Calculus 8th Edition

$\cos ^{-1}\dfrac{1}{\sqrt 6}\approx 66^\circ$
As we have from the given curves $r_1(t)=\lt t,t^2,t^3 \gt$ and $r_2(t)= \lt \sin t,\sin 2t,t \gt$ Tangent vector $r'(t)$ can be found by taking the derivative of $r_1(t)$ and $r_2(t)$ with respect to $t$. This yields,$r_1'(t)=\lt 1,2t,3t^2 \gt$ ,$r_2'(t)= \lt \cos t,2 \cos t,1 \gt$ $r_1'(0)=\lt 1,0,0 \gt$ and $r_2'(t)= \lt 1,2,1 \gt$ Consider $\theta$ be the angle between the two curves. Thus, $\cos \theta=\dfrac{\lt 1,0,0 \gt \cdot\lt 1,2,1 \gt}{\sqrt {1+0+0}\sqrt {1+4+1}}$ or, $\theta=\cos ^{-1}[\dfrac{\lt 1,0,0 \gt \cdot\lt 1,2,1 \gt}{\sqrt {1+0+0}\sqrt {1+4+1}}]$ or, $=\cos ^{-1}\dfrac{1}{\sqrt 6}\approx 66^\circ$