r(t)→=tan−1ti+sintj+t2k\overset{\rightarrow}{r(t)}=tan^{-1}ti+sintj+t^2kr(t)→=tan−1ti+sintj+t2k ആയാൽ r′(t)→t=0=\overset{\rightarrow}{r'(t)}_{t=0}=r′(t)→t=0= Ai-jB1/2i+j+kCi+j-kDi+jAnswer: D. i+j Read Explanation: r′(t)→=11+t2i+costj+2tk\overset{\rightarrow}{r'(t)}=\frac{1}{1+t^2}i+costj +2tkr′(t)→=1+t21i+costj+2tkr′(t)→t=0=11+0i+cos0+2×0k\overset{\rightarrow}{r'(t)}_{t=0}=\frac{1}{1+0}i+cos0+2 \times 0kr′(t)→t=0=1+01i+cos0+2×0k=i+j=i+j=i+j Read more in App