HW07-solutions - HW07 neitzke(55325 This print-out should...

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– HW07 – neitzke – (55325)1Thisprint-outshouldhave17questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.00110.0pointsWhenCis parametrized byc(t) = (sin 4t)i+ 3tj+ (cos 4t)k,find its arc length betweenc(0) andc(2).1.arc length = 10correct2.arc length = 43.arc length = 64.arc length = 125.arc length = 8Explanation:The length of the curve betweenc(t0) andc(t1) is given by the integralL=integraldisplayt1t0bardblc(t)bardbldt .Now whenc(t) = (sin 4t)i+ 3tj+ (cos 4t)k,we see thatc(t) = (4 cos 4t)i+ 3j(4 sin 4t)bk .But then by the Pythagorean identity,bardblc(t)bardbl= (16 + 9)1/2= 5.ThusL=integraldisplay205dt=bracketleftBig5tbracketrightBig20.Consequently,arc length =L= 10.00210.0pointsFind the arc length of the curver(t) = (4 +2t)i+etj+ (2et)kbetweenr(0) andr(3).1.arc length = 2e32.arc length = (e3e3)23.arc length =e3+e34.arc length = (e3+e3)25.arc length =e3e3correct6.arc length = 2e3Explanation:The length of a curver(t) betweenr(t0)andr(t1) is given by the integralL=integraldisplayt1t0bardblr(t)bardbldt .Now whenr(t) = (4 +2t)i+etj+ (2et)k,we see thatr(t) =2i+etj+etk.But thenbardblr(t)bardbl= (2 +e2t+e2t)1/2=et+et.ThusL=integraldisplay30(et+et)dt=bracketleftBigetetbracketrightBig30.Consequently,arc length =L=e3e3.00310.0pointsFind the unit tangent vectorT(t) to thegraph of the vector functionr(t) =(3 sint,+4t,+3 cost).
– HW07 – neitzke – (55325)21. T(t) =(3 sint,4,3 cost)2. T(t) =(bigg35cost,45,35sint)bigg3. T(t) =(bigg35sint,45,35cost)bigg4. T(t) =(bigg35cost,45,35sint)biggcorrect5. T(t) =(3 cost,4,3 sint)6. T(t) =(3 sint,4,3 cost)Explanation:The unit tangent vectorT(t) tor(t) is givenbyT(t) =r(t)|r(t)|.Now whenr(t) =(3 sint,+4t,+3 cost),we see thatr(t) =(3 cost,+4,3 sint)while|r(t)|=radicalBig32(cos2t+ sin2t) + (4)2= 5.Consequently,T(t) =(bigg35cost,45,35sint)bigg.keywords:00410.0pointsDetermine the curvature,κ, of the curver(t) =t2i+ 5tk.1.κ(t) =10t4t2+ 252.κ(t) =10t(4t2+ 25)3/23.κ(t) =(4t2+ 25)3/24.κ(t) =10(4t2+ 25)1/25.κ(t) =10(4t2+ 25)3/2correctExplanation:The curvatureκof the curver(t) =t2i+ 5tkis given byκ(t) =bardblr(t)×r′′(t)bardblbardblr(t)bardbl3.Now in this case,r(t) = 2ti+ 5k,r′′(t) = 2i.Thusr(t)×r′′(t) =10kandbardblr(t)×r′′(t)bardbl=radicalbig102= 10.Sincebardblr(t)bardbl= (4t2+ 25)1/2,it follows thatκ(t) =10(4t2+ 25)3/2.00510.0pointsFindthecurvature,κ(t),ofthecurveparametrized byr(t) = 2ti+ 5 sintj+ 5 costk1.κ(t) =5292.κ(t) =529correct3.κ(t) = 29
– HW07 – neitzke – (55325)34.κ(t) =295.κ(t) = 5Explanation:Nowr(t) = 2i+ 5 costj5 sintkwhiler′′(t) =5 sintj5 costk.

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