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Kevitzinn
Average Member

USA
14 Posts

Posted - 10/06/2007 :  20:43:16  Show Profile
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
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skeeter
Advanced Member

USA
5634 Posts

Posted - 10/06/2007 :  21:14:16  Show Profile
you sure its not ...

|y''/[1+((y')]^(3/2)| = y/r

?
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Kevitzinn
Average Member

USA
14 Posts

Posted - 10/06/2007 :  21:45:23  Show Profile
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?
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galactus
Advanced Member

USA
1464 Posts

Posted - 10/07/2007 :  09:34:55  Show Profile
An image can be inserted by using ImageShack.
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skeeter
Advanced Member

USA
5634 Posts

Posted - 10/07/2007 :  12:22:35  Show Profile
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

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Kevitzinn
Average Member

USA
14 Posts

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

USA
2399 Posts

Posted - 10/09/2007 :  17:41:04  Show Profile
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' = (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



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.
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