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

  • Thread starter tifa8
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  • #1
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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=[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
 

Answers and Replies

  • #2
14
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nobody ?
 

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