Assignment 3 Solutions

Assignment 3 Solutions - Soweto» Assignment #3 DUE DATE:...

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Unformatted text preview: Soweto» Assignment #3 DUE DATE: This assignment is to be submitted entirely on paper on Friday, February 13th by 9pm in your TA’s drop box. Hand in one copy per pair. Learning Dbiectit'es: This assignment eoyers applications of integration that we have not studied in class. The concept of' 1Iii-"erk is a concept that you are familiar with from physics. and 1his assign menl will allow you to es plore what happens when a force is not constant. and integration mu st be done to calculate work. The assignment will require you to read the test on your own. and apply your knowledge ot‘integration and its appliemions in a new ecu-nest! ASSIGKMEKT 9 marks total : Read sections {1.2 and 5.4 of the test {pp 422—431] and 433—44 1} and answer the following questions [MUTE lQuestions must be completed by hand]: Section +3.2 — {3 marks] IIC‘tJnsider the curve Its} — 1'25 — 1': between I = 2 and .t' = 4. Ili'r'hat is the 1.rolume obtained by resolvingI the curve around the iii-axis? 2. Section +3.2 — {3 marks] IIC‘tJnsider the area enclosed between the curves fir} — I: — l and pix; : s r: w‘nm is me volume obtained by res-e1 sin 3. the area between these two curves arcamd the line _i.- : W 3. Section as — {5 marks] A bucket lhs1 WE‘lgltS Slb and a rope ol‘negiigible weight are used to draw water from a well that is Tfli't deep. The bucket starts with Sfllb of water and is pulled up at a rate of was, but water leaks out of a hole in the bucket at a rate ot‘tlebts. Find the work clone in pulling the bucket le1he top of 1he well a a.) C9 {bah $19? 3 Xe [is] )6 L61: kl . \lGQJ" g Atxwtx waste Abe) is Am. 0?— msk we» Moos izs-xl 9. *1 AM: Ti (F's—e 3": “W (2: say “ ‘t So dat- & T(l€—><1\alx :— «mm—sex] s 0%“. 2 2; © £60: X'L_H > {goatgrxl‘ commas \FJ-vengecr WAEU &(x\s-5(x-3 J 1.5. weep XIH ="\-><1' <=> le“—%)=O (=9 xvi-.1, we MATE Tax; Ang 15an 'H-YE 9.64st mean tgc—l Line. 7. 006' = 3 A00 09.76 , “MERE AOC) is AMA ca: “WW2.” (Amouus) swung ma. mius vim. Mblus “l (b‘EL—Ou) A; 8; Au) .-.- “W (km-\- fg" - {QM H11 ) = 2&1“ 04— 23') So doe-=8 QH‘W (H-x”)&x.= QHTY CKX'$X}S&1= lfie‘fi’ '2 'Z @ Lei: M) = umaém’ as: (soch & UJ-KTEIL ) x3 —_ msoLkgMzuT (GELTlZfiL usR-ma) o - _ MDKKra S “30952? _ "\F 0,56 eats up Time To? 09 we urau, As THE chem CQPX :20 m p951: uoOa) (macaw AS A FUNK—Lox) cc ‘3) .kpou) g'échiL: ~05 & («man #50) w=§o+§¢§5 L—> so wee): —o.§Jc,+C, W m(o\—.—. —o.s.o+C=SS =9 (1:51; . so mtg): -os-b +51; . ALsca mm) %= \o & wan) 'E=QJ 13—70 (—39 VAC-m: \O‘E—k’é M H pm) (“53:40: \0-0 +6 =9 6::10 so «L153: Wk -70 302. 957-7‘5'3457 THEM um ‘20: —o.§ - ere‘sg = “fit-Ha $73933 £03 =...\_ 103k 2. 0 wow = 3 “2‘8? +--‘§})G°Ag= ("3631" 13:3?) = 15:5"7°Z+ "2370 137275 ’70 ...
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Assignment 3 Solutions - Soweto» Assignment #3 DUE DATE:...

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