University Calculus: Early Transcendentals (3rd Edition)

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