## Calculus (3rd Edition)

$$r(t)=\lt \frac{\sqrt 3}{2}\cos t, \frac{1}{2}, \frac{\sqrt 3}{2}\sin t \gt.$$
The intersection of the plane $y = 1 /2$ with the sphere $x^2 + y^2 + z^2 = 1$ is given by $$x^2 + (1/4)^2 +z^2 = 1\Longrightarrow x^2 + z^2 = 3/4.$$ To find a parametrization, we put $x=\frac{\sqrt 3}{2}\cos t$ and $z=\frac{\sqrt 3}{2}\sin t$; that is, we have the parametrization $$r(t)=\lt \frac{\sqrt 3}{2}\cos t, \frac{1}{2}, \frac{\sqrt 3}{2}\sin t \gt.$$