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When gradient is parallel to position vector

  1. Apr 15, 2007 #1
    1. The problem statement, all variables and given/known data

    suppose that grad of f(x,y,z) is always parallel to the position vector xi+yj+zk. show that f(0,0,a)=f(0,0,-a) for any a.


    3. The attempt at a solution
    grad of f= fx(x,y,z)i+fy(x,y,z)j+fz(x,y,z)k ; then gradf (dot) pos.vector = |gradf|*|pos.vector| (since cos(teta)=1 ) ===> however ı could not do anything with this equation and ı have not any other idea...
     
  2. jcsd
  3. Apr 15, 2007 #2
    dot product, or cosine of theta

    positive if theta is 0, negative if theta is 180 or pi

    that tells your max and min slope
     
  4. Apr 15, 2007 #3
    what should ı do with the max and min slope value?
     
  5. Apr 15, 2007 #4
    what is the vector from f(0,0,a) to f(0,0,-a)?

    find the unit vector of this vector,

    and then do the dot product of the unit vector and the gradient vector.
     
  6. Apr 15, 2007 #5
    since we do not know f how can we find the vector from f(0,0,a) to f(0,0,-a)?
    please be more clear. I do not know perfectly this subject.
     
  7. Apr 15, 2007 #6
    lets call f(0,0,a) point P, and f(0,0,-a) point S

    What is vector PS?
     
  8. Apr 15, 2007 #7
    is it -2ak (ı am not sure)
     
  9. Apr 15, 2007 #8
    find the unit vector of that and do the dot product of that and the gradiant and see if you get the answer you are looking for
     
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