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
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
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.