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# particle moving on the astroid at constant frequency

In parametric Cartesian equations, the astroid can be represented by

$x=a\cos^{3}\omega t,\quad y=a\sin^{3}\omega t,$ |

where $a>0$ is a known constant, $\omega>0$ is the constant angular frequency, and $t\in[0,\infty)$ is the time parameter. Thus the position vector of a particle, moving over the astroid, is

$\mathbf{r}=a\cos^{3}\omega t\,\mathbf{i}+a\sin^{3}\omega t\,\mathbf{j},$ |

and its velocity

$\mathbf{v}=-3a\omega\sin\omega t\cos^{2}\omega t\,\mathbf{i}+3a\omega\sin^{2}% \omega t\cos\omega t\,\mathbf{j},$ |

where $\{\mathbf{i},\mathbf{j}\}$ is a reference basis. Hence for the particle speed we have

$v=3a\omega\sin\omega t\cos\omega t.$ |

From the last two equations we get the tangent vector

$\mathbf{T}=-\sin\omega t\,\mathbf{i}+\cos\omega t\,\mathbf{j},$ |

and by using the well known formula ^{1}^{1}By applying the chain rule,
∥dTdt∥=∥dTds∥—dsdt—=
∥Nρ∥v=vρ,
by Frenet-Serret. $\mathbf{N}$ is the normal vector.

$\bigg\|\frac{d\mathbf{T}}{dt}\bigg\|=\frac{v}{\rho},$ |

$\rho>0$ being the radius of curvature at any instant $t$, we arrive to the useful equation

$v=\omega\rho.$ |

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## Comments

## Ultra minor correction: period

Could you remove the period at the end of the title? (I think this is way too minor to bother you with a formal correction).

## Re: Ultra minor correction: period

Thanks a lot my friend.