Homework_1_Spring_2010_SOLUTIONS

Homework_1_Spring_2010_SOLUTIONS - 5W" 4913— 1 Egg...

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Unformatted text preview: 5W" 4913— 1 Egg MAE 105: Introduction to Mathematical Physics Homework 1 Posted: April 2, 2010, DUE: April 9, 2010 (Friday, 1 pm), 1. In class, we derived the heat equation for a one-dimensional rod of constant cross-sectional area, A. If the rod has a variable cross-Sectional area, say A(x), derive the modified form of the heat equation. Assume no internal heat sources. Hint: Use similar rationale as the class derivation. dx 2. A brief review of basic math. and Math 20D (A) Write down the values of (i) sin 0°, (ii) cos 180°, (iii) tan 30°, (iv) cos 90°, (v) sinh 1, (vi) cosh 1 m V1... 1 i 1451. I l The area A(x + Ax) can be approximated as A(x) + mama) etc. E (B) Solve the following integrals, by using the method of “integration by parts”: (5011:2 c053: 011', (b) Jaggsin axeos bx dz, for (i) a = b, and (ii) a 75 b, l O (W (c)f sin ax sin bx (12:, for (i) a = b, and (ii) 61 ¢ :5 (C) Solve the following ordinary differential equations (ODEs) for y(x): d1... ¢¥_i d_3"_.2 £_ — day: d3y=- (a)d,—1’Cb);_y’(°)as_{was y—x,(e)dma 932,00 m, 9y (D GD (2;) (a; @ é) (g):—:=2,x:>o,y(0)=3 6331* C29 ( 2.0 [MW ( (h) M, =3x2, 2 :> x > 0,3703) = swig) = 2 C39 HGMENOR :4 1L - SoLm'IoNS fix a,“ N0 mug IMF Emmi—E; 510%):0 M . (L) M (a We} Jf‘t'hna.‘ My )OvaoSs a. SLCqug Muss Ax = _9__ amt) ACX)A'£l “()0qu M 2y.» ,m Mex-Ag qu‘caung‘ waxy» b4» she/0L 5} bk om (Hakim! Wu) 7" $030941) -~ CH1 TA1,t‘) M1161) Q, {we} Aw %[fi(‘llb) k(1)£ht]: (#61, b) A61) .. (bf {1+A1!t) ’(bl-l-A‘A) giant) ACQ— f¢Cma +Ax CPCX, t3 EAR) +A1A(m)} mm dab-me) 33¢(n.e) Mm (fit) 32. Hm 10"2 Many/Afi) ~i-Ax ¢'(’1,t) M1) +¢(1)L.)A1A'(u)+ @331 Nagr‘hrwj M AM ’3 6W7) 2— —- M [5&0 Au) +£{>(1,b) Aw] “W“ at flgm class aC-xrt) I CCK) 19(1)TC;E)J fl [4?“ t): It iglf). ASSWU (3,1? am what (wJLfWQK)_ y... .. K W a‘Téut) + 9M) 015(3) 9x” 37. d1, (b) m. We. MW‘ cm 55%!” 0:5 I” J“ S»; (Ad 3): EMA (253+ CosASMB a 5‘ SmCA ‘6) .: SLA- C‘Sfit— CaSA-Sung ') swCA-esfisMA—B): 25MAro56 "I j LsMMbh M + raw-w we] 0 0* (on-b) , “a“ “I - m" mt COSLK (10L "— .1, £0 1 Sw- J‘ES (M b) = 0M} Q g 20:45) ,5) wot-L0" 0 = b w é 68—5" tfi'e't (Arab l 1‘72.th lax (1M TIL gm 6-1 £3be a“ = Z 0 ° “/2. .‘:: I _ [@ 1&1. a 3:( 2;) S J . [gaxwww1ja5gfiomu] I -_—_. __L _§¢M(a-—bi)7t -— I $1n(0\‘fb)3(. (a) 0%”5} [a mm our—r] Assam a {NJ 50mm J} 6‘1““ 2140'!“ firm—)mmhmy iLA w "19‘“ 2W3 =6'3 =) (my—‘1)a;0 =9m-113 /> 1 TL"). W. (SMWL (3 “4M 3(1): A1631 +Ale—3x M (am also Le owl-Rn an ”9 [1) :. A: CQSLLCBX) + A: Mb") 00 “e +9 I..': @§L(3I) 51%;“: 6(1); A. e?“ + A2, 53"" mmalh 1e. (,3me I a (1),»: A" Co; 31+ A; Sm31 ’ 4‘31 CLO e LSxfe :- 60} 3x fid— ‘ 45¢ eL51_e 1&3“. ______________.__... (1 2 C ) (E2! ‘) 2(6)‘EC 3: )-.- 19d” afio ‘3 'Ui (Jump L z 311' (a) A 1, mw ...
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