Author 
Topic 

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' = (2r2x)/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/y
y" = 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/r
since 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 Kevitzinn
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' = (2r2x)/2y or may be something else? Any help anyone can give me would be very much appreciated. Thank you for your time.
KG
The 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. 



Topic 
