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Partial derivatives & gradient

  1. Oct 31, 2007 #1
    [​IMG]

    Let f(x,y)=depth.
    What I've seen in the model solutions is that they used the estimate that
    the partial dervaitve of f with respect to x evaluate at (0,0) is equal to [f(100,0) - f(0,0)] / 100 = 1/4,
    & the partial dervaitve of f with respect to y evaluate at (0,0) is equal to [f(0,100) - f(0,0)] / 100 =-1/2

    Gradient of f at (0,0) = (1/4, -1/2)

    So we suggest going in the direction (-1/4, 1/2) [answer] which is in the direction opposite to the gradient.

    ======================
    Now there are three subtle points that I don't understand:

    1. WHY can you use the estimate that partial dervaitve of f with respect to x evaluate at (0,0) is equal to [f(100,0) - f(0,0)] / 100? This is like taking the secant line to be equal to the tangent line, but intution tells me that this estimate can be way way off...

    2. Is there any possible way to find the formula for the function of the dome in quesiton (half-sphere)?

    3. (-1/4, 1/2) is in the direction opposite to the gradient, why does this give the answer? Pointing in the direction of maximum rate of decrease of f doesn't necessary mean that it's pointing to the absolute minimum of f (i.e. top of dome), right? If I am right, then (-1/4, 1/2) can't be correct...


    Thanks for explaining! I really appreciate your help!
     
  2. jcsd
  3. Oct 31, 2007 #2

    HallsofIvy

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    Yes, it might be- but do you have any more accurate information? That's the best you can do with the information given.

    You know that the general equation of a sphere is [itex](x-x_0)^2+ (y-y_0)^2+ (z- z_0)^3= R^2[/itex]. You are given (x,y,z) for three points. Is that enough to determine the four unknown constants, [itex]x_0[/itex], [itex]y_0[/itex], [itex]z_0[/itex], and R?

    No one said it was "correct" (would give the minimum depth)- it's the best you can do given the information. Derivatives can only give "local" information (what is true close to the given point) and can not directly give the "global" maximum or minimum. However, "following" the direction of the gradient at different points will, eventually, lead to a local maximum.


     
    Last edited: Nov 1, 2007
  4. Nov 1, 2007 #3
    2. So is it true that...
    we need 4 points to determine a unique sphere, and that 3 points can NEVER determine a unique sphere?


    3. "However, "following" the direction of the gradient at different points will, eventually, lead to a local maximum" <-----I don't get this point. In our case, it's a dome, so there is one and only one local/absolute maximum, and therefore the local max is the absolute max...same thing...


    Thanks a lot!
     
  5. Nov 1, 2007 #4

    HallsofIvy

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    Yes, that's true. Just as you need 3 points to determine a unique circle in the plane, you need 4 points to determine a unique sphere in three dimensions.


    Yes, in this particular case, any "local" maximum is a "global" maximum. I wasn't referring to this situation only. You can "follow" the gradient by: choose a point. Calculate the gradient at that point. Move in the direction of the gradient a short distance. At this new point, repeat. Eventually (it can be implemented on a computer but tends to be slow) this will lead you to a local maximum but not necessarily a global maximum.
     
  6. Nov 1, 2007 #5
    Thanks for explaning! I am definitely understanding it better now.
     
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