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# UT M 408D - Homework 10

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abdallah (haa2348) – Homework 10 Due by 11:59pm on Sun 5/3 – reyes – (52535) 1This print-out should have 31 questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.001 10.0 pointsA rectangular piece of cardboard is 3 timesas long as it is wide. If the length of theshorter side is y inches and an open box isconstructed by cutting equal squares of side-length x inches from the corners of the piece ofcardboard and turning up the sides as shownin the figurexxx xxxxxDetermine the volume, V , of the box as afunction of x and y.1. V (x, y) = 3xy2+ 8x2y + 2x3cu. ins2. V (x, y) = 3xy2− 8x2y − 4x3cu. ins3. V (x, y) = 8xy2+ 3x2y − 2x3cu. ins4. V (x, y) = 8xy2− 3x2y + 2x3cu. ins5. V (x, y) = 3xy2−8x2y + 4x3cu. ins cor-rect6. V (x, y) = 8xy2+ 3x2y − 4x3cu. insExplanation:The rectangular sheet of cardboard willhave dimensions y × 3y. Thus the lengthof the shorter side of the base of the box isy − 2x, w hi le the length o f the longer side is3y − 2x. Since the height of the box is x, thevolume will thus beV ( x ) = (length) × (width) × (height)= x(3y − 2x)(y − 2x).Consequently,V (x) = 3xy2− 8x2y + 4x3cu. ins .002 10.0 pointsWhich of the following surfaces could havecontour map-4-3-2-101230-1121.zyxcorrect2.xzyabdallah (haa2348) – Homework 10 Due by 11:59pm on Sun 5/3 – reyes – (52535) 23.zxy4.zxy5.zxyExplanation:The graphs in the contour map are parallellines so the horizontal cross-sections of t hesurface are all lines as in planes and paraboliccylinders. On the other hand, the values of thecontours and the grid show that these cross-sections grow at a constant rate. Now for aplane the contours are equally spaced lines,whereas for a parabolic cylinder the cont ourswill be increasingly close together.Consequently, of the surfaces shown, theonly one having the given conto ur map is azyxkeywords:003 10.0 pointsWhich one of t he following surfaces is thegraph off(x, y) =12(8 − x2− y2) ?1.-2-1012x-2-1012y01234z-2-1012x-2-101201234correct2.-2-1012x-2-1012y-4-2024z-2-1012x-2-1012-4-2024abdallah (haa2348) – Homework 10 Due by 11:59pm on Sun 5/3 – reyes – (52535) 33.-2-1012x-2-1012y01234z-2-1012x-2-1012012344.-2-1012x-2-1012y02468z-2-1012x-2-1012024685.-2-1012x-2-1012y01234z-2-1012x-2-101201234Explanation:When y = 0 the function becomes z =(1/2)(8 − x2). Its graph is a parabola in thex-z plane opening downwards , i.e., a parabolaopening downwards in the plane y = 0. Thusthe cross-section of t he graph of f by theplane y = 0 should give a parabola openingdownwards . Similarly, the cross-section ofthe graph by the plane x = 0 gives a parabolaopening downwards.This is exactly the case with-2-1012x-2-1012y01234z-2-1012x-2-101201234004 10.0 pointsUse a table of numerical values of f (x, y)for (x, y) near the origin to make a conjec-ture abo ut the value of the limit of f(x, y) as(x, y) → ( 0, 0).f(x, y) =x2y3+ x3y2− 93 − xy.1. Cannot determine the value from thegiven information.2. −3 correct3. −94. 35. 9Explanation:005 10.0 pointsDetermine fx− fywhenf(x, y) = 3x2+ 2xy − 3y2− x − 3y .1. fx− fy= 8x − 4y − 42. fx− fy= 8x + 8y − 43. fx− fy= 4x + 8y + 2 correct4. fx− fy= 4x + 8y − 4abdallah (haa2348) – Homework 10 Due by 11:59pm on Sun 5/3 – reyes – (52535) 45. fx− fy= 8x − 4y + 26. fx− fy= 4x − 4y + 2Explanation:After differentiation we see thatfx= 6x + 2y − 1 , fy= 2x − 6y − 3 .Consequently,fx− fy= 4x + 8y + 2 .006 10.0 pointsDetermine fx+ fywhenf(x, y) = x2− 3xy + 3y2+ 3x − y .1. fx+ fy= −x + 3y + 2 correct2. fx+ fy= 5x − 9y + 43. fx+ fy= −x − 9y + 24. fx+ fy= −x + 3y + 45. fx+ fy= 5x + 3y + 46. fx+ fy= 5x − 9y + 2Explanation:After differentiation we see thatfx= 2x − 3y + 3 , fy= −3x + 6y − 1 .Consequently,fx+ fy= −x + 3y + 2 .007 10.0 pointsDetermine whether the parti a l derivativesfx, fyof f ar e positive, negative or zero at thepoint P on the gra ph of f shown inPxzy1. fx> 0 , fy= 02. fx< 0 , fy< 03. fx= 0 , fy> 04. fx< 0 , fy= 0 correct5. fx= 0 , fy< 06. fx< 0 , fy> 07. fx> 0 , fy> 08. fx= 0 , fy= 0Explanation:The value of fxat P is the slope of thetangent line to graph of f at P in the x-direction, while fyis the slope of the tangentline in the y-direction. Thus the sign of fxindicates whether f is increasing or decreasingin the x-direction, or w hether the tangent linein that direction at P is horizontal.Similarly, the value of fyat P is the slopeof the tangent line at P in the y-direction,and so the sign of fyindicates whether f isincreasing or decreasing in the y-direction, orwhether the tangent line in that direction atP is horizontal.From the graph it thus follows that at Pfx< 0 , fy= 0 .abdallah (haa2348) – Homework 10 Due by 11:59pm on Sun 5/3 – reyes – (52535) 5keywords: surface, partial derivative, first or-der partial derivative, graphical interpreta-tion008 10.0 pointsDetermine fxwhenf(x , y) = (x2+ y)(2y2− x) .1. fx= 2y + 2xy2+ 3x22. fx= 4xy2+ y − 3x23. fx= 2xy2− 2y + 3x24. fx= 4xy2− y − 3x2correct5. fx= y − 4xy2− 3x26. fx= 2y − 2xy2+ 3x2Explanation:From the Product Rule we see thatfx= 2x(2y2− x) − (x2+ y) .Consequently,fx= 4xy2− y − 3x2.009 10.0 pointsDetermine fxwhenf(x, y) =2x + yx + 2y.1. fx= −3x(x + 2y)22. fx=5y(x + 2y)23. fx=3y(x + 2y)2correct4. fx= −5x(x + 2y)25. fx= −4x(x + 2y)26. fx=4y(x + 2y)2Explanation:From the Quotient Rule we see thatfx=2(x + 2y) − (2x + y)(x + 2y)2.Consequently,fx=3y(x + 2y)2.010 10.0 pointsDetermine fxwhenf(x, y) = x sin(2y − x) − cos(2y − x) .1. fx= −cos(2y − x) − x sin(2 y − x)2. fx= x sin(2y − x)3. fx= −x cos(2 y − x) correct4. fx= −2 sin(2y − x) − x cos( 2 y − x)5. fx= 2 sin(2y − x) − x cos(2y − x)6. fx= −x sin( 2y − x)7. fx= x cos(2y − x)8. fx= x cos(2y − x) − sin(2y − x)Explanation:From the Product Rule we see thatfx= sin(2y −x) −x cos(2y −x) −sin(2y −x) .Consequently,fx= −x cos(2y − x) .abdallah (haa2348) – Homework 10 Due by 11:59pm on Sun 5/3 – reyes – (52535) 6011 10.0 pointsDetermine fxwhenf(x, y) = x sin(x + y) − cos(x + y) .1. fx= −x sin(x + y)2. fx= 2x cos(x + y)3. fx= 2 sin(x + y) − x cos(x + y)4. fx= x cos(x + y)5. fx= 2 sin(x+ y) + x cos(x + y) correct6. fx= 2 cos(x + y) − x sin(x + y)7. fx= 2 cos(x + y) + x sin(x + y)8. fx= −2x sin(x + y)Explanation:From the Product Rule we see thatfx= sin(x + y) + x cos(x + y) + sin(x + y) .Consequently,fx= 2

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