13_2.pdf

# 13_2.pdf - 3 r(t =(t 2 t2 1(b r(t =(1,2t r(l =(3 2 r(l =(1...

• 2

This preview shows pages 1–2. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

This is the end of the preview. Sign up to access the rest of the document.

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. ...
View Full Document

• Summer '14

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern