## Thomas' Calculus 13th Edition

$\dfrac{\pi}{2}$
We know that the velocity and acceleration is defined as: velocity is given as: $v(t)=r'(t)$ and acceleration is given as: $a(t)=v'(t)$ Now, $v(t)=r'(t)=\dfrac{1}{3}\lt 2(1+t)^{(1/2)},-2(1-t)^{(1/2)},1 \gt \\ \implies v(0)=\dfrac{1}{3}\lt 2,-2,1 \gt$ and $a(t)=\dfrac{1}{3}\lt (1+t)^{-(1/2)},(1+t)^{-(1/2)},0 \gt \\ \implies a(0)= \dfrac{1}{3}\lt 1,1,0 \gt$ As we know that $\theta =\cos ^{-1}\dfrac{v(0) \cdot a(0)}{|v(0)||a(0)|}$ Thus, $\theta=\cos ^{-1}(\dfrac{0}{(1)(\dfrac{\sqrt{2}}{3})})=\dfrac{\pi}{2}$