Piecewise 3D function
asterisco500 shared this question 2 years ago
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?
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== means "are they equal?" rather than "set them equal"
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
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.
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