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Curvature of plane curves

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$begingroup$

My text defines the curvature of a plane curve as $<ddotx,N>$ where $N$ is the normal to the normalized tangent of $x$ and $x$ is the curve. I thought the $ddotx$ also was perpendicular to $dotx$, making this projection kind of odd. Can someone see where I go wrong? Isnt projection of parallel lines a wierd thing?

differential-geometry

share|cite|improve this question

edited Mar 21 at 13:44

Twist

asked Mar 21 at 13:26

TwistTwist

636

$endgroup$

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  $begingroup$
  $ddotxperpdotx$, not necessarily to $x$, for example take the curve $tmapsto (cost,sint)$
  $endgroup$
  – PSG
  Mar 21 at 13:43
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @PSG Right sorry, that what I ment.
  $endgroup$
  – Twist
  Mar 21 at 13:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  usually curvature is defined as $|ddotx|$, $because ddotx || N$, assuming $ddotxneq 0, N=fracddotxddotx$. So, $<ddotx,N>=|ddotx|^2/|ddotx|$
  $endgroup$
  – PSG
  Mar 21 at 13:49
  
  

  
  
  
  

add a comment | 

0

$begingroup$

My text defines the curvature of a plane curve as $<ddotx,N>$ where $N$ is the normal to the normalized tangent of $x$ and $x$ is the curve. I thought the $ddotx$ also was perpendicular to $dotx$, making this projection kind of odd. Can someone see where I go wrong? Isnt projection of parallel lines a wierd thing?

differential-geometry

share|cite|improve this question

edited Mar 21 at 13:44

Twist

asked Mar 21 at 13:26

TwistTwist

636

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $ddotxperpdotx$, not necessarily to $x$, for example take the curve $tmapsto (cost,sint)$
  $endgroup$
  – PSG
  Mar 21 at 13:43
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @PSG Right sorry, that what I ment.
  $endgroup$
  – Twist
  Mar 21 at 13:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  usually curvature is defined as $|ddotx|$, $because ddotx || N$, assuming $ddotxneq 0, N=fracddotxddotx$. So, $<ddotx,N>=|ddotx|^2/|ddotx|$
  $endgroup$
  – PSG
  Mar 21 at 13:49
  
  

  
  
  
  

add a comment | 

$begingroup$

My text defines the curvature of a plane curve as $<ddotx,N>$ where $N$ is the normal to the normalized tangent of $x$ and $x$ is the curve. I thought the $ddotx$ also was perpendicular to $dotx$, making this projection kind of odd. Can someone see where I go wrong? Isnt projection of parallel lines a wierd thing?

differential-geometry

share|cite|improve this question

edited Mar 21 at 13:44

Twist

asked Mar 21 at 13:26

TwistTwist

636

$endgroup$

share|cite|improve this question

edited Mar 21 at 13:44

Twist

asked Mar 21 at 13:26

TwistTwist

636

share|cite|improve this question

edited Mar 21 at 13:44

Twist

asked Mar 21 at 13:26

TwistTwist

636

asked Mar 21 at 13:26

TwistTwist

636

asked Mar 21 at 13:26

TwistTwist

636