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

puttingsquare 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