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Question from reading Div Grad Curl and All That

  1. Mar 17, 2012 #1
    on pages 14-15, in deriving the normal vector to a surface, they use a plane to cut the surface (the plane is parallel to the xz plane) then use the curve 'c' in the xz plane (this curve being where the plane intersects the surface), draw a tangent vector 'u' and want to use the components of 'u'. Now, they use an arbitrary length called 'u sub x' in the x direction (this again is in the xz plane) but WHY is the component in the z direction designated 'partial f/partial x times u sub x'? My question is NOT "why is the derivative of z with respect to x replaced with the partial derivative of f with respect to x", that I understand. But why in the first place is the change in the z direction a derivative times change in x?
  2. jcsd
  3. Mar 17, 2012 #2


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    "on pages 14-15" of what???

    If ux is a vector in the (positive) x direction, then "the component in the z direction" cannot be "partial f/partial x time ux" because that is a vector in the x direction, not the z direction. Are you sure you are quoting correctly?
  4. Mar 17, 2012 #3
    Pgaes 14-15 of 'div grad curl and all that', second edition however.
    Let me rephrase:
    there is a surface S in 3-d (x, y, z). This is cut by a plan parallel to the xz plane. This intersection of the plane and surface creates a curve C.
    Now there is a picture showing the x and z axes with this curve C, and a tangent vector called U. U is decomposed into U sub X and U sub Z HOWEVER, U sub Z is 'equal' here to as 'partial f/ partial x' times U sub X.
    So my question is why is U sub Z equal to partial f/partial x time U sub X.
  5. Mar 17, 2012 #4
    Oh, and f refers to the 3-d surface: z=f(x,y). The second picture I referred to is a 2-d picture with the x and z axes, just to clarify.
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