With : $ r=$ position vector , $v=r'=$ velocity vector , $a = v'=$ acceleration vector , and $T=\frac{v}{|v|}= $ unit tangent vector, one formula for curvature $ \kappa$ is :
$\kappa = \frac{|T'|}{|r'|}= \frac{|T'|}{|v|}$.
I'd like to know whether the above formula is equivalent to :
$\kappa = \frac{|a|}{|v|^2}\space\space $?
I tried this derivation:
$T'= [\frac {v} {|v|}]' = \frac{1}{|v|} [v]' = \frac{1}{|v|}a$.
$\kappa = \frac{|T'|}{|v|}= \frac{\big|\frac{1}{|v|}a\big|}{|v|}= \frac{ \big|\frac{1}{|v|}\big||a|} {|v|}= \frac{|a|}{|v|^2}$.
Could you please tell me whether this attempt at deriving the alledged formula does contain any algebra mistake, particularly mistakes regarding absolute values.