ha (lvh262) – Homework 13.1 – karakurt – (56295) 1This print-out should have 16 questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.001 10.0 pointsFind the domain of the vector functionr(t) = ht3,√t − 2, ln(9 − t) i.1. 2 < t ≤ 92. 2 ≤ t < 9 correct3. 2 < t < 94. t < 2, t > 95. 2 ≤ t ≤ 9Explanation:A vector functionr(t) = hf(t), g(t), h(t) iis defined when each of f (t) , g(t) and h(t) isdefined. Now, for the given function,f(t) = t3is defined for all t, whileg(t) =√t − 2is defined only when t ≥ 2. On the otherhand,h(t) = ln(9 − t)is defined only 9 − t > 0. Consequently, thedomain of r(t) consists of all t,2 ≤ t < 9 .keywords: vector function, domain, powerfunction, square root function, log function,002 10.0 pointsFind limt → 0+r(t) whenr(t) = h6 cos t, 8et, 4t ln t i.1. limit = h0, 8, 0 i2. limit = h6, 0, 4 i3. limit = h0, 0, 0 i4. limit = h6, 0, −4 i5. limit = h6, 8, 4 i6. limit = h6, 8, 0 i correctExplanation:For a vector functionr(t) = hf (t), g(t), h(t) i,the limitlimt → 0+r(t)=Dlimt → 0+f(t), limt → 0+g(t), limt → 0+h(t)E.But for the given vector function,limt → 0+f(t) = limt → 0+6 cos t = 6 ,whilelimt → 0+g(t) = limt → 0+8et= 8 ,andlimt → 0+h(t) = limt → 0+t ln t = 0 ,using L’Hospital’s Rule. Consequently,limt → 0+r(t) = h6, 8, 0 i .keywords: vector function, limit, trig functionlog function, exponential function003 10.0 pointsha (lvh262) – Homework 13.1 – karakurt – (56295) 2A space curve is shown in black on thesurfacexyzWhich one of the following vector functionshas this space curve as its graph?1. r(t) = ht cos t, t, t sin t i, t > 0,correct2. r(t) = hsin 2t, cos t, sin t i3. r(t) = hcos t, sin t, sin 2t i4. r(t) = hcos t, sin t, sin 4t i5. r(t) = hsin 4t, cos t, sin t i6. r(t) = ht cos t, t sin t, t i, t > 0,7. r(t) = ht cos t, t sin t, t i8. r(t) = ht cos t, t, t sin t iExplanation:The surface is a cone such that verticalcross-sections perpendicular to y-axis inter-sect the surface in a circl e. On the otherhand, the space curve exists only for y > 0.But of the given vector functions, onlyr(t) = ht cos t, t, t sin t i, t > 0,satisfies the equationsy(t)2= x(t)2+ z(t)2, y(t) > 0 .keywords: surface, space curve, vector func-tion, 3D g raph, circular cylinder,004 10.0 pointsA space curve is shown in black on thesurfacexyzWhich one of the following vector functionshas this space curve as its gr a ph?1. r(t) = hsin t, cos t, −t i2. r(t) = hcos t, sin t, t i3. r(t) = hcos t, sin t, sin 4t i correct4. r(t) = hsin t, cos t, cos 2t i5. r(t) = hcos t, sin t, cos 4t i6. r(t) = hcos t, sin t, cos 2t iExplanation:If we writer(t) = hx(t), y(t), z(t) i,ha (lvh262) – Homework 13.1 – karakurt – (56295) 3thenx(t)2+ y(t)2= 1for all the given vector functions, showing thattheir graph will always lie on t he cylindricalcylinderx2+ y2= 1To determine which particular vector functionhas the given graph, we have to look moreclosely at the graph itself. Notice that thegraph oscillates with peri od 4, so r(t) is one ofhcos t, sin t, sin 4t i, hcos t, sin t, cos 4t i.On the other hand, it passes it through(1, 0, 0) and (0, 1, 0). Consequently, thespace curve is the graph ofr(t) = hcos t, sin t, sin 4t i .keywords:005 10.0 pointsThe vector functionr(t) = (2 − 5 sin t) i + (3 + 5 cos t) j + 4 ktraces out a ci rcle in 3-space as t varies. De-termine the radius and center of this circle.1. radi us = 5 , center = (2, 3, 4) correct2. radius = 5 , center = (3, 4, 2)3. radius = 5 , center = (4, 3, 2)4. radius = 3 , center = (4, 3, 2)5. radius = 3 , center = (−2, 3, 2)6. radius = 3 , center = (4, 3, −2)Explanation:Writingr(t) = x(t) i + y(t) j + z(t) k ,we see that z(t) = 4 for all t, whilex(t) = 2 − 5 cos t , y(t) = 3 + 5 sin t .Thus r(t) traces out a curve in the plane z = 4and(x(t) − 2)2+ ( y(t) − 3)2= 25 .Consequently, r(t) traces out a circle withradius = 5 , center = (2, 3, 4) .keywords: vector function, space curve, circle,plane, radius, center circle,006 (part 1 of 2) 10.0 pointsThe vector functionr(t) = (1 + 3 cos t) i + 5 j + (2 − 3 sin t) ktraces out a circle in 3-space as t varies. Inwhich plane does this circle lie?1. plane y = 5 correct2. plane y = −53. plane x = −54. plane z = −55. plane z = 56. plane x = 5Explanation:Writingr(t) = x(t) i + y(t) j + z(t) k ,we see that y(t) = 5 for all t. Consequently,r(t) traces out a curve in theplane y = 5 .007 (part 2 of 2) 10.0 pointsDetermine the radius and center of the cir-cle traced out by r(t).ha (lvh262) – Homework 13.1 – karakurt – (56295) 41. radius = 3 , center = (2, 1, 5)2. radius = 3 , center = (5, 1, 2)3. radius = 1 , center = (5, 1, −2)4. radi us = 3 , center = (1, 5, 2) correct5. radius = 1 , center = (5, 1, 2)6. radius = 1 , center = (−2, 1, 2)Explanation:Writingr(t) = x(t) i + y(t) j + z(t) k ,we see also thatx(t) = 1 + 3 cos t , z(t) = 2 − 3 sin t .Thus r(t) traces out a curve in the plane y = 5and(x(t) − 1)2+ (z(t) − 2)2= 9 .Consequently, r(t) traces out a circle withradius = 3 , center = (1, 5, 2) .Consequently, r(t) traces out a circle withradius = 3 , center = (1, 5, 2) .keywords: vector function, space curve, ci r cl e,plane, radi us, center ci r cl e,008 10.0 pointsFind a parametrization for the circle havingradius 4 and center (3, −2, −5) that lies in aplane parallel to the yz-plane.1. r(t) = h−3, 2 − 4 cos t, 5 − 4 sin t i2. r(t) = h−3 + 4 sin t, 2 − 4 cos t, 5 i3. r(t) = h3 − 4 cos t, −2, −5 + 4 sin t i4. r(t) = h−3 + 4 cos t, 2, 5 − 4 sin t i5. r(t) = h3 − 4 sin t, −2 + 4 cos t, −5 i6. r(t) = h3, −2 + 4 cos t, −5 + 4 sin t icorrectExplanation:If the vector functionr(t) = hx(t), y(t), z(t) itraces out the circle having radius 4 and center(3, −2, −5) that lies in a plane parallel …