= Solution
Use the convention $x=r\sin\theta\cos\phi$, $y=r\sin\theta\sin\phi$, $z=r\cos\theta$, with $\theta$ measured from the positive vertical axis. The squared <radius> is
$$
r^2=\frac{\cos^2t+\sin^2t+t^2}{1+t^2}=1.
$$
Since the horizontal radial distance is $1/\sqrt{1+t^2}>0$ for finite $t$,
$$
\cos\theta=\frac{t}{\sqrt{1+t^2}},\qquad
\sin\theta=\frac1{\sqrt{1+t^2}}.
$$
The horizontal angle advances with $t$. Thus the <spherical coordinates> are
$$
\boxed{r(t)=1,\qquad\theta(t)=\frac\pi2-\arctan t,\qquad
\phi(t)=t\pmod{2\pi}.}
$$
An unwrapped continuous azimuth is simply $\phi=t$; a principal-angle convention introduces harmless jumps by $2\pi$. The polar angle stays in $(0,\pi)$ and approaches the two poles only as $t\to\pm\infty$.
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