page 1 of 10 Math 1B First Exam Tuesday 24 July 2009 Instructor Theo Johnson Freyd http math berkeley edu theojf 09Summer1B Name Problem Number Score Maximum ANSWERS 1 2 20 20 3 4 5 20 20 20 Total 100 Please 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 exam by myself without any help during the exam from other people or from sources other than my allowed one page hand written cheat sheet Moreover I have not provided any aid to other students in the class during the exam I understand that cheating prevents me from learning and hurts other students by creating an atmosphere of distrust I consider myself to be an honorable person and I have not cheated on this exam in any way I promise to take 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 10 1 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 compute both coordinates x and y using the following formulae Z 1 1 x x x x2 dx A 0 Z 11 1 2 y x x2 2 dx A2 0 R1 x x2 dx 23 13 13 We have used Here A is the area of the figure equal to A 0 R1 the fact that the two curves intersect at 0 0 and at 1 1 and that 0 xn dx 1 n 1 Then Z 1 1 9 2 1 3 2 3 x x x dx 3 1 3 0 5 4 20 Z 1 3 1 1 9 1 x x4 y 2 3 0 2 2 5 20 9 9 x y 20 20 b 5 pts Recall the Theorem of Pappus that the volume of the solid of revolution formed by rotating a region R around a line is the area of R times the distance that the centroid of R travels as it revolves around the line Use this theorem and some geometry to find the volume of the solid of revolution formed by rotating the region bound by the curves y x and y x2 around the line y x 9 9 The shortest path from the centroid 20 20 to the line y x is to the origin the q 9 2 9 2 9 distance from the centroid to the origin is 20 20 20 2 Thus the distance 9 9 traveled by the centroid is 20 2 2 10 2 and so the volume is V Ad 1 9 3 2 2 3 10 10 Name Math 1B First Exam 7 July 2009 page 3 of 10 2 A trough is constructed with vertical ends and flat sides that meet at a 45 point so that each cross section is a right triangle as shown in the picture One end of the trough can move by sliding along the trough slider h x a 10 pts If the trough is filled with water to a depth h find the hydrostatic force on one end of the trough If the trough is filled to a depth h then the submerged part of the end of the tough is a right h h triangle Thus letting y denote the depth of a generic point the width of the triangle at depth y is h y as y changes from 0 to h So the hydrostatic force is Z F 0 h y2 y3 gy h y dy g h 2 3 h gh3 6 0 Here 1 g cm3 1000 kg m3 and 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 that the trough is one meter long the water comes to a depth of ten centimeters If you move the 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 a constant Therefore the depth h is given by p h 2V x where V 1 m 0 1 m 2 2 0 005 m3 100 cm 10 cm 2 2 5000 cm3 Since we know the volume we can simplify a bit h p 1 10 000 cm3 x 100 cm x cm 1 2 m x m 1 2 10 c 5 pts The R a work required to move the slider from a length a to a length b is defined to be W b F x dx where F x is the hydrostatic force on the slider when the slider is at length x Find the work required to shrink the trough from a length of one meter to a length of fifty centimeters In part a we computed the p force when the depth is h to be F h gh3 6 In part b we computed h h x 2V x Thus we have F x g p 3 g23 2 V 3 2 3 2 2V x 6 x 6 Then the work is a Z a g23 2 V 3 2 3 2 1 1 g23 2 V 3 2 g23 2 V 3 2 1 1 2 x dx x 6 6 1 2 3 a b b b where V 0 005 m3 5000 cm3 g 10 000 kg m2 s2 1000 g cm2 s2 and a 1 m 100 cm and b 0 5 m 50 cm In terms of the numbers we have g 1 1 2 1 3 3 2 W 10000 cm 108 g cm2 s2 3 3 50 cm 100 cm This is a much better number in mks units whence W 2 1 kg m2 s2 0 07 kg m2 s2 3 where we have rounded 2 to 1 42 to make the numbers nice This is about the accuracy of the estimate g 10 m s2 Name Math 1B First Exam 7 July 2009 page 5 of 10 3 The class sizes at a certain large university are distributed exponentially in a given semester there are N x Ae x B classes of size x where A and B are positive constants Since universities are large we approximate x and N x which actually can only take integer values by continuous variables …
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