showing curvature?

Question

when you have the cornu spiral expressed parametrically as
x= integral[cos{(pi*u^2)/2}du from 0 to t]
y =integral[sin{(pi*u^2)/2}du from 0 to t]

how do you use k(s) = ||r''(s)|| to show that the curvature of the cornu spiral evaluated at any arc length is proportional to the absoute value of that act lenght? would s = t? i know that you have to find s in terms of t and then plug s into the formula, but I can't seem to figure out how to.
is the constant you acquire here the constant of proportion?

help would be great!! thanks!

also, if I am trying to prove that the parameter in the vector valued function that i found is an arc-length parameter with a reference point where t=0, is this correct --> r'(t)=
now
|r'(t)|=sqrt[ cos{(pi*t^2)/2} ^2 + sin{(pi*t^2)/2} ^2 ] =1

therefore arc length = s(t) = integral form 0 to t of |r'(u)| du
= integral from 0 to t of 1 du
= t : therefore, it is an arc-length parameter with a reference point where t=0

also, if I am trying to prove that the parameter in the vector valued function that i found is an arc-length parameter with a reference point where t=0, is this correct --> r'(t)=
now
|r'(t)|=sqrt[ cos{(pi*t^2)/2} ^2 + sin{(pi*t^2)/2} ^2 ] =1

therefore arc length = s(t) = integral form 0 to t of |r'(u)| du
= integral from 0 to t of 1 du
= t : therefore, it is an arc-length parameter with a reference point where t=0

Answer 1

HA! if you can figure out to give me about 990 extra points on top of the usual 10, it might be worth my while...