I want to prove a theorem about a vector curve ${\bf c}(t): \mathbb{R} \to \mathbb{R}^3$ (for $t \in [a, b]$) which lies on the surface of a sphere in $\mathbb{R}^3$. It is in my understanding that this curve should be expressed in general. One can take ${\bf c}(t)$ to satisfy the equation $x^2+y^2+z^2=d$, for some constant $d$, such as $\sqrt{d}(\cos(t), \cos(t)\sin(t), \sin^2(t)$ since $\cos^2(t)\sin^2(t) + \sin^4(t) + \cos^2(t)) =1 $. But this curve looks like a spherically curved infinity sign, which is understandable - this curve is not general.
But how can one express this curve in general?
We can parametrize your sphere by: $${\bf x}(\theta,\phi) = (\sqrt{d}\cos \theta \cos \phi, \sqrt{d}\cos \theta \sin \phi, \sqrt{d}\sin \theta)$$so that a generic curve is: $${\bf c}(t) = {\bf x}(\theta(t),\phi(t)) = (\sqrt{d}\cos \theta(t)\cos \phi(t), \sqrt{d}\cos \theta(t) \sin \phi(t), \sqrt{d}\sin \theta(t)),$$ for maps $\theta = \theta(t)$ and $\phi = \phi(t)$.