If gamma(t) = ( x(t), y(t) ) is a 2 dimensional curve, the curvature

k(t) at point gamma(t) is this:

k(t) = A/B

A = [ (dx/dy)*(d^2y/dt^2) - (d^2x/dt^2)*(dy/dt) ]

B = [ (dx/dt)^2 + (dy/dt)^2 ]^(3/2)

can someone help me prove that?

thank you

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- Feb 24th 2007, 11:15 PMtea217Help with curvature proof
If gamma(t) = ( x(t), y(t) ) is a 2 dimensional curve, the curvature

k(t) at point gamma(t) is this:

k(t) = A/B

A = [ (dx/dy)*(d^2y/dt^2) - (d^2x/dt^2)*(dy/dt) ]

B = [ (dx/dt)^2 + (dy/dt)^2 ]^(3/2)

can someone help me prove that?

thank you - Apr 22nd 2007, 06:14 AMRebesques
Remember that the curvature is given by k=dQ/ds, where Q is the angle between the tangent vector (x'(t),y'(t)) and (1,0), and ds the arclength element. Begin by

k=dQ/ds=(dQ/dt)/(ds/dt),

and use:

-> (ds/dt)^2= (x')^2+(y')^2

but also

-> Q=Arctan(dy/dx)=Arctan([dy/dt]/[dx/dt]),

etc.