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Partial Derivative Signs Through Level Curves

  1. Sep 30, 2013 #1
    1. The problem statement, all variables and given/known data

    Question 2 from http://math.berkeley.edu/~mcivor/math53su11/solutions/hw6solution.pdf here.

    I do not understand b) and e). How do I think of the slope with respect to y?

    2. Relevant equations



    3. The attempt at a solution

    I do know that the partial derivatives are the slopes of the tanget lines. I do know that fx < 0 as the slope is negative there (at point P), but I do not understand how the slop in terms of y looks.
     
  2. jcsd
  3. Sep 30, 2013 #2

    Office_Shredder

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    When you say "the slope", it's not clear what you mean because there is a slope in every direction from P for this function. If you did the correct procedure to figure out the sign of fx you should be able to do the exact same thing to figure out fy - why don't you elaborate on how you did fx?
     
  4. Sep 30, 2013 #3
    Well, the way I solved for fx was as follows: draw a cross section of the function's 2D hill image (similar to y=1/x). Then place my point p. As the line tangent through P has a negative slope, fx is negative. Actually, I would appreciate help with getting fxx as well - I don't understand how "b/c the x-slope becomes less negative as we move right" correlates with fxx. If fx is the tangent to the curve, then what is fxx geometrically? Answering that would help me a lot.

    EDIT: Just googled the second derivative. Now I remember that it tells us the concavity of the quadratic. In this case, is that why fxx is positive? Is that the entire solution to part c) ?

    Thanks.
     
    Last edited: Sep 30, 2013
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