If I have a cubic bezier curve $B(t)$, I know that to get the tangent of the curve $B(t)$, just take the derivative of it. However, what if I'm looking for the tangent at one of the endpoints, how do I go about getting it?
Thanks.
[Math] Tangent of Cubic Bezier curve at end point
bezier-curvederivatives
Related Solutions
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In your example, the coefficient of $t^3$ turns out to be $100$.
In the general case $$X_0(1-t)^3 + 3X_1(1-t)^2 t + 3X_2(1-t) t^2 + X_3t^3,$$ the coefficient of $t^3$ will turn out to be $-X_0+3X_1-3X_2+X_3$. The constant term is $X_0$. The coefficient of $t^2$ is $3X_0-6X_1+3X_2$, and the coefficient of $t$ is $-3X_0+3X_1$.
For the expansion, all you need is $(1-t)^3=1-3t+3t^2-t^3$ and $(1-t)^2=1-2t+t^2$.
If (as is likely) you will be using a numerical method to solve the cubic, it is not even necessary to expand, since the numerical method will take care of things.
The simplest heuristic approximation method is the one proposed by Tiller and Hanson. They just offset the legs of the control polygon in perpendicular directions:
The blue curve is the original one, and the three blue lines are the legs of its control polygon. We offset these three lines, and intersect/trim them to get the points A and B. The red curve is the offset. It's only an approximation of the true offset, of course, but it's often adequate.
If the approximation is not good enough for your purposes, you split the curve into two, and approximate the two halves individually. Keep splitting until you're happy. You will certainly have to split the original curves at inflexion points, if any, for example.
Here is an example where the approximation is not very good, and splitting would probably be needed:
There is a long discussion of the Tiller-Hanson algorithm plus possible improvements on this web page.
The Tiller-Hanson approach is compared with several others (most of which are more complex) in this paper.
Another good reference, with more up-to-date materials, is this section from the Patrikalakis-Maekawa-Cho book.
For even more references, you can search for "offset" in this bibliography.
Best Answer
The line through the first two control points is tangent to the curve, as is the line through the last two control points, i.e., the curve is tangent to the first and last segments of the Bézier polygon. This is true for any order Bézier curve.