If one saw a parametric equation $r(u,v)=<\cos^3u\cos^3v,\sin^3u\cos^3v,\sin^3v$ > on an exam, how would they be able to visualize that this is an astroid shape without prior knowledge of that fact? Or at least how would one match the equation to an astroid?
I notice that if you set u constant you get y=x which would make me think to look for a grid line on the surface that is a straight line in the xy plane. But besides that I see no way to know how to visualize the other parts of it.
I would appreciate a veteran approach to this.
Thanks.
$$((\cos(u)\cos(v))^3,(\sin(u)\cos(v))^3,(\sin(v))^3) \tag{1}$$
A graphical representation :
If the cubes are suppressed in (1):
$$(\cos(u)\cos(v),\sin(u)\cos(v),\sin(v))$$
one finds the classical representation of the unit sphere ($u$ = longitude, $v$ = latitude) providing a key of understanding, as a kind of "deflation" of the unit ball (important : it suffices to understand what happens in the first octant (all coordinates positive).
It will be simpler to see it on the implicit equations.
The implicit equation of the unit sphere is
$$x^2+y^2+z^2=1,$$
whereas the implicit equation of the 3D astroid surface is
$$x^{2/3}+y^{2/3}+z^{2/3}=1,$$
appearing in this way as a natural extension of the 2D case, ($x^2+y^2=1$ vs. $x^{2/3}+y^{2/3}=1$) or with parametric representations $(\cos(t),\sin(t))$ for the unit circle vs. $(\cos^3(t),\sin^3(t))$ for the astroid.
A side remark: this action (cancellation of $z$ coordinate) explains why the intersection of the surface represented above with coordinate plane $xOy$ appears as an astroid. The same for coordinate planes $yOz$ and $xOz$.