tapia (jat4858) – HW06 – clark – (52990) 1This print-out should have 22 questions.Multiple-choice questions may continue o nthe next column or page – find all choicesbefore answering.001 10.0 pointsIf the constant C is chosen so that the curvegiven parametrically byCt,C28t2, 0 ≤ t ≤ 4 ,is the arc of the parabola8y = x2from (0, 0) to (4, 2), find the coor di nates ofthe point P on this arc corresponding to t = 2.1. P =1 ,182. P =2 ,12correct3. P =1 ,124. P =12, 25. P =18, 26. P =18, 1Explanation:The point P has coordinatesCtt=2,C28t2t=2=2C,C22,so we need to find C. But we are told that thegraph ofCt,C28t2passes through (4, 2) when t = 4. Thus4C = 4 , i .e., C = 1 .Consequently,P =2C,C22=2 ,12.keywords: parametric curve, parabo la002 10.0 pointsFind a Cart esia n equation for the curvegiven in parametric form byx(t) = 3 cos 4t , y(t) = 2 sin 4t .1.x29−y24=1362.x24−y29=1363.x24+y29=1364. 9x2+ 4y2= 365. 4x2+ 9y2= 36 correct6. 4x2− 9y2= 36Explanation:We have to eliminate the parameter t fromthe equations for x and y. Nowcos2θ + sin2θ = 1 .Thusx29+y24= 1 .But then after simplification, the curve hasCartesian form4x2+ 9y2= 36 .003 10.0 pointsDescribe the motion of a parti cle w ith posi-tion P (x, y) whenx = 5 sin t , y = 3 cos tas t varies in the interval 0 ≤ t ≤ 2π.tapia (jat4858) – HW06 – clark – (52990) 21. Moves along the linex5+y3= 1 ,starting at (0, 3) a nd ending at (5, 0).2. Moves once counterclockwise along theellipse(5x)2+ (3y)2= 1 ,starting and ending a t (0, 3).3. Moves once clockwise along the ellipse(5x)2+ (3y)2= 1 ,starting and ending a t (0, 3).4. Moves along the linex5+y3= 1 ,starting at (5, 0) a nd ending at (0, 3).5. Moves once counterclockwise along theellipsex225+y29= 1 ,starting and ending a t (0, 3).6. Moves once clockwise along the ellipsex225+y29= 1 ,starting and ending a t (0, 3). correctExplanation:Sincecos2t + sin2t = 1for all t, the particl e travels along the curvegiven in Cartesian form byx225+y29= 1 ;this is an ellipse centered at the origin. Att = 0, the particle is at (5 sin 0, 3 cos 0), i.e.,at the po int (0, 3) on the elli pse. Now as tincreases from t = 0 to t = π/2, x(t) increasesfrom x = 0 to x = 5, while y(t) decreases fromy = 3 to y = 0 ; in particular, the particlemoves from a point on the posi tive y-axis toa point on the po sitive x-ax is, so it is movingclockwise.In the same way, we see that as t increasesfrom π/2 to π, the particle moves to a pointon the negative y-axis, then to a point on thenegative x-axis as t increases from π to 3π/2,until finally it returns to its sta rting po int onthe posi tive y-axis as t increases from 3π/ 2 t o2π.Consequently, the particle moves clockwiseonce around the ellipsex225+y29= 1,starting and ending at (0, 3).keywords: motion on curve, el lipse004 10.0 pointsWhich one of the following could be thegraph of the curve given parametrically by(x(t), y(t))when the graphs of x(t) and y(t) are shown in010111tx(t) : y(t) :tapia (jat4858) – HW06 – clark – (52990) 31.010111xy2.010111xy3.010111xy4.010111xy5.010111xy6.010111xycorrectExplanation:Sincex(1) = y(1) = 0 ,x(0) = 1 , y(0) = 0 ,the gra ph passes through the origin and thepoint (1, 0). This al ready eliminates three ofthe graphs. T here are two ways of decidingwhich of the remaining three it is:(i) sincex12=34, y12= 1 ,the graph must pass also through (3/4, 1),(ii) the graph of y(t) is increasing near t = 0at the same rate as it is decreasing near t = 1,while the graph of x(t) is decreasing fasternear t = 1 than it is near t = 0. So the graphof (x(t), y(t)) is decreasing faster near t = 1than it is increasing near t = 0.Consequently, the graph of (x(t), y(t)) istapia (jat4858) – HW06 – clark – (52990) 4010111xykeywords: parametric curve, graph,005 10.0 pointsFind parametric equati ons representing theline segment joi ning P (−5, 3) to Q(1, −5) as(x(t), y(t)) , 0 ≤ t ≤ 1 ,whereP = (x( 0), y(0)) , Q = (x(1), y(1))and x(t), y(t) are linear functions of t.1.−5 + 6 sin2πt2, −5 + 8 cos2πt22. (−5 + 6t, 3 − 8t) correct3. (1 − 6t, −5 + 8t)4.1 − 6 sin2πt2, 3 − 8 cos2πt25. (1 − 6t2, −5 + 8t2)6. (−5 + 6t2, 3 − 8t2)Explanation:All six possible answers represent the l inesegment joining P (−5, 3) to Q(1, −5) in para-metric form(x(t), y(t)) , 0 ≤ t ≤ 1 .But only in the cases(−5 + 6t, 3 − 8t)and(1 − 6t, −5 + 8t)are x(t), y(t) linear functions of t. This al-ready eliminates four of the possible answers.On the other hand, the first of these starts atP and ends at Q, while the second starts at Qand ends a t P . Consequently, only(−5 + 6t, 3 − 8t)represents the line segment in parametricform by linear functions x(t), y(t) so thatP = (x( 0), y(0)) , Q = (x(1), y(1)) .006 10.0 pointsFind the path (x(t), y(t)) of a particle thatmoves once clockwise around the curvex2+ (y − 5)2= 49 ,starting at (7, 5).1. (7 cos t, 5 + 7 sin t), 0 ≤ t ≤ π2. (7 cos t, 5−7 sin t), 0 ≤ t ≤ 2π correct3. ( − 7 cos t , 5 − 7 sin t), 0 ≤ t ≤ π4. (7 cos t, 5 + 7 sin t), 0 ≤ t ≤ 2π5. ( − 7 cos t , 5 − 7 sin t), 0 ≤ t ≤ 2π6. (7 cos t, 5 − 7 sin t), 0 ≤ t ≤ πExplanation:The graph ofx2+ (y − 5)2= 49is the circl etapia (jat4858) – HW06 – clark – (52990) 5O ADCB(not drawn to scale) centered at (0, 5) havingradius 7 . As the particle moves clockwisearound this circle starting at (7, 5), it startsat A, then moves to B, then to C, and finallyto D.The Pythagorean identitysin2θ + cos2θ = 1 ,suggests settingx(t) = 7 cos t , y(t) = 5 ± 7 sin t .For thenx2(t) + (y(t) − 5)2= 49for either choice of sign ±. The choice of signdetermines whether the path moves clockwiseor counterclockwise.Indeed, supp ose we takex(t) = 7 cos t , y(t) = 5 + 7 sin t .Then (x(0), y(0)) = (7, 5), so the pat h startsat the right point. But as t increases from 0 t oπ/ 2, x is decreasing while y is increasing; infact, the point (x(t), y(t)) moves from (7, 5)to (0, 12), and it does so counter-clockwise.On the other hand, if we takex(t) = 7 cos t , y(t) = 5 − 7 sin t .Then (x(0), y(0)) = (7, 5), so the pat h startsat the right point. But as t increases from0 to π/2, both x and y are decreasing; infact, the point (x(t), y(t)) moves from (7, 5)to (0, −2), a nd it does so clockwise.Consequently, the path of the particle isdescrib ed parametrically by(7 cos t, 5 − 7 sin t),