Solve Implicit Function Question: Derivatives, Jacobi Matrix, Diff.

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The discussion focuses on solving the implicit function problem defined by the equation f(x,y,z)=0 near the point (1,1,1). The user employs two methods to compute the derivative of the implicit function g(z) = (x,y) using the implicit function theorem. The first method involves calculating the Jacobi matrix of f and checking the invertibility of its minor, while the second method uses implicit differentiation directly. The user encounters a sign discrepancy between the two methods and seeks clarification on the geometric interpretation of the results, emphasizing the importance of understanding the orientation of solution vectors.

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littleHilbert
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Hi! I've got a question about implicit functions.

I have to solve a system f(x,y,z)=0 in the neighbourhood of (1,1,1). I have a problem computing the derivative of an implicit function (x,y)=g(z), whose existence is given by the implicit function theorem when applied to the given function f(x,y,z) which goes from R^3 to R^2. I use as it seems two equivalent standard methods.

I compute the Jacobi matrix of f and check whether the minor - a square matrix - is invertible at the given point. Then I invert it and with one more step get the derivative of g at z=1.

The other method is: I use implicit differentiation, i.e. I differentiate the system f(x,y,z)=0 directly and get differentials dg_1/dz, dg_2/dz (because g is two-component).

Now I get two different answers with either method. The difference is only the sign, but I cannot figure out WHY there is sign change! Is there anything I should pay attention to when I use these two methods? I mean they are basically the same! I checked each step in both of them. What could be the reason for the difference?
 
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well thiunk about it, you havee a map from R^3 to R^2, so you have a family of curves roughly parallel to one another in three space, eachc urve reprewenting the points collapsing under your map to one point in R^3.

and you have mnaged somehow to view the curve passing through (1,1,1) as a graph of a parametrized path, parametrized by the coordinate z. so the natural velocity vector would be the one pointing roughly in the directiion of the positive z axis.

it would seem this is the one you get by yur second method.

but you have not described your first method clearly enough to reveal why you may be getting the other direction from it. maybe you are just taking a cross product of two vectors and not choosing them so it has the same orientation as the other one.
 
maybe what i said is wrong, maybe the second way gives the projection into the x,y plane of the velocity vector to the curve?

i haven't quite understood it. but anyway if you understand the geometry of your two solution vectors you wil see why they point different directions.
 

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