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WUSTL MATH 132 - m132_E3sF10

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MATH 132 EXAM III ( ) FALL 2010solutions1) An aquarium 6 ft long , 1 ft wide and 4 ft high is full of water weighing 62.5 lb/ft Find the work needed , in ft-lbs , to pump out$Þ half the water of the aquarium.A) 749 B) 749.5 C) 750 D) 750.5 E) 751 F) 751.5 G) 752 H) 752.5 I) 753solution:Let the bottom be 1 and the top 4. For each the crossCœ Cœ !ŸCŸ%ß section so the weight of a layer is 6 (62.5) 375. Half the water meansEÐCÑ œ 'ß Ð Ñ! œ 2 4, and the distance lifted is Then work done is given byŸCŸ %! CÞ '#%$(&Ð% ! CÑ .C œ (&! 0> ! 6,=Þ ÐGÑ-------------------------------------------------------------------------------------------------2) A chain 10 ft long, uniformly weighing 32 lb, is lying on the ground. Find the work (in ft-lbs) required to raise the chain up vertically so that the bottom of the chain will be 2 ft above the ground . (Note: The top will end up,12 feet above the ground.)A) 100 B) 150 C) 314 D) 512 E) 1664 F) 1600 G) 384 H) 320 I) 224 J) 160solution: If is the ground, then we want to lift the top of the chain to TheC œ ! C œ "#Þ weight is 32/10 = 3.2 lb/ft. For when the top is lifted to level y, the! Ÿ C Ÿ "!ßweight is 3.2 y lb. For 1 2, the weight is always 32 lb. So work is!ŸCŸ" '!"!$Þ# C .C " Ð#чÐ$#Ñ œ "'! " '% œ ##% 0 > ! 6,=Þ ÐMÑ----------------------------------------------------------------------------------------------------3) Suppose the waiting time for a customer's call to be answered is related to a probability density function which has an average waiting time of 3 minutes. Find the probability that a customer waits more than 6 minutes .A) eeeeee!"Î# !"Î# !"Î$ !"Î$ !$Î# !$Î#"" "$$ $FÑ GÑ HÑ IÑ JÑKÑ LÑ MÑ / NÑee e!#Î$ !#Î$ !# !#Î$""$$solution: :Ð>Ñ œ / ß T Ð> # 'Ñœ / .>œ/ œ/ Þ ÐMÑ""$$!>Î$ !>Î$ !'Î$ !#'_'-------------------------------------------------------------------------------------------------4) Find the for the probability density functionmean f( ) = .12xBÐ" ! BÑ 30 ! Ÿ B Ÿ "!9>2/<A3=/!#A) 1 B) C) """#$233537558881010HÑ IÑ J Ñ KÑ LÑ MÑ N Ñsolution: . œ B 0ÐBÑ .B œ "#B Ð" ! BÑ .B œ "# B ! B.Bœ ÞÐIÑ'' '!_! !_" "$$%$&2.5) Find the solution of the differential equation , with initial condition, y(0) = 1..C.B %C/œ#B$A) e B#FÑ # / ! "GÑ/ HÑÐ/ " ÑIÑÐ#/! "Ñ#B B #B "Î% #B "Î%""##JÑ / " KÑ Ð/ ! "Ñ " "LÑÐ/ " "Ñ MÑ Ð/ ! "Ñ " """ "##BÎ# B "Î% %B #B#È ""##solution: Given y(0) 1 we get''%C .C œ / .B Ê C œ / " GÞ œ$#B %#B"# 1 then Then we have So we getœ " Gß G œ Þ C œ / " Þ"" ""## ##%#B y ) It can't be y = ) , since y(0) 1. œ/" Þ ! / " ÑœÐHÑ( ( "" ""## ###B "Î% #B "Î%-------------------------------------------------------------------------------------------------------------6) How many years will it take an investment to if the interest rate isdouble 6 % , compounded continuously ?A) 2 B) 4 C) 6 D) 8 E) 10 F) 12 G) 14 H) 16 I) 18 J) 20solution: Solve EÐ>Ñ œ EÐ!Ñ / Þ EÐ>Ñ œ #EÐ!Ñ Ê EÐ!Ñ / œ # EÐ!Ñ ÊÞ!'> Þ!'>/ œ # Ê > œ œ ""Þ&&Þ "#Þ!'>68Ð#ÑÞ!'Need years. (F)-------------------------------------------------------------------------------------------7) A tank contains 20 lb of salt dissolved in 6000 L of water . Brine that contains 0.01 pounds of salt per liter, enters the tank at the rate of 50 L/min. The solution is kept thoroughly mixed and drains out of the tank at the rate of 50 L/min . How much salt remains in the tank after 2 hours? (Find the answer to 2 decimals.) A) 42.84 B) 43.52 C) 44.68 D) 45.28 E) 46.34 F) 47.12 G) 48.62 H) 49.74 I) 50.26 J) 51.64 Note: 2 hours is 120 minutes.)Ðsolution: This is seperable, so we get.C C C '!!C.> '!!! "#! "#!œ ÐÞ!"Ñ † Ð&!Ñ ! Ð&!Ñ œ !Þ& ! œÞ ''.C'!!C "#! "#!.> >G !œÊ! 68l'! ! Cl œ " GÊl'!! Cl œ / / Þ>"#!From y(0) = 20 we get Since the right side is/ œ %! Ê l'! ! Cl œ %! / ÞG !>"#! never zero and is sometime positive, we get '! ! C'!! Cœ%!/ Ê!>"#! CÐ>Ñ œ '! ! %! / Þ CÐ"#!Ñ œ '! ! %! / œ '! ! µ %&Þ#)Þ ÐHÑ!!%!/>"#! "#!120-------------------------------------------------------------------------------------------8) Find the values of which would make the function y = e a solution to the<<> differential equation yww w" C ! 'C œ !ÞEÑ ß FÑ"ß GÑ#ß% HÑ'ß"# IÑ"ß$ JÑ!ß' KÑ"ß# LÑ! 'ß ! $"" "#' $MÑ ! $ß # NÑ ! 'ß #solution: y = e , y = e y = e Then y<> w <> ww # <> ww w<ß < Þ " C ! 'C œe So we get solution <> #Ð< " < ! 'Ñ œ !Þ < œ ! $ß #Þ ÐMÑ$Þ9) A bacteria culture starts with 500 bacteria and grows with a constant relative growth . After 1 hour there are 1000 bacteria. In , from the start,rate how many hours will the population be 13,500 ? ( Find the answer to 2 decimals )A) 1.25 B) 2 C) 2.50 D) 3.25 E) 4.50 F) 4.75 G) 5.25 H) 5.75 I) 6.5 J) 7.25solution: We have and So EÐ>Ñ œ &!! / EÐ"Ñ œ "!!!Þ / œ # Ê 5 œ 68Ð#ÑÞ5> 5Then Now, solve for t given 13,500EÐ>Ñ œ &!! Ð#Ñ Þ œ &!!Ð#Ñ Ê # œ #(>>>Ê > œ œ %Þ(&%))) µ %Þ(&Þ ÐJ Ñ68Ð#(Ñ68Ð#Ñ---------------------------------------------------------------------------------------------------10) Determine whether the sequence converges orœ "#8 " "$8 " "_8œ" diverges . If it converges then find the limit .A) 0 B) "#" $&$$# ##GÑ$ HÑ IÑ JÑ# KÑ LÑ MÑ% NÑ.3@/<1/=solution: lim lim8Ä_8Ä_#8""#$8""$#"$"œœÞÐHÑ"8"8---------------------------------------------------------------------------------------------11) Find the sum of the series if it converges.$ !" ! " !ÞÞÞÞÞ ß$$ $ $&#&"#&'#&A) "# "# $ & * $ & * #&& ( "! # % & $ #& *FÑ GÑ HÑ I Ñ JÑ KÑ LÑ MÑ NÑ .3@/<1/=solution: This is a Geometric Series with ,$Ð"!"ÐÑ! ÐÑ" ÞÞÞÞ ÑÞ + œ $"" "&& &#$<œ ! ÞœÞÐHÑ"$&&#""So it converges to "&--------------------------------------------------------------------------------------------------12) Which of the following 3 series is are convergent ?Î I) II) III) !!!8œ 8œ___"8 68Ð8Ñ 868Ð8Ñ =38 Ð8Ñ8œ"8822#ÈA) I B) II C) III D) I &


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WUSTL MATH 132 - m132_E3sF10

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