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

  • Thread starter rbnvrw
  • Start date
  • #1
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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 [itex]f(x,y) = (x^2+y^2-4)(x+y)[/itex] and their nature.

Homework Equations


[itex]\vec{\nabla} f(x,y) = \vec{0}[/itex]

The Attempt at a Solution


[itex]\frac{\partial f(x,y)}{\partial x} = 3x^2+2xy+y^2-4 = 0[/itex]
[itex]\frac{\partial f(x,y)}{\partial y} = 3y^2+2xy+x^2-4 = 0[/itex]
These are both ellipses with center (0,0) and angles [itex]\frac{5\pi}{8}[/itex] and [itex]\frac{\pi}{8}[/itex] 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 [itex]2xy[/itex] 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!
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
41,794
925
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 [itex]f(x,y) = (x^2+y^2-4)(x+y)[/itex] and their nature.

Homework Equations


[itex]\vec{\nabla} f(x,y) = \vec{0}[/itex]

The Attempt at a Solution


[itex]\frac{\partial f(x,y)}{\partial x} = 3x^2+2xy+y^2-4 = 0[/itex]
[itex]\frac{\partial f(x,y)}{\partial y} = 3y^2+2xy+x^2-4 = 0[/itex]
The first thing I would do is subtract one equation from the other to get
[tex]3(x^2- y^2)+ (y^2- x^2)= 2(x^2- y^2)= 0[/tex]
so that either y= x or y= -x.

These are both ellipses with center (0,0) and angles [itex]\frac{5\pi}{8}[/itex] and [itex]\frac{\pi}{8}[/itex] 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 [itex]2xy[/itex] 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!
 
  • #3
ehild
Homework Helper
15,409
1,813
Subtract the second equation from the first one and factorise.

ehild
 
  • #4
10
0
Thanks for your advice.
However, this doesn't give me the four points of intersection. If I take the solutions [itex]y=\pm x[/itex], I get the solutions [itex] (0,0) \wedge (\pm \sqrt{2},\pm \sqrt{2})[/itex].
However, the points [itex] (\pm \sqrt{\frac{2}{3}},\pm \sqrt{\frac{2}{3}})[/itex] are also solutions of the equation. And the point [itex](0,0)[/itex] is not a solution at all.
How would I get the other two points?
Thanks in advance!
 
  • #5
ehild
Homework Helper
15,409
1,813
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
 
Last edited:
  • #6
10
0
I got this:
[itex]y = \pm x[/itex]
[itex](x^2+x^2-4)(x+x)=0[/itex]
[itex]x(x^2-2)=0[/itex]
[itex]x = 0 \vee x=\pm \sqrt{2}[/itex]
According to the solutions manual, which only states the values, not the method, the solutions are:
[itex]x = \pm \sqrt{2} \vee x = \pm \sqrt{\frac{2}{3}}[/itex]
 
  • #7
10
0
Oops... I see it now. I used f(x,y) instead of the partial derivatives.
I will try again!
 
  • #8
10
0
Yes! I got it!
Thank you for your help, I appreciate it! :D
 
  • #9
ehild
Homework Helper
15,409
1,813
You see, it was worth to write it again...:smile:

ehild
 

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