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**Unformatted text preview: **3. r(t) = (t — 2, t2 + 1), (b) r’(t) = (1,2t),
r(—l) = (—3, 2). r'(—l) = (1, —2)
Since(9:+2)2=t2=y-l =>
y = (a:+2)2 + 1,thecurveisa
parabola. 5. r(t) = e” i + e‘j, r(0) = i+j. (b) r’(t) = 2e” i + e‘j,
Since .1: = e2t = (6)2 = ya, the r’(0) = 2i+j
curve is part of a parabola- Note
that here a: > 0, y > 0. 9. r(t) = (Vt — 2, 3, l/t2) :>
r’(t) = <% [m] ,% [31% [1/t2]> = <%(t — 2)-1/2,o, —2t‘3> 13. r(t) =tsinti+etcostj+sintoostk => r’(t) = [t . cost + (sint) - l] i+ [e‘(—sin t) + (oost)e*] j + [(sint)(— sin t) + (cost)(cost)] k = (toast+sint)i+e‘ (cost-sint)j + (cos2t-sin2 t) k 19. r'(t)=—sinti+3j+4oos2tk => r'(0)=3j+4k. Thus T(0) = 27. First we parametrize the curve C of intersection The projection of 0 onto the xy-plane is contained in the circle :52 +1412 = 25,2 = 0,sowecanwn'tea:= 5cost, y: 5sint. Calsoliesonthecylindery’ +2:2 = 20, andz Z 0
nearthepoint (3,4,2),sowecanw1itez= «20—312 = V20 —253in2t. AvectorequationtholforCis
r(t)=<Seost,53int,\/20-253in2t> :> r'(t)=<-53int,5cost,%(20—253in2t)_1/2(—5Osintcost)>. The point (3,4, 2) corresponds to t = cos—1 (g), sothe tangentvectorthere is
r’(cos‘1(%)) = (-5(%) ﬁe) , ;(2o — 2s (92)“ <—5o<%)(%))> = <—4,s, —e>. The tangent line is parallel to this vector and passes through (3, 4, 2), so a vector equation for the line
is r(t) = (3 — 4t)i + (4 + 3t)j + (2 — 6t)k. 35. f: (ti—t3j+3t5k)dt= (f: tdt)i— (f: t3dt)j+ (f: 3t5dt)k = [W]: i- Et‘]: 5 + [#13 k =%(4—0)i—%(16-0)j+%(64—0)k=2i—4j+32k 37 1;” 1 '+ t kdt— 1Lott” 1 1 dt'+ 1 t dtk ' o t+1 t“’+1J t2+1 _ o t+1 o t2+1 J o t2+1
=[ln|t+l|](1,i+[tan—1t];j+[%ln(t2+l)];k
=(1n2—1n1)i+(§—0)j+%(1n2—1n1)k=1n2i+gj+§1n2k 41. r'(t) = 2ti+3t2j + Jik => r(t) = t2i+t3j + §t3/2k+C, where Cisaconstantvector. Buti+j=r(1)=i+j+%k+C.ThusC=—§kandr(t)=t2i+t3j+(%t3l2_%)k_ 4s. % [u(t) x v(t)] = u’(t) x v(t) + u(t) x v’(t) [by Formula 5 ofTheorem 3]
= (cost, —sint, l) x (t, cost, sint) + (sint,coet,t) x (l, —sint,cost) = (—sinzt—costJ—cost sint,cos2t+tsint> + (c032t+tsint,t—cost sint,—sin2t—cost) = (ooszt—sin2t—cost+ts'mt,2t—2oost sint,c032t—sin2t—cost+tsint> l 55. % Iran = % [r(t) -r(t)]1/2 = are) -r(t)1"”l2r(‘) "W = Ir(t)| r(t) - r’ (t) 56. Since r(t) - r'(t) = 0, wehave 0 = 2r(t) - r'(t) = [r(t) - r(t)] = a |r(t)|2. Thus |r(t)|2, and so |r(t)|, is a constant, and hence the curve lies on a sphere with center the origin. ...

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- Summer '14
- Calculus