12. If a particle moves in the \( xy \)-plane so that at time \( t > 0 \) its position vector is \( (t^3 - 1, \ln{\sqrt{t^2 + 1}}) \), then at time \( t = 1 \), its velocity vector is
Answer is: option3
Solution:
\( (3, \frac{1}{2}) \)
