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?

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- Dec 15th 2010, 12:43 PMPlaythiousClassifying 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? - Dec 15th 2010, 01:44 PMchiph588@
- Dec 15th 2010, 08:38 PMPlaythious
- Dec 15th 2010, 09:17 PMdwsmith
$\displaystyle \displaystyle f_x=2xe^{x^2-y^2}+(x^2+y^2)e^{x^2-y^2}(2x)=2xe^{2x-y^2}(1+x^2+y^2)=0$

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

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

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

By substitution,

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

Critical point:

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

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

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

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

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

Inconclusive. Those, being $\displaystyle f_{xx}, \ f_{yy}, \ f_{xy}$, derivatives were pretty nasty though so you might want to double check them. - Dec 16th 2010, 10:07 AMPlaythious
so (0,0,0) is inconclusive

but aren't there other possibilities with 1 = x^2 + y^2 - Dec 16th 2010, 01:22 PMdwsmith
No, would $\displaystyle (\pm 1,0)$ or $\displaystyle (0,\pm 1)$ satisfy $\displaystyle 1+x^2+y^2=0\mbox{?}$

- Dec 16th 2010, 01:38 PMdwsmith
http://www.mathhelpforum.com/math-he...6&d=1292539051

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