# Thread: Classifying Critical Points in R3

1. ## Classifying Critical Points in R3

f(x,y) = (x^2 + y^2)e^(x^2-y^2) = z
I have found that the part ∂z/∂y = 0 when y = 0 or +/- sqrt(1 - x^2)
and that ∂z/∂x = 0 when x=0 or x^2 + y^2 + 1 = 0 which is not real
so now how do I classify them? i.e. what is min, what is max, what is saddle?

2. which points am I testing though? I'm not sure how to write them

3. $\displaystyle f_x=2xe^{x^2-y^2}+(x^2+y^2)e^{x^2-y^2}(2x)=2xe^{2x-y^2}(1+x^2+y^2)=0$

$\displaystyle x=0, \ 1+x^2+y^2=0$

$\displaystyle f_y=2ye^{x^2-y^2}+(x^2+y^2)e^{x^2-y^2}(-2y)=2ye^{x^2-y^2}(1-x^2-y^2)=0$

$\displaystyle y=0, \ 1=x^2+y^2$

By substitution,

$\displaystyle x^2+y^2+x^2+y^2=0\Rightarrow 2x^2+2y^2=0\Rightarrow x^2+y^2=0$

Critical point:

$\displaystyle f(0,0)=0\Rightarrow (0,0,0)$

$\displaystyle f_{xx}(0,0)=2$

$\displaystyle f_{yy}(0,0)=2$

$\displaystyle f_{xy}(0,0)=2$

$\displaystyle d=f_{xx}f_{yy}-[f_{xy}]^2=4-4=0$

Inconclusive. Those, being $f_{xx}, \ f_{yy}, \ f_{xy}$, derivatives were pretty nasty though so you might want to double check them.

4. so (0,0,0) is inconclusive
but aren't there other possibilities with 1 = x^2 + y^2

5. No, would $(\pm 1,0)$ or $(0,\pm 1)$ satisfy $1+x^2+y^2=0\mbox{?}$

6. I decided to upload the image since it was interesting when I graphed.