P 5 3 4 3 2 p 5 1 4 4 correct 3 p 5 4 4 1 4 p 5 3 4 3

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11Determine the area ofA.1.P= (5,3),(-4,-3)2.P= (-5,1),(4,4)correct3.P= (-5,4),(4,1)4.P= (5,-3),(-4,3)5.P= (5,1),(-4,4)6.P= (-5,-3),(4,3)Explanation:Sincex(t) = 3t2-6,y(t) = 2t ,the slope of the tangent line to the curve isgiven bydydx=y(t)x(t)=2t3t2-6.But if this tangent line is to be parallel to theline defined parametrically by (-3t,2t), then2t3t2-6=-23.After cross-multiplication and simplification,this becomes the equation6t2+ 6t-12 = 6(t-1)(t+ 2) = 0.Consequently,Pis the point corresponding tothe valuest= 1 andt=-2, in which caseP= (-5,1),(4,4).01810.0pointsThe regionAenclosed by the liney= 5 andthe graph of the curve given parametrically byx(t) =t-1t,y(t) = 2t+2tis similar to the shaded region inxy1.area(A) =172-4 ln 42.area(A) =172+ 4 ln 43.area(A) =152+ 4 ln 44.area(A) =152-4 ln 4correct5.area(A) =132+ 4 ln 46.area(A) =132-4 ln 4Explanation:If the graph ofx(t) =t-1t,y(t) = 2t+2tintersectsy= 5 whent=t0andt=t1, thenAis similar to the shaded region in5xyabwitha=x(t0) andb=x(t1). In this casearea(A) =integraldisplayba(5-y)dx=integraldisplayx(t1)x(t0)parenleftBig5-2t-2tparenrightBigdx ,
hayes (tsh489) – HW01 – milburn – (55800)12wherex=x(t) =t-1t,dx=parenleftBig1 +1t2parenrightBigdt .Thus after a change of variable fromxtotwereduce the integral describing area(A) tointegraldisplayt1t0parenleftBig5-2t-2tparenrightBigparenleftBig1 +1t2parenrightBigdt=integraldisplayt1t0parenleftBig5-2t-4t+5t2-2t3parenrightBigdt=bracketleftBig5t-t2-4 lnt-5t+1t2bracketrightBigt1t0.But the graph ofx(t) =t-1t,y(t) = 2t+2tintersectsy= 5 when2t+2t= 5,i.e., when2t2-5t+ 2 = (2t-1)(t-2) = 0.Thust0= 1/2 whilet1= 2, soarea(A) =parenleftBig154-4 ln 2parenrightBig-parenleftBig-154-4 lnparenleftBig12parenrightBigparenrightBig.Consequently,area(A) =152-4 ln 4.keywords: parametric curve, area,01910.0pointsWhich one of the following integrals givesthe length of the parametric curvex(t) = 2t2,y(t) =t ,0t8.1.I=integraldisplay40|16t2+ 1|dt2.I=integraldisplay80radicalbig16t2+ 1dtcorrect3.I=integraldisplay40radicalbig16t2+ 1dt4.I= 2integraldisplay80radicalbig16t2+ 1dt5.I= 2integraldisplay80|16t2+ 1|dt6.I= 2integraldisplay40|16t2+ 1|dtExplanation:The arc length of the parametric curve(x(t), y(t)),atbis given by the integralI=integraldisplaybaradicalBig(x(t))2+ (y(t))2dt .But whenx(t) = 2t2,y=t ,we see thatx(t) = 4t ,y(t) = 1.Consequently, the curve hasarc length =integraldisplay80radicalbig16t2+ 1dt.

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