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Gradient vector

  1. Feb 20, 2007 #1
    I want to find the gradient vector of f(x,y,z)=2*sqrt(xyz) at the point ((3,-4,-3).

    I find the partials and set in for the x-, y-, and z-values, and find the grad. vector (2, (1,5), 2). The right solution is (2, (-1,5), -2), so I have obviously made a mistake with the sqrt. How do I know whether it's + or -?

    And another question: Have I used the chain rule correctly when I have calculated the gradient of f(x,y,z) at (-5,1,3) to be (160,-240,400)?
    Last edited: Feb 20, 2007
  2. jcsd
  3. Feb 21, 2007 #2
    what did you get for your gradient??
  4. Feb 21, 2007 #3


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    Call your variables [itex]x_{i}, i=1,2,3[/tex]
    Then, [itex]f(x_{1},x_{2},x_{3})=2\sqrt{x_{1}x_{2}x_{3}}[/itex]
    And the partials are:
    [tex]\frac{\partial{f}}{\partial{x}_{i}}=\frac{2x_{j}x_{k}}{f}, i,j,k=1,2,3,j\neq{i}\neq{k},j\neq{k}[/tex]
    Mind your signs..
  5. Feb 21, 2007 #4


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    ?? You know whether it is + or - (- is correct here) because that's what you get when you do the arithmetic!

    What did you get for the gradient vector of [itex]f(x,y,z)= 2\sqrt{xyz}[/itex]?
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