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Why is the gradient vector normal to the level surface?

  1. Sep 6, 2011 #1
    In functions involving only two variables the gradient is supposed to be the instantaneous rate of change of one variable with respect to the other and this is usually TANGENT to the curve. So then why is the gradient NORMAL to the curve at that point, since it is supposed to represent the direction of maximum increase?

    Same thing for 3 dimensions. Shouldn't the direction of maximum increase be some vector tangent to the level surface at a point rather than orthogonal? Yet the gradient vector is the direction of maximum increase and is orthogonal. Also I would like to know what exactly the gradient vector represents, since I have usually understood derivatives to be of one variable with respect to another eg, "dy/dx". Than what exactly does "d/dx" or "d/dy" represent (ie. what is their physical/geometric interpretation, how can I visualize this?). And by extension, what does a gradient vector in 3d represent, an increase in exactly WHAT? I know that when the equation of a plane is given in the form:

    z = x + y

    The gradient vector has no z component and represents the direction in the xy plane corresponding to maximum increase in "z". But what about when an equation is given in the form

    K = x + y + z

    Then the gradient vector is calculated in terms of x, y and z, but what does it now represent? Since K is a constant and neither increase nor decreases?

    Thanks
     
  2. jcsd
  3. Sep 6, 2011 #2

    HallsofIvy

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    No, you are misunderstanding what function we are talking about. You seem to be thinking of f as changing value on a surface. f is a function of three variables, NOT restricted to that surface. A "level surface" (or "level curve" in two dimensions) for function f(x,y,z) is a surface upon which f is constant. The derivative of f in a direction tangent to a level surface would have to be 0, not a maximum. Since the derivative of f in a direction pf unit vector [itex]\vec{v}[/itex] tangent to the level surface is [itex]\nabla f\cdot\vec{v}= 0[/itex], and [itex]\vec{v}[/itex] is not 0, it follows that [itex]\nabla f[/itex] is perpendicular to [itex]\vec{v}[/itex]. Since it is perpendicular to every vector tangent to the surface, it is perpendicular to the surface.

    [itex]\nabla f[/itex] is the rate of change of f relative to a change of position in two or three dimensions. That change of position is a vector, not a number.

    The gradient vector of what function? Are you thinking of this as z= f(x,y)= x+ y or as f(z,y,z)= x+ y- z= 0?
    The gradient of the first is [itex]\nabla z= \nabla f= \vec{i}+ \vec{j}[/itex] which has "no z-component" because you really talking about a function of x and y only. If you are thinking of z= x+ y as a level surface of f(x,y,z)= x+ y- z, then [itex]\nabla f= \vec{i}+ \vec{j}- \vec{k}[/itex] which certainly does have a z compoent.

    Again, you are confused as to what your function is. If you are thinking of z as a function of x and y, z= x+ y+ k so the gradient is [itex]\nabla z= \vec{i}+ \vec{j}[/itex] which has no z component just as before. But if you are thinking of this as a level surface of the function f(x,y,z)= x+ y+ z, then [itex]\nabla f= \vec{i}+ \vec{j}+ \vec{k}[/itex].

    You are thinking there is a difference between the two because you changed your interpretation of the function. In the first you were thinking of z= x+ y with z a function of the two variables x and y and in the second you were thinking of k= x+ y+ z as a level surface of the function f(x,y,z)= x+ y+ z, of the three variable x, y, and z.

     
  4. Sep 6, 2011 #3
    Ok, I understand that much now.

    I was talking about the gradient vector of, f(x,y,z)= x+ y- z. Because this equation still represents a plane so does that mean that shifting the plane in the direction of the gradient will yield the largest increase in f ?

    Does this mean that we can rewrite f(x,y,z)= x+ y+ z as:

    K = x + y + z + q where say, q = f(x, y, z) and K = some arbitrary constant?
     
  5. Sep 6, 2011 #4

    HallsofIvy

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    Well, no,it would be x+ y+ z- q= 0.
     
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