Directional derivative and gradient vector

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SUMMARY

The discussion focuses on calculating the directional derivative of the function f=sqrt(xyz) at the point P(2,-1,-2) in the direction of the vector v=i+2j-2k. The correct gradient vector at point P is determined to be ∇f = <1/2, -1, -1/2>, leading to a directional derivative of -1/6. The initial calculation error stemmed from an incorrect gradient vector, which was highlighted by participants in the discussion.

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kasse
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Homework Statement



Find the directional derivative of f=sqrt(xyz) at P(2,-1,-2) in the direction of v=i+2j-2k

The Attempt at a Solution



I calculate the gradient vector and obtain grad(f) at P= <1/2, 1, 1/2>

Then I find the unit vector of v, which is <1/3, 2/3, -2/3>

The directional derivative is the dot product of these two: 1/6 + 2/3 -2/6, which is 1/2. However, the answer is supposed to be -1/6 according to my textbook. Where's my mistake?
 
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Your mistake lies in your calculation of [tex]\vec\nabla f[/tex]

Its should be: [tex]\vec\nabla f = <1/2, -1, -1/2>[/tex]

Check your work for that part. Other than that everything else looks good. I get -1/6 for the directional derivative with that gradient. Good Luck!

G01
 

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