Π 3 5 4 cosθ dθ correct 5 1 2 integraldisplay π 1

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π/35-4 cosθ dθcorrect5.12integraldisplayπ0(1-2 cosθ)26.integraldisplayππ/25-4 cosθ dθ7.integraldisplay2π/305-4 cosθ dθ8.12integraldisplayππ/2(1-2 cosθ)2Explanation:The arc length of a polar curver=f(θ)betweenθ=θ0andθ=θ1is given by theintegralL=integraldisplayθ1θ0radicalBigr2+f(θ)2dθ .Now whenr= 1-2 cosθ .its graph will(i) pass through the origin whenr= 0,i.e.,atθ=-π/3, π/3,(ii) cross thex-axis also atθ= 0, π,
cai (atc667) – HW18 - Sect. 10.4 – chavez-dominguez – (55235)6(iii)andcrossthey-axisalsoatθ=π/2,3π/2,as the graph above indeed shows.Thusθ0=π/3 whileθ1=πandr2+f(θ)2= (1-2 cosθ)2+ 4 sin2θ= 1-4 cosθ+ 4 cos2θ+ 4 sin2θ= 5-4 cosθ .Consequentlyarc length =integraldisplayππ/35-4 cosθ dθ.keywords:arc length,polar graph,polarcurve, trig function, radical, double angle for-mula00810.0pointsFind the arc length of the portion of thegraph shown as a solid curve inof the polar curver= 2(1-cosθ).1.arc length = 422.arc length = 223.arc length = 2(2-2)4.arc length = 45.arc length = 8correct6.arc length = 4(2-2)Explanation:The arc length of a polar curver=f(θ)betweenθ=θ0andθ=θ1is given by theintegralL=integraldisplayθ1θ0radicalBigr2+f(θ)2dθ .Now in the graph above,θ0= 0 whileθ1=π.Forr= 2(1-cosθ)the indicated portion of its graph thus has arclengthL=integraldisplayπ0radicalBig4(1-cosθ)2+ 4 sin2θ dθ .But(1-cosθ)2+ sin2θ= 1-2 cosθ+ cos2θ+ sin2θ= 2(1-cosθ),soL= 22integraldisplayπ01-cosθ dθ .To evaluate this last integral we need to elim-inate the radical:1-cosθ=radicalBig2 sin2(θ/2).ConsequentlyL= 4integraldisplayπ0sinθ2=-8bracketleftBigcosθ2bracketrightBigπ0,in which casearc length =L= 8.keywords:arc length,polar graph,polarcurve, trig function, radical, double angle for-mula

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