Now when c t t i t 2 j t k t 1 we see that c t i 2 t

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Now whenc(t) =ti+t2j+tk,0t1,we see thatc(t) =i+ 2tj+k,0t1,whileF(c(t)) = 3t3i+ 4t2j+ 2t3k.In this case,F·c(t) = 13t3,so thatI= 13integraldisplay10t3dt= 13bracketleftBig14t4bracketrightBig10.Consequently,Work Done =134units.

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