Answer
$\dfrac{\pi}{2}$
Work Step by Step
Since we know velocity is:
$v(t)=r'(t)=\lt 2t(t^2+1)^{-1},(t^2+1)^{-1},t(t^2+1)^{-1/2} \gt$
and $v(0)=\lt 0,1,0 \gt$
And we know the acceleration is:
$a(t)=v'(t)=\lt 2t(t^2+1)^{-1}-4t^2(t^2+1)^{-2},-2t(t^2+1)^{-2},(t^2+1)^{-1/2} -t^2(t^2+1)^{-3/2}\gt$
and $a(0)= \lt 2,0,1 \gt$
Hence, $\theta =\cos ^{-1}\dfrac{v(0) \cdot a(0)}{|v(0)||a(0)|}=\cos ^{-1}(\dfrac{0}{\sqrt{1}\sqrt{5}})=\cos ^{-1} (0)$
or, $\theta=\dfrac{\pi}{2}$
