## Bezier Curves

Suppose we don't want to deal with derivatives when interpolating polynomial curves. Instead of interpolating through Hermite curves, we can use Bezier curves to represent an interpolated cubic polynomial.

Pierre Bezier invented the Bezier curve when working at Renault.

We have four control points $P_0$, $P_1$, $P_2$, and $P_3$. The curve starts at $P_0$, and ends at $P_3$.

The matrix multiplication expression to interpolate a Bezier curve at any parameter $t$ is given by:

$P_0$, and $P_3$ are points in which the curve ultimately pass through. $P_1$, and $P_2$ serve the purpose in that they control the curvature of the curve. Moving these points around individually yields a different curve.

So how do we obtain the coefficient matrix $\bold{M}$?

The formal definition of the Bezier curve is given by the expression:

Where ${n \choose i} = \frac{n!}{i!(n-i)!}$.

We only need to worry about $i = 0 \rightarrow 3$ here, so $n = 3$. If we expand the above, and sum all the polynomials, we obtain the basis polynomial:

Expanding each term:

Taking the coefficients and forming the basis vectors for each polynomial we can then obtain the coefficient matrix $\bold{M}$:

### Example

Given $P_0 = (-0.4, 0)$, $P_1 = (-0.2, 0.4)$, $P_2 = (0.2, 0.5)$, $P_3 = (0.3, -0.3)$, we obtain the following curve where it passes through the first ($P_0$), and last $P_3$ control points.

Here is a video example showing adjustment of each control point yielding different curve: