Math_138__Assignment_3_Solutions

Math_138__Assignment_3_Solutions - MATH 138 SOLUTIONS T0...

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Unformatted text preview: MATH 138 SOLUTIONS T0 ASSIGNMENT 3 PAGE 1 Pﬁﬁ—L Qdfmfigg OLD Typet m "[39:12, cund dJIanm'ccanu-maf- m Magma e—é mch_hnpnom win.‘ @104. b ‘ 1 3 S f dJ: IS Tﬂpti , amd. Lb Hm Ana ilin. IE Cit - _ lo I l-l-t‘ b |+£a an m at :- 4‘3’“ - C] I ' 5 1+4? it log-w - 5" 2t t , beﬁumk. I ‘ J tt “1-: {+tz’ = A; b . W" *2- E’mlueﬂ = g—ﬂmwm Aim. b—aoo s‘r‘ﬂﬁ- Mll’rb'" —9 +09 ab baa, OJ} maes‘ an . 90‘ t = ell—cit- e') JL__J——-9+tz |9 Type i. 33 W1 £9+£1°L 25\$ 9H9. ¢‘ CL-b’: 2 l _ ‘ k ..——-—- . d g 4‘. .L. ._L._-— ‘M in 9+9 inan 9w;z t .2 5‘31 9! 1+ git mmm u ‘ 11- I 2 alnu onc‘ZuhC')-—vg-°D“]L"'°°' P554“ ' . . . . 1 JxQM-szm. “3 Tape. 7.: 5mm. :6 meh—émd. ‘ .-. benxabg z: QL‘m‘r 813nm” ‘ afar ushvadU-gxd”. o =Pda-t=§;d»<, v:- 3&3, _ {\m { XI l I‘ L 2 ’ —- em 1 — S 2‘. .L. a-ym [2 'x a a. 2 x 0le _ Km {0 Q7- 1?- ‘ CL—rO" " E’Q‘V‘O- "' #:50- = _le|m afwa _ __|_ en“ 0: {Eek}? swam) z aﬁo tin-40* 4 . 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It, 5 J—abx awe» , no mo °° ‘ on oxm l 2' w {{J‘l'x‘zl W a) Fan Smgrdt , we. mot: ezt+tl>z e“ W t e “2‘ n e +t‘ A at \F at 4 at = e—‘t o‘e:5+’c"e§E_+t1 “ 31.1: 55‘ tho- ID \a c: S‘ﬁ'a Ee-‘t51t:: Se-tdfc :.- Ee- Jab": Q‘L'm "e- + e =1 O 5-” a 5-H» éiaLr—g Le. 00.1 ‘9 1: ED ' ’ 8 ABC Paw-MQLB, 5‘0 “Q30 j it {:1 ch: aonumano. e + D MATH 138 SOLUTIONS TO ASSIGNMENT 3 i ‘E‘ Q, U “5 £2 0’ .L a 3 (3“ j I 5 f5. “fj—N \:e e. clt- bi,“ a m c __ . 5 =3'2.\ ) Tim Na: mm = M ,T‘CWWZS ‘2 "W‘H..V.0rcms-(WD‘- 3 D e :1 9 MATH 133 SOLUTIONS T0 ASSIGNMENT 3‘ 'PAGE 5 \ 455d) ‘ PS 4. musk f0 &Cn¢i 11. voﬁw, summattd. Ema Momma M "Pu n3 Gum's “u. Cunu-L 5:34‘92 J 85-. -aD-C'N: <63. We. choose CKKQLLV‘CLIR’COJ skews as Thx'ckmm Ax} QwigH' I5, amé. ﬁnds» x) AN 2 a" 1:3 Ax Swmm'mcz M ME 361 O\$x<o°) W12:ng 'Vvu. lbw-LT a.» Ax.,o; c0 . 1 V: ZQTTXLAAX = :urjtxe‘x/zcbx , 0m immopcn. 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MATH 138 SOLUTIONS TO ASSIGNMENT 3 PAGE 6 CnALLEQGE: L> # Assignment 3 - Problem 8 - A String of Beads E> restart; E> # Plot the graph of y=exp(-0.1*x)*sin(x) on [0,6Pi] > plottexp(-O.l*x)*sin(x),x=0..6*Pi); > # Evaluate the volume of the solid of revolution created by revolving the graph about the x—exis. > int(Pi*exp(-O.2*x)*(sin(x))“2,x=0..6*Pi); 7596945615 > # Define th) as the volume obtained by revolving y=exp(-k*x)*sin(x) ahot the x-axis for {(n—l)Pi,nPi]. PHI > V := n —> int(Pi*exp(—2*k*x}*(sin(x))‘2,x=(n—1)*Pi..n*Pi); V2: n —) 7: e (‘2 kx) sirloc)2 dx menu Mm: 138 SOLUTIONS TO ASSIGNMENT 3 PAGE 3]- l [> # Find V(n) and V(n+1); some simplification is necessary. > V(n); _(-2k2-1+2k2cos(nrr)2-2ksin(nn:)cos(nn'))z(e(2k")-1) calm”) 41:09“) > subs( {cos (n*Pi))"2=1,sin (n*Pi)=O,V(n)); Me(2kx)_l)e(.2knu) 4k(k2+1) > V(n+1); 7r(—2k2-1+21:2cos(mr)2-2ksin(n:r)cos(n7r))(e(2”)-1)“JUN—2”) 4k(k2+l) > subs ( (cos (n*Pi) ) *2=1 ,sin (n*Pi)_=0 ,V(n+1)); 3(e(2kn’)_1)e(-2knx—2kn) 4k(k2+1) [> # Find. the ratio of the volumes of two successive beads, and simplify. T ratio = V(n+1)/V(n); [ > [> # That is, the ratio of successive volumes is constant, which is e(—2nwk—2k1r) rati0= e(—2n:rk) # Note that, after cancelling the two terms exp(-2nkPi) , ratio=exp(-2kPi) which is independent of n. unexpected. > # Thus this ration equals 1/2 if k=ln(2)/(2Pi) . [> # Finally, we find the total volume of an infinite string of such beads. 13 4k(k2+1) [ sum (th) ,n=l . . infinity) ; [> [> ...
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This note was uploaded on 05/23/2010 for the course MATH 138 taught by Professor Anoymous during the Fall '07 term at Waterloo.

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Math_138__Assignment_3_Solutions - MATH 138 SOLUTIONS T0...

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