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Mathematic Methods HW Solutions 52
Course: MHF 2312, Fall 2011School: UNF
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10 9.18 9.19 9.20 9.21 52 Chapter r 1 e , r 1 e r , 3 2e , er cos e sin , 3 r 1 , r 3 , 0 2r1 , 6, 2r4 , k2 eikr cos 11.4 Vector 11.5 ds2 = du2 + h2 dv 2 , hu = 1, hv = u(2v v 2 )1/2 , v dA = u(2v v 2 )1/2 du dv, ds = eu du + hv ev dv , eu = i(1 v ) + j(2v v 2 )1/2 , ev = i(2v v 2 )1/2 + j(1 v ) au = = eu au , av = hv ev , av = ev /hv uv 2 V = = Fu v (2 v ) u uv 2 (v 1) V uv + 2uv = Fv + = u1...
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10
9.18
9.19
9.20
9.21
52
Chapter r 1 e , r 1 e r , 3
2e , er cos e sin , 3
r 1 , r 3 , 0
2r1 , 6, 2r4 , k2 eikr cos
11.4 Vector
11.5 ds2 = du2 + h2 dv 2 , hu = 1, hv = u(2v v 2 )1/2 ,
v
dA = u(2v v 2 )1/2 du dv, ds = eu du + hv ev dv ,
eu = i(1 v ) + j(2v v 2 )1/2 , ev = i(2v v 2 )1/2 + j(1 v )
au = = eu au , av = hv ev , av = ev /hv
uv 2
V
=
= Fu
v (2 v )
u
uv 2 (v 1)
V
uv + 2uv
= Fv
+
= u1 [v (2 v )]1/2
1/2
v
[v (2 v )]
[v (2 v )]3/2
11.6 m u
m
11.7
U = eu U /u + ev u1 v (2 v ) U /v
V = u1 (uVu )/u + u1 v (2 v ) Vv /v
U
1
1
U
2
u
+ 2 v (2 v )
U=
v (2 v )
u u
u
u
V
V
11.8 u1 ,
u1 k,
0
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UNF - MHF - 2312
Chapter 113.23.53.83.113.143/232/35(5/3)1(4/3)3.33.63.93.123.159/1072(5/4)(2/3)/3(2/3)/417.12 B (5/2, 1/2) = 3/1617.33 B (1/3, 1/2)7.5 B (3, 3) = 1/3017.72 B (1/4, 1/2)7.10 4 B (1/3, 4/3)37.12 (8/3)B (5/3, 1/3) = 32 2 3/27
UNF - MHF - 2312
Chapter 115412.15 2 E 23 E12.14 12E 35 15.86=12.16 2 2 E 1/ 2 3.820=a12.23 T = 8 5g K 1/ 5 ; for small vibrations, T 2 2a=3g13.713.913.1113.1313.1513.1713.1913.2113.2213.24(4) = 3!3/22.4222E135 F arc sin 4 , 4/5 = 0.1834 sn u d
UNF - MHF - 2312
Chapter 121.11.31.51.71.9yyyyy2.1= a1 xex= a1 x= a0 cosh x + a1 sinh x= Ax + Bx3= a0 (1 x2 ) + a1 x3= a0 e x= a0 cos 2x + a1 sin 2x= (A + Bx)ex= a0 (1 + x) + a2 x22= (A + Bx)ex1.21.41.61.81.10yyyyySee Problem 5.32.4Q0 =
UNF - MHF - 2312
Chapter 125610.5 sin (35 cos3 15 cos )/210.4 sin 10.6 15 sin2 cos 11.111.311.511.611.711.811.911.10y = b0 cos x/x211.2 y = Ax3 + Bx332y = Ax + Bx11.4 y = Ax2 + Bx31/21/2y = A cos(2x ) + B sin(2x )y = Aex + Bx2/3 [1 3x/5 + (3x)2 /(5 8
UNF - MHF - 2312
Chapter 1257= Ax + B x sinh1 x x2 + 1= A(1 + x) + Bxe1/x22= A(1 x ) + B (1 + x )exx= Ax Be= A(x 1) + B [(x 1) ln x 4]= A x + B [ x ln x + x]xx= A 1x + B [ 1x ln x + 1+x ]2= A(x2 + 2x) + B [(x2 + 2x) ln x + 1 + 5x x3 /6 + x4 /72 + ]= Ax2 +
UNF - MHF - 2312
Chapter 132.1T=2.2T=2.32.42.7T=nx(1)n+1 ny/10esinn101200 41odd nn22even n22n=2+4knx 1 ny/20sinen20neny sin nx1x 1 3y/30120 y/303x1 5y/305xesinesin+esin2309302530n4sinh n(1 y ) sin nxT= 2 (n2 1) sinh
UNF - MHF - 2312
Chapter 13592.14 For f (x) = x 5: T = 4021odd nnx ny/101cose2n10For f (x) = x: Add 5 to the answer just given.2.15 For f (x) = 100, T = 100 10y/3401For f (x) = x, T = (30 y) 263.23.33.43.53.63.7u=4001odd nu = 100 u=40 1od
UNF - MHF - 2312
Chapter 134.64.74.8y=y=y=604hl2 v9hl3 v4l2 v4.91odd n1nnwnxnvt1sinsinsinsin2n2lllnnxnvt1sinsinsinn33ll1sin(n/2)xvt2x2vtnxnvtsinsin+sinsinsinsin3ll16lln(n2 4)lln=31. n = 1, = v/(2l)2. n
UNF - MHF - 2312
Chapter 136116, n odd, the six solutions on (0, ) aren (4 n2 )1. T = bn eny sin nxbnsinh n(H y ) sin nx2. T =sinh nH23. u =bn e(n) t sin nxh 2 n24. =bn sin nx eiEn t/h , En =2m5. y = bn sin nx cos nvtbn6. y =sin nx sin nvtnv12(1)n+1
UNF - MHF - 2312
Chapter 135.14 u =6250 ln r200+ln 2o dd nln r5.15 u = 50 1 ln 2r n r nsin nn(2n 2n )200o dd nn (2n1 2 n )r2nr2nsin n6.2The rst six frequencies are 10 , 11 = 1.59310 , 12 = 2.1351020 = 2.29510 , 13 = 2.65210 , 21 = 2.91710.6.4
UNF - MHF - 2312
Chapter 136322227.21 n (x) = e (x +y +z )/2 Hnx (x)Hny (y )Hnz (z ), = m/ ,h1113En = (nx + 2 + ny + 2 + nz + 2 ) = (n + 2 ) .hh(n + 2)(n + 1), n = 0 to .Degree of degeneracy of En is C (n + 2, n) =2M e4+12r7.22 (r, , ) = R(r)Ylm (, )
UNF - MHF - 2312
Chapter 1310.12 u =40064o dd n1rna2nsin 2n10.14 Same as 9.1210.15 u =1rn 104006nsin 6n =2(10r)6 sin 6200arc tan1012 r12o dd n10.16 v 5/(2 )10.17 mn , n = 0; the lowest frequencies are:11 = 1.5910 , 12 = 2.1410 , 13 = 2.6510, 21 = 2.
UNF - MHF - 2312
Chapter 141.11.31.51.71.81.91.101.111.121.131.151.171.181.191.201.21u = x3 3xy2 , v = 3x2 y y 31.2 u = x, v = yu = x, v = y1.4 u = (x2 + y 2 )1/2 , v = 0u = x, v = 01.6 u = ex cos y , v = ex sin yu = cos y cosh x, v = sin y sinh xu
UNF - MHF - 2312
Chapter 143.13.53.93.123.173.203.236612+i11(a) 5 (1 + 2i)3(a) 0(a) 072i3.2 (2 + i)/33.6 1, 13.10 (2i 1)e2i(b) 1 (8i + 13)3(b) i(b) 17i/43.24 17i/963.3 03.7 (1 i)/83.11 2i3.43.8i/2i/23.18 i 63/3.22 i 3/1083.16 03.19 16
UNF - MHF - 2312
Chapter 14676.286.306.326.346.146.186.226.266.276.316.3511R(3i) = 16 + 24 i6.29R(0) = 1/6!6.31R(2i) = 3ie2 /326.33R(0) = 7, R(1/2) = 76.35 i/46.15 06.1606.19 06.2006.23 sinh 2 6.2433 2 i(1 + cosh 3 + i 3 sinh 3 )316.29
UNF - MHF - 2312
Chapter 146810.4 T = 200 1 arc tan(y/x)10.5 V = 200 1 arc tan(y/x)2210.6 T = 100y/(x + y ); isothermals y/(x2 + y 2 ) = const.;ow lines x/(x2 + y 2 ) = const.10.7 Streamlines xy = const.; = (x2 y 2 )V0 , = 2xyV0 , V = (2ix 2jy )V010.9 Streamlines
UNF - MHF - 2312
Chapter 151.11.31.51.71.91/10, 1/91/3, 5/91/4, 3/4, 1/3, 1/29/26, 1/2, 1/133/10, 1/31.21.41.61.81.103/8, 1/8, 1/41/2, 1/52, 2/13, 7/1327/52, 16/52, 15/529/100, 1/10, 3/100, 1/103/82.122.142.152.17(a) 3/4(b) 1/5(c) 2/3(d) 3/4(e
UNF - MHF - 2312
Chapter 15704.94.124.174.2125/102, 25/77, 49/101, 12/254.11 0.097, 0.37, 0.67; 1354.14 n!/nnMB: 16, FD: 6, BE: 104.18 MB: 125, FD: 10, BE: 35C (n + 2, n)4.22 0.1354.23 0.305.1 = 0, = 35.2 = 7, = 35/65.3 = 2, = 25.4 = 1, = 21/25.5 = 1, =
UNF - MHF - 2312
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Q,-p r . E ,ot.r'' T r n xa ( rt.t : - h n t"!cfw_:"ili. tIl= )rE,"riti",1,of,._ ' / r cfw_ s r - 4 ' t ''(l( -t I- oo,^ l,= H r.\.odEth-t)*xrX;Xt.,i)G-XG) vS/rllci-L)Z' ;tt achooodnleo,.o.^: kl-(ttt,i--.Y:' . oF- 7t-.i-^ .
York University - MATH - 2131
York University - MATH - 2131
York University - MATH - 2131
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York University - MATH - 2131
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York University - MATH - 2131
- */7).-!r!rg-'u'l*,- - lr'zJ=;'.;E1t+ri,C-, -, l)-!)t tL>\/ +Jy,l, - 6 < - s tz Io<-tet'^lt^ta t y r ,( y l . ) = t r . , z* e , E . o + v < r^l4.aza+riL ,lz !+Vt-.)At-rvj=.'"=? -tt_cfw_L dt\r,*.-4,-.If a,t.mar-r.t'L
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| -.i ncfw_it, 1.6'I_ -Q_+ 1 o P "\t-o o3IIt,t r, t y L tc), ' a 4 e r y t a v . ! . . t - a a . f L I,v3f1a r'rILUs L -7 z tJt ( 9 t t =z) - 3i-"-i=) ? .rIiQ tt - 4.9o1:t.eIiy t11rt/lr) dY= 3'tcfw_ \La tq:Ia\ 2,xr al jI
York University - MATH - 2131
EF:' t"I'1r-*A"nn?-Y-t\) l -u't+ q lrr)l,ILoV I X,YIt
York University - MATH - 2131
3/IIt=+laa'L'W.6(4tll,z t(Vrl r .(N F tt ) - v1,. 1_=- I ^l4r) + /&re + /tt
York University - MATH - 2131
+tl1-1"PIX=o) =l=oA P/r,zllo- olt z + orP / x ^ 4 n ) - t - . P * . - t ) 1 , . ." t r , ^It1.y,S-J,ILllel,l!+/v-\\,/D' vsuJltos-8,.-y11-t Y ,\ 1 - f e qch .a "e t .1.d,-)-!l,X^ + -x- ) - f | / sx)^\.%x\e.uene xL P (f ^ - - tL,p
York University - MATH - 2131
York University - MATH - 2131
>tt P(11-Xl>tl) = P(b\-^,.t. ,.n)&hi- I- tI >ntI-XJh ., t
York University - MATH - 2131
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York University - MATH - 2131
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York University - ECON - 2350
York University - ECON - 2350
York University - ECON - 2350
Price Discrimination Firm charges dierent prices to dierent consumers. Prices are tailored to extract greater surplus from consumers with greaterwillingness to pay and do not reect (only) cost dierences.Prerequisites for Price Discrimination Market
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The Assumptions of Perfect Competition1. Large Numbers: No individual consumer buys and no individual rm produces asignicant proportion of the total output.(No economies of scale at the rm level; minimum ecient scale is a small portion ofthe total amo
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York University - ECON2350 - V. BardisPractice Set 11. Find the aggregate demand function, X (p), if the demand of person 1 is given by x1 = 50 pand the demand of person 2 is given by x2 = 25 (1/2)p. Is any of the three demands moreelastic than the ot
York University - ECON - 2350
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York University - ECON - 2350
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York University - ECON - 2350
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York University - ECON - 2350
York University - ECON2350 - V. BardisPractice Set 31. Using the concepts of consumer and producer surplus argue that the equilibrium in a perfectlycompetitive market is ecient.2. Provide a set of assumptions that describes a perfectly competitive mar