     
  2. jcsd
  3. May 21, 2008 #2

    arildno

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    Suppose that a surface S can be written as a level surface of a scalar function f, that is the equation
    [tex]f(x,y,z)=K[/tex]
    describes the surface S.

    But, a surface may also be parametrized, that is the the set of points (x,y,z) in S may exactly be described by a paremetrization:
    [tex]\vec{x}=(x,y,z)=\vec{S}(u,v)[/tex]

    Therefore, we have that for all u,v, the identity holds:
    [tex]f(\vec{S}(u,v))=K[/tex]
    Differentiating with respect to u yields:
    [tex]\nabla{f}\cdot\frac{\partial\vec{S}}{\partial{u}}=0[/tex]
    But [tex]\frac{\partial\vec{S}}{\partial{u}},\frac{\partial\vec{S}}{\partial{v}}[/tex] are TANGENT vectors to surface S, hence, the gradient of f must be the normal vector to S.
     
  4. May 21, 2008 #3
    Thanks for the reply.

    I understand what you're saying about the gradient being normal to the surface in space. My textbook follows the same reasoning that you've delineated above. I just don't understand why my book shows the position vector r(t) as a curved double-arrow. Maybe it's just a bad representation of the position vector..I don't know..


    Also if you (or someone) could help me understand a related topic, I would greatly appreciate it. I've been studying Lagrange multipliers, where it's important to understand the grandient as a normal vector. In the context of functions of two variables, where the maximums and minimums of a function f(x, y) are subject to the constraint g(x, y) = 0, the extrema occur where grad(f) is parallel to grad(g) in the xy-plane, that is, where the level curves of f(x, y) = (some constant) and g(x, y) = 0 are tangent to one another. I understand how the geometry works out, but I don't understand why the two gradients being parallel implies extrema...
     
  5. May 21, 2008 #4

    HallsofIvy

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    The picture your book is showing is the range of the r(t) function. For each t, r(t) is a vector from the origin to some point (x(t),y(t)). The curve your book is showing is the set {(x(t),y(t))} for all t in the domain.
     
  6. May 21, 2008 #5
    I see, that makes much more sense, thank you :)

    Any thoughts on why parallel gradients grad(f) and grad(g) imply extrema for f(x, y)?..
     
  7. May 23, 2008 #6

    arildno

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    Sure.

    In the region where g(x,y)=0, then the function F(x,y,k)=f(x,y)+kg(x,y) coincides with f(x,y).

    Extrema for F requires [itex]\nabla{F}=0[/itex]; note in particular that we get:
    [tex]\frac{\partial{F}}{\partial{k}}=g(x,y)=0[/tex]
    That is, ALL extrema of F will lie in that region in which F coincides with f, and therefore solve the problem of extremizing f under the constraint g=0
     
  8. May 23, 2008 #7
    Thanks for the reply. I have a few questions..so F(x,y,k) is essentially a function that evaluates all of the points that satisfy g(x,y) = 0, correct? Is F(x,y,k) called the Lagrangian? And finally, how can one show that F(x,y,k) coincides with f(x,y) for points on and near the level curve?

    Thanks again.
     
    Last edited: May 23, 2008
  9. Aug 26, 2011 #8
    i'm using a textbook that says nothing at all about a unit normal vector,except that the gradient is normal to the surface at the pt given. as arildno said, can i say [itex]\partialS[/itex]/[itex]\partialx[/itex] and [itex]\partialS[/itex]/[itex]\partialy[/itex] can be considered as the components of the unit normal vector?? but wouldn't that be just those of the gradient itself? i'm trying to get back to my books after a 3 year gap without an instructor, so please bear with my doubts, as stupid as they maybe. Pls help too .
     
  10. Aug 26, 2011 #9
    sorry , my understanding of what arildno said was wrong. he actually meant to say how the gradient is normal to the surface, not anything about the unit normal vector. Can anybody tell me what the unit normal vector to a surface is ??
     
  11. Aug 26, 2011 #10

    LCKurtz

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    Typically when you take the gradient at a point you don't get a unit normal, so you need to divide it by its length to get one. One example where it is used is in calculating flux integrals. There, you need the component of the flux vector which is normal to the surface in the integral, so you dot it with the unit normal, as in

    [tex]\iint_S \vec F \cdot \hat n\, dS[/tex]
     
  12. Jan 18, 2013 #11
    Though this is now more than a year old, I would like to add an intuitive approach which I hope should make sense. The reason why the gradient vector seems hard to visualize to be a normal vector is because of our taking-it-for-granted attitude in introductory calculus that gradient and tangent are synonymous where as gradient and normal are not.
    Ok... Imagine you have a bowl which is a surface as you might expect... take any arbitrary point and try to tell me the tangent... Well, you can't because you don't know the direction... Imagine a pencil as the tangent line... It is obvious that because you can rotate the pencil in any direction while the pencil is still a tangent to the bowl at that point... this implies that the gradient can have a lot of values... As a remainder, gradient of z=f(x,y) will tell us how much is z changing with respect to both x and y. The gradient is given by dot product ∇f⋅u where u is the unit vector in any direction. From the idea of a dot product ∇f⋅u will be maximum when both vectors are parallel. Since -1≤cos([itex]\theta[/itex])≤1 then there can only be 2 directions where z is having a maximum ascent or maximum descent. From geometry we know that the shortest distance is a perpendicular between the two lines. What this means is that at any point we can have only one direction of max ascent or descent and that occurs along a perpendicular from the level curve where you are to the nearest level curve. Hence the gradient in the direction of the gradient is the shortest path to take to move from one level curve to the other which should be a perpendicular (normal) otherwise it is not the shortest distance.
    I hope I was able to convey the way I understand it...
     
  13. Jan 19, 2013 #12

    lavinia

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    The pointss in space where the function takes on a particular constant value from a surface. If you agree with this - proof? - the the function must have zero derivative along any curve on the surface (because the derivative of a constant is zero).

    This means that the inner product of the gradient with any tangent vector to the surface is zero - since any tangent vector is tangent to some curve on the surface. This follows from the Chain Rule.
     
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