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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

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2 Answers 2

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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.

enter image description here

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  • $\begingroup$ Thank you.Does parameter "t" go from 0 to 1? Which value from 0 to 1 should I choose while constructing "x" and "y"? $\endgroup$
    – user504068
    Jan 24, 2021 at 20:34
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    $\begingroup$ Please check at what $t$ value cusp occurs. Extend the domain to include it and view it by graphing. $\endgroup$
    – Narasimham
    Jan 24, 2021 at 20:38
  • $\begingroup$ Isn't it enough to check for which $t$ that $\dot{x}^2+\dot{y}^2=0$? $\endgroup$
    – Jap88
    Jan 24, 2021 at 22:12
  • $\begingroup$ The statement "zero curvature gives a cusp" is quite questionable. A straight line has zero curvature everywhere but there is no cusp in a straight line. $\endgroup$
    – fang
    Jan 25, 2021 at 2:14
  • $\begingroup$ @Jap88 Sorry about the typo. $\endgroup$
    – Narasimham
    Jan 25, 2021 at 6:46
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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.

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  • $\begingroup$ Thanks a lot. At least what you say is understandable. So in order to detect cusps I need to plot the graph of the given function and inspect it? $\endgroup$
    – user504068
    Jan 30, 2021 at 10:36
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    $\begingroup$ You can look for places where the first derivative vector is zero. If there aren’t any, then there are no cusps. That’s the case with your particular curve. A place where the derivative is zero may or may not be a cusp. To find out, you have to look at the direction of the derivative vector (I.e. the tangent vector) slightly before and after the zero point. $\endgroup$
    – bubba
    Jan 31, 2021 at 10:12

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