Answer
$\theta=\dfrac{\pi}{2}$
Work Step by Step
$r_1'(t)=-\sin t i+\cos t j+k$ when $t=0$ then we have $ r_1'(0)=0 i+ j+k$
$r_2'(t)=i+2t j+3t^2 k$ when $t=0$ , then we have $r_1'(0)=i+0 j+0 k$
The angle can be found by using the dot product as:
$\theta=\cos^{-1} [\dfrac{a \cdot b}{|a||b|}]$
Now, we have
$\theta=\cos^{-1} [\dfrac{(0) \times (1)+(1) \times (0)+(1) \times (0)}{|1||1|}]$
$\theta=\cos^{-1}(0)$
Hence, we have $\theta=\dfrac{\pi}{2}$
