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Mathematics
General Math
Moving center of coordinates in the polar graph
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[QUOTE="Mark44, post: 6561413, member: 147785"] Or as a bit more readable, $$t(\rho, \phi) = \frac{H^2}{H^2 + \rho^2}$$ If you click what I wrote above, you can see my LaTeX script. There's more information about LaTeX in the LaTeX Guide line at the lower left of the input pane. It looks to me like you are omitting the ##L^2## term in the denominator. After the substitution -- in equation (2) -- I get this: $$t(\rho, \phi) = \frac{H^2}{H^2 + (x - L)^2 + y^2} = \frac{H^2}{H^2 + x^2 - 2Lx + L^2 + y^2} $$ $$ = \frac{H^2}{H^2 +\rho^2 - 2L\rho\cos(\phi) + L^2}$$ Does my change make a difference? [/QUOTE]
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Mathematics
General Math
Moving center of coordinates in the polar graph
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