Question

Multivariable Calc Question:

Hi all! I'm having issues finding the critical points of the following function. I've figured out the partial derivatives and set them to zero, but I don't know where to go from there. I'll attach the problem and my attempt at a solution in the comments.

find and classify the critical points of\[\large f(x,y)=(x^2-y^2)e^{-x^2-y^2}\]

Answer 1

Here the problem:

Answer 2

For the partial derivatives I got that:
\[\frac{ \delta f }{ \delta x } = x(x^2 - y^2 + 1) e^(-x^2) \]
\[\frac{ \delta f }{ \delta y } = y(y^2-x^2+1)e^(-y^2)\]

Answer 3

those partials aren't right...

Answer 4

oops I guess that's why I'm getting nowhere...

Answer 5

\[f(x,y)=(x^2-y^2)e^{-x^2-y^2}\]\[{\partial f\over\partial x}=2xe^{-x^2-y^2}+(x^2-y^2)(-2x)e^{-x^2-y^2}\]

Answer 6

do you see your mistake? you didn't let u=-x^2-y^2 effectively

Answer 7

yeah I think so. Let me try that again with df/dy.

Answer 8

good idea

Answer 9

ok here goes:\[\frac{ \delta f }{ \delta y } = -2ye ^{-x^2 - y^2} - 2y(x^2 - y^2)e^{-x^2-y^2}\]

Answer 10

yep :)
so there are two solutions to each, what are they?

Answer 11

well, since e^(anything) can never be zero...

Answer 12

x=0 is a solution

Answer 13

and \[x = \sqrt{y^2 - 1}\] is the other solution?

Answer 14

\[x=\sqrt{y^2+1}\], no?

Answer 15

oh right right I was looking at df/dy sorry

Answer 16

so yeah x is 0 or sqrt(y^2 + 1)

Answer 17

and if you look at dy then y = 0 or sqrt(x^2 + 1) ?

Answer 18

right, now here I have to do some refreshing, but basically we need to start solving the system\[f_x=0\\f_y=0\] because those values of x,y will be our critical points

Answer 19

I know that part. I'm just not sure how to solve the system at this point.

Answer 20

{0,0} we can say is a critical point right away, for the others we have to solve the system using sustitution

Answer 21

should we plug in one value of x and then plug in the other value?

Answer 22

and yeah the origin is definitely a critical point

Answer 23

yes, plug \[x=\sqrt{1+y^2}\]into \(f_x=0\)

Answer 24

wait but didn't we originally derive that value of x from f subx? Shouldn't we plug it into f sub y?

Answer 25

oh yeah, sorry :P

Answer 26

ah ok in that case...

Answer 27

we get.... y = 0?

Answer 28

x = 1?

Answer 29

let me check, one sec

Answer 30

and the other solution if you plug the other way around is (0,1)

Answer 31

I think I've confused myself... I'm getting only y=0 again

Answer 32

I took the value we got for x and plugged it into fy then I did the opposite, plugging in the value for y into fx

Answer 33

the first resulted in (1,0) and the other resulted in (0,1)

Answer 34

\[f_x=0\implies x=\{0,\sqrt{1+y^2}\}\]\[f_y=-2ye^{-x^2-y^2}(1+x^2-y^2)\]

Answer 35

\[f_y(\sqrt{1+y^2})=-2ye^{-1-2y^2}(1+1+y^2-y^2)=-4ye^{-1-2y^2}\]did I make a mistake?

Answer 36

oh I see what you are saying.... so this gives (1,0) and the other gives (0,1)
I'm slow today :P

Answer 37

yeah I think so. Unless I screwed up royally.

Answer 38

no you are right

Answer 39

ok sweet. So the critical points are (0,0), (1,0), and (0,1)

Answer 40

now to determine the max/min/saddle of each...

Answer 41

there is that formula involving the second derivatives....

Answer 42

need the determinant thing

Answer 43

fxx*fyy - (fxy)^2

Answer 44

if that determinant is negative, it's a saddle

Answer 45

yeah that one
pretty formulaic from here

Answer 46

otherwise, if fxx is positive it's a local min and if it's negative it's a local max. thank you so much!

Answer 47

oh hey wait, we missed two solutions...

Answer 48

wha....

Answer 49

>.<

Answer 50

wolfram says that (0,-1) and (-1, 0) are also critical

Answer 51

where did we lose a sign ambiguity?

Answer 52

well, almost everything in this problem is squared so....that's probably why.

Answer 53

yeah I know it's some small mistake like that, but I guess since it's your problem I'll leave it top you to hunt down specifically, unless it jumps out at me

Answer 54

Thanks anyway! This was VERY helpful

Answer 55

Thanks for the review :)

Answer 56

it jumped out at me :D
if you're still there

Answer 57

\[f_y(0)=-2ye^{-y^2}(1-y^2)\]now we have already considered the solution y=0, x=0 so here we assume that y is NOT zero, and wind up solving\[1-y^2=0\]from where we get our sign ambiguity

Answer 58

same argument goes for \(f_x(0)\)

Answer 59

Oh there we go makes sense. Molte grazie! Allons-y!