# Quest 9 - ha(kh26295 M408D Quest Homework 9 pascale(54550...

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ha (kh26295) – M408D Quest Homework 9 – pascaleff– (54550)1Thisprint-outshouldhave21questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.00110.0 pointsFindlimt0+r(t) whenr(t) =3 cost,4et,6tlnt.1.limit =0,0,02.limit =0,4,03.limit =3,0,64.limit =3,4,0correct5.limit =3,0,-66.limit =3,4,6Explanation:For a vector functionr(t) =f(t), g(t), h(t),the limitlimt0+r(t)Which one of the following vector functionshas this space curve as its graph?=limt0+f(t),limt0+g(t),limt0+h(t).But for the given vector function,limt0+f(t) =limt0+3 cost= 3,whilelimt0+g(t) =limt0+4et= 4,andlimt0+h(t) =limt0+tlnt= 0,using L’Hospital’s Rule. Consequently,limt0+r(t) =3,4,0.1. r(t) =tcost, tsint, t2. r(t) =sin 4t,cost,sintcorrect3. r(t) =sin 2t,cost,sint4. r(t) =cost,sint,sin 2t5. r(t) =tcost, tsint, t ,t >0,6. r(t) =tcost, t, tsint ,t >0,7. r(t) =tcost, t, tsint8. r(t) =cost,sint,sin 4tExplanation:The surface is a cylinder such that verti-cal cross-sections perpendicular to thex-axis
keywords: vector function, limit, trig functionlog function, exponential function00210.0 pointsA space curve is shown in black on thesurfacexyz
ha (kh26295) – M408D Quest Homework 9 – pascaleff– (54550)2intersect the surface in a circle.Now of thegiven vector functions onlyr(t) =sin 2t,cost,sintr(t) =sin 4t,cost,sintlie on such a cylinder; in particular, the sur-face is the graph ofy2+z2= 1,and the space curve passes through the pointr(0) = (0,1,0)on they-axis as well as the pointrπ2= (0,0,1)on thez-axis.But thex-coordinate of thespace curve takes both positive and nega-tive values between these two points, whilesin 2t >0 for 0< t <π/2. Consequently, thespace curve is the graph ofr(t) =sin 4t,cost,sint.keywords: surface, space curve, vector func-tion, 3D graph, circular cylinder,00310.0 pointsFind a vector function whose graph is thecurve of intersection of the circular cylinderx2+y2= 4and the hyperbolic paraboloidz= 2xy .1. r(t) = 2 sinti-2 costj+ 4 sin 2tk2. r(t) = 2 costi+ 2 sintj+ 4 cos 2tk4.r(t)=2 costi+ 2 sintj+ 4 sin 2tkcorrect5. r(t) =12costi+12sintj+14cos 2tk6. r(t) = 2 cos 2ti+ 2 sin 2tj+ 2 costk7. r(t) = 2 sin 2ti+ 2 cos 2tj+ 2 sintk8. r(t) =12costi-12sintj+14sin 2tkExplanation:The graph of the vector functionr(t) =x(t)i+y(t)j+z(t)kis the intersection of the circular cylinderx2+y2= 4and the hyperbolic paraboloidz= 2x2ywhenx(t)2+y(t)2= 4,z(t) = 2x(t)y(t).Nowcos2θ+ sin2θ= 1,which already eliminates two of the functions.On the other hand,2 sinθcosθ= sin 2θ,which is satisfied only byr(t) = 2 costi+ 2 sintj+ 4 sin 2tk.keywords: vector function, cylinder, circularcylinder, hyperbolic paraboloid, curve of in-tersection004
3. r(t) = 2 sinti+ 2 costj+ 4 cos 2tk
10.0 points
ha (kh26295) – M408D Quest Homework 9 – pascaleff– (54550)