Feet from the mortar since sin θ cos θ 1 2 sin 2 θ

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feet from the mortar sincesinθcosθ=12sin 2θ .But sin 150= 1/2. Consequently,v0=64×4900 = 560 ft/sec.keywords:trajectory,parabola,distance,range, initial velocity00510.0pointsA ladder 11 feet in length slides down a wallas its bottom is pulled away from the wall asshown in11xyθPUsing the angleθas parameter, find theparametric equations for the path followed bythe pointPlocated 3 feet from the top of theladder.1.(3 secθ,8 tanθ)2.(8 cosθ,3 sinθ)3.(8 sinθ,3 cosθ)4.(8 tanθ,3 secθ)5.(3 sinθ,8 cosθ)6.(8 secθ,3 tanθ)
hayes (tsh489) – HW01 – milburn – (55800)57.(3 tanθ,8 secθ)8.(3 cosθ,8 sinθ)correctExplanation:By right triangle trigonometry, the coordi-nates (x, y) are given respectively byx= 3 cosθ ,y= (11-3) sinθ .Consequently, the curve traced out byPhasthe parametric form(3 cosθ,8 sinθ)for 0θπ/2. Eliminatingθ, we see thatPtraces out the portion of the ellipsex29+y264= 1in the first quadrant.keywords: parametric form, circle, ellipse,00610.0pointsFind the path (x(t), y(t)) of a particle thatmoves once clockwise around the curvex2+ (y-2)2= 49,starting at (7,2).1.(7 cost,2-7 sint),0t2πcorrect2.(-7 cost,2-7 sint),0t2π3.(-7 cost,2-7 sint),0tπ4.(7 cost,2-7 sint),0tπ5.(7 cost,2 + 7 sint),0t2π6.(7 cost,2 + 7 sint),0tπExplanation:The graph ofx2+ (y-2)2= 49is the circleOADCB(not drawn to scale) centered at (0,2) havingradius 7.As the particle moves clockwisearound this circle starting at (7,2), it startsatA, then moves toB, then toC, and finallytoD.The Pythagorean identitysin2θ+ cos2θ= 1,suggests settingx(t) = 7 cost ,y(t) = 2±7 sint .For thenx2(t) + (y(t)-2)2= 49for either choice of sign±. The choice of signdetermines whether the path moves clockwiseor counterclockwise.Indeed, suppose we takex(t) = 7 cost ,y(t) = 2 + 7 sint .Then (x(0), y(0)) = (7,2), so the path startsat the right point. But astincreases from 0 toπ/2,xis decreasing whileyis increasing; infact, the point (x(t), y(t)) moves from (7,2)to (0,9), and it does so counter-clockwise.On the other hand, if we takex(t) = 7 cost ,y(t) = 2-7 sint .Then (x(0), y(0)) = (7,2), so the path startsat the right point.But astincreases from0 toπ/2, bothxandyare decreasing; in
hayes (tsh489) – HW01 – milburn – (55800)6fact, the point (x(t), y(t)) moves from (7,2)to (0,-5), and it does so clockwise.Consequently, the path of the particle isdescribed parametrically by(7 cost,2-7 sint),0t2π.00710.0pointsIf the constantCis chosen so that the curvegiven parametrically byparenleftBigC28t2, CtparenrightBig,0t5,is the arc of the parabolay2= 8xfrom (0,0) to (2,4), find the coordinates ofthe pointPon this arc corresponding tot= 2.1.P=parenleftBig225,85parenrightBig2.P=parenleftBig45,825parenrightBig3.P=parenleftBig825,85parenrightBigcorrect4.P=parenleftBig85,825parenrightBig5.P=parenleftBig225,45parenrightBig6.P=parenleftBig45,225parenrightBigExplanation:The pointPhas coordinatesparenleftBigC28t2vextendsinglevextendsinglevextendsinglet=2, Ctvextendsinglevextendsinglevextendsinglet=2parenrightBig=parenleftBigC22,2CparenrightBig,so we need to find

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