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Kevitzinn
Average Member
USA
14 Posts |
Posted - 10/06/2007 : 20:43:16
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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
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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
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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
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| 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
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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
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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
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| 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
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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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