Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A failing in 3d gradients?

  1. Sep 8, 2014 #1
    So my question is regarding the gradient of a function. Suppose we have a function expressed In cylindrical coordinates. Its expressed as z=rcos2@

    I expressed the equation in cylindrical, but for the sake of my logic I'll now talk about it in cartesian. It appears that dz/dx and dz/dy at the origin are both equal to one, and so the gradient would imply (back into cylindrical coordinates) dz\Dr at the 45 degree angle is not what the graph implies, What am I missing here?

    It seems to me that d(dz/ dr) \d@ is still valid as a gradient?

    Isn't the argument of the gradient in cartesian that dz/dy should not change much over small changes in x, and yet this is not true in this example.
     
    Last edited: Sep 8, 2014
  2. jcsd
  3. Sep 12, 2014 #2

    FactChecker

    User Avatar
    Science Advisor
    Gold Member

  4. Sep 12, 2014 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    The gradient, whether in one or several variables, is a derivative, not a second derivative, like your "d(dz/dr)/d@".
     
  5. Sep 12, 2014 #4

    WWGD

    User Avatar
    Science Advisor
    Gold Member

    If I understood correctly, this is true only when the gradient itself is continuous.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: A failing in 3d gradients?
  1. Gradient of a vector (Replies: 4)

  2. Gradient of a surface (Replies: 2)

  3. Gradient of r (Replies: 1)

  4. Gradient Confusion (Replies: 22)

  5. Gradient Intuition (Replies: 7)

Loading...