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Old 02-22-2006, 09:33 AM   #1 (permalink)
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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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Old 02-23-2006, 09:02 AM   #2 (permalink)
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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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Old 03-08-2006, 01:04 PM   #3 (permalink)
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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...5-B2HW3Sol.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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