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**Unformatted text preview: **1. r(t)=(t,3cost,3sint) => r’(t)=(1,—3sint,3cost) => lr'(t)l = 1’ + (—3sint)2 + (3cost)2 = ‘/1 + 9(sin’ t + 0032 t = m,
Thenusing Formula 3, wehaveL = f; |r'(t)|dt = [fax/Fldt = x/ﬁt]:5 = 10 m. 3. r(t)=s/§ti+e‘j+e“k => r’(t)=\/§i+e‘j—e‘*k => |r’(t)| = v («5)2 + (e“)2 + (—e—‘)2 = v2 + e2t + 8"“ = (e‘ + e“)2 = e‘ + 8" [since e‘ + e" > 0]- ThenL = I: Ir’(t)ldt = folk: +e-t)dt = [3‘ — e4]; = e _ 8'1. 5. r(t)=i+t2j+t3k => r'(t)=2tj+3t2k => |r’(t)|=\/4t2+91‘.4=t\/4+9t2 [sincetZO]. 1 Then L = fol Ir’(t)| dt = j: “/4 +9t! dt = ﬁ . %(4+ 9t2)3/2] o = %(133/2 _ 43/2) = 2.1703372 _ s). . 'I‘heprojectionofthecurveContothezany-planeisthecm'vea:2 = 2yory = %22,z = 0. Thenwecanchoosetheparameter
a: = t => y = %t2. Since Calsolies onthe surface 32 = my, we havez = §xy = %(t)(%t2) = %t3. 'Ihenparametric
1 - , z = Et3 andthe corresponding vector equation is r(t) = (t, 1t2, 1is). The oﬁgin conespondstot=0andthepoint(6, 18,36) conespondstot=6,so L=I§ Ir'<t>ldt=1: mm a=1:‘\/12+t2+<1t2)’dt=1:\/1+—t2+1t7dt = f: \/(1+ %t’)’dt= f:(1 + %t”)dt = [t+ %t"]§ = 6+36= 42 23. r(t) = ¢ét2i+2tj +2t3k => r'(t) = 2J§ti+2j +6t21g r”(t) = 2Jéi+ 12tk,
|r'(t)| = m = W = W = 2(3):2 + 1),
r’(t) x r”(t) = 24:1— 12J§t2j — 4mg
|r’(t) x r”(t)| = «m = W = m = 4% (31:2 + 1). |r’(t) x r”(t)| _ 4J§(3t2 + 1) _ J6 mean“): |r'(t)|3 _ 8(3t2+1)3 _W' 25. r(t) = <t,t2,t3) => r’(t) = (1,2239). The point (1, 1, l) compondstot = 1, and r’(1) = (1,2,3) => |r’(l)| =m=Jﬁ r”(t)= (0,2,6t) 2» r”(l)= (0,2,6). r’(l) xr”(1)=(6,—6,2),so ...

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- Summer '14
- Calculus