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Stompy · 02-23-2006, 09:02 AM

Ahhh, I spent SO much time on it. Figured it out (I think, anyway).

Set the 2/((1+1/x2)3/2x3) - 3/((1+1/x2)5/2x5) = 0

x = sqrt(1/2)

Point at max curvature would be [ sqrt(1/2) , ln(sqrt(1/2)) ]

02-23-2006, 09:02 AM	#2 (permalink)
Stompy Banned from being Banned Location: Donkey	Ahhh, I spent SO much time on it. Figured it out (I think, anyway). Set the 2/((1+1/x<sup>2</sup>)<sup>3/2</sup>x<sup>3</sup>) - 3/((1+1/x<sup>2</sup>)<sup>5/2</sup>x<sup>5</sup>) = 0 x = sqrt(1/2) Point at max curvature would be [ sqrt(1/2) , ln(sqrt(1/2)) ] __________________ I love lamp.