Can the Implicit Function Theorem be Applied to Solve this System of Equations?

minderbinder
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Homework Statement



Can the equation x^2 + y^2 + z^2 = 3, xy + tz = 2, xz + ty + e^t = 0 be solved for x, y, z as C^1 functions of t near (x, y, z, t) = (-1, -2, 1, 0)?

Homework Equations





The Attempt at a Solution



The mixed-partial derivatives matrix I got was:
[2x, 2y, 2z, 0]
[y, x, t, z]
[z,t,x, y+e^t]

Plugging in the numbers I get:
[-2, -4, 2, 0]
[-2, -1, 0, 1]
[1, 0, -1, -1]

I know the theorem states that when this matrix is invertible, it means explicit functions exist, however, how should I proceed as the matrix is not square?
 
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The point in question, (-1, -2, 1, 0) is not on the sphere x2 + y2 + z2 = 3. IOW, the first three coordinates of this point don't satisfy the sphere's equation. That seems problematic to me.
 
I just looked up the errata for the textbook I was using. It looks like the author made a mistake and the first equation is actually supposed to be x^2 + y^2 + z^2 = 6.

So with that, and re-reading the theorem, I think the answer should be as follows:

The mix-partial derivatives matrix of the variables I want to solve for is:
[-2, -4, 2]
[-2, -1, 0]
[1, 0, -1]

Determinant of this is non-zero. Therefore, the conclusion is that explicit functions are possible.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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