What is the definition of a cusp of a plane curve and how to find them?

Question

I am pretty confused about what the definition of a cusp of a plane curve is (let's say that the curve admits the parametrization $\gamma(t)=(f(t), g(t))$). On Wikipedia I found two possible definitions: one that says that it is a singular point (that is a point where $\gamma'(t)=0$) satisfying some condition on a directional derivative that I do not understand and one that says that it is a point where the curve is not differentiable.

This confuses me because I was looking at this post How to make a sharp 5-pointed astroid in parametric coordinates? and from the comments I gather that the following curve $\gamma(\varphi)=(\frac{n - 1}{n}\cos(\varphi) + \frac{1}{n}\cos\left( (n-1) \varphi \right), \frac{n - 1}{n}\sin(\varphi) - \frac{1}{n}\sin\left( (n-1) \varphi \right)$ has $n$ cusps. So I think that cusps are indeed singular points, since this curve is everywhere differentiable. However, what other conditions should I impose for a cusp and how would I show that this particular curve has $n$ cusps?

Answer 1

I am posting an (attempt at an) answer to my own question on how to find the cusps of that $$\gamma(\varphi)=\left(\frac{n - 1}{n}\cos(\varphi) + \frac{1}{n}\cos\left( (n-1) \varphi \right), \frac{n - 1}{n}\sin(\varphi) - \frac{1}{n}\sin\left( (n-1) \varphi \right)\right)$$ curve.

Firstly, $$\gamma'(\varphi)=\left(-\frac{n-1}{n}\sin(\varphi)-\frac{n-1}{n}\sin((n-1)\varphi), \frac{n-1}{n}\cos(\varphi)-\frac{n-1}{n}\cos((n-1)\varphi)\right),$$ so it is easy to see that $\gamma'(\varphi)=0 \iff \varphi \in \{\frac{2m\pi}{n}|m\in \mathbb{Z}\}$.

Now, $||\gamma'(\varphi)||=\sqrt{\frac{4(n-1)^2\sin^2\left(\frac{n\varphi}{2}\right)}{n^2}}=\frac{2(n-1)}{n}|\sin\left(\frac{n\varphi}{2}\right)|$ and this shows me that when $\varphi=\frac{2m\pi}{n}$ for some $m\in \mathbb{Z}$ we will have limiting unit tangent vectors of opposite signs, so all these point will be cusps.
Due to the fact that $\gamma$ is periodic of period $2\pi$ (I think this means that the curve is closed), we only have to count $m\in \{0, 1, ..., n-1\}$ and this is why there will be $n$ cusps.
EDIT: I will add some more details now. We have $\frac{\gamma'(\varphi)}{||\gamma'(\varphi)||}=-(\frac{\sin(\frac{n\varphi}{2})\cos(\frac{(n-2)\varphi}{2})}{|\sin(\frac{n\varphi}{2})|}, \frac{-\sin(\frac{n\varphi}{2})\sin(\frac{(n-2)\varphi}{2})}{|\sin(\frac{n\varphi}{2})|})$.
So, now it is clear that for $\varphi \nearrow \frac{2m\pi}{n}$ and $\varphi \searrow \frac{2m\pi}{n}$ we get vectors of opposite sign. If we want to write them down explicitly we should discuss this after $m$'s parity, but I think this is pretty straightforward now.

Answer 2

At a cusp the radius of curvature is infinite. Its reciprocal, the curvature of curve is zero. This happens in the cases you mentioned.

The curvature locally switches/swaps sign between $ \pm \infty$ .

Recall that at the cusp we have radius of curvature expression in Cartesian coordinates

$$ R=\frac{(x^{'2}+y^{'2})^{3/2}}{xy^{''}- y x^{''}}\to 0 $$

We have at cusp $$ \frac{dy}{dx}=const, R=0 $$

In polar coordinates

$$ R=\frac{{(r^{'2}+ r^{'2})^{3/2}}}{{(r^2+2r^{'2}-r r^{''})}}\to 0 $$

At cusp we have $$\psi=0, \psi'=0, R=0. $$

Other familiar examples are a cardioid, a multi cornered star, trochoid or plane cycloid with cusps created when a point on a rolling circle touches the straight line or circle it is rolling upon with no slip. $\psi$ is angle made between the arc and radius vector.