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Restrict a function in 3D view
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Hello,
I need restrict this function:
 z=1x^2
in the intervals
 0<=x<=1
 0<=y<=2
But, I get a error when I use the function if.
please, help me
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Hi,
z=If[0<=x<=1&&0<=y<=2,1x^2]
works for me... what exactly did you type in?
Cheers,
Zbynek
Will the If() command work in 3D only when the function is stated explicitly? I have tried
If(0<=x<=85, (xz)^2 + y^2 = z^2) and
If(0<=x<=85 && z>=0, (xz)^2 + y^2 = z^2)
but get an input error. I don't require any restriction on y.
you can not use a condition on z for 3D surface at this moment
in your case you can try if(0<=x,(x^2+y^2)/(2x))
Thanks mathmagic. So z has to be expressed explicitly in terms of x & y. Is it possible to restrict the surface by restricting the values of z, say 76<z<114?
first define a(x,y)=(x^2+y^2)/(2x) then if(76<=a(x,y)<=114,a(x,y))
I think better if you can parametrize the function and use surface() because you can have a problem with discontinuity
in your case you have horizontal circles centred in (z,0,z)
surface(u+ucos(v),u sin(v),u,76,114,v,0,2 pi)
You can do it like me, restricting first in an axis (could be x) and then you need to restrict again the function but on the other axis.
For example:
e(x, y) = If(0 ≤ x ≤ 2π, sen(x) + sen(y) + sen(x + y))
f(x,y) = If(0 ≤ y ≤ 2π, e(x, y))This will restrict the domain on X and Y from 0 to 2pi in both axis.
es más corto, y creo que más fácil
If(0 ≤ x ≤ 2π && 0 ≤ y ≤ 2π , sen(x) + sen(y) + sen(x + y))
&& es .AND. es decir el y logico
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