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=$$\int$$cos(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 ?