Solving for variables as functions of other variables

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SUMMARY

The discussion focuses on solving the system of equations defined by xy² + zu + v² = 3, x³z + 2y - uv = 2, and xu + yu - xyz = 1 for (x, y, z) as functions of (u, v) near the point (1, 1, 1, 1, 1). Participants emphasize the importance of calculating the Jacobian matrix and ensuring its determinant is non-zero to confirm the existence of solutions. Additionally, they explore substituting variables to simplify the equations and verify the conditions for solvability.

PREREQUISITES
  • Understanding of multivariable calculus differentiation
  • Familiarity with Jacobian matrices and their determinants
  • Knowledge of implicit function theorem
  • Experience with algebraic manipulation of equations
NEXT STEPS
  • Study the implicit function theorem in detail
  • Learn how to compute Jacobian matrices for multivariable functions
  • Practice solving systems of nonlinear equations
  • Explore numerical methods for approximating solutions to nonlinear systems
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Students and professionals in mathematics, particularly those studying calculus and differential equations, as well as anyone involved in solving complex systems of equations.

gummz
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Homework Statement



Show that the equations

xy^2+zu+v^2=3
x^3z+2y-uv=2
xu+yu-xyz=1
can be solved for (x,y,z) as functions of (u,v) near the point (x,y,z,u,v)=(1,1,1,1,1) and find dy/du at (u,v)=(1,1)

Homework Equations


Multivariable calculus differentiation

3. The Attempt at a Solution

I want to know if I'm doing this correctly (I don't think I am). What I did was find the Jacobian and plug the point into that to get a numerical matrix. Then I take determinants of matrices consisting of [x/y/z u v] (or is it [x y z]?) and show that none of them are zero.
Then I find the inverse of [u v] and do [u v]^-1 * [x y z]
I'm not very good at this.
 
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gummz said:
find the Jacobian and plug the point into that to get a numerical matrix.
What did you get ? Why do this ?

gummz said:
take determinants of matrices consisting of [x/y/z u v] (or is it [x y z]?) and show that none of them are zero.
What did you get ? Why do this ? Is that enough to show that the original three equations can be solved simultaneously for (x,y,z) as functions of (u,v) near the point (x,y,z,u,v)=(1,1,1,1,1) ? Aren't you mixing up with things that apply to linear equations only ? And why the question mark ?

gummz said:
find the inverse of [u v] and do [u v]^-1 * [x y z]
What did you get ? How do you do this at all ?

How whould you show that the original equations can be solved for (u, v) = (1,1) ?
 
gummz said:

Homework Statement



Show that the equations

xy^2+zu+v^2=3
x^3z+2y-uv=2
xu+yu-xyz=1
can be solved for (x,y,z) as functions of (u,v) near the point (x,y,z,u,v)=(1,1,1,1,1) and find dy/du at (u,v)=(1,1)
The last equation, xu+ yu- xyz= 1 is the same as (x+ y)u= xyz+ 1 so u= (xyx+ 1)/(x+ y). Putting that into the second equation gives
x^3z+ 2y- [(xyz+ 1)/(x+ y)]v= 2 or v= (x^2z+ 2y- 2)(x+y)/(xyz+1).
You should to see if, with those, xy^2+ zu+ v^2 is equal to 3. If not, the equation cannot "be solved for (x,y,z) as functions of (u,v)".

2. Homework Equations
Multivariable calculus differentiation

3. The Attempt at a Solution

I want to know if I'm doing this correctly (I don't think I am). What I did was find the Jacobian and plug the point into that to get a numerical matrix. Then I take determinants of matrices consisting of [x/y/z u v] (or is it [x y z]?) and show that none of them are zero.
Then I find the inverse of [u v] and do [u v]^-1 * [x y z]
I'm not very good at this.
 

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