Problem : Represent the intersection of $x^2+y^2+z^2=5$ and $x=2y$ as a vector-valued function.
(Hint: Let $y=\cos t$)
My Attempt $$z^2 = 5- 4\cos^2t - \cos^2t = 5- 5\cos^2t$$
$$z = \sqrt{5(1-\cos^2t)} = \sin(t) \sqrt5$$
and that is where I am stuck. I cannot figure out whether I should use the positive or negative sign after taking the square root of $z$. Please let me know which one to choose and why.
Through the work that you've shown, you've come up with two possible answers:
$$ \left(2\cos(t),\cos(t),\sqrt{5}\sin(t)\right) $$ and $$ \left(2\cos(t),\cos(t),-\sqrt{5}\sin(t)\right). $$ both with a domain of $0\leq t<2\pi$.
Since both satisfy the given equations, both are correct. But also notice that if you replace $t$ with $-t$ in the first parametric form, you get $$ \left(2\cos(-t),\cos(-t),\sqrt{5}\sin(-t)\right)=\left(2\cos(t),\cos(t),-\sqrt{5}\sin(t)\right). $$
Therefore, the two parameterizations trace out the same curve just in different directions!
In general, there are many parameterizations for the same curve. For example, $$ \left(2\cos(20t),\cos(20t),\sqrt{5}\sin(20t)\right) $$ with a domain of $0\leq 2<\frac{\pi}{10}$ is also a parameterization for the intersection.