Calc 3 directional derivative question

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SUMMARY

The discussion focuses on calculating the directional derivative of the function f(x, y, z) = z ln(x/y) at the point (1, 1, 2) in the direction toward the point (2, 2, 1). The user correctly identified the directional vector as (1, 1, -1) and normalized it to obtain the unit vector (1/sqrt(3), 1/sqrt(3), -1/sqrt(3)). After computing the gradient of the function, the user performed the scalar product with the unit vector and arrived at a directional derivative of 0, confirming the approach was correct.

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meadow
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The question asks:
Find the directional derivative of f (x, y, z) = z ln (x/y) at (1, 1, 2) toward the point (2, 2, 1).

What I did was find the distance between the two points to be the directional vector (i+j-k) and then I took the norm of the direction vector. so my unit vector = 1/sqrt(3) * u; then I found the gradient. From there, I found the scalar product of my unit vector and the gradient to get 0. Did I approach this problem right? Does that answer seem correct to you?
 
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Seems all right to me :smile:
 

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