I need to parametrize the intersection between the cylinder $ x^2 + y^2= \frac{1}{4}$ and the sphere $(x+ \frac{1}{2})^2 + y^2 +z^2 = 1$.
I tried parametrizing the first equation which gives $r(t) = (\frac{cos(t)}{2}, \frac{sin(t)}{2})$ since the radius is 1/2.
Then, I plugged in the values in the second equation which yields $(\frac{cos(t) +1}{2})^2 + \frac{sin^2(t)}{4} + z^2 = 1$. We isolate $z$ and get that $z = \sqrt{1-\frac{cos(t)}{2}}$. I'm stuck here because the parametrization would be incomplete if we choose the positive or negative root, am I doing something wrong? If so, what would a correct parametrization be?
EDIT: I've just recalled that $\sqrt{\frac{1-cos(t)}{2}} = sin(\frac{x}{2})$ or $cos(\frac{x}{2})$. Is this helpful?
$x^2+y^2=\frac{1}{4}$
$\left( x+\frac{1}{2} \right)^2 + y^2 + z^2 = 1$
Expand
$x^2 + x + \frac{1}{4} + y^2 + z^2 = 1$
Collect
$\left(x^2+y^2\right)+x+\frac{1}{4}+z^2=1$
Substitute first equation
$\frac{1}{4}+x+\frac{1}{4}+z^2=1$
The key equations now being$\ldots$
$x+z^2=\frac{1}{2}$
$x^2+y^2=\frac{1}{4}$
We pick $z=t$ hoping for a cleaner solution...
$x=\frac{1}{2}-t^2$
$x^2+y^2=\left( \frac{1}{2}-t^2 \right)^2+y^2=t^4-t^2+\frac{1}{4}+y^2=\frac{1}{4}$
$y^2=t^2-t^4$
$y^2=t^2\left(1-t^2\right)$
$y=\pm t\sqrt{1-t^2}$
And we write...
$\left(\frac{1}{2}-t^2,\pm t\sqrt{1-t^2},t\right)$ for $t \in \left[-1,1\right]$
and I expect someone else to point out the algebra mistake if there is one because there's alot of expansion in my approach, but I checked it and I'm pretty sure it's consistent.
Because the cylinder is of lesser radius, the intersection is a single point at the xy plane. The vertical (xy) projection of the curve is a circle. The lateral (xz) cross section of the curve is a parabola.