Compare the following two parametrized curves for $k \in \mathbb{N}^+$:

$$x_r(t) = \cos(t)(1 + r\sin(kt))$$ $$y_r(t) = \sin(t)(1 + r\sin(kt))$$

with $0 \leq t < 2\pi$ and $0 \leq r \leq 1$ (being the plot of the sine function with amplitude $r$ over the circle instead of the real axis) and

$$X_R(t) = R\cos(t/2)\sin(kt/2)$$ $$Y_R(t) = R\sin(t/2)\sin(kt/2)$$

(which give Grandi roses) with $R = 2\sqrt{r}$.

Find depicted the outer ("sine") curve $(x_r,y_r)$ thinner, the inner ("rose") curve $(X_R,Y_R)$ thicker, and the argument circle (for $t$) only looming.

The curves are coloured according to the argument $t$ that gives rise to them, the color ranging from **black** (for $t=0$) over **red** (for $t=\pi/2$), **white** (for $t=\pi$) and **blue** (for $t=3\pi/2$) back to **black** (for $t = 2\pi$).

Here for $k=1,3,5,7$:

My questions are:

Why doesn't this work for even $k$?

What happens when $r = 1$, that is when also $(x_r,y_r)$ exhibits an $k$-fold intersection point – like $(X_R,Y_R)$ always does?

Especially: How are the curves $(x_1,y_1)$ and $(X_2,Y_2)$ related (topologically)?

Which other pairs of curves $(x,y)$, $(X,Y)$ behave in a similar way?

For $r < 1$ the curves$(x_r,y_r)$ and $(X_R,Y_R)$ are obviously not homeomorphic. On the other hand $(x_1,y_1)$ and $(X_2,Y_2)$ are homeomorphic as point sets – but not as parametrized curves, because there is no continuous bijection $f: [0,2\pi] \rightarrow [0,2\pi]$ such that $x_1(t) = g(X_2(f(t)))$, $y_1(t) = g(Y_2(f(t)))$ with $g$ the homeomorphism that maps the two curves as points sets.

Is this the right way to say it – "homeomorphic as point sets, but not as parametrized curves" – and is that all there is to say?

To see what goes wrong for even $k$, find here the cases $k=2,4,6$. I didn't try to "fix" them: