ece302hw2 - may ’ Catareetiia Stare...

Info iconThis preview shows pages 1–10. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 6
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 8
Background image of page 9

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 10
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: may ’ Catareetiia Stare ‘eetsreeemeo‘sweeem, egmfimf Elam’lfil Er Camellia? Engineering Department ergusen ‘ ' ’ I; I v ESE 392-. ' . - Hessewoea a It; 7 i a. section as assessenmeeeabye 522 s 2', a gas so“, and _ mange-£79m; (a) the surface area ofthe section, (is) manslasedvolnme.‘ I ,_ ‘ “ Also Search the causes of the section. Evaluate the integral Elfin, Where l? = r one :35 and Fla a sphere ofunit'radius centered at the origin. . ~ 1 ' ® (:1) Express the field I) = (x2 + y2)—1(;cax + yay) in cylindrical components and cylindrical variables; (21) Evaluate D at the point Where p = 2, 45 = 0.27:, . and z = 5, expressing the result in cylindrical and rectangular components. LE x Express in cylindrical components:' (a) the vector from C6, 2, —7) to D(-—l, v U —4, 2); (b) a unit vector at D directed toward C; (c) a unit'vectorat D - directed toward the origin. - ‘ ® The surfaces ,0 = 3, p = 5, 45 =100°,¢ = 130°, 2: = 3, and'z = 4.5 define a ’ closed surface. (a) Find the enclosed volume; (29) Find the-total area of the , enclosing surface; (c) Find the total length of the twelve edges of the surfaces; (d) Find the length of the longest straight line that lies entirely Within the volume. . 1 ' @ Express the vector field, G : 8 sin¢ in (a) rectangular components; (2)) cylindrical components. ' ' Express the unit vector a; in spherical components at the point: (a) r := 2, 6 =‘lracl1 ¢,=0.8rad; (17):: = 3, y = 2,2: —-1;(c),o :25, 45 =O.7racl., r \z = 1.5. . , ' . ' (89 Convert the vector :1, + a), —— Via, at the point (1,1, V?) to one in spherical coordinates. , ~ Determine if the vector (ap— V? :19, + 332) at the point (3, 7r/3, 5) in cylin- drical coordinates is equal to the vector (Say— V? aa -" aqb) at the point . (1, 77/3, 7r/6) inspherical coordinates. ark/ea: ‘ 8‘ A A I' A a d6 dc? A 5) ir— =© "“ S smgq) n = s and E— = -p 66 3‘? q) 9’ A as r 39 = 4 *- = 005% as 3‘? (33:0 93:: —sin9f#00566 as 3‘? Givenr=ag+by+sawheremcare constants. Is A a constant vector? Find the cylin- drical and spherical components of A, expressing them in terms of p, (p, z and r, 6, q) resPectively. Points .41 and b are ‘both located on the surface of a sphere. The s‘phe‘re has a radius of 4.5 m. At fioint a, 6 = WW, 4) = 1.517. At point b, B = 17/2, 4) = 17/8. . v - ex 1 (2) Calculate the angle (in rachans) between a straight line extending from the Center of “V‘s” the sphere to point a and another straight line extending from the center of the sphere to 0‘ 3 L 7% Wu point b. (b) Calculate the shortest distance, as measured on’ the surface of the sphere, between points P 1 8 Z W A _ a and b. ‘ ‘ tmswm~ r; l 0x.) : 1i ' 2' _ 3 r" A @ (a) (1/p)a,, (b) 0.5%, or 0.412% + 0.2925, @1 (a) —6,66ap — 2.77% + 932 (a) 0.593r +. 0.3339 — 0.72% 1 (b) 4.59.29 + 0.21 _'0.7ga (b1 0.30:1r — ohm — 0.55% . (C) -0.90ap —— 0.44:: z (c) 0.66ar + 0.3939 -- 0.642qu1 {5] (a) 6.23 (1020.7 (6) 22.4 (d) 3.21 ._,. XE ‘3? .N A. 6606*: L z 2158‘ lx+ '2— waz 1* +v +2. *3 k *t L) , 3 {w _ . . > G \T-gsfzfil P r 2 @ VMWS W Jaw (3 Gt) (3 :(QCOS<% 4+— A—Q-msm4: + + fl : (5x§u~9061<{5 «((BS'MQ Squs + 00949); ._A + (QCe’SQC/m+ +£Case§a~cde " CSM@)Q + (*fisww‘f: + _ A A 6” A d €3__A 591:6 —-r—=31n6q3 an dq) p 33 3(1) A A 86 fl = _f ~— = 0036(3) 66 392 fizg €51: —sm6f—c0565 as 6c? V Given A = a5: + by + c2 where a, b, c are constants. Is A a constant vector? Find the cylin- dricalvand spherical components of A, expressing them tenf’ns of p, (p, z and r, 6', (p re3pecfive1y. ® deihfitslla and b are 'both Ioeated on the surface of a sphere. The sphere has a radius of 4.5 m. At 150th a, 6 = ear/7,1!) =’ 1.5% At point b, 6 = *n‘l2, d) = 17/8. _ v r (2) Calculate the angle (in radians) between a straight h'he extending from the center of ‘3‘];— the sphere to point a and another straight line extending from the Center of the sphere to 0‘ 3 '- 7” we: point 17. f (1)) Calculate the shortest distance, as measured 011' the surface of the sphere, between points D =- 8 7 W a and b. ' N» o» - ,2 M)S:1—3tT-V;Mz E) v: %K 3 @ H A G) (a) (I/p)ap (b) 05:19, or 0.412;: + 0,29%, 61. (a) fififiaé 523% £935 a L ' (a) 0.59:1r 420.3829 — gig? -. (b) —0;59ap, + 0.21 v _ 0.78 ' (1710.802r — 0:32:49 -— . 21¢ . (a) —0.9«0ap — 0.44:: 3" (c) 0.66ar + 0.39219 — 0.64% , Av '35 I 'xe _A 3% s A 6"»)G}= 1/2V2,;‘i‘—:X+L Lbrz "‘i’v'rw‘. L ' A, ' . cs- _ A E) Q: , i. AF—‘E ,. A s 2 e a h f +12— L. @Vedw; WM H a) fi‘:(&cesc% J.— LCnagm++boe¢+><§> “PC-"ti. _ _. I E) 3:1:‘(agwecetc454’hs‘mesiwcb'v003/19)? . , _ ._A + (fiCa’Séceg+ +55956§M+“ CSMQJQ ‘ .1. («a5w454f game); nun—a.— ‘ _ O A section of a sphere is giesczibedtby 0 g R _<_ 2, 0 g 9 g 90", and 30° 5 q) s 90°. Find: j ' (a) the surface atea of the spherical section, z ,ch _sez Him we 7.. “khan (b) the enclosed volume. ' ' Also sketch the outline of the section. 11/2 rc/Z ' S = R2 ' e _ (pm-:16 e=o gm d9 dwhz _ 7: TC 4 . :4 [—COSQIS/z].=4xg=§ (1112)1 ‘, 2 “It/2 n/Z » 1 V = i stin samequ i 12:0 time/6 03:0 R3 2 7: 7; lat/'7 8 T: 8'31: ‘ ‘3— 0 (a‘g) [-cosfilo 1: g5; = ~9— (m3). The Spherical co ordinate veri-ablee'ere inns'feonvenient for this problem. To cover all points ' Within the volume, the rangeof the-coorihzate: must be 0 < f < 1,0 < 9 < 77', O < q’; < 211-. ' A150, @171 .= Fsinfl circledq‘: at themiume. Subsfiiufinginto the integal, we have Zn- r.- 1 . A f?_cie~=f f geese-Sinaidr'dadqfi. ' v 0 0 0- - This integral is not as easy in; evaluate as it mzyfh-st appear, Because of the prescnce of the unit Vector fir, which varies the position van'ables 9 and q': . - ~ WE repreSent :3, in Cartesian components as hr = snacdse a: ,+ sinfisinééy + «20535:. Subsfimfing, we obtain . 2n- n' I... ' - _ _217 1r 1 " ~ . [Pdv=§:f ffimzacosze'drded¢+a,f Psin29sin95cos¢drd6d¢' IV ~ 0 0 .0 - 0 o D ‘ ‘ ' ' - l . . I“ 21:- ‘rr 1 D I +azf [ [PsinacosecOqudrdequ a 0 0 v.- . ' T333 second and third integrals on the right-hand side of this expression are zero, since cTrain 45 cos 45005 = O and If?” c0545 6195 = O, respecfively, leaving ‘ I .. 211' 1r 1 “ 241- 1r " ' ‘ [Pd0=§=f frisinzacosz¢drdad¢aefiif f sinlfldedqb- C552? - V 0 "o 0 _ 4 0 0 _’ . "mi: 1w.' ' ~1Tzi. _ _EL. 3. a- 3. : a) Exprééé“£hé'fie1d D 2&2 '¥§;’~”)‘3'1"(23‘§;+gag)"- iii"'5y11‘ndn"" "éél‘fiéfiifidfléfltfi:Hfié'fififiéfiféfiffiiiablesz“ ‘ I Have a; = pcos qS, y : psin <15, and 1:2 + y2 = p2. Therefore ' = % (cos gbaz + sin 45.21,) Then ' V ’ - . I 1‘ _ , 1 DP = D rag: [cos flax - 21,) + sin (May - ap)]‘ = 3 [cos2 915 + sin2 ([51:13; and V , 1 . Dqg = D - a¢ = E [c205145(’am - agg) +sin¢(ay - a¢,)] = [cos qb(— sinqb) + sin‘qbcos q5] = O Theréfore I . 1 . p b) Evaluate D at the pbinj: Where p = 2, q5 = 0.27, and z = 5, expressing the result cylindrical and I, cartesian coordinates: At the given point, and in cylindrical coordinates, D. = 0.5ap. To express this in cartesian,'we use ' r ' / D = 0.5(ap - a¢)a¢ + 0.5(ap . away ;= 0.5 cos 36°am + 0.5 sin36°ay ‘= 0.4121; + 0.29% Express in cylindrical components: _ ' . ‘Jlfector T 2' C(3, 2, —7) —+ C(p = 3.61, ,5 = 337°, 2 = —7) and j D(—1, ‘—4, 2) —+ 130; = 4.12, 3 = -—104.0°,z = 2). , _ Now Rap (—4, ~43, 9) and RP = ROD - 31,, —4 cos(33.7) — 6sin(33.7) = —-6.66. Then R¢,= ROD - a¢ =4sin(33.7) — 6cos(33.7) = ~2.77. So ROD = —6.66ap — 2.7730,, + 9&3 V a. unit vector at D directed toward 0: RCD = (4,6, —9) and RP 2 BBC - ap = 4cos(—104.0) + 631n(—104.0) = —6.79. Then Egg 2 R130 - .395 = 4[— sin(—104.0)] + 5COS(—104.0) = 2.43. SO RDC = —6.79ap + 2.43a¢ -- 9313 Thus ape = —-0.59ap + 0.21a¢ — 0.78511z a unit vector at D directed toward the origin: Start with rD = (-1, —4,2), and so the vector toward the origin will be —rD = (1,4, ——2). Thus in cartesian the unit vector is a = (0.22, 0.87, ——0.44). Convert to cylindrical: , - up = (0.22, 0.87, —0.44) -a,, = 0.22.cos(—104.0) + 0.87sin(—-104.0) —0.90, and 0,, = (0.22,0.87,—0.44) -a¢ = 0.22[— sin(—104.0)] + 0.87cos(-104.0) = 0, so that finally, a = —O.90ap —- 0.44az. Tr+~>» 05 = 130°, 2 =3, and 2 =5 4.5 define a closed surface. @ The surfaces ,0 = 3, p = 5, 05 = 100°, 3) Find the enclosed volume: - 4.5 130° 5 Vol=/ / / pdpd¢dz=628 3 100° 3 NOTE: The limits on the 05 integration must be converted to radians (as was done here, but not 1 ‘ shown). 13) Find the total area of the enclosing surface: ' 130° 5 ~45 130° Area=2/ / pdpdcé +/ / 1 3 3 100° 00° . 4.5 4.5 5 + / Edodz —i- 2/ / dpdz=20.7 3 3 3 _ 3 do dz 130° c) Find the total length of the twelve edges of the surfaces: . x3 + 3600 x27rx5] —2_2_._§ Length=4>§l.5 +4.22 + 2x[ d) Find the length of the longest straight line that lies entirely within the volume: This will be between the points A(p = 3, 45 = 100°, 2 = 3) and B(p = 5, 45 = .130°,’z = 4.5). Performing point transformations to cartesian coordinates, these become AC1: = —0.52, y = 2.95, z = 3) and B(:c = —3.21, y = 3.83., z = 4.5). Taking A and B as vectors directed from the origin, the requested length is ‘ 3.2 Length = ts — AI = |(—2.69, 0.88, 1.5); finpressmthes’rector field; G ‘= 85inq§ a9 in p V a.) rectangular components: ‘ x ' Smyz = (51:2 +212)\/a:2 +yz+z2 ‘ A » . . 8y z y Gy=Ssm¢ag-%=Ssm¢cosflsm¢=mmm - , ' 8:122 _. M..t,c_,_____»_~n_> ~ ’ (x2 + yZ)‘ /$2 + y2 + 22 = m2+y2+22 Finally, ’ ' ' 8y . 1:2 . yz G i = ’ - “ z ' (“my 2) x/m2+y2+z2 [$2+-yza“+m2+yzay a] b) cylindrical components! The as direction Will transform to cylindrical components in the ap andraz directions only, where ‘ Z W? 7 The 2 component will be the same as found in part a, so we finally obtain ' i G(p’z)=M[za _,_ a2] WE” @ Express the unit vector am in spherical components at the point: a)r=2,6=lrad,¢=0.8rad:Use ’ , as = (a2 - ar)a¢ + (as - ae)aa + (am -a¢)a¢ 9—- sin(1) cos(0.8)a., + cos(i) cos(0.8)ag + (— sin(0r8))a¢ : 0.593... + 0.38219 —- 0.72% b) a: =., 3, y = 2, ’z = —1: First, transform the point to spherical coordinates; Have r = m 9 = cos-1(—1/fiZ) =5 1055", and ¢ = tan-1(2/3) = 33.7°. Then a9: sin(105.5°) cos(33.7°)a.r + cos(105.5°) cos(33.7°)ag + (—- sin(33.7°))a¢ 2' = 0.80%. -— 0.22ag —- 0.552195 GP = Ssimfiag -ap =85in¢cos€= Ssinqb 9 'c) p = 2.5, ¢ = 0.7 red, 2 = 1.5: Again, convert the point to spherical coordinates. 7' = \/ ,0? + 22 -= V8.5, 0 = cos-1(z/r) = cos‘1(l.5/\/8.5) = 59.02am ¢ = 0.7rad .= 40.1°. Now = 0.66.3, + 0.39m, — 0.64%: ' ' __ ' ' ....., hr. .___.....v,. i .. — - ‘ 8y ' 27 __w_.__ Gm: 8511145a0'aw =85in¢c°SGC°S¢= ‘/'¢'2+y2 1/m'2+y2+z2 1/$2+1/2 Ac spherical coordinates of (1, 1, WE) are I ‘ - ~ ' Ci [ai’ ‘fia‘i’ + 3az](3. m, 5) r~=1/1+1+2 =2 . .=(co.s ¢+flf§sin ¢)ax+(5in¢' NEWS ¢)ay+3az snail—I—‘fiiiwm r 7r 7: h . I 2 > e ‘ = [cos 73‘- + 3 sin r3—jax + (sin ~3- — J5 cos 3—)513, + SaZ ¢ = tan-1 %= 7r/4 r ‘ Then [3315. — x539 - 3¢] ax: sifi%cos%§ Cosg-ag—éio%a¢ - (Ly/3,115) =%af+%ag_%a¢ {I} =(35in-735coo—765— 3cos§cos§+sin%)ax . x I lay=sini£sin€1£ar +cos§sm%o9+cos%a¢ +(3smgsm%_fiws%sm%_cosg)ay =.;_ar:+%aa+71§a¢ + (3c§s§+¢§sm§)az . I. ' _-\/-2_az=—-\/-2— (cos-gar ~Sin-Ea9] =(—Z——%+%)ax+[%§——@—g)ay . Ii ’ . ‘ =—ar+ae I I +Q+§ z ., [ax +2}. —J2_a;](LLJ:) =2a9 - r .=22x+332 The two vectors are equal. ' A ' . . 4 - n. r: «xyli'ihwgscwffiqj +weqz a6: C616 Carri: 3.x +9519 sg+ab~m~g «12 A _ ' A ' A, - a¢y=~sm+axfwéqj . : ‘ .4. AA - . "; _ «*1: (34:19 C‘s-34>va +Cr°15$w§>tibf SM,3QE - 39' A ' A 7' A I , DC” —_- - SwG 5.4.4.45 4.x {— Sane ens—q, 23 : smeflqb (54> g 4 A A : ~55Hem+q “Smégm43c; soared :..a D 2. be, . DA fa. ,. A A A v e - "M9 SWCFQ +W9c91+i : Cased: a‘f‘ ' 3 - ‘F ‘ .4 ‘\ fl . A B d? = "9“:4’ 4' ‘" 5““ ‘E a " fighgwgr; _ _.. - « -_ - w _y e 54: . DA <14, - o G: , A = —1.95259'+ 4.05442 At point I): " :1: = 4.5sin(1r/2) cos (1r/8) = 4.1575 2/ = 45 sin (qr/2) sin (7r/8) = 1.7221 » z = 4.5 cos (1r/2) = 0 B = 4.15755- + 1.72211} (a) A - B L ' I - : > _ 4.52 ,m- = 4.5551(1/7) cos (1.5”) = 0 I I . a = 1.7375 red 9‘: 4.51551 (ar/T)sin (1.51:) = 4.9525 , - ' (b) _ z = 4.5 cos (1r/7) = 4,0544 ‘ cos (a) = = —0.16604 Distance = 4.5 x 1.7376 = 7.8192 In Z i??? S b 3.535% 3:6: >1 L > Qt- %_- 4-.‘0 we 5.5m}: <4. 9:4»..33)‘. Q5 ...
View Full Document

This note was uploaded on 08/12/2010 for the course ECE ECE 302 taught by Professor Ferguson during the Spring '10 term at Cal Poly Pomona.

Page1 / 10

ece302hw2 - may ’ Catareetiia Stare...

This preview shows document pages 1 - 10. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online