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I have a function $z = B \sin x \ \sin y+\cos x \ \cos y$. Where $0 \leq x \leq \pi$ and $0 \leq y \leq \pi$. I need to find the length of the curve that describes a level set for any value of $B$. That is, if I set $z = A$ (where $A$ is some scalar constant), what is the length of the level curve for any value of the parameter $B$. Obviously some symmetry can be exploited to solve the problem, but I'm having trouble figuring out how to derive the length.

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Are you sure the variables $x,y$ have not to satisfy some bound (e.g., $-\pi\leq x,y\leq \pi$)? For otherwise the problem is meaningless... –  Pacciu Mar 29 '11 at 16:55

1 Answer 1

The length is either 0 or infinity, depending on A and B, because the function z is periodic in x and in y.

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Because the level sets are made of infinite closed curves (which can degenerate into points), I guess. –  Pacciu Mar 29 '11 at 16:53
    
I should have specified, the domain is 0 <= x <= pi and 0 <= y <= pi. The length is not always zero. You can plot the level sets and geometrically determine the length for a few simple cases of A and B. For example: when A = B = -1 the length (in the given domain) is pisqrt(2), when A = 0, and B = -1 the length is also pisqrt(2); however, when A = 0, and B -> 0 the length approaches 2*pi. So in general the length is finite and nonzero, but there are a few special cases where the length will be zero. –  okj Mar 29 '11 at 19:45
    
@Pacciu –  okj Mar 29 '11 at 19:56
    
@okj: Then, why don't you edit your post and write "with $0\leq x,y\leq \pi$" somewhere in it? ;-) –  Pacciu Mar 29 '11 at 22:26
    
@Pacciu: Thanks, I didn't realize I could edit the post, this is my first time on stackexchange. I think the problem should be more easily understood now. –  okj Mar 29 '11 at 23:07

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