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Grad vector

  1. Mar 20, 2007 #1
    If you are to find the normal vector at a point P of a 3D curve, you have to fina the gradient vector.

    But if you are to find the direction in which the curve is steepest, you also have to find the gradient vector.

    Is this wrong? I find it ambiguous.
  2. jcsd
  3. Mar 20, 2007 #2


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    "Direction in which a curve is steepest"???! You SHOULD find it wrong and ambiguous. It is. The normal of a 3D curve has nothing to do with a gradient. You may be thinking of 2D level curves (like f(x,y)=C).
    Last edited: Mar 20, 2007
  4. Mar 20, 2007 #3
    There are an infinity of vector normal to a curve in 3D.
    Two particular kind of normal vector are useful.
    The first, is perpendicular to the plane tangent to the curve in that point.
    The second is perpendicular to the tangent vector but inside the tangent plane (like if the curve was 2D on a short scale).
  5. Mar 20, 2007 #4
    First, check out the interactive illustration at:

    http://www.math.umn.edu/~rogness/multivar/dirderiv.shtml" [Broken]

    If you drag the graph around you can change your view of the contour. Dragging the dot changes the facing direction. Notice that facing eastward shows a "level" (slope = 0) tangent vector at the green dot. If you started moving from P (the dot) you would be neither going up or down. If you turn 90 degrees (i.e. perpendicular or "normal") to the north from this level possibility, you will be facing the steepest ascent - the gradient. Turn to the south and your facing the steepest down hill decent. Of course, you could imagine this same possiblity at every point on the mountain. The important point is that at every point, the gradient is perpendicular to the level path directions.

    If you think of a contour map of a hill, each contour line represents a path that is level - no slope - for various elevations. A single contour line would be constructed by all the adjacent points around the hill where the tangent vector is level (referring to the above web site illustration). You could call it a "level tangent path". So, as you move from point to point around this contour line you would neither be going up the hill or down along the path. But, at each point on the path there is a perpendicular direction to the level tangent path that is the steepest path you could chosse at that point. A gradient map of the hill, superimposed over the contour map, would have representitive arrows (gradient vectors), pointing toward the steepest slope at their head points and each vector would be seen as perpendicular (normal) to the level tangent path contour lines.

    Here's an example of such a superimposed contour and gradient map. Darker shades represent downhill, lighter uphill.

    http://home.comcast.net/~perion_666/Files/index_149.gif" [Broken]

    So, you can see how the gradient represents the steepest rate of increase of some function for each of the points where the function is defined, but each gradient vector is also perpendicular to the level tangent directions (like the eastward direction in the illustration) at a point. Note that these gradient vectors are not "normal" to the surface of the hill - as in pointing up into the sky. They are normal to level directions.
    Last edited by a moderator: May 2, 2017
  6. Mar 20, 2007 #5


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    It's ambiguous because you are dealing with two differnt kinds of problems.

    Also, you first statement is slightly wrong- a "3D curve" doesn't have a normal vector, it has a normal plane. What is true is that a surface f(x,y,z)= constant has a normal vector.

    Yes, it is true that if you calculate grad f, you get the direction in which it increases fastest. That is, of course, perpendicular to the directions in which it does not change all- the directions in which it is donstant.

    It is perfectly reasonable that grad f, the vector pointing in the direction in which f(x,y,z) increases fastest, is perpendicular to f(x,y,z)= constant.
  7. Mar 20, 2007 #6
    I really should delete my first post because rather than deal with the confusion it actually illustrates. So, I'll leave it.

    Sometimes we confuse the gradient when dealing with a scalar function of two variables, f(x,y). Even though we might decide to assign the values that f produces to distances in the z direction from the x-y plane and thus produce a pretty surface, it is incorrect to think that the gradient vectors for f are somehow attached to that surface and point "out into space" or are somehow tangent to that surface. Grad f produces a vector function of x and y. Any vector that graf f produces relates entirely to the x-y plane - there is no z direction for grad f, just as there is no z direction for f. The gradient for some point (x,y) points in a direction entirely in the x-y plane and indicates what direction and magnitude that f increases maximally. It's perpendicular to the direction in which f doesn't vary in moving some infinitesimal distance.

    The same is true for a 3 dimensional scalar field. If T(x,y,z) is a scalar field function that gives us the temperature at any point (x,y,z), grad T doesn't point out of 3-space into some hyper-space. It points in the direction in which the temperature rise will be greatest.
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