From zeilberg@math.rutgers.edu Tue Dec 12 18:44:51 2006 Received: (from zeilberg@localhost) by math.rutgers.edu (8.11.7p2+Sun/8.8.8) id kBCNiUA18002; Tue, 12 Dec 2006 18:44:30 -0500 (EST) Date: Tue, 12 Dec 2006 18:44:30 -0500 (EST) From: Doron Zeilberger Message-Id: <200612122344.kBCNiUA18002@math.rutgers.edu> To: zeilberg@math.rutgers.edu Cc: lpudwell@math.rutgers.edu, scottsch@math.rutgers.edu Subject: MathIsFun; Yesterday's "quiz"; Clarifications and announcements Status: O Content-Length: 5069 Dear Students, Below are the solutions to the last attendance "quiz". First a few reminders and clarifications. 1) Regarding the deal, I am currently going over the envelopes to see who qualified. I hope and expect most people to qualify, so I will only send E-mail (tomorrow evening) to those poor souls who did not qualify. So if you don't get an E-mail message from me by tomorrow (Wed.) 8:00pm, you can assume that you qualified. 2) Let me clarify again how the "deal" works. I compute the grade in TWO ways. The first way is the usual way, described in the syllabus page that I handed in at the first class and that can be read on-line at http://www.math.rutgers.edu/~zeilberg/math251.html Here is the relevant section: -------------------------- GRADE: The max. number of points is 500, with the following breakup. * Input from Recitation Leader (on his/her discretion, some combination of Maple workshops and quizes) 100 points. * 2 Midterm Exams: 100 points each. * Final: 200 points. 460-500: A ; 430-459: B+ ; 400-429: B ; 370-399: C+ ; 325-369: C ; 280-324: D ; 0-279: F . No Curve!. ------------------------------------ If you didn't participate in the deal, that's how your grade gets computed. If you did participate in the deal (and qualified), then I ALSO look at your FINAL grade (just by itself) which is out of 200, so for the grade according to the final: we have 184-200: A ; 172-183: B+ ; 160-182: B ; 148-159: C+ ; 130-159: C ; 112-129: D ; 0-111: F . No Curve!. -------------------------------------- Now whatever is HIGHER is your grade for the course. Example 1 : John Smith got the following scores: >From The TAs: 80 (out of 100) Exam 1: 70 (out of a 100) Exam 2: 80 (out of a 100) Final: 173 (out of 200) The total is: 80+70+80+173= 403 According to the first scheme this is a B. According to the second scheme it is a B+ (barely) so he will get a B+ and he benefited by the deal. Example 2 : Nancy Brown got the following scores: >From The TAs: 80 (out of 100) Exam 1: 70 (out of a 100) Exam 2: 80 (out of a 100) Final: 110 (out of 200) The total is: 80+70+80+110= 340 According to the first scheme this is a C. According to the second scheme it is an F. so she will get a C (and the deal didn't help her but didn't hurt her either). Please note: The Final counts for AT LEAST 40% of the grade, so if you got full score from the TAs and scored 100 on both exams 1 and exam 2, but got 0 on the Final then you would get a D for the course (since the total is 300). ------------------------------------------------------------ 3) On the Final you are allowed a 5x7 index card written on BOTH sides with AT MOST TEN non-math words 4) As I hope you know, there is going to be Optional Review for the Final on Thurs. Dec. 14, 2006, 8:40am-1:20pm (with breaks, of course) at ARC 103. You are welcome to ask me any questions then. 5) The attendance quizes can be downloaded from the directory http://www.math.rutgers.edu/~zeilberg/calc3/quizes 6) As usual, absolutely no calculators allowed on the Final. 7) The Final for Sections 1-3 is this coming Friday, Dec. 15, 2006, 8:00-11:00am in the Classroom (SEC 118). The Final for Sections 4-6 is Thurs., Dec. 21, 2006, 12:00-3:00pm in the Classroom (SEC 117). 8) Bring your notebooks to the final, I'll be checking a random sample. ----------------------------------------- ----------------------Solution to the "quiz" of Dec. 11, 2006 Most people got yesterday's quiz (that was about basic skills) right, but as usual, a few people messed up the algebra. Q: a) Draw the triangle bounded by the lines y=x, y-2x, and x+y=1 b) determine its vertices c) Find its circumference Solution: a) You do it. Note that the line x+y=1 can be written as y=1-x b) The point where y=x means y=2x (lets' call it P) is obtained by setting x=2x, solving (x=0) and plugging into either y=x or y=2x and getting y=0. So the point P is (0,0). The point where y=x means y=1-x (lets' call it Q) is obtained by setting x=1-x, which means 2x=1. Dividing by 2, we get x=1/2 and plugging into either y=x or y=1-x and getting y=1/2. So the point Q is (1/2,1/2). The point where y=2x means y=1-x (lets' call it R) is obtained by setting 2x=1-x, so 3x=1, and dividing by 3 we get x=1/3 and plugging into either y=2x or y=1-x and getting y=2/3. So the point R is (1/3,2/3). c) The circumference (which is the same as the arclength) of the triangle is dist(P,Q)+dist(P,R)+dist(Q,R)= sqrt((1/2)^2+(1/2)^2)+ sqrt((1/3)^2+(2/3)^2)+sqrt((1/6)^2+(-1/6)^2)= sqrt(1/2)+ sqrt(5/9)+sqrt(1/18)= sqrt(2)/2+ sqrt(5)/3+sqrt(2)/6. That's the answer to (c). ------------------------------------