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Course: M 408D 408D, Spring 2012
School: University of Texas
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gilbert HW09 (55035) This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points Find an equation for the plane passing through the origin that is parallel to the tan-gent plane to the graph of 2 2 z = f (x, y) = 4x 2y 3x + y at the point (1, 1, f (1, 1)). 1 keywords: 002 10.0 points Find an equation for...

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gilbert HW09 (55035) This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points Find an equation for the plane passing through the origin that is parallel to the tan-gent plane to the graph of 2 2 z = f (x, y) = 4x 2y 3x + y at the point (1, 1, f (1, 1)). 1 keywords: 002 10.0 points Find an equation for the tangent plane to the graph of p 2 2 f (x, y) = 5 x + 3y at the point P (2, 1, f (2, 1)). 1. 3x + 2y 2z 5 = 0 1. z + 5x + 3y 8 = 0 2. 2x 3y + 2z 5 = 0 correct 2. z 5x + 3y = 0 correct 3 . 2 x 3y 2z + 3 = 0 3 . z 5 x 3y = 0 4. 3x 2y + 2z + 3 = 0 4 . z + 5x 3y 2 = 0 5. 2x + 3y + 2z 11 = 0 5. z + 5x + 3y = 0 6. 3x + 2y 2z 11 = 0 6 . z 5 x 3y + 8 = 0 Explanation: Parallel planes have parallel normals. On the other hand, the tangent plane to the graph of z = f (x, y) at the point (a, b, f (a, b)) has normal n = hfx(a, b), fy(a, b), 1 O . Explanation: The equation of the tangent plane to the graph of z = f (x, y) at the point P (a, b, f (a, b)) is given by z But when 2 2 f (x, y) = 4x 2y 3x + y = f (a, b) + f x (a, b)(x a) + f we see that fx = 8x 3 , fy = 4y + 1 , z 5x + 3y = 0 . (a, b)(y b) . f (x, y) = we see that Thus an equation for the plane through the origin with normal parallel to n is hx, y, z On = hx, y, z Oh5, 3, 1 O = 0 , which after evaluation becomes Now when and so when a = 1, b = 1, n = h5, 3, 1 O . y p 5 x 2 + 3 y2 , f x x while = p f y 5 x2 + 3y 2 3y =p 5 x + 3y 2 . 2 , HW09 gilbert (55035) Thus at P , But the partial derivatives of f (2, 1) = 2 , while f x (2, 1) 2 f = 1 , y (2, 1) f (x, y) = tan = are 3 2. fx = z = 2 (x 2) + f 1 y f 2(y 1) , , x (1, 0) =2 (1, 0) and f . y L(x, y) = 10.0 points 1 (x + 4y) 3. L(x, y) = 1 2x + 4. L(x, y) = x + 2y 5. L(x, y) = x + 2y 6. L(x, y) = 1 004 1 4 1. (a,b) y (a,b) z 0.72 correct 4. . z 0.82 3. L(x, y) = f (a, b) f f + (x a) +(y b) z 0.42 2. Explanation: The linearization of f = f (x, y) at a point (a, b) is given by 10.0 points as (x, y) changes from (2, 4) to (2.04, 4.03). 1 2x + 2y + 4 2 correct x . 2 2 z = 3x + y + 1 1 2x + 2y + 4 2 Use Linear Approximation to estimate the change, z, in 2 1 2y 2 1 + 4 2 1 2 2. L(x, y) = x + 2y + 1 keywords: Find the linearization, L(x, y), of the func- f (x, y) = tan at the point (1, 0) . 1 + (x + 4y) Consequently, keywords: tangent plane, partial derivative, radical function, square root function, tion 2. = 2. 003 4 which after rearrangement becomes 2x 3y + 2z 5 = 0 = 2 1 + (x + 4y) It follows that So at P the tangent plane has equation 3 1 (x + 4y) z 0.52 5. z 0.62 Explanation: The Linear Approximation to z = f (x, y) at a point (a, b) is L(x, y) = f (a, b) + fx(a, b)(x a) + fy(a, b)(y b) , HW09 gilbert (55035) and so the change, z in z as (x, y), changes from (a, b) to (a + x, b + y) can be estimated by z = L(a + x, b + f x (a, b) = y) f (a, b) f x + y (a, b) = 4. dw f in which case at (2, 4), f f x (2, 4) = 12 , y (2, dw dt 4) = 8 . 4 100 , x= w = xe 2 + 8 100 3 = 0.72 . dt = 2t 006 dw dt when y/z dw dt dw 2. dt y = 1t, = t = t x z + x z 4 xy y/z z 2e . x 4 xy z z 2 ey/z z = 1 + 4t . xy y/z ze + xy y/z ze . 10.0 points z u when and r = uv , 2 y/z r z = e cos and x=t, ze Use the Chain Rule to find 10.0 points w = xe x Consequently, keywords: dierentials, multi-variable esti-mate, polynomial function Use the Chain Rule to find z = 1 + 4t , therefore, dw 005 y/z y = 1t, dw y/z dt = 2te 4 4xy When Consequently, z 12 100 + 4xy z ey/z z dx w d y w d z dt + y dt + z dt . x x=t, 3 100 . x= x x and On the other hand, when (x, y) changes from (2, 4) to (2.04, 4.03), 1. w = + 2 2t 2t + z 2 = 2y , y x ey/z dt z z Explanation: By the Chain Rule for Partial Di erentia-tion, we see that = 6x , = = + dt z z dw dt 6. x xy y/z z2 e x 4xy 2t ey/z t = correct 5. 2 2 z = f (x, y) = 3x + y , f dw dt dw 3. y. Now when 3 = p u2 + v2 . z sin r = e v cos 2 u2+ v2 1 . u z u sin r = e v cos u2 + v2 correct 2 . u z sin 2 2 3. u = ue v cos + u + v r HW09 gilbert (55035) z sin r 2 2 4. u = e v cos + u + v z u sin while z = e Thus sin , x dx = + F F dy y dx x . u2 + v2 z u sin r = e v cos u2 + v2 . u + F y dx F dy =v u u = F dx Thus r u r 2y sin(x 4y) + e 6. = 2y 4xe 2 sin(x 4y) dx Explanation: Applying the Chain Rule to both sides of the equation F (x, y) = 0, we see that dy 2 2 r 5. u = e v cos + 2 u + v Explanation: By the Chain Rule for Partial Di erentia-tion, z z r z u = r u + u . But z r r = e cos , 4 dx = F x y 007 10.0 points Use partial di erentiation and the Chain Rule applied to F (x, y) = 0 to determine dy/dx when F (x, y) = cos(x 4y) xe 2y dx 1. 2y sin(x 4y) + e = 2y 2xe 4 sin(x 4y) 2y = sin(x 4y) + 2xe 2y 4 sin(x 4y) e dy = dy dx 2. dy 3. dx 4. 5. = 0. dy dx dy dx sin(x 4y) + e F (x, y) therefore, 4 sin(x 4y) 2xe 2y sin(x 4y) 2e = 2y 2 sin(x 4y) 4xe 2y sin(x 4y) 2xe = 2y 2 sin(x 4y) 4e Fx = Fy . = cos(x 4y) xe 2y sin(x 4y) e dy = 4 sin( x dx 4y) = 0, 2y 2x e2y . Consequently, 2y sin(x 4y) + e = 2y . 4 sin(x 4y) 2xe dy dx 008 10.0 points The temperature at a point (x, y) in the plane is T (x, y) C. If a bug crawls on the plane so that its position in the plane after t minutes is given by (x(t), y(t)) where x= 5+ t , y = 3+ 3 4t , determine how fast the temperature is rising on the bugs path at t = 4 when 2y 2y = 0. When keywords: partial dierentiation, Chain Rule, dy correct Tx(3, 6) = 18 , Ty(3, 6) = 4 . 1. rate = 6 C/min correct 2. rate = 4 C/min 3. rate = 5 C/min HW09 gilbert (55035) 4. rate = 3 C/min 5. rate = 48 cu. ins/min 5. rate = 2 C/min Explanation: Explanation: By the Chain Rule for partial dierentiation, the rate of change of temperature on the bugs path at (x(t), y(t)) is given by dT dt = dT (x(t), y(t)) dt = 5 T dx T dy x dt + y dt . The volume of a right circular cone having base r radius and height h is given by V 1 dy dt = 2 5 + t , dt = dV = dt But when t = 4, the bug is at the point (3, 6), while 1 dy 1 dx 3 dt t=4 = 4 , t=4 = 25 + t t = 4 = 6 , dt and we are told that Tx(3, 6) = 18 , Ty(3, 6) = 4 . Consequently, after 4 minutes the temperature on the bugs path is changing at a rate = 1 3 18 6 + 4 4 009 V dr + V r dt 3 4. = 6 C/min . 10.0 points The radius of the base of a right circular cone is increasing at a rate of 4 ins/min while its height is increasing at a rate of 2 ins/min. 2 h. 3r When h and r are changing with t, then by the Chain Rule the rate of change of V with respect to t is given by Now dx 1 = = 2 h dt 3rh But dr dt = 4, in which + 1 case dV dh dr dt 2 dh 3r dt . dh dt = 2 , 8 2 2 dt = 3rh + 3 r . Consequently, when r = 3 and h = 5, dV dlt (r=3, h=5) = 46 cu.in./min. . 010 10.0 points The length and width w of the closed box shown in At what rate is the volume of the cone changing when the radius is 3 inches and the height is 5 inches? h 1. rate = 46 cu. ins/min correct 2. rate = 50 cu. ins/min w 3. rate = 47 cu. ins/min 4. rate = 49 cu. ins/min are increasing at a rate of 4 ft/min while its height h is decreasing at a rate of 6 ft/min. HW09 gilbert (55035) 6 Find the rate at which the volume of the box is increasing when = 2, 011 10.0 points w = h = 3 feet . Find the directional derivative, fV, of p f (x, y) = 4x + 2y at 1. rate = 30 cu ft/min 2. rate = 26 cu ft/min the point (3, 3) in the direction 3. volume is decreasing, not increasing v = i+j. 4. rate = 28 cu ft/min 5. rate = 24 cu ft/min correct 1 . fV V = w h . Thus by the Chain Rule for partial di eren-tiation, the rate at which the surface area is changing is given by V t V V w V h t + w t + h t . = = 16 2 . fV Explanation: The box has volume = 13 f 3. V 1 4. fV = f Now V V = wh , w = h , V 5. h = w , and so we see that =0 V 2 correct = 23 Explanation: For an arbitrary vector v, V w h t = wh t + h t + w t . fV = f , v || When = 2, w = h = 3, and v where we have normalized the direction vector so that it has unit length. Now the partial derivatives of t = therefore, V t w t p h = 4 , t = 6 , f (x, y) = 4x + 2y are given by =334 f x = 4x + 2y , + (3 2) 4 (3 2) 6 = 24 > 0 . Consequently, the volume is increasing at a rate = 24 cu ft/min . 2 and f 1 y = 4x + 2y . HW09 gilbert (55035) Thus 7 we see that f (x, y) = f x i + f 2 2 3 f = y + 9x y, 2xy + 3x . y j 2 1 j , = 4x + 2y i + keywords: 4x + 2y 013 and so Determine the gradient of 2 f (x, y) = xy at the point P (3, 2). 1 2 1 f (3, 3) = 2 3 i + 3 j . On the other hand, 1 v = v = i+j But then f v v = 21 = |v| 2 1 3 i+3 j 1. f |P = 4 i 12 j correct 2 (i + j) . 2. f |P = 12 j 4 k (i + j) . 3. f |P = 12 i + 4 k 4. f |P = 4 i + 12 j || Consequently, f V =2 1 5. f |P = 4 i 12 k 2 3 + 3 1 = 2 1 . keywords: 012 10.0 points 6. f |P = 12 j + 4 k Explanation: Since 2 f = fx i + fy j = y i + 2xy j , and P = (3, 2), we see that f |P = 4 i 12 j Find the gradient of 2 3 f (x, y) = xy + 3x y . 1. f 2. f 3. f 4. f = 10.0 points y2 + 9x2y, 014 3 2xy + 3x correct 2 2 = 9x y y , 3 = 3x 2xy, 2 2 = y + 9x y, 3 = 2xy + 3x , 2xy + 3x 3 f (x, y) = The contour map given below for a function f shows also a path r(t) traversed counter-clockwise as indicated. 3x 2xy 2 9x y y 5. f 32 2 6. f = 2xy + 3x , y + 9x y Explanation: Since f 10.0 points 2 2 y + 9x y 3 2 2 1 0 3 Q -3 -2 P R x f , y , . -1 0 HW09 gilbert (55035) Which of the following properties does the derivative d dt f (r(t)) II FALSE: at Q we are ascending - the con-tours are increasing in the counterclockwise direction. III TRUE: at P we are descending - the con-tours are decreasing in the counterclockwise direction. have? I zero at R, II negative at Q, III 8 keywords: contour map, contours, slope, curve on surface, tangent, Chain Rule, multi-variable Chain Rule, negative at P . 1. none of them 015 2. all of them Find the directional derivative, fV, of the function 3. II and III only f (x, y) = 3 + x y 4. I only 5. I and III only correct 6. III only at the point P (1, 4) in the direction of the vector v = h 3, 4 O . f 7. I and II only 1. = Explanation: By the multi-variable Chain Rule, d f (r(t)) = (f )(r(t))r (t) . 2. fV 10 3. d dt will be the sign of the slope of the surface in the direction of the tangent to the curve r(t), and we have to know which way the curve is being traversed to know the direction the tangent points. In other words, if we think of the curve r(t) as defining a path on the graph of f , then we need to know the slope of the path as we travel around that path - are we going uphill, downhill, or on the level. That will depend on which way we are walking! From the contour map we see that I TRUE: at R we are on the level - we are following the contour. 11 10 = 5 V f = 7 10 V 5 . fV f (r(t)) = 4 f 4. Thus the sign of 9 V 8. II only dt 10.0 points = 1 correct Explanation: Now for an arbitrary vector v, fV = f v v , || where we have normalized so that the direc-tion vector has unit length. But when f (x, y) = 3 + x y , then 1 x f = ( y) i + 2 y j. HW09 gilbert (55035) At P (1, 4), therefore, f P = 2 i + 1 4 9 so that at P (3, 3), 1 1 (f )(3, 3) = 6 i + 6 j j. = D 1, 1 E . 6 6 Consequently, when v = h 3, 4O , On the other hand, v = fV(1, 4) = 2, 41 vv = 1 . || h 4, 3O which as a vector of unit length becomes 43 v =D5, 5 E. v || keywords: Consequently, 016 10.0 points f |P = Find the directional derivative, fV, of f (x, y) = y 017 x P Q when Q = (7, 6). 2. f V 1 = 10 3. f V 1 = 30 correct 1 6 2. max slope = 5 3. max slope = 1 2 4. max slope = 20 = 15 1 = 15 5. max slope = 20 correct 6. max slope = The directional derivative of f (x, y) at P in the direction of v = P Q is given by the dot product fV|P = f |P |v v | f (x, y) = xy 1/2 8. max slope = 5 , || we see that 2 7. max slope = 4 Explanation: At P (0, 5, 0) the slope in the direction of v is given by f (0,5) vv . . But when = 10.0 points 1. max slope = 4 Explanation: f x But when 1 y 1/2 x3 1 = 30 . at the point P (0, 5). = 5 . fV 5 3E f (x, y) = 4 sin(xy) 1. f V V 5, Find the maximum slope on the graph of 4. 4 1/2 at P = (3, 3) in the direction of the vector f D 1, 1 E.D 66 , f y = 2 1 1 1/2 xy , f (x, y) = 4 sin(xy) , HW09 gilbert (55035) of the partial derivative fy at P = (1, 2). the gradient of f is f (x, y) = 4y cos(xy) i + 4x cos(xy) j , so at P (0, 5) f (0,5) 10 On the other hand, the slope at P in the direction of any vector v = ai+bj = 20 i . Consequently, the slope at P will be maxi-mized when v = i in which case is given by f (1,2) max slope = 20 . = keywords: slope, gradient, trig function, max-imum slope 018 10.0 points If the graph of z = f (x, y) at P = (1, 2) has slope = 2 in the x-direction and slope = 3 in the y-direction, find the slope at P in the direction of the vector v = 4 i + 3 j . 1 1. slope = 5 correct 2 2. slope = 5 3. slope = 1 4 4. slope = 5 3 5. slope = 5 Explanation: The slope of the graph of z = f (x, y) in the x-direction at P = (1, 2) is the value f fx(1, 2) = x (1, 2) = 2 of the partial derivative fx at P , while the slope of the graph of z = f (x, y) in the ydirection at P is the value f fy(1, 2) = y (1, 2) = 3 vv || 1 |v | fx(1, 2) i + fy(1, 2) j (a i + b j) . When v = 4 i + 3 j , therefore, the graph of f has slope = 1 5(2 i 3 j)(4 i + 3 j) = at P in the direction of v. keywords: 1 5
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Summary of Key Learning Objectives forIndianapolis City Services (A)The Indianapolis City Services case continues the theme that we started with Kanthal, of using accountingcost data for purposes other than product costing. Specifically, in this case,
UC Davis - MGT ` - 170
Summary of Key Learning Objectives forKanthal and Cooperative Bank CasesThe Kanthal Case extends previous lessons on cost system design to the analysis of selling, general andadministrative expenses (SG&amp;A), which typically fall outside of product costi
UC Davis - MGT ` - 170
Summary of Key Learning Objectives forClass 4: Wilkerson &amp; Cooperative Bank CasesThese cases continue the theme of consequences of growth started with the Seligram case. Cost systemdesign is used to align product and customer costing information with t
UC Davis - MGT ` - 170
Summary of Key Learning Objectives forClass 3: Seligram CaseThe Seligram case provides an example of how traditional cost systems that meet financial accountingneeds for inventory valuation, fail in dynamic competitive environments characterized by rap
UC Davis - MGT ` - 170
Summary of Key Learning Objectives forClass 2: Bridgeton CaseThe Bridgeton Case introduces cost accounting by providing a simple example of the design andfunctioning of traditional product cost systems. The case setting and managerial decisions covered
UC Davis - MGT ` - 170
MGT 170: Class 1Summary of Key Learning Objectives forP&amp;G Polska CaseThe P&amp;G Polska case introduces the topic of management accounting, including both costmanagement and management control (i.e., performance measurement and evaluation). The caseillus
UC Davis - MGT ` - 170
5-10 Year Plan DevelopmentJanF ebM arA prM ayJ unJ ulA ugS epO ct2 pageB u r k e L e t t erCE O J &amp; JDis c u s sw it hE xec .C om mStr ateg yReviewExec . C om mStolz erDeb atean dA dju s tCod m anBoar dFunc . D epts .P r ior Yr
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JanFebMarAprMayJunJulAugSepO ctCEO J&amp;JExec. CommStolzerDebateandAdjustCodmanBoardFunc. Depts.Prior Yr2nd YearFcstDebateandAdjustMissionState.5/10 Plan-by area-bus. seg-strategies-mktg plan-other deptsInitialPrepApproved
NUCES - FINANCIAL - 101
CHAPTER 10ACQUISITION AND DISPOSITION OFPROPERTY, PLANT, AND EQUIPMENTIFRS questions are available at the end of this chapter.TRUE-FALSEConceptualAnswerFTFTFTFFFTTTTFFTTFFTNo.Description1.2.3.4.5.6.7.8.9.10.11.12.13
McMaster - CELL BIO - 2b03
Cancer biology: tumorigenesis can stem from errors in signaling pathways and cell cycle regulationmechanisms Cancer is result of uncontrolled cell divisiono Leading cause of premature death in Canadao In young animal, cell proliferation exceeds cell
McMaster - CELL BIO - 2b03
McMaster - CELL BIO - 2b03
Final Exam Practice QuestionsThese questions can be used as a guide to help you focus your studying to cover key points that willlikely be on the final exam. Two disclaimers: 1) This guide should not be used as a substitute for thelecture notes, althou
Utah Valley University - HIST - 2710
AmistadAmistad We are all born into something, a certain belief or ideology. Others create their ownideologies, either out of conviction or for the purpose of convenience. Our lives revolve aroundthese ideas, these ideologies, for it is through them th
Utah Valley University - HIST - 2710
Brazil TourismSubject: Tourism in Brazil Statement: A brief overview of Brazil. Introduction: With its 8,512thousands km2, Brazil is the fifth largest country in the world and it is today America Latin'smost important country, if considering the produc
Utah Valley University - HIST - 2710
Brazil, Japan, UkExecutive Summary A cross analysis between three countries around the world were arrangedto determine which of those countries present the best investment opportunities. An evaluationamong the countries was structured through the compa
Utah Valley University - HIST - 2710
BRAZIL For years Brazil has been a featured country in South America. Not only is Brazil avery pretty country, it is filled with an abundance of natural resources. For three centuries Brazilwas ruled by Portugal. The country finally gained its independe
Utah Valley University - HIST - 2710
Brazilians and North AmericansBrazilians have a strikingly distinct culture which favors physical contact to show love andaffection compared to the North American culture. The personal space that Brazilians have isnot as great as the North Americans ha
Utah Valley University - HIST - 2710
The world's most luxurious hotel, Burj Al-Arab or the Arabian Tower, is the second modernwonder of the world. Ever since its completion in November 1997, the hotel has become an iconfor the technologically expanding city of Dubai. The jewel of the Middl
Utah Valley University - HIST - 2710
Child Prostitution EssayThe word &quot;prostitution&quot; dates back to the year 1553. Webster's dictionary defines it as the act orpractice of indulging in promiscuous sexual relations especially for money or the state of beingprostituted. Prostitution probably
Utah Valley University - HIST - 2710
China 1Abstract the purpose of this paper is to present China as a nation that has gone through severaleconomic phases until it reaches a point that has made several consider this country as apromising future power in the region. While Japan is falling
Utah Valley University - HIST - 2710
AMERICANCIVILIZATIONUNITEDSTATESHISTORY170RhettS.JamesMarcioCarvalheiroHistory170Chapter1September16,2008ThemeetingofCultures1.MeltingPotOverthecourseofmanycenturies,migrantsfromRussiaandprobablyalso fromotherareasofAsiaandfromthePacificIslandss
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AMERICAN CIVILIZATIONUNITED STATES HISTORY 170Rhett S. JamesMarcio CarvalheiroHistory 170Chapter 2September 16, 2008Transplantations and borderlands1.Great wealthAs Jamestown struggled to survive, the London Company (now renamed the VirginiaComp
Utah Valley University - HIST - 2710
AMERICANCIVILIZATIONUNITEDSTATESHISTORY170RhettS.JamesMarcioCarvalheiroHistory170Chapter3September16,2008SocietyandCultureinProvincialAmerica1. FairpaymenttopeopleandnationsforterritorywoninwarsThesystemoftemporaryservitudedevelopedoutofpractices
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AMERICANCIVILIZATIONUNITEDSTATESHISTORY170RhettS.JamesMarcioCarvalheiroHistory170Chapter5September22,2008TheAmericanRevolution1. TheU.S.AisafederalistsystemThreeweeksafterthebattlesofLexingtonandConcord,whentheSecond ContinentalCongressmetinPhil
Utah Valley University - HIST - 2710
AMERICANCIVILIZATIONUNITEDSTATESHISTORY170RhettS.JamesMarcioCarvalheiroHistory170Chapter6September30,2008TheConstitutionandtheNewRepublic1. TheU.S.AisafederalistsystemFiftyfivemen,representingallthestatesexceptRhodeIsland, attendedoneormoresessi
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AMERICANCIVILIZATIONUNITEDSTATESHISTORY170RhettS.JamesMarcioCarvalheiroHistory170Chapter7September30,2008TheJeffersonianEra1. Greatwealth/workethicSomewomenaspiredtomore.In1784,JudithSargentMurraypublishedanessaydefendingtherightofwomentoeducati