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Kevitzinn
Average Member

USA
14 Posts

 Posted - 10/06/2007 :  20:43:16 OK, the question is:For the circle x + y = r show that|y''/[1+((y')]^(3/2)| = 1/r.What I'm confused about right now is what do with the r. Should I say that the, y' = (-2x/2y) or is it, y' = (2r-2x)/2y or may be something else? Any help anyone can give me would be very much appreciated. Thank you for your time.-KG

skeeter
Advanced Member

USA
5634 Posts

 Posted - 10/06/2007 :  21:14:16 you sure its not ...|y''/[1+((y')]^(3/2)| = y/r?

Kevitzinn
Average Member

USA
14 Posts

 Posted - 10/06/2007 :  21:45:23 Nope, the text book says 1/r, and just to clarify to the max, it's |(y'')/([1+(y')[square]]^(3/2))| = 1/r.Why do you believe it would be equal to y/r?

galactus
Advanced Member

USA
1464 Posts

 Posted - 10/07/2007 :  09:34:55 An image can be inserted by using ImageShack.

skeeter
Advanced Member

USA
5634 Posts

 Posted - 10/07/2007 :  12:22:35 o.k. ... missed a factor of "y" in my initial attempt, found my error.y' = -x/y 1 + (y') = 1 + x/y = r/yy" = -r/y-----------------------------------------y"/[1 + (y')]3/2 =(-r/y)/(r/y)3/2 =[(-1/y)*r/y]/(r/y)3/2 =(-1/y)/(r/y)]1/2 = -1/rsince r > 0 ... |-1/r| = 1/r

Kevitzinn
Average Member

USA
14 Posts

 Posted - 10/07/2007 :  13:22:51 Wow, thank you so much, I really appreciate it. You guys are the coolest of the cool!

sahsjing
Advanced Member

USA
2399 Posts

 Posted - 10/09/2007 :  17:41:04 quote:Originally posted by KevitzinnOK, the question is:For the circle x + y = r show that|y''/[1+((y')]^(3/2)| = 1/r.What I'm confused about right now is what do with the r. Should I say that the, y' = (-2x/2y) or is it, y' = (2r-2x)/2y or may be something else? Any help anyone can give me would be very much appreciated. Thank you for your time.-KGThe left side |y''/[1+((y')]^(3/2)| is called the curvature, which is always equal to 1/R, where R is called the radius of the curvature.
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