Level curves (1 Viewer)

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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.
 

HallsofIvy

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Do you have a particular reason for converting to polar coordinates?

Level curves are points (x,y) that give f(x,y) a constant value:
[tex]f(x,y)= \frac{2xy}{x^2+y^2}= c[/tex].

That is the same as 2xy= c(x2+ y2 or
x2+ (2/c)xy+ y2= 0. Obvious point: if c= 1 that is
x2+ 2xy+ y2= (x+ y)2= 0 which is the single straight line y= -x. It should be easy to show that for other values of c, it can still be factored and the graph will be two straight lines intersecting at (0,0).
 

benorin

Homework Helper
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[itex]f(x,y)= \frac{2xy}{x^2+y^2}= c\Rightarrow y^2-\frac{2xy}{c}+x^2= 0[/itex], which is quadratic in y, hence

[tex]y=\frac{ \frac{2x}{c} \pm\sqrt{ \left( \frac{2x}{c}\right) ^2 - 4x^2} }{2}[/tex]
 
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Converting to polar coordinates just seemed to simplify the expression. I'll have another look at it. Thanks for the help.
 

benorin

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A little more detail...

[itex]f(x,y)= \frac{2xy}{x^2+y^2}= c\Rightarrow y^2-\frac{2xy}{c}+x^2= 0[/itex], which is quadratic in y, hence

[tex]y=\frac{ \frac{2x}{c} \pm\sqrt{ \left( \frac{2x}{c}\right) ^2 - 4x^2} }{2} = \frac{x}{c}\left( 1\pm\sqrt{1-c^2}\right)[/tex]

note that this family of level curves given by [itex]f(x,y)=c[/itex] is only meaningful for [itex] -1\leq c\leq 1[/itex]. This is because, for every [itex](x,y)\in \mathbb{R} ^2[/itex], we have

[tex] 0\leq (x - y)^2 =x^2 - 2xy + y^2 \Rightarrow 2xy\leq x^2 + y^2 \Rightarrow f(x,y)= \frac{2xy}{x^2+y^2} \leq 1[/tex]

and

[tex] 0\leq (x + y)^2 =x^2 + 2xy + y^2 \Rightarrow -2xy\leq x^2 + y^2 \Rightarrow -1\leq\frac{2xy}{x^2+y^2} =f(x,y) [/tex]

thus

[tex]f(x,y)= \left| \frac{2xy}{x^2+y^2} \right| \leq 1[/tex]

so no other values of c are even possible (except c=0) and we thereby gain assurence that our family of curves is complete. The plot below shows 7 level curves superimposed on the plot of f(x,y) over -3 < x < 3, -3 < y < 3 .

For c=0, we get 2xy=0, so either x=0 or y=0, but not both, as that would involve division by zero, and so among our level curves, count also the x-axis & the y-axis less the origin.
 

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Thanks again for your help. I'll go over your work after I get some sleep.:zzz:
 

benorin

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I too will go over my work after I get some sleep. So be careful if you get up before me.
 

HallsofIvy

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benorin said:
[itex]f(x,y)= \frac{2xy}{x^2+y^2}= c\Rightarrow y^2-\frac{2xy}{c}+x^2= 0[/itex], which is quadratic in y, hence
[tex]y=\frac{ \frac{2x}{c} \pm\sqrt{ \left( \frac{2x}{c}\right) ^2 - 4x^2} }{2} = \frac{x}{c}\left( 1\pm\sqrt{1-c^2}\right)[/tex]
note that this family of level curves given by [itex]f(x,y)=c[/itex] is only meaningful for [itex] -1\leq c\leq 1[/itex]. This is because, for every [itex](x,y)\in \mathbb{R} ^2[/itex], we have
[tex] 0\leq (x - y)^2 =x^2 - 2xy + y^2 \Rightarrow 2xy\leq x^2 + y^2 \Rightarrow f(x,y)= \frac{2xy}{x^2+y^2} \leq 1[/tex]
and
[tex] 0\leq (x + y)^2 =x^2 + 2xy + y^2 \Rightarrow -2xy\leq x^2 + y^2 \Rightarrow -1\leq\frac{2xy}{x^2+y^2} =f(x,y) [/tex]
thus
[tex]f(x,y)= \left| \frac{2xy}{x^2+y^2} \right| \leq 1[/tex]
so no other values of c are even possible (except c=0) and we thereby gain assurence that our family of curves is complete. The plot below shows 7 level curves superimposed on the plot of f(x,y) over -3 < x < 3, -3 < y < 3 .
For c=0, we get 2xy=0, so either x=0 or y=0, but not both, as that would involve division by zero, and so among our level curves, count also the x-axis & the y-axis less the origin.
In other words, if c= 0, the level curve is a single straight line through the origin and for other values of c, the level curve is two straight lines passing through the origin, as I said above.
 

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