# ReviewHW09-solutions - huynh(lth436 ReviewHW09...

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huynh (lth436) – ReviewHW09 – gilbert – (56690)1Thisprint-outshouldhave15questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.0010.0pointsThe astroid shown inxyis the curvec(t) = 4 cos3ti+ 4 sin3tj.Determine the arc length of the astroid.1.arc length = 24correct2.arc length = 123.arc length = 24π4.arc length = 45.arc length = 12π6.arc length = 4πExplanation:The arc length of a curvec(t),atb, isgiven by the integral expressionI=integraldisplaybabardblc(t)bardbldt .But whenc(t) = 4 cos3ti+ 4 sin3tj,we see thatc(t) =12 sintcos2ti+ 12 costsin2tj.In this casebardblc(t)bardbl=radicalBig(12 costsint)2(cos2t+ sin2t),which simplifies tobardblc(t)bardbl=|12 costsint|.On the other hand,c(t) traces out the astroidastranges from 0 to 2π. ThusI= 12integraldisplay2π0|costsint|dt= 48integraldisplayπ/20costsint dt= 24bracketleftBigsin2tbracketrightBigπ/20.Consequently, the astroid hasarc length = 24.keywords:arc length, parametric curve, as-troid trig functions, definite integral0020.0pointsFind the arc length of the curver(t) = (42t)i+etj+ (1et)kbetweenr(0) andr(3).1.arc length = (e3+e3)22.arc length = (e3e3)23.arc length = 2e34.arc length =e3e3correct5.arc length = 2e36.arc length =e3+e3Explanation:The length of a curver(t) betweenr(t0)andr(t1) is given by the integralL=integraldisplayt1t0|r(t)|dt .
huynh (lth436) – ReviewHW09 – gilbert – (56690)2Now whenr(t) = (42t)i+etj+ (1et)k,we see thatr(t) =2i+etj+etk.But then|r(t)|= (2 +e2t+e2t)1/2=et+et.ThusL=integraldisplay30(et+et)dt=bracketleftBigetetbracketrightBig30.Consequently,arc length =L=e3e3.0030.0pointsThe curveCis parametrized byc(t) = (4 + 2t)i+ ln(2t)j+ (1t2)k.Find the arc length ofCbetweenc(1) andc(4).1.arc length = 15 + ln 4correct2.arc length = 15ln 43.arc length = 8ln 44.arc length = 152 ln 45.arc length = 16 + 2 ln 46.arc length = 4 + ln 8Explanation:The arc length ofCbetweenc(t0) andc(t1)is given by the integralL=integraldisplayt1t0|c(t)|dt .Now whenc(t) = (4 + 2t)i+ ln(2t)j+ (1t2)kwe see thatc(t) = 2i+1tj2tk.But then|c(t)|=parenleftBig4 +1t2+ 4t2parenrightBig1/2=2t2+ 1t.ThusL=integraldisplay41parenleftBig2t+1tparenrightBigdt=bracketleftBigt2+ lntbracketrightBig41.Consequently,arc length =L= 15 + ln 4.0040.0pointsFind the unit tangent vectorT(t) to thegraph of the vector functionr(t) =(3 sint,+4t,+3 cost).