Viviani’s curve for a cylinder and sphere intersection in parametric form

Question

How to find Vivianis curve of the following intersection in parametric form \begin{align*} x^2+y^2&=\frac14\tag{1} \\ \left(x+\frac12\right)^2+y^2+z^2&=1\tag{2} \end{align*}

We are given $-1\leq z\leq 1$ and $z=\sin t$

My understanding is that we have to eliminate $x,y,z$ but it’s not clear where the above line comes from...

I have attempted to proceed by solving for z in equation $(2)$ given $x=a\cos t$ and $y=a\sin t$

$$1=(x+a)^2+y^2+z^2$$

I arrive at $z^2=1-\cos^2(t/2)$... not sure how to proceed further...

Answer 1

You just can take the spherical coordinates: $$x= \cos(\theta)\cos(\varphi)-0.5 , y=\cos(\theta)\sin(\varphi), z=sin(\theta). $$ Then set $\varphi=\theta$ and you get $$x= \cos(\theta)\cos(\theta)-0.5 , y=\cos(\theta)\sin(\theta), z=sin(\theta). $$ You can then check that they fulfil the equation of the cylinder (see Wikipedia).