Gradient vector

  • Thread starter kasse
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  • #1
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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)?
 
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  • #2
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what did you get for your gradient??
 
  • #3
arildno
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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..
 
  • #4
HallsofIvy
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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)?
?? 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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