Define z as a function of x and y

  • Thread starter Thread starter MeMoses
  • Start date Start date
  • Tags Tags
    Function
Click For Summary

Homework Help Overview

The problem involves defining z as a function of x and y through the equations x=uv, y=u+v, and z=u^2-v^2. The original poster seeks to find the partial derivative of u with respect to x.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the process of expressing z in terms of x and y, with the original poster questioning how to properly represent z as a function of x and y. Clarifications are sought regarding the derivative and the relationships between the variables.

Discussion Status

The discussion is ongoing, with some participants providing insights into the relationships between u, v, x, and y. There is a divergence in understanding the relevance of z to the problem, indicating multiple interpretations are being explored.

Contextual Notes

There is a mention of the chain rule and the need to solve for u and v before substituting into z, highlighting potential complexities in the relationships among the variables.

MeMoses
Messages
127
Reaction score
0

Homework Statement


The equations x=uv, y=u+v and z=u^2-v^2 define z as a function of x and y. Find [tex]\frac{\partial u}{\partial x}[/tex]

Homework Equations



Chain rule

The Attempt at a Solution


Once I get z as a function of x and y the solution seems pretty straight forward, but how exactly is that done. Would it be along the lines of z(x(u,v),y(u,v))? That doesn't seem right though. Thanks for any help in advance.
 
Physics news on Phys.org
Did you mean to say you're trying to find [itex]\partial u/\partial x[/itex] and not [itex]\partial z/\partial x[/itex]?
 
No, I meant what I stated. I realized you solve for u and v and then plug into z=u^2-v^2. You get u=x/v and substitute that into y=u+v and then multiply both sides by v to get a quadratic equation. You do that for both u and v and plug them into z and its smooth sailing from there
 
Frankly, I don't see why z has anything to do with the problem then.
 

Similar threads

  • · Replies 27 ·
Replies
27
Views
2K
Replies
8
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
Replies
20
Views
5K
  • · Replies 19 ·
Replies
19
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 16 ·
Replies
16
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K