Envelope of one parametric family of lines $y=mx+f(m)$ is claimed to touch the given family of lines. I find/list five cases of lines and their respective envelopes below:
\begin{matrix} [1]: y=mx+e^{-m} & \rightarrow y=-x(\ln{x}-1) \\ [2]: y=mx+\sin m & \rightarrow y=x \cos^{-1}(-x)+\sqrt{1-x^2} \\ [3]: y=mx+m^3 & \rightarrow y=\frac{2x \sqrt{-x}}{3\sqrt{3}} \\ [4]: y = mx+m^2 &\rightarrow y = \frac{-x^2}{4} \\ [5]: y = mx+\cos m & \rightarrow y = \sqrt{1-x^2}+x\sin^{-1}(x) \end{matrix}
I find that in [1] and [4] $m$ can take any real value for a line to touch its envelope. In other cases $m$ is restricted: In case [2], $m \in [0,\pi]$. In case [3], $m \in [0,\infty).$ In case [5], $m\in [-\pi/2, \pi/2].$ I would like to know why and when $m$ gets restricted to a smaller domain than that of the real set. Please help, Thanks in advance.
The envelope of a family of lines given by $y=mx+f(m)...(1)$ is found as $x+f'(m)=0...(2)$ where $m=y'$ so one solves the ODE $x+f'(y')=0...(3)$ and gets the envelope as $y=g(x)$...(4). Its is crucial to note that the domain of $g(x)$ and Eq. (2) simultaneously determine all the possible values of $m$ for which the line (1) will touch the fixed curve (envelope) given by Eq. (4).
Let us analyse five interesting examples of yours, one by one.
[1]: $x=e^{-m}$ and the domain of $g(x)$ is $x \in (0, \infty)$. Hence $m$ can take any real value.
[2]: $x+\cos m=0 \Rightarrow m= \cos^{-1} (-x) \Rightarrow m \in [0,\pi]$ as the domain of $g(x)$ is $[-1,1]$.
[3]: $x+3m^2=0\Rightarrow m=\pm \sqrt{\frac{-x}{3}}$. In this case one should get two envelopes $y=\pm g(x), ~\mbox{for}~ x\le 0.$ So, for $m \ge 0$ the envelope is $y=+g(x)$ and for $m\le 0$ it is $y=-g(x).$ In your listing you have missed out the second part.
[4]: $x+2m=0$ and the domain of $g(x)$ is $(-\infty, \infty).$ Hence $m \in (-\infty,\infty).$
[5]: $x-\sin m=0 \Rightarrow m=\sin^{-1} (x) \Rightarrow m \in [-\pi/2,\pi/2]$ as the domain of $g(x)$ is $[1,1].$
You have made an interesting observation, keep it up. In this regard, one may see Clairaut equation (ODE):
https://en.wikipedia.org/wiki/Clairaut%27s_equation