Gradient delta f of f= z^-1 * (sqrt((9x^2*y^2))

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SUMMARY

The discussion focuses on calculating the gradient delta f of the function f = Z^-1 * sqrt(9x^2*y^2) at the point (1, 4, 10). The partial derivatives were computed as fx = 0.18, fy = 0.08, and fz = -0.05. The resulting gradient vector ∇f at the specified point is (0.18, 0.08, -0.05). The conversation also clarifies the distinction between finding the gradient vector and evaluating the function at perturbed coordinates.

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Unemployed
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Homework Statement



Find delta f of f=Z^-1 * Sqrt(9x^2*y^2)
at point (1,4,10)

Homework Equations


f =f+(fx*delta x )+(fydelta y)+(fz*delta z)

The Attempt at a Solution


fx = 9*2x/(z*(2*sqrt(9x^2*y^2)) =.18 plugging in (1,4,10)
fy=2y/(z*(2*sqrt(9x^2*y^2)) =.08
fz=(sqrt(9x^2*y^2) ) *-z^-2=-.05

f=.5

Where do you get the delta x, y, and z from. I plugged in 1,4,10 in these equations?
 
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Hi Unemployed! :smile:

(have a grad: ∇ and a delta: ∆ and a square-root: √ and try using the X2 and X2 icons just above the Reply box :wink:)

I'm not sure what the question is :redface:

are you being asked to find the vector ∇f at (1,4,10),

or to find the number f(1+∆x,4+∆y,10+∆z) ? :confused:

If it's the former, then (assuming your figures are correct) ∇f is simply the vector (0.18,0.08,-0.05) :wink:
 

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