Finding Tangent of Bezier Curve - Is Maths or Programming the Problem?

[Bucky]

Hi, I'm programming a visual implimentation of a bezier curve for a coursework. It would be beneficial for me to find the tangent at any point t on the curve. I can calculate the position of a point t, and so can hash the problem somewhat by finding the gradient between t+- a small value, but I was hoping for a more mathmatically accurate method.

So I have the basic equation

nCr * t^r * (1-t)^(n-r) * Vector for point

which I tried to differentiate to...

nCr * r * t^(r-1) * -(n-r) * (1-t) * Vector for point


where
t is progress along curve (between 0 and 1)
n is the number of control points
r is which iteration of the list of points we're on.


is this correct? I'm not sure if it's the programming or the maths that are causing the problem.

 

[lippyka]

Any help would be appreciated. Thanks!Yes, your differentiation is correct! To find the tangent at a point t, you would use the derivative of the equation you provided above and evaluate it at the point t. The resulting vector is the tangent vector at point t.