Piecewise 3D function

asterisco500 shared this question 1 week ago
Answered

I'm trying to graph this function


f(x,y) = 0 if y=-x^2 and f(x,y) = 2xy/(x^2+y^2) otherwise


I tried to use the If command this way


f(x,y) = If( y=-x^(2), 0, (2xy)/(x^(2)+y^(2)) )


But it didn't work. Is there a way to graph the function?

Comments (3)

photo
1

Try this

If( y==-x^(2), 0, (2xy)/(x^(2)+y^(2)) )

== means "are they equal?" rather than "set them equal"

photo
1

I think not possible because y=-x^2 is a curve and the graph has not thickness

the aspect must be equal to f(x)=-x^2 graphed in 2D and 2x y/(x^2+y^2) graphed in 3D in graphics 3d window

it happens same with if(x==0,0,x^2+1) in 2D

photo
1

Hi,


here my 2 Cents.

Michaels definition should work. Try e. g. f(2,-4) or f(3,-9).

But, as mathmagic mentioned, you can't see this in graphics.

Maybe it's sufficient to mark/cut off the "difference" between the "normal surface" and the exception for y=-x².


82a30a9718585f865da37bdef9e74198

Kind regards

mire2

© 2020 International GeoGebra Institute