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Directional derivative at a point

  1. Oct 30, 2015 #1
    1. The problem statement, all variables and given/known data
    upload_2015-10-29_21-55-55.png

    2. Relevant equations


    3. The attempt at a solution
    part a) finding partial derivatives:
    upload_2015-10-29_22-5-49.png

    and plugging in (2,0,1) into each, I get the gradient which is <0,-2,0>

    to find the directional derivative, it is the dot product of the gradient and unit vector of (3,1,1):

    upload_2015-10-29_22-1-26.png

    part b) isn't the direction of the greatest increase just the gradient, which I found in part A?

    upload_2015-10-29_22-4-50.png

    and the greatest increase at (2,0,1) is the magnitude of the gradient there, which is

    upload_2015-10-29_22-7-4.png
     
  2. jcsd
  3. Oct 30, 2015 #2

    andrewkirk

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    Yes.
    A pedant might insist that a direction should be given as a normalized vector. But we are not pedants.
     
  4. Oct 30, 2015 #3

    Ray Vickson

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    Some of your partial derivatives are incorrect (although not by very much).
    One of your partial derivatives is wrong.
     
  5. Oct 30, 2015 #4
    Also, you don't say anything about the continuity of the first partial derivatives in a neighborhood of ##(2,0,1)##. It is under that condition that the formula ##(D_{\vec v}f)(2,0,1) = (\nabla f. \vec v) (2,0,1) ## is valid.
     
  6. Oct 30, 2015 #5
    wow, I can't believe I didn't spot that +x at the end.

    it should be (-yz e^-xyz) + 1
     
  7. Oct 30, 2015 #6
    the correct answers after fixing the partial derivative with respect to x

    upload_2015-10-30_4-27-18.png
     

    Attached Files:

  8. Oct 30, 2015 #7
    do you mean like saying the domain is from negative infinity to positive infinity?
     
  9. Oct 30, 2015 #8
    Not at all and I wonder why you ask that ?
     
  10. Oct 30, 2015 #9

    Ray Vickson

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    Well, it IS possible to have a well-defined directional derivatives in some directions ##\vec{d}## at a point ##\vec{r_0}= (x_0,y_0,z_0)##, but to not have well-defined, finite partial derivatives at ##\vec{r_0}##. However, the present example is not so pathological.
     
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