Question

Please explain all steps.

1. (25 pts) Let F(x, y, z) = (2xy + 25)i + (4.r?y3 + 2yz?)j + (5.624 + 3y222)k and let C be the curve given parametrically by r(t) = (3t+1)i + tºj + 5tk for 0 <t<1. Evaluate the line integral (Fd

Thanks!

Answer 1

for ostsi solution F19,4.3) = (229* + 29) 1+ (98y?+ 2yz!)j + (5224+ 3y?z? Jé 04+) = (3t+1) ++?; + 5tk First we need the the vector field evaluated along the leave. F(814) = { (3t+1) C29%+2547%)s'+[413t+1)*; [ty+ 2(49)(5213); + (5(37+1)&(st)+3142)? (5+)?) fešlus) - [2(3t+1)+8 + 3125+*]i + [4t" (?t?+1+6+) + 25 [3125 *(3+1) + 15 = 3k = [6t+2+8+31257575+ [36+®+4+*+24+7)+250tJit (9375073125*25JA Now the set to derivative of the parametrization + 250+5); +

+ p ' (t) = 36+2+ ſ +5X Finally let's get the dot product taken care of (olp) .3'142 18+'+678 + 937575 + 72+9+8+*+48 +*+ 500t6 +4687575 +15625t" + 37576 Flole)°24 = 9049 +54487847487546 + 5625075 +150257" b + The line integral is then Pal ridra 1562545 5) 7 faida -}(90+ (90t+54 +8+8+7 +875467 5625075 +15625t") dt Sedm = ( 9ot "t sy eet ett 8754* + 562 50 t, tror = (970 +679 + 78 +125+*+9875 ++312545) JF de 29+6+1+1257 987 5+3125) - 0) Side = 1314 6 5 > 9. 54)/