- #1
Benny
- 584
- 0
Hi, I'm having trouble drawing the level curves of a function because I can't really visualise what's going on.
[tex]
f\left( {x,y} \right) = \frac{{2xy}}{{x^2 + y^2 }}
[/tex]
In polar coordinates (x,y) = (rcos(phi),rsin(phi)) I get:[tex]f\left( {r\cos \phi ,r\sin \phi } \right) = \sin \left( {2\phi } \right)[/tex].
Level curves are obtained by setting f equal to a constant. In the case of a circle for example f(x,y) = x^2 - y^2 - 9 if I set f(x,y) = k where -9 < k <= 0, in the x-y plane I get level curves which are circles with radii between 0 and 3 (3 inclusive).
I'm not sure which happens here though. I have f(polar) = sin(2phi) where phi is the polar angle. Since, 1- <= sin(x) <= 1 then does that mean I can only set f equal to values in the range [-1,1]?
I'm still not really sure what the level curves should be like. Any help would be good thanks.
[tex]
f\left( {x,y} \right) = \frac{{2xy}}{{x^2 + y^2 }}
[/tex]
In polar coordinates (x,y) = (rcos(phi),rsin(phi)) I get:[tex]f\left( {r\cos \phi ,r\sin \phi } \right) = \sin \left( {2\phi } \right)[/tex].
Level curves are obtained by setting f equal to a constant. In the case of a circle for example f(x,y) = x^2 - y^2 - 9 if I set f(x,y) = k where -9 < k <= 0, in the x-y plane I get level curves which are circles with radii between 0 and 3 (3 inclusive).
I'm not sure which happens here though. I have f(polar) = sin(2phi) where phi is the polar angle. Since, 1- <= sin(x) <= 1 then does that mean I can only set f equal to values in the range [-1,1]?
I'm still not really sure what the level curves should be like. Any help would be good thanks.