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You are not logged in. #1 20060505 12:16:06
arc length of a spiralMy tetherball thread hit a brick wall as it turned out finding the length of a simple spiral is not as simple as I initially thought. So I'm going to look at finding the length of polar curves, particularly spirals, instead. (would a mod please delete the tetherball thread) Last edited by mikau (20060505 14:02:09) A logarithm is just a misspelled algorithm. #2 20060505 14:23:27
Re: arc length of a spiralwhoah! dang! I was just looking for pictures of archemedian spirals and accidently came arcoss the formula for their arclength. They're using the exact same formula I derived! I guess its right then. lol... A logarithm is just a misspelled algorithm. #3 20060506 23:16:19
Re: arc length of a spiralYes, polar coords can be really helpful and a good alternative! X'(yXβ)=0 #4 20060507 08:29:15
Re: arc length of a spiralAll right, here's another one. A hyperballic spiral of the form rθ = a (or r = a/θ), this spiral approaches the line y = a as an asymptote as θ approaches zero. Using a/θ for for r and a/θ^2 for dr in the expression sqrt ( (dr)^2 + (r dθ)^2 ) you can rearrange it into the form a/θ sqrt ( 1 + 1/θ^2 ) dθ. Can it be integrated? Yes! I used a trig sub and let θ = cot x, so sqrt ( 1 + 1/θ^2) becomes sqrt ( 1 + tan^2 x) = sec x and dθ becomes csc^2 x, put it all together and you have a ∫ sec^2 x csc x which is easy to solve using integration by parts. In the end after resubstituting: A logarithm is just a misspelled algorithm. 