Calc 3 Question on Curvature - Tilted Forum Project Discussion Community

Stompy · 02-22-2006, 09:33 AM

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/x2 / (1 + 1/x2)3/2

To find the min max, you need to take the derivative of the curvature, which ends up being:

2/((1+1/x2)3/2x3) - 3/((1+1/x2)5/2x5)

...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/x2 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!

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)) ]

BadNick · 03-08-2006, 01:04 PM

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.

02-22-2006, 09:33 AM	#1 (permalink)
Stompy 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! __________________ I love lamp.

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.

03-08-2006, 01:04 PM	#3 (permalink)
BadNick 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...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.