# Critical points of multivariable equations

1. Jun 7, 2012

### ihateblackbox

1. The problem statement, all variables and given/known data

Find all critical points of

u(x,y)=(x-y)(x^2+y^2-1)

2. Relevant equations

-

3. The attempt at a solution

Partial differentials:

ux=3x^2+y^2-1-2xy=0

uy=-x^2-3y^2+2xy+1=0

I know the critical points are the solutions to the above two equations. But how do I solve these equations? Or am I going about it the wrong way?

2. Jun 7, 2012

### Vorde

Well I have no experience but by reasoning it seems that the critical points should be when both of those partial derivatives equal zero. So it seems that if you solved each equation for y (or x, I guess), then the intersection of the plots of the two functions should be the points you are looking for.

You don't even need to solve it for a single variable, but then you would need an implicit grapher like wolfram alpha.

Last edited: Jun 7, 2012
3. Jun 7, 2012

### algebrat

Might want to be a little clever, try adding the two equations... I get y=±x. A little more work should tell you more.

4. Jun 7, 2012

### Vorde

It seems like adding the two equations would just give you the sum of the partial derivatives, which will give you zero points both when there is a critical point and when each partial derivative is the opposite of the other.

5. Jun 8, 2012

### algebrat

The sum of the equations gives you:

$2x^2-2y^2=0.$

6. Jun 8, 2012

### Ray Vickson

A standard trick whenever you see an equation with terms like y^2 +- 2xy in it is to look for something of the form (x+-y)^2. We have ux = 2x^2 -1 + [x^2 + y^2 - 2xy] = 2x^2-1+t, where t = (x-y)^2. Also, uy = -2y^2 +1 - x^2 - y^2 + 2xy = -2y^2 + 1 - t, so we have x^2 = 1/2 - t/2 and y^2 = 1/2 - t/2. We can get x and y in terms of t (although there will be multiple roots), then can express ux and uy in terms of t, so can solve for t. Then we can get x and y.

In this way you will get 4 stationary points.

RGV