TL;DR

Using Derivatives for BĂ©zier curves to find the extreme candidates useful for the bounding box.

In our quest to find the bounding box for BĂ©zier curves (i.e. the minimum area rectangle that has sides parallel to the coordinate axes and that fully contains the curve) we did a step ahead in Derivatives for BĂ©zier curves, because we found a way to easily find out the parameters of the derived polynomial.

Why the derivation? From calculus, we know that all values of $t$ where the derivative of a function is zero are candidate positions for either a minimum or a maximum. Hence, it suffices to find where the derivative components are equal to zero and we will have candidate values of the parameter $t$ to find the extreme points for either axes. We will of course also make sure that these values of $t$ are admissible (i.e. fall in the regular $[0,1]$ interval for it) and also check the values at the two sides of the interval for $t$ itself.

We left Derivatives for BĂ©zier curves with this formula to calculate the coefficients of the polynomials for the derivative in the two axes:

$M_{D,n} \cdot \hat{P}$

As anticipated, these matrices can be pre-calculated and used when necessary upon the specific set of control points $\hat{P}$, which of course depends on the specific BĂ©zier curve we are considering.

The generic solution would involve finding the zeros of a polynomial of any degree. For quadratic and cubic BĂ©ziers, instead, we are lucky because itâ€™s very easy to find the roots.

$M_{D,2} = M_1 \cdot D_2 = \begin{bmatrix} 1 & 0 \\ -1 & 1 \end{bmatrix} \cdot 2 \cdot \begin{bmatrix} -1 & 1 & 0 \\ 0 & -1 & 1 \end{bmatrix} \\ M_{D,2} = 2 \cdot \begin{bmatrix} -1 & 1 & 0 \\ 1 & -2 & 1 \end{bmatrix}$

Multiplying this matrix by the matrix $\hat{P}$ obtained putting the three control points as row vectors we end up with the following:

$\begin{bmatrix} \mathbf{q} \\ \mathbf{m} \end{bmatrix} = \begin{bmatrix} q_x & q_y \\ m_x & m_y \end{bmatrix} = 2 \cdot \begin{bmatrix} - \mathbf{P}_1 + \mathbf{P}_2 \\ \mathbf{P}_1 - 2 \cdot \mathbf{P}_2 + \mathbf{P}_3 \end{bmatrix}$

The two row vectors $\mathbf{q}$ and $\mathbf{m}$ represent the intercept and the angular coefficient of the derivatives for the two components over the X and Y axes. Hence, it is trivial to find the two candidates for the bounding box:

$t_x = - \frac{q_x}{m_x} \\ t_y = - \frac{q_y}{m_y} \\$

$M_{D,3} = M_2 \cdot D_3 = \begin{bmatrix} 1 & 0 & 0 \\ -2 & 2 & 0 \\ 1 & -2 & 1 \end{bmatrix} \cdot 3 \cdot \begin{bmatrix} -1 & 1 & 0 & 0 \\ 0 & -1 & 1 & 0 \\ 0 & 0 & -1 & 1 \end{bmatrix} \\ M_{D,3} = 3 \cdot \begin{bmatrix} -1 & 1 & 0 & 0 \\ 2 & -4 & 2 & 0 \\ -1 & 3 & -3 & 1 \\ \end{bmatrix}$
$\begin{bmatrix} \mathbf{c} \\ \mathbf{b} \\ \mathbf{a} \end{bmatrix} = \begin{bmatrix} c_x & c_y \\ b_x & b_y \\ a_x & a_y \end{bmatrix} = 3 \cdot \begin{bmatrix} - \mathbf{P}_1 + \mathbf{P}_2 \\ 2 \cdot (\mathbf{P}_1 - 2 \cdot \mathbf{P}_2 + 1 \cdot \mathbf{P}_3) \\ - \mathbf{P}_1 + 3 \cdot \mathbf{P}_2 -3 \cdot \mathbf{P}_3 + \mathbf{P}_4 \end{bmatrix}$
Here, the three row vectors $\mathbf{a}$, $\mathbf{b}$, and $\mathbf{c}$ hold the parameters for the second-degree equations that allow us to find the zeros in the two dimensions X and Y. Note that $\mathbf{b}$ can be easily divided by $2$, so we can find the roots as follows:
$t_{x,12} = \frac{- \frac{b_x}{2} \pm \sqrt{\left(\frac{b_x}{2} \right)^2 - a_x \cdot c_x}}{a_x} \\ t_{y,12} = \frac{- \frac{b_y}{2} \pm \sqrt{\left(\frac{b_y}{2} \right)^2 - a_y \cdot c_y}}{a_y}$