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 $$\bigg\Vert\frac{d\mathbf{T}}{dt}\bigg\Vert=\frac{v}{\rho},$$ $\rho>0$ being the radius of curvature at any instant $t$ , we arrive to the useful equation $$v=\omega\rho.$$

Footnotes

...¹

By applying the chain rule, $$\bigg\Vert\frac{d\mathbf{T}}{dt}\bigg\Vert=\bigg\Vert\frac{d\mathbf{T}}{ds}\bigg\Vert\bigg\vert\frac{ds}{dt}\bigg\vert= \bigg\Vert\frac{\mathbf{N}}{\rho}\bigg\Vert v=\frac{v}{\rho},$$ by Frenet-Serret. $\mathbf{N}$ is the normal vector.