tapia (jat4858) – HW11 – 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 pointsFind the domain of the functionf(x, y) = ln(8 − x2− 5y2) .1.n(x, y) :15x2+18y2< 1o2.n(x, y) :18x2+58y2< 1ocorrect3.n(x, y) :18x2+58y2> 1o4.n(x, y) :15x2+18y2≤ 1o5.n(x, y) :18x2+58y2≤ 1o6.n(x, y) :15x2+18y2> 1oExplanation:Since ln x is defined only for x > 0, thefunctionf(x, y) = ln(8 − x2− 5y2)is defined only forn(x, y) :18x2+58y2< 1o.keywords: function several variables, ln func-tion, domain002 10.0 pointsA rectangular piece of cardboard is 4 timesas long as it is wide. If the length of theshorter side is y inches and an open box isconstructed by cutting equal sq uares of side-length x inches from the corners of the piece ofcardboard and t urning up the sides as shownin the figurexxx xxxxxDetermine the volume, V , of the box as afunction of x and y.1. V (x, y) = 4xy2+ 10x2y + 2x3cu. ins2. V (x, y) = 4xy2− 10x2y + 4x3cu. inscorrect3. V (x, y) = 10xy2+ 4x2y − 2x3cu. ins4. V (x, y) = 10xy2− 4x2y + 2x3cu. ins5. V (x, y) = 4xy2− 10x2y − 4x3cu. ins6. V (x, y) = 10xy2+ 4x2y − 4x3cu. insExplanation:The rectangular sheet of cardboard willhave dimensions y × 4y. Thus the lengthof the shorter side of the base of the box isy − 2x, whi le the l ength of the longer side is4y − 2x. Since the height of the box is x, thevolume will thus beV (x) = ( length) × (width) × (height)= x( 4y − 2x)(y − 2x).Consequently,V (x) = 4xy2− 10x2y + 4x3cu. ins .003 10.0 pointsWhich one of the following surfaces is thegraph off(x, y) = 2x2?tapia (jat4858) – HW11 – clark – (52990) 21.-2-1012x-2-1012y-4-2024z-2-1012x-2-1012-4-20242.-2-1012x-2-1012y-4-2024z-2-1012x-2-1012-4-20243.-2-1012x-2-1012y01234z-2-1012x-2-1012012344.-2-1012x-2-1012y02468z-2-1012x-2-101202468correct5.-2-1012x-2-1012y02468z-2-1012x-2-101202468Explanation:Note that the function z does not dependon the variable y. Hence the cross-sections ofthe graph by any plane, parallel to x-z-planewould give a parabola with branches directedupwards.This is exactly the case with-2-1012x-2-1012y02468z-2-1012x-2-101202468004 10.0 pointsWhich of the following surfaces could havecontour map-6-5-4-3-2-10123456012345tapia (jat4858) – HW11 – clark – (52990) 31.zxy2.zxy3.xzy4.zxycorrect5.zyxExplanation:The graphs in the contour map show thatthe horizontal cross-sections of the surface areall circles as in cones and paraboloids. On theother hand, the values of the contours and thegrid show that these cross-sections increase ata constant rate. This occurs for a cone, butnot for a paraboloid.Consequently, of the surfaces shown, theonly one having the given contour map is azxykeywords:005 10.0 pointsWhich one of the following functions hastapia (jat4858) – HW11 – clark – (52990) 4-2-1012x-2-1012y-4-2024z-2-1012x-2-1012-4-2024as its graph.1. f(x, y) = y2− x2correct2. f(x, y) = x2− y23. f(x, y) = 2 x24. f(x, y) =12(x2+ y2)5. f(x, y) =12(8 − x2− y2)Explanation:The cross-section of t he gra ph by the planey = 0 gives a parabola opening downwardswith vertex at t he point O(0,0,0). Similarly,the cross-section of t he graph by the planex = 0 gives a parabola opening upwards withvertex at the origin O. Hencef(x, y) = y2− x2.keywords:006 10.0 pointsFind lim(x,y)→(4,−3)x5+ 2x3y − 3xy2.1. 4242. 7483. 532 correct4. 13005. 1516Explanation:007 10.0 pointsFind lim(x,y)→(6,2)xy cos(x − 3y) .1. 62. 12 correct3. 24. 05. −2Explanation:008 10.0 pointsFind lim(x,y)→(0,0)4xy2x2+ y2, if it exists.1. 42. 83. 24. The limit does not exist.5. 0 correctExplanation:009 10.0 pointsFind lim(x,y)→(0,0)7xypx2+ y2, if it exists.1. 142. 7tapia (jat4858) – HW11 – clark – (52990) 53. 0 correct4. The limit does not exist.5. 3.5Explanation:010 10.0 pointsFind lim(x,y)→(0,0)4xy4x2+ y8, if it exists.1. 82. The limit does not exist. correct3. 04. 25. 4Explanation:011 10.0 pointsFind lim(x,y,z)→(0,0,0)xy + 6yz2+ 5xz2x2+ y2+ z4, if itexists.1. The limit does not exist. correct2. 03. 54. 115. 6Explanation:012 10.0 pointsDetermine the set of points at which thefunctionF (x, y) = arctan5x + 6py − 6is continuous.1.n(x, y)|x ≥ 0, y ≥ 0o2.n(x, y)|y ≥ 6,5x + 6py − 6≤ 1o3.n(x, y)|x ≥ 0o4.n(x, y)|y ≥ 0o5.n(x, y)|y ≥ 6ocorrectExplanation:013 10.0 pointsDetermine the set of points at which thefunctionf(x, y, z) =xyz3x2+ 2y2− zis continuous.1.n(x, y, z)|z 6= 3x2+ 2y2, xyz < 0o2.n(x, y, z)|z 6= 3x2+ 2y2ocorrect3.n(x, y, z)|xyz > 0o4.n(x, y, z)|z 6= −3x2− 2y2o5.n(x, y, z)|z 6= 3x2+ 2y2, xyz > 0oExplanation:014 10.0 pointsFrom the contour map of f shown below de-cide whether fxand fyare positive, negative,or zero at P .tapia (jat4858) – HW11 – clark – (52990) 600224466Pxy1. fx> 0 , fy< 02. fx= 0 , fy< 03. fx> 0 , fy> 0 correct4. fx= 0 , fy> 05. fx< 0 , fy> 06. fx< 0 , fy< 0Explanation:When we walk in the x-direction from P weare walking uphill, so fx> 0. On the otherhand, w hen we walk in the y-direction from Pwe are again walking uphill, so fy> 0 also.Consequently, at Pfx> 0 , fy> 0 .keywords: contour map, contours, partialderivative, slope,015 10.0 pointsDetermine fywhenf(x, y) = y cos(4x − y) − sin(4x − y) .1. fy= y sin(4x − y)2. fy= −y cos(4x − y)3. fy= 2 cos(4x− y) +y sin(4x−y) correct4. fy= −y sin(4x − y)5. fy= −2 sin(4x − y) + y cos(4x − y)6. fy= −2 cos(4x − y) − y sin(4x − y)7. fy= 2 sin(4x − y) − y cos(4x − y)8. fy= y cos(4x − y)Explanation:From the Product Rule we see thatfy= cos(4x−y)+ y sin(4x − y) +cos(4x− y) .Consequently,fy= 2 cos(4x − y) + y sin(4x − y) .016 10.0 pointsDetermine fxwhenf(x, y) = (3 − 2xy)e−xy.1. fx= x( 5 − 3xy)e−xy2. fx= x( 3xy − 5)e−xy3. fx= y(2xy − 5)e−xycorrect4. fx= −x(1 + 2xy)e−xy5. fx= y(5 − 2xy)e−xy6. fx= −y(1 − 2xy)e−xy7. fx= −y(1 + 2xy)e−xy8. fx= −x(1 + 3xy)e−xyExplanation:From the Product Rule we see thatfx= −2ye−xy− y( 3 − 2xy)e−xy.tapia (jat4858) – HW11 – clark – (52990) 7Consequently,fx= y(2xy − 5)e−xy.017 10.0 pointsFind the va lue of fxat (−2, 1) whenf(x, y) =5x2+ 3xyx + y.Correct answer: 3.Explanation:After differentiation using the quotient rulewe see thatfx=(x + y)(10x + 3y) − 5x2− 3xy(x + y)2=5x2+ 10xy + 3y2(x + y)2.At (−2, 1), therefore.fx(−2, 1)= 3.018 10.0 pointsFind fywhenf(x, y) =Zxycost5dt