Find dy/dx and d^2y/dx^2 for a spiral of cornu in funtion of t

[tifa8]

Hello,
I need help for this problem on my calculus chapter curves and motion on curves

Homework Statement

a) for the spiral cornu defined by the parametric equations
x=\intcos(pi*u^{2}/2)du and

y=\int sin(pi*u^{2}/2)du

obtain the length of the curve s(t) from 0 to t and hence reparametrize the curve in term of s

b)Obtain dy/dx, d^{2}y/dx^{2} and k the curvature.

Homework Equations

The Attempt at a Solution

a) I have found that ds/dt =1 thus s(t)=t and t(s)=s (chain rules)

b) using chain rules,
dy/dx=dy/dt *dt/dx=(dy/dt)/(dx/dt)= sin(pi*t^{2}/2)/cos(pi*t^{2}/2)=Tan(pi*t^{2}/2)

and
d^{2}y/dx^{2}=pi*t/cos^{2}(pi*t^{2}/2)
by differentiating dy/dx a second time

However I can't seem to find K

My first attempt was by using K=\left\|acceleration x velocity\left\|/speed^{3}

I have found K=pi

Then to verify that i used the second formula i have which is
k=(d^{2}y/dx^{2})/(1+(dy/dx)^{2})^{3/2}

which gave me k= pi*t*abs(cos(pi*t^{2}/2))

Thank you for your help

 

[tifa8]

nobody ?