Question

I have a question regarding critical points

Answer 1

Find the critcal points of :

\[f(x,y)=(x^2-y^2)e ^{(-x^2-y^2)/(2)}\]

Answer 2

for partial x i got:

\[xe ^{(-x^2-y^2)/(2)}(-x^2+y^2+2)\]

Answer 3

for partial y i got:

\[ye ^{(-x^2-y^2)/(2)}(-x^2+y^2-2)\]

Answer 4

I now i am suppose to set them to zero, but how do i solve for the critical points.

Answer 5

@Zarkon , could you help me out, is the critical point (0,0)

Answer 6

or anybody

Answer 7

you didnt use the product rule?

Answer 8

oh yes sorry

Answer 9

but, what i am looking for is the critcal points

Answer 10

the question is how do we find the zeroes i understand

Answer 11

right, because, to find the critical points we set partial x and partial y to zero

Answer 12

idk :(

Answer 13

help @hero (sorry for botherin )

Answer 14

@estudier

Answer 15

@lalaly could you help

Answer 16

From your first partial , we see that x=0 makes fx=0
similarly, y=0 for the 2nd equation, so (0,0) looks critical
if we take x=0 from the first, we can make the second equation zero by solving for
y^2-2 = 0

Answer 17

oh @eigenschmeigen :D

Answer 18

or vice versa, y=0 from the 2nd equation and x^2=2 makes the first 0

Answer 19

i think phi has this covered :D

Answer 20

oh yes ur latee :D

Answer 21

what do you mean take x=0 from the first, you mean plug x=0 into the partial derivative of y?

Answer 22

yes

Answer 23

because that point will make fx=0 and fy=0

Answer 24

i see

Answer 25

okay, so i plugged in x=0 into the partial derivative of y and got:
\[ye ^{(-y^2)/(2)}(y^2-2)\]

Answer 26

is this what i should have gotten, and how do i procced from here

Answer 27

You have 3 factors, if any = 0, the expression = 0.
so try each in turn

Answer 28

i dont understand

Answer 29

It is the same idea used to solve
(x-1)(x+2)=0 either (x-1)=0 or (x+2)=0
here you have 3 factors
y
e^stuff
(y^2-2)

Answer 30

okay, so i got y=0, and y=plus or minus sqrt(2), how do i deal with that expoenet?

Answer 31

It only approaches 0 as y-> inf, so it isn't a player

Answer 32

but so then what is the meanig of the y=plus or minus sqrt(2)

Answer 33

y= sqrt(2) makes fy=0 when x=0
x=0 makes fx=0
so (0,sqrt(2)) makes fx=0 and fy=0

Answer 34

i understand, so then the ctircal points would be (0,0) and (0,sqrt(2))

Answer 35

and (0,-sqrt(2))

and don't forget y=0 from the 2nd eq. solve for x in the first to get even more

Answer 36

oh man, we still need to plug in the y=o into the f sub x. Hmm, okay, i will do that. Thanks

Answer 37

now when i plugged in y=o into partial x, i got 3 terms, and set them to zero. i got x=0, and then i also got:x=plus or mius sqrt(2), so, then i more critical points right? they would be (sqrt(2),0) and (-sqrt(2),0)

Answer 38

Yes. It looks like 5 critical points.

Answer 39

http://www.wolframalpha.com/input/?i=%28x2%E2%88%92y2%29e%5E%28%28%E2%88%92x2%E2%88%92y2%29%2F%282%29%29

Answer 40

it has 4

Answer 41

(0,0) is also a solution of partial derivatives.

Answer 42

i got another one