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164 (1 CHAPTER 10 VECTOR FUNCTIONS 40.3. Are Length and Curvature 4. rt) = (Qeost.5.—2sint) = ir'(t}| = cost)? +? + (—2sin#)? = V2H. Then using Formula 3, we have Le 2% rola = £9, Vat = V251]",, = 20 v3, 2. r’(t) = (2t.cost + tsint — cost. — sind + Leos! + sind) = (24. tsint. tcost} => je'(O)| = /QE+ sin)? + (eos ty? = \/at? + P(sin? t+ cos?) = V5 It] = V5t for 0 < ( < 7. Then using Formula 3. we have L = ff r'()}dt = fo Vatdt = as] ° e (Qt) = 2tj+3tk = [| = MTF = tb VISOR (since t 2 0). Then La fi r(jlats fiivis or dt = % 54 ory]! = 3 (199? - 4/7) = (13° — 8) 4 ef(t) = is Vij 4 Gtk = [e'()| = Vd ts GF = /B0F + 2)? = 6k + 2] = G(t +2) ford St <) Then L = f2 [r'(ehat = fo 6(¢ + 2) dt = [ar + 12e]) = 15 . The point (2.4.8) corresponds to f = 2. so by Equation 2. b= 2 /()? + (2)? + (G7) de. 1 f(t) = VT aE FF, then Simpson's Rule gives L & 2—2 [f(0) + 4f(0-2) + 2f(0.4) ++ + 4f(1.8) + f(2)] © 9.8706. 10-3 if . Here are two views of the curve with parametric equations x = cost. y = sin3t, z = sint ‘ ‘ 1 - 6. o ' + ~ +1 “1 vt - 9 o @ ‘ ¥ Ox ¥ rl ea The complete curve is given by the parameter interval (0. 271 fo" (sin f? + (eos di)? + (cost)? dt = fo” VI+ G cos? Bt dt = 13.9744. _#(t) = 24-35 44k and & = [PW] = VEs OF 18 = VB Then s = s(t) = fp Ie'(w)du = fy V28du = V29t Therefore. t = hg. and substituting for t in the original equation, we have r(t(s)) = gsiz (1 - w)i + (5+ s+) k 8 r’(t) = 2e"(cos 2t — sin 2t) i + 2e™ (cos 2é + sin 2t) k. 4: = |r’ (t)] = 26% (cos 2t — sin 2)? + (cos Dt + sin At}? = 2e* \/2 cos? 2 + 2sin? 2 = 2/2 a s= a(t) = fl fe'(ulldu = fi 2VFe™ du= Ve], = VBE" —-1) = wn a ~ yeti = r= 3m(y+ 1) Substitusing. we have r(t(s)) = 23 #(84)) cose G in( Jp + 1)) ipajeetl (5t1)) gino G ns, = (+1) cos(in( =2))i 4 2j+ (e+ 1)sin(In( jy +1) Here r (#) = (3sin f, 4¢. 3cos#). so r/(t) = (3.cosé. 4. —3sin#) and fr’ (t)] = /9cos?t + 16 + 9sin?t = V/25 = 5. The point (0. 0. 3) corresponds to t = 0, so the arc length function beginning at (0. 0. 3) and measuring in the positive +1))k ° direction is given by s(t) = ff Ir’ (wWidu = ff ddu= St s()=5 = St=5 = ¢= 1. thus your location after moving 5 units along the curve is (3sin 1.4. 3.cos 1) 
