How to find the unit vector 'u' tangent to the curve and its component 'ux' and 'uy'...?

I want to know how to find the unit vector u tangent to the curve and its component ux and uy. I need it for the measurment of the orientation. please help me to solve this problem, thanks in advance

[harshal j]

Actually i want measure the orientation at each point on the boundary the formula is Ψ(b)={360-arcos(ux(b)) if uy<=0 and Ψ(b)={arcos(ux(b)) if uy>=0 for that i required the unit tangent vector and its direction cosines ux and uy, please help if possible, thanks in advance

[Torsten]

How is your curve given ?

If it's a parametric curve given by

C=(x(s),y(s))

the unit tangent vector at a given point is simply

1/sqrt(xdot^2(s)+ydot^2(s))*[xdot(s) ; ydot(s)]

where xdot, ydot are the derivatives with respect to the parameter s.

Best wishes

Torsten.

[harshal j]

Thanks for the comments, I have obtain the curve or the boundary of 2-D image by using the bwboundaries function in MATLAB, i want to obtain the unit tangent vector for the each point on the boundary, please help me to solve this problem

[harjeet singh]

kindly upload the boundary image which you have after bwboundaries

[harshal j]

[harshal j]

Please help me to find the unit tangent vector (U) and its component (ux & uy) along the boundary.