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r(t)=tan1ti+sintj+t2k\overset{\rightarrow}{r(t)}=tan^{-1}ti+sintj+t^2k ആയാൽ r(t)t=0=\overset{\rightarrow}{r'(t)}_{t=0}=

Ai-j

B1/2i+j+k

Ci+j-k

Di+j

Answer:

D. i+j

Read Explanation:

r(t)=11+t2i+costj+2tk\overset{\rightarrow}{r'(t)}=\frac{1}{1+t^2}i+costj +2tk

r(t)t=0=11+0i+cos0+2×0k\overset{\rightarrow}{r'(t)}_{t=0}=\frac{1}{1+0}i+cos0+2 \times 0k

=i+j=i+j


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