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Homework 7 Solutions

# Homework 7 Solutions - APPM 2350 HOMEWORK)7 FALL 2010 Due...

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Unformatted text preview: APPM 2350 HOMEWORK ()7 FALL 2010 Due Friday, October 15, 2010 at 4 pm under your TA’s door. 1. Previously in the semester, we derived equations for the distance between a point and a line, and a point and a plane. Your task is to derive the same results, but using the method of Lagrange multipliers. (a) Use Lagrange multipliers to establish the formula for the distance D from the point (11:0, yo) to the line can + by = d. (b) Use Lagrange multipliers to establish the formula limo + 53/0 + 030 — dl \/ a2 + b2 + c2 for the distance D from the point (:30, yo, 2,0) to the line am + by + oz 2 d. D: 2. Suppose you have three identical boxes, and some total number of particles, say N. Let n1, ng, and n3 denote the number of particles in each box. Of course, we have that 3 Zn,- 2 N. (1) j=1 3 We can then deﬁne the probability of a particle being in the jth box as = N. Then clearly, P,- = 1. (2) 3 =1 .7 Now we deﬁne the entropy, S, of our system of boxes and particles as 3 S=—ZP,-1nP,. (3) j=1 Entropy, roughly speaking, is a measure of the disorder of a system, and it turns out, according to the Second Law of Thermodynamics, that any closed system wants to maximize its entropy. (3.) Using the method of Lagrange multipliers, prove that the entropy S of our system is maximized when P1 = P2 = P3 = Thus, to maximize entropy, how should the particles be distributed across the three boxes? Now, suppose that the three boxes are not exactly alike. Suppose that if a particle is in box 1, then it has energy E1, if it is in box 2, it has energy E2, etc... Without loss of generality, we assume 0 < E1 < E2 < E3. Finally, suppose that we have a ﬁxed total amount of energy we are trying to distribute across the three boxes, say E. Therefore, we now have the constraints N, (4) 'M 3. 11 (5) ‘ M \$9 ‘53 ll be If the total number of particles, N , is very large, it turns out that we can still say the probability 71' . ”’ . —3 Thus if we deﬁne the constant E :2 —, we can rewrite our 7 fb' 'th "hb '13-: . o eingin e] oxrs 5, N N constraints as (6) ‘M £9 H II D3 3 Z 3ij ". (7) (b) Using the method of Lagrange multipliers with more than one constraint, ﬁnd the probabilities P]- such that the entropy S is maximized. Let )r denote your Lagrange multiplier for the con— straint (6) and ,8 denote the Lagrange multiplier for the constraint Your answer for each P]- should be in terms of A, ,8, and E}. (c) Using (6), determine each Pj only in terms of the Lagrange multiplier 3. Assuming B < 0, and given that 0 < E1 < E2 < E3, which energy state is a particle most likely to be in? Note: when a system has a ﬁxed number of particles and a ﬁxed amount of energy, it is called a canon- ical ensemble. What you have now shown is how the Second Law of Thermodynamics determines the allocation of a ﬁxed amount of energy for a canonical ensemble. Appm ZESO CQ‘CuzLLS jg; HOMWOV‘L 7 Nix} 0A We \N 3k #9 vvﬁhlwnu M4 A§s+b~v~oe fmw‘ +u Pong; €X°\\/o) “\$386+ *9 +M LonS+vu,w+ IHAOA’ (X,y) are “A ,‘kz 56+ 04: rah/U6 [Dy OKX-F by C A . ﬁ‘nui oMY Etum+\.0V\S area 1 \/7' BMW) [(X’Xo’ Aty’Yﬁ/K j(x,y): ax+by «A S'mw D(x)y)z L“; a mlwivvxum wkev‘ever D(\<,y) \Aos a MM;W\MVV\) \Ne wa MYK ‘\W;’\ “RN, D(Ki\,)l) wLH'CL‘ {5 V‘AoY-C vadobe ~ (9 , up Kc) ’ 30‘ v01 : 1v? (1(K~Xo))z(\/lx/°)) v X<<M 55 v» r )b CD :5 A ' Z(‘/’y<>) QLXMWO ax+by - aX*b\/ CA (,9 , L“ , B ' me‘ "fM ‘34 1 (’iua4{o¥~5 \A/Q [Sné (X’XD) ’ z ) (j/V‘) ’ 1 f L ’Hnus D(X)\/)1: {><~><o)l'¥(//‘/»)1 3 ﬁ(0‘1*'o) MN,..M.WMW ‘ {A X WQ V‘ “WI guy X qwé rHM/vx f A a \, Ox - (9%“) " ("m ’ W V’ M a; W X :méamym/ww g me @1 y: “1’ :3 X“ o 0\c§+ We «Move «1 a} “11" i) x : W 3 X("*?) : If ' L M“ a “a 7. M» We «we vxdcfﬂs") W a — a I J.” ~ : / X X allHai M“ obkLm l/(X’m , ( \ SUM/t X '5 (A 3 'Z~ “Adam 3 we Com budghkk g”, x W‘ Dug/)1 ﬂ—ﬂ/Wﬂmw‘mw I ‘ oval 5) ka‘j) : (mug) :: Dbcwl): a”? b) we froCbeé a: 39%?ng I ux-Xol )W‘ D : W3 v z) mm): W a :0 1(Z’ZO/i XL C\)<4"’/"'CZ:é FNMA e‘gi 1’“ wt p\"\é (Y’XHC I 7. S0 D1: %(O‘L*I,1¢L) NW «(9* 1260* “(3) :3 :> :3 Tk \ Ll( Avax (LyaeCZo) AM, 1 E} a) We Ax) vaK;\,\/\;%e K S<fwﬂxpl> ( it! ' ' \ « 3.5/7, 63<P“F11P3) : Pv“pz* f3’\ 3'0 VS : ”V(P\‘WP14 P1\$WP1*F”\$I”P3\} : —<)m{)’+)> 1W101+I’)y‘{)3111> \ ,/ 3:0 ;3 p‘;}azk\93::pl:{ :3 P’,3 ' 'é EP\ : {>1 1" f’ \/ M5X§m;z,es S Wrx¥k Cor\$+ragw+ a ‘ k) Now we w\L--S”'-' maxlwﬁge W\5<\« Mspc'ci, )ro “Z (ows’ha’wxi ? ~ .g Qwé (LC/3?}:E’ 3:! J 3:! iff ‘3‘(9HP7,P3\)Z PIAPZ/yfg 1‘ C J r / 31(P,)P1\1€g\)fE,Pl’uzPl4/E3(\g C [ILXW cu} wwme’La/Hom e?ua+(qy\g arc," » «E (9 VS : \.v3\}’ ‘- A V \' : w 3E1 @ ﬂ 32 C \ ~ , I E‘AP2*?2 ~/ ’ ‘ f‘ f? V P C P 1/ 3 C d 1’ b2 2 4, 5 , Max] meé’xwﬁeky ~ 1K3 QHMY) (éé (AH Hwee mwaErs L‘H‘vw Favjf )3 , 3 \\ ...
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