Envelope of family of curves $x(u,v)=\cos^2(u)\cos(v)+\cos(u)\sin(u)\sin(v)$, $y(u,v)=\cos^2(u)\sin(v)-\cos (u)\sin(u)\cos(v)$

Question

How to generally find singular solution or envelope of a two parameter family of curves $ x(u,v),y(u,v) $ in the plane?

The parametric equations

$$x(u,v) = \cos^2 (u) \cos (v) + \cos (u) \sin (u) \sin (v),\\ y(u,v) = \cos^2 (u) \sin (v) - \cos (u) \sin (u) \cos (v),\\ (0 < v < 2 \pi), (0 < u < \pi ),$$

represent rotating circles of unit diameter passing through the origin and rotating about the origin. How do we obtain their envelope $ x^2+y^2=1 ? $

EDIT1:

The envelope is $x^2+y^2==1 $ as shown below.

EDIT 2:

We can wlog write $v = c $ for any rotated position.

$$x(u,c) = \cos^2 (u) \cos (c) + \cos (u) \sin (u) \sin (c),\\ y(u,c) = \cos^2 (u) \sin (c) - \cos (u) \sin (u) \cos (c).\\ $$

I think that I know the C-discriminant method with one parameter, but would like to know how to extend it to two parameters.

Can it be extended to 3D space to find a surface envelope?

Answer 1

So if I understood you correctly, we have the curves $\gamma_v(u):(0, \pi)\to\mathbb R^2$, given by: $$\gamma_v(u)=\begin{pmatrix}x_v(u)\\y_v(u)\end{pmatrix} = \begin{pmatrix}\cos^2 (u) \cos (v) + \cos (u) \sin (u) \sin (v)\\ \cos^2 (u) \sin (v) - \cos (u) \sin (u) \cos (v)\end{pmatrix}$$ and we are looking for the envelope of the family $\{\gamma_v(u)\mid v\in [0,2\pi)\}$.

I believe that you already found that each $\gamma_v(u)$ describes the circle $$\left( x-\frac12 \cos(v)\right)^2+\left( y-\frac12\sin(v)\right)^2=\frac14.$$

We now convert to the implicit form $$F(u,v)=\left( x-\frac12 \cos(v)\right)^2+\left( y-\frac12\sin(v)\right)^2-\frac14=0.$$ Note that $(0,0)$ lies on $F(u,v)=0$, for every $v$.

We can now find our envelope by solving: $$\begin{cases} F(u,v)=0\\ \\ \dfrac{\partial F(u,v)}{\partial v}=0\end{cases}$$ You can easily check that $\dfrac{\partial F}{\partial v}=x\sin(v)-y\cos(v).$ So we find that points lie on our envelope, if and only if, it is an intersection point of the circle $F(u,v)=0$ and the line $x\sin(v)-y\cos(v)=0$, for some $v$.

Since we can easily see that $\left(\frac12 \cos(v), \frac12 \sin(v)\right)$ (the centre of $F(u,v)=0$) always lies on the line $x\sin(v)-y\cos(v)=0$, we infer geometrically (see picture) that the envelope is: $$(0,0)\cup \left\{ (x,y)\mid x^2+y^2=1\right\}.$$ Note how the origin should be included in the envelope.

^{$\color{red}{\text{The circles $F(u,v)=0$ (red)}}$;
The line $x\sin(v)-y\cos(v)=0$;
$\color{blue}{\text{The envelope (blue)}}$.}

The same result can be derived using algebraic methods, but this is a lot more writing.