02222006, 09:33 AM  #1 (permalink) 
Banned from being Banned
Location: Donkey

Calc 3 Question on Curvature
My main question is on curvature and how to find the min/max. I haven't had Calc 1 in about 2 years, so some things are foggy to me at this point.
"At what point does y = ln(x) have maximum curvature? What happens to the curvature as x > infinity" You end up with the curvature equation: k = 1/x<sup>2</sup> / (1 + 1/x<sup>2</sup>)<sup>3/2</sup> To find the min max, you need to take the derivative of the curvature, which ends up being: 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>) ...and this is where I'm stuck. How do I get the min/max from that equation? I remember something about setting that equal to 0 and testing points but... yeah, that's an algebraic mess to figure out. I know based on limits that the curvature, as x approaches infinity, gets closer to 0 (1/x<sup>2</sup> gets closer to 0 as x gets bigger) .. at least I hope that part is right. Based on that, I took a stab in the dark and said that the max curvature is at (1, 0). Somehow, without proof, I don't feel that's right. Any help would be GREATLY appreciated!
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02232006, 09:02 AM  #2 (permalink) 
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)) ]
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03082006, 01:04 PM  #3 (permalink) 
Riding the Ocean Spray
Location: S.E. PA in U Sofa

Congrats if you did. I was searching around to refresh myself and found some links, this one "might" help, maybe look at the bottom of page 1 here:
http://www.public.iastate.edu/~aaall...5B2HW3Sol.pdf PS: And you might be famous by now. In case you haven't received tons of calls and requests for your autograph yet, when I googled "maximum curvature of ln (x)" your post here in this thread on TFP was the fourth find at google. 
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