page 1 of 10Math 1B: First ExamTuesday, 24 July 2009Instructor: Theo Johnson-Freydhttp://math.berkeley.edu/~theojf/09Summer1B/Name: ANSWERSProblem Number 1 2 3 4 5 TotalScoreMaximum 20 20 20 20 20 100Please do not begin this test until 2:10 p.m. You may work on the exam until 4 p.m.Please do not leave during the last 15 minutes of the exam time.You must always justify your answers: show your work, show it neatly, and when in doubt,use words (and pictures!) to explain your reasoning. Please box your final answers.Calculators are not allowed. Please sign the following honor code:I, the student whose name and signature appear on this midterm, have completed the examby myself, without any help during the exam from other people, or from sources other thanmy allowed one-page hand-written cheat sheet. Moreover, I have not provided any aid toother students in the class during the exam. I understand that cheating prevents me fromlearning and hurts other students by creating an atmosphere of distrust. I consider myselfto be an honorable person, and I have not cheated on this exam in any way. I promise totake an active part in seeing to it that others also do not cheat.Signature:Name: Math 1B First Exam: 7 July 2009 page 2 of 101. (a) (15 pts) Find the centroid of the region bound by the curves y =√x and y = x2.By symmetry, the centroid definitely lies on the line y = x. Nevertheless, we computeboth coordinates ¯x and ¯y, using the following formulae:¯x =1AZ10x√x − x2dx¯y =1A12Z10(√x)2− (x2)2dxHere A is the area of the figure, equal to A =R10√x − x2dx =23−13=13. We have usedthe fact that the two curves intersect at (0, 0) and at (1, 1), and thatR10xndx = 1/(n+1).Then:¯x =11/3Z10x3/2− x3dx = 325−14=920¯y =12/3Z10x − x4=3212−15=920(¯x, ¯y) =920,920(b) (5 pts) Recall the Theorem of Pappus, that the volume of the solid of revolution formedby rotating a region R around a line ` is the area of R times the distance that the centroidof R travels as it revolves around the line `. Use this theorem and some geometry tofind the volume of the solid of revolution formed by rotating the region bound by thecurves y =√x and y = x2around the line y = −x.The shortest path from the centroid (920,920) to the line y = −x is to the origin; thedistance from the centroid to the origin isq(920)2+ (920)2=920√2. Thus, the distancetraveled by the centroid is920√2 × 2π =910√2π, and so the volume is:V = Ad =13910√2π =310√2πName: Math 1B First Exam: 7 July 2009 page 3 of 102. A trough is constructed with vertical ends and flat sides that meet at a 45◦point, so thateach cross section is a right triangle, as shown in the picture. One end of the trough can moveby sliding along the trough.sliderxh(a) (10 pts) If the trough is filled with water to a depth h, find the hydrostatic force on oneend of the trough.If the trough is filled to a depth h, then the submerged part of the end of the tough isa right h × h triangle. Thus, letting y denote the depth of a generic point, the width ofthe triangle at depth y is h − y, as y changes from 0 to h. So the hydrostatic force is:F =Zh0ρgy(h − y)dy = ρghy22−y33h0= ρgh3/6Here ρ = 1 g/cm3= 1000 kg/m3and g = 10 m/s2= 1000 cm/s2.Name: Math 1B First Exam: 7 July 2009 page 4 of 10(b) (5 pts) Enough water is poured into the trough so that when the sliding end is such thatthe trough is one meter long, the water comes to a depth of ten centimeters. If you movethe slider so that the trough has length x, how deep will the water now be?The volume is a constant. The volume is equal to the length x times the area of an end.The end is an h × h triangle, so it has area h2/2. Thus, the number V = xh2/2 is aconstant. Therefore, the depth h is given byh =p2V/xwhere V = (1 m)(0.1 m)2/2 = 0.005 m3= (100 cm)(10 cm)2/2 = 5000 cm3. Since weknow the volume, we can simplify a bit:h =p10, 000 cm3/x = 100 cm (x/cm)−1/2=110m (x/m)−1/2(c) (5 pts) The work required to move the slider from a length a to a length b is defined tobe W =RabF (x)dx, where F (x) is the hydrostatic force on the slider when the slider isat length x. Find the work required to shrink the trough from a length of one meter toa length of fifty centimeters.In part (a), we computed the force when the depth is h to be F (h) = ρgh3/6. In part(b), we computed h = h(x) =p2V/x. Thus, we have:F (x) = ρgp2V/x3/6 =ρg23/2V3/26x−3/2Then the work is:Zabρg23/2V3/26x−3/2dx =ρg23/2V3/261−1/2x−1/2ab=ρg23/2V3/231√b−1√awhere V = 0.005 m3= 5000 cm3, ρg = 10, 000 kg/m2s2= 1000 g/cm2s2, and a = 1 m =100 cm and b = 0.5 m = 50 cm.In terms of the numbers, we have:W =ρg3(10000 cm3)3/21√50 cm−1√100 cm=√2 − 13× 108× g cm2/s2This is a much better number in mks units, whence:W =√2 − 13kg m2/s2= 0.07 kg m2/s2where we have rounded√2 to 1.42 to make the numbers nice. This is about the accuracyof the estimate g = 10 m/s2.Name: Math 1B First Exam: 7 July 2009 page 5 of 103. The class sizes at a certain large university are distributed exponentially: in a given semesterthere are N (x) = Ae−x/Bclasses of size x, where A and B are positive constants. Sinceuniversities are large, we approximate x and N(x) (which actually can only take integervalues) by continuous variables.(a) (5 pts) How many classes does the university offer in a given semester?We add up the total number of classes by integrating:total number = N =Z∞0N(x) dx =Z∞0Ae−x/Bdx = ABThe integral starts at 0 because we cannot have a negative number of classes.(b) (5 pts) Write a probability distribution expressing the probability that a given class hassize x.The probability distribution ρ(x) should be proportional to N(x) but normalized so thatR∞0ρ(x)dx is 1. Thus, we haveρ(x) =1NN(x) =1Be−x/BName: Math 1B First Exam: 7 July 2009 page 6 of 10(c) (5 pts) What is the average class size at this university?The average class size is:¯x =1NZ∞0x N(x) dx =Z∞0x ρ(x) dx =1BZ∞0xe−x/Bdx =1BB2= B(d) (5 pts) Jimmy Stewart is a student at this university. Since the probability that he endsup in any particular class is proportional to the number of students in that class, theprobability that his first-period class has size x is proportional to xAe−x/B. Find theexpected size of Jimmy’s first-period class.From Jimmy’s perspective, the probability of being in a class of size x is proportional toM(x) = xe−x/B. Thus, the average class