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A contradiction? - Calculus III Question

  1. Oct 16, 2011 #1
    yz = ln(x+z)
    So I'm trying to find the tangent plane to the surface at a particular point (x0,y0,z0).
    Here's the general formula:
    Fx(x0,y0,z0)(x-x0) + Fy(x0,y0,z0)(y-y0) + Fz(x0,y0,z0)(z-z0) = 0,
    where Fx, Fy, and Fz are the partial derivatives of the below F(x,y,z):

    1. F(x,y,z) = ln(x+z) - yz
    2. F(x,y,z) = yz - ln(x+z)

    When I take the partial derivatives Fx,Fy,Fz for both of these functions, they turn out to be different:

    For (1.), I get:
    Fx = 1/(x+z)
    Fy = -z
    Fz = 1/(x+z) - y

    For (2.), I get:
    Fx = -1/(x+z)
    Fy = z
    Fz = y - 1/(x+z)
    , which all have the opposite signs of (1.)

    So my question is: how do I know which F(x,y,z) to use when trying to find the tangent plane through a particular point?
     
  2. jcsd
  3. Oct 16, 2011 #2

    SammyS

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    Your first version of F(x,y,z) is the negative of your second version, so of course that gives you a different result.

    Added in Edit:

    Of course I should have said, ... a different result for the gradient. The result for the overall problem should be the same for both versions of F(x,y,z).
     
    Last edited: Oct 16, 2011
  4. Oct 16, 2011 #3

    Ray Vickson

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    It doesn't matter: an equation of the form [itex] \sum_{i} a_i x_i = 0[/itex] is the same as [itex] \sum_{i} (-a_i) x_i = 0.[/itex]

    RGV
     
  5. Oct 16, 2011 #4
    Yeah, that's true, which is my point.
    Differentiating
    F(x,y,z) = ln(x+z) - yz and F(x,y,z) = yz - ln(x+z) should give me the same result for the tangent plane and normal line of the surface at a particular point, but it doesn't. It gives me the negative of what it should be (by should be, I mean the correct answer to the homework problem).

    The normal line has parametric equations:
    x = x0 + Fxt
    y = y0 + Fyt
    z = z0 + Fzt, where <Fx,Fy,Fz> is the gradient.
    So if both functions give me a negative result of the gradient, the parametric equations are also negative of each other for both situations, and there can be only one answer.
     
  6. Oct 16, 2011 #5

    SammyS

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    It may look like a different line, but any point on one of the lines is also on the other. So, both forms define the same line.
     
  7. Oct 16, 2011 #6
    But don't they point in opposite directions?
     
  8. Oct 16, 2011 #7
    Then the lines only differ by a constant -t. If you remember the equation of a line.
     
  9. Oct 16, 2011 #8

    SammyS

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    Vectors point.

    A line goes to infinity in both directions.
     
    Last edited: Oct 16, 2011
  10. Oct 16, 2011 #9

    Ray Vickson

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    Vectors point in directions, but lines do not. Anyway, you are getting a *plane* and that does not point in any direction at all (at least as normally understood by the word "point"). Locally, the floor you are standing on is the plane perpendicular to a vertical line, and is the same plane whether you regard it as perpendicular to an upward-pointing vector or a downward-pointing one.

    RGV
     
  11. Oct 16, 2011 #10
    It makes sense, but when the normal line to a plane is mentioned, I automatically think of a vector that points in a specific direction and not a line that goes on forever.
    So in summary, both answers should be correct? My online homework only accepts one of the answers as correct, for both the plane and the line.
     
  12. Oct 17, 2011 #11

    SammyS

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    It's ridiculous for the online homework to take only one specific answer for either of these. There are an infinite number of ways to write a correct equation for each of the line & the plane.
     
  13. Oct 17, 2011 #12
    K, thanks. I guess it was just programmed to take that one answer.
     
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