Finding irregular points of a homotopy between two parametrisations of the unit circle with varying speed

Question

The curve $d: [0, 2 \pi] \to \mathbb R$, $d(t) := \left(\cos\left(t \cdot e^{t - 2 \pi}\right), \sin\left(t \cdot e^{t - 2 \pi}\right)\right)$ is a parametrisation of the unit circle of non-constant velocity.

I suspect (from graphing) that the following homotopy between the reguarly parametrised unit circle and $d$ is irregular for $c \approx 0.27$ and $c \approx 0.83$, i.e. $\frac{\partial h(t,c)}{\partial t} = 0$, but I haven't been able to do the algebraic manipulations: $$ h: [0, 2 \pi] \times [0, 1] \to \mathbb R, \ (t, c) \mapsto \begin{pmatrix} (1 - c) \cos\left(t \cdot e^{t - 2 \pi}\right) + c \cdot \cos(t) \\ (1 - c) \sin\left(t \cdot e^{t - 2 \pi}\right) + c \sin(t) \end{pmatrix}. $$ We have $$ \frac{\partial h(t,c)}{\partial t} = \begin{pmatrix} (c - 1) \sin(t) - c (t + 1) e^{t - 2 \pi} \sin\left(t \cdot e^{t - 2 \pi}\right) \\ (1 - c) \cos(t) + c (t + 1) e^{t - 2 \pi} \cos\left(t \cdot e^{t - 2 \pi}\right) \end{pmatrix} $$ Thus we have to solve $$ (c - 1) \sin(t) = c (t + 1) e^{t - 2 \pi} \sin\left(t \cdot e^{t - 2 \pi}\right) \quad \text{and} \quad (c - 1) \cos(t) = c (t + 1) e^{t - 2 \pi} \cos\left(t \cdot e^{t - 2 \pi}\right). $$ The method for solving them will be analogous, so lets focus on the first one, which can be rearranged to $$ \frac{c - 1}{c} e^{2 \pi} = (t + 1) e^{t} \frac{\cos\left(t \cdot e^{t - 2 \pi}\right)}{\cos(t)}, $$ but I have no idea how to continue. I suspect we have to use the Lambert W-function i.e. the product log function but I don't know how because of the cosines involved.

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A similar problem arises when I try to find the intersection of the ray extending from the origin and going through $d'(t)$ and the curve $d$ itself. We are thus trying to find the function $\lambda$ satisfying $\lambda(t) d'(t) = d(t)$, i.e. $$\lambda(t) \begin{pmatrix} -e^{t - 2 \pi} (t + 1) \sin(e^{t - 2 \pi} t) \\ e^{t - 2 \pi} (x + 1) \cos(e^{t - 2 \pi} t)\end{pmatrix} = \begin{pmatrix} \cos(t e^{t - 2 \pi}) \\ \sin(t e^{t - 2 \pi})\end{pmatrix},$$ which can be rearranged to $$ \lambda(t) = \frac{\cos(e^{t - 2 \pi} t)}{-e^{t - 2 \pi} (t + 1) \sin(e^{t - 2 \pi} t)} = - \frac{e^{2 \pi}\cot(t e^{t- 2\pi})}{(t + 1)e^{t}}$$ and $$ \lambda(t) = \frac{\sin(e^{t - 2 \pi} t)}{e^{t - 2 \pi} (t + 1) \cos(e^{t - 2 \pi} t)} = \frac{e^{2 \pi} \tan(t e^{t - 2 \pi})}{(t + 1) e^t}$$

Answer 1

Thus we have to solve $$ (c - 1) \sin(t) = c (t + 1) e^{t - 2 \pi} \sin\left(t \cdot e^{t - 2 \pi}\right) \quad \text{and} \quad (c - 1) \cos(t) = c (t + 1) e^{t - 2 \pi} \cos\left(t \cdot e^{t - 2 \pi}\right). $$

It follows

$$(c - 1) \sin(t) c (t + 1) e^{t - 2 \pi} \cos\left(t \cdot e^{t - 2 \pi}\right)= (c - 1) \cos(t) c (t + 1) e^{t - 2 \pi} \sin\left(t \cdot e^{t - 2 \pi}\right)$$

Assuming $c\ne 0,1$, it follows

$$\sin(t)\cos\left(t \cdot e^{t - 2 \pi}\right)= \cos(t)\sin\left(t \cdot e^{t - 2 \pi}\right),$$

that is $$\sin(t(1- e^{t - 2 \pi}))=0,$$

so $f(t)=t(1- e^{t - 2 \pi})=\frac {\pi n}{2},\, n\in\Bbb Z\tag{1}$

The graph of the function $f(t)$

suggests that for $t\in [0,2\pi]$ Equation (1) has six solutions $(t,n)$, namely, $(t_0=0,0)$, $(t_1,1)$, $(t_2,2)$, $(t_3,2)$, $(t_4,1)$, and $(t_6=2\pi,0)$. Equation (1) is transcendental, and I expect that roots of such equations have simple analytic expressions only in exceptional cases. On the other hand, the values of $t_i$ for $i=1,\dots, 4$ can be approximated numerically. Next for each $t_i$ we have a linear equation for $c$, for instance, $$(c - 1) \sin(t_i) = c (t_i + 1) e^{t_i - 2 \pi} \sin\left(t_i\cdot e^{t_i - 2 \pi}\right).$$

Answer 2

Too long for comments.

Starting from @Alex Ravsky's answer, we can have explicit decent approximations for the six solutions of $$f(t)=t(1- e^{t - 2 \pi})=\frac {\pi n}{2}$$ For sure, we shall not work the trivial cases $t_1=0$ and $t_6=2\pi$.

Using the simplest $[1,1]$ Padé approximants built around the bounds, we have $$t_2=\frac{2 e^{\pi } \pi \sinh (\pi )}{8 \sinh ^2(\pi )-\pi }\approx 1.57838$$ while the exact solution is $1.58524$.

Similarly, $$t_5=2\pi-\frac{6(7 \pi-1) }{146 \pi -45}\approx 5.97872$$ while the exact solution is $5.97836$.

More interesting are the two other roots since $f'(t)$ cancels at $$t_*=W\left(e^{1+2 \pi }\right)-1$$ from which $$t_3=t_*-\sqrt{2} \sqrt{t_*+\frac{4+\pi }{t_*+2}-(\pi+2)}\approx 3.55408$$ $$t_4=t_*+\sqrt{2} \sqrt{t_*+\frac{4+\pi }{t_*+2}-(\pi+2)}\approx 5.57878$$ while the exact solutions are $3.31111$ and $5.41539$.