Curvature and Vectors?

I have been trying to work the idea behind curvature, as in #kappa = (d hat T)/ (ds) #, where #hat T# is the unit tangent vector, and have been hoping to do so using purely vector algebra

i can get this far....

position vector #vec r(t)# has tangent vector #vec v = vec dot r# and unit tangent vector #hat v = (vec v)/(abs ( vec v))#

using the chain rule

#kappa = (d hat v(t))/ (ds) = (d hat v(t))/ (dt)* (dt)/(ds) #

# = (d hat v(t))/ (dt)* 1/(dot s)= (d hat v(t))/ (dt)* 1/(abs (vec v)) #

as #dot s = abs (vec v)#

now

#(d hat v(t))/ (dt) = d/(dt) (vec v)/(abs(vec v))#

by quotient rule...is this too aggressive(??)

#= ( d/dt(vec v) abs(vec v) - vec v d/dt(abs(vec v)) ) /abs(vec v)^2#

#= ( vec a abs(vec v) - vec v d/dt(abs(vec v)) ) /abs(vec v)^2 qquad triangle#

and
#d/dt(abs(vec v)) = d/dt(sqrt(vec v * vec v)) #

# =1/2 * 1/(sqrt(vec v * vec v)) d/dt (vec v * vec v) #

# =(vec a * vec v)/(sqrt(vec v * vec v)) = (vec a * vec v)/(abs( vec v)) qquad square#

putting#square# into #triangle#:

#= ( vec a abs(vec v) - vec v (vec a * vec v)/(abs( vec v)) ) /abs(vec v)^2#

#= ( vec a (vec v * vec v) - vec v (vec a * vec v) ) /abs(vec v)^3 qquad circ#

that's still shy of the usual result ie

#(abs (vec v times vec a))/(abs(vec v)^3)#

and to further complicate matters, i make #circ# out to be:

#= ( vec v times( vec a times vec v ) ) /abs(vec v)^3 #