Integraldisplay t 1 t parenleftbig 5 4 t 5 t 5 t 2 1

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integraldisplayt1t0parenleftBig5-4t-5t+5t2-1t3parenrightBigdt=bracketleftBig5t-2t2-5 lnt-5t+12t2bracketrightBigt1t0.But the graph ofx(t) =t-1t,y(t) = 4t+1tintersectsy= 5 when4t+1t= 5,i.e., when4t2-5t+ 1 = (4t-1)(t-1) = 0.
qazi (kaq87) – HW01 – berg – (55290)10Thust0= 1/4 whilet1= 1, soarea(A) =parenleftBig-32-5 ln 1parenrightBig-parenleftBig-878-5 lnparenleftBig14parenrightBigparenrightBig.Consequently,area(A) =758-5 ln 4.keywords: parametric curve, area,01610.0pointsThe curve traced out by a pointPon thecircumference of a circle as the circle rollsalong a straight line shown inPxyis called aCycloidand the shaded region isthe region,A, below anarch.If the circle has radiusR, the cycloid isgiven parametrically byx(t) =R(t-sint),y(t) =R(1-cost).-cost)dx ,wherex=x(t) = 5(t-sint),dx= 5(1-cost)dt .Thus after a change of variable fromxtotwesee thatarea(A) = 25integraldisplay2π0(1-cost)2dt= 25integraldisplay2π0(1-2 cost+ cos2t)dt= 25integraldisplay2π0parenleftBig32-2 cost+12cos 2tparenrightBigdt .Consequently,area(A) = 75π.keywords: parametric curve, area, cycloid01710.0pointsWhich one of the following integrals givesthe length of the parametric curvex(t) =t2,y(t) = 2t ,0t12.1.I=integraldisplay
Find the area ofAwhenR= 5.125(1-cost)dx ,wherex=x(t) = 5(t-sint),dx= 5(1-cost)dt .Thus after a change of variable fromxtotwesee thatarea(A) = 25integraldisplay2π0(1-cost)2dt= 25integraldisplay2π0(1-2 cost+ cos2t)dt= 25integraldisplay2π0parenleftBig32-2 cost+12cos 2tparenrightBigdt .Consequently,area(A) = 75π.keywords: parametric curve, area, cycloid01710.0pointsWhich one of the following integrals givesthe length of the parametric curvex(t) =t2,y(t) = 2t ,0t12.1.I=integraldisplay

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