- #1

tifa8

- 14

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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=[tex]\int[/tex]cos(pi*u[tex]^{2}[/tex]/2)du and

y=[tex]\int[/tex] sin(pi*u[tex]^{2}[/tex]/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[tex]^{2}[/tex]y/dx[tex]^{2}[/tex] 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[tex]^{2}[/tex]/2)/cos(pi*t[tex]^{2}[/tex]/2)=Tan(pi*t[tex]^{2}[/tex]/2)

and

d[tex]^{2}[/tex]y/dx[tex]^{2}[/tex]=pi*t/cos[tex]^{2}[/tex](pi*t[tex]^{2}[/tex]/2)

by differentiating dy/dx a second time

However I can't seem to find K

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

I have found K=pi

Then to verify that i used the second formula i have which is

k=(d[tex]^{2}[/tex]y/dx[tex]^{2}[/tex])/(1+(dy/dx)[tex]^{2}[/tex])[tex]^{3/2}[/tex]

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

Thank you for your help