Find the critical points of (x^2+y^2-4)(x+y) and their nature

[rbnvrw]

Hi,

I am studying for my Analysis final and came across this problem I just can't get my head around:

Homework Statement

Find the critical points of f(x,y) = (x^2+y^2-4)(x+y) and their nature.

Homework Equations

\vec{\nabla} f(x,y) = \vec{0}

The Attempt at a Solution

\frac{\partial f(x,y)}{\partial x} = 3x^2+2xy+y^2-4 = 0
\frac{\partial f(x,y)}{\partial y} = 3y^2+2xy+x^2-4 = 0
These are both ellipses with center (0,0) and angles \frac{5\pi}{8} and \frac{\pi}{8} respectively.
According to Wolfram Alpha, there are four intersection points. (Click here to view) However, I need to solve this problem analytically and I wish to understand how it is done.
I have tried to eliminate a variable from the equations, but it is not possible due to the 2xy term. Another solution would be to write the ellipses in a parametric form, but this I feel would be overly complicated.

Could anyone please shed some light on this?
Thanks in advance!

 

[HallsofIvy]

Hi,

I am studying for my Analysis final and came across this problem I just can't get my head around:

Homework Statement

Find the critical points of f(x,y) = (x^2+y^2-4)(x+y) and their nature.

Homework Equations

\vec{\nabla} f(x,y) = \vec{0}

The Attempt at a Solution

\frac{\partial f(x,y)}{\partial x} = 3x^2+2xy+y^2-4 = 0
\frac{\partial f(x,y)}{\partial y} = 3y^2+2xy+x^2-4 = 0

The first thing I would do is subtract one equation from the other to get
3(x^2- y^2)+ (y^2- x^2)= 2(x^2- y^2)= 0
so that either y= x or y= -x.

These are both ellipses with center (0,0) and angles \frac{5\pi}{8} and \frac{\pi}{8} respectively.
According to Wolfram Alpha, there are four intersection points. (Click here to view) However, I need to solve this problem analytically and I wish to understand how it is done.
I have tried to eliminate a variable from the equations, but it is not possible due to the 2xy term. Another solution would be to write the ellipses in a parametric form, but this I feel would be overly complicated.

Could anyone please shed some light on this?
Thanks in advance!

 

[ehild]

Subtract the second equation from the first one and factorise.

ehild

 

[rbnvrw]

Thanks for your advice.
However, this doesn't give me the four points of intersection. If I take the solutions y=\pm x, I get the solutions (0,0) \wedge (\pm \sqrt{2},\pm \sqrt{2}).
However, the points (\pm \sqrt{\frac{2}{3}},\pm \sqrt{\frac{2}{3}}) are also solutions of the equation. And the point (0,0) is not a solution at all.
How would I get the other two points?
Thanks in advance!

 

[ehild]

Your values are not correct. Give the solutions as pairs.
(√2, √2) is not a solution, but (√2, -√2) is.
(0,0) is not a solution. How did you get it? If you substitute x=y into one of the original equation, you will get the pair with √(2/3)

ehild

 

[rbnvrw]

I got this:
y = \pm x
(x^2+x^2-4)(x+x)=0
x(x^2-2)=0
x = 0 \vee x=\pm \sqrt{2}
According to the solutions manual, which only states the values, not the method, the solutions are:
x = \pm \sqrt{2} \vee x = \pm \sqrt{\frac{2}{3}}

 

[rbnvrw]

Oops... I see it now. I used f(x,y) instead of the partial derivatives.
I will try again!

 

[rbnvrw]

Yes! I got it!
Thank you for your help, I appreciate it! :D

 

[ehild]

You see, it was worth to write it again...

ehild