How can I prove in general form that the tangent at the start point of a Bézier curve goes through control point 1?

Question

I need to prove that the tangent to the start point of any Bézier curve goes through the control point. I have proven this for specific Bézier curves but I am struggling to do it in general, thank you.

Answer 1

The control points of a Bézier curve are considered to be sequentially linked by straight lines. These lines have geometrical significance; they are not there just for the sake of appearance. The segment $P_0P_1$ , of course passes through $P_1$ : It was meant to do so. $P_0$ is a terminus of the curve and $P_0P_1$ is tangent to the curve at $P_0$ .

Answer 2

The equation of a Bézier curve with control points $\mathbf{P}_0, \mathbf{P}_1, \ldots, \mathbf{P}_m$ is $$ \mathbf{C}(t) = \sum_{i=0}^m \phi_i^m(t)\mathbf{P}_i $$ where $\phi_i^m(t)$ is the $i$-th Bernstein polynomial of degree $m$. Differentiating and doing some re-arranging, we get $$ \mathbf{C}'(t) = \sum_{i=0}^{m-1} \phi_i^{m-1}(t)\left(\mathbf{P}_{i+1} - \mathbf{P}_i\right) $$ Setting $t=0$, we get $$ \mathbf{C}'(0) = m\left(\mathbf{P}_{1} - \mathbf{P}_0\right) $$ So $$ \mathbf{P}_{1} = \mathbf{P}_0 + \frac{1}{m}\mathbf{C}'(0) $$ This shows that the point $\mathbf{P}_{1}$ lies on the line through $\mathbf{P}_{0}$ in the direction of the tangent vector $\mathbf{C}'(0)$.

For another explanation, look at this page.