Cusps on quadratic Bezier curve

Question

A quadratic Bezier curve is given in parametric form by: $$C(t) = (1-t)^2P_0 + 2(1-t)tP_1 + t^2P_2.$$

My points are: $(1,1)$, $(2,2)$ and $(3,3)$. How do I show that this curve has cusps?

Best regards, Sergey

Answer 1

Compute curvature with standard parametric form. Zero radius of curvature gives a cusp.

Primed with respect to $t$,

$$\kappa= \dfrac{(x'y''-y'x'')}{ (x'^2+y'^2)^{\frac32} }\to \infty $$

At a cusp slope $\dfrac{y'}{ x'} $ remains stationary or maximum/minimum.

Answer 2

I’m assuming that a “cusp” is defined to be a place where the curve’s tangent direction is discontinuous; i.e. there’s a sharp corner.

Your particular curve is a straight line, so it certainly doesn’t have a cusp.

In general, a quadratic Bézier curve is a parabola, and parabolas don’t have cusps, either.

It might seem that cusps occur at points where the first derivative vector has zero length. But this isn’t true. Take the quadratic Bézier curve with control points $(0,0)$, $(0,0)$, $(1,1)$, for example. It has zero first derivative at the origin, but it’s a straight line, so it doesn’t have a cusp.

Generally, you can’t learn about cusps by looking at derivatives and their magnitudes, because these depend on how the curve is parameterized, whereas the presence of a cusp is a purely geometric thing that does not depend on any particular parameterization.