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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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.
Narasimham
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Thank you.Does parameter "t" go from 0 to 1? Which value from 0 to 1 should I choose while constructing "x" and "y"? – Jan 24 '21 at 20:34
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1Please check at what $t$ value cusp occurs. Extend the domain to include it and view it by graphing. – Narasimham Jan 24 '21 at 20:38
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Isn't it enough to check for which $t$ that $\dot{x}^2+\dot{y}^2=0$? – Jap88 Jan 24 '21 at 22:12
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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. – fang Jan 25 '21 at 02:14
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@Jap88 Sorry about the typo. – Narasimham Jan 25 '21 at 06:46
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@SergeyDukman Sorry there was a typo Please read Zeeman's (and Rene Thom's ) Cusp catastrophe if possible, Swallow tail cusp etc. – Narasimham Jan 25 '21 at 06:56
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Alright. Will do – Jan 25 '21 at 13:12
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Ah, the points are on a line. Thanks @Narasimham. – Jap88 Jan 25 '21 at 18:11
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The curve is $x=1+2t$ and $y=1+2t$ and does not have any cusp. – Jap88 Jan 25 '21 at 18:28
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Not clear to me. $( x'^2+y'^2)^{\frac32} =2 \sqrt 2 \ne 0.$ A straight line has no cusp. – Narasimham Jan 25 '21 at 18:48
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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.
bubba
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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? – Jan 30 '21 at 10:36
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1You 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. – bubba Jan 31 '21 at 10:12