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Assignment#1 - Model Answers Mathematical Methods for Mechanical Sciences ASSIGNMENT 1 Problems 2A 2A-1 Find the value of:1 given that a 2 3i —

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Unformatted text preview: Model Answers Mathematical Methods for Mechanical Sciences ASSIGNMENT 1 Problems 2A 2A-1: Find the value of :1: given that a 2 3i —- 2j and b 2 4i + (vj are perpendicular. Solution: If a J. b then a - b = 0, (3i — 23) - (4i + xj) 12 — 21' u n 9.0 (I) 2A-3: Solve for x the vector equation x + a(b - x) = c, where a, b, c are constant vectors. What happens when a-b = —1?[x = c— a(b-c)/(1 +a-b)]. Solution: To solve x+a(b-x)=c. (1) First find b - x. Take the scalar product of b with (1): b-x+(b-a)(b-x) =boc (b'x)[l+b-a] =b-c b - c m Substitution of (2) into (1) and rearrangement yields _ a(b - c) 1 + a - b ’ (b-X)= X=C which is the desired result. If a . b = —1 and b - c aé 0, there is no solution. The system of three equations represented by (1) is inconsistent. If h - c = 0, there is an infinity of solutions. 2A-5: Solve the simultaneous equations x + y x p = a, y + x x p = b. [x = {(P-a)p+a—b X p}/(1 +102), y = {(p-b)p+b—a x p}/(1 +172) ]- Solution: Write the equations: x + y X p = a (1) xxp+y=b (2) 1 AM 400 Engineering Mathematics Model Answers Mathematical Methods for Mechanical Sciences Take the cross product of (1) with p, x><p+(y><p)><p=a><p, and use (y x p) X p = p(p-y) -p2y- Then 2 _ xxp-p y+p(p-y) —a><p- Elimination of x x p between (2) and (3) results in 3'(1 +p2)=p(p-Y)+b-(a><1_>)- To find p - y take the scalar product of p and (2): p - y = p ~ b. Substitute into (4) to get _ (p-b)p+b—aXp 1 + p2 To find x, substitute for y in The result will be (p-a)p+a—b><p x: 1+1)2 (3) (4) (5) (Alternative solution for x: Because ( 1) and (2) are symmetric in x and y, the solution for x can be written down by inspection.) 2A-7: Show that (axb)-(axc) xd= (a-d)(a-bxc). Solution: (axb)-(axc)xd = (axb)'[—a(c-d)+c(a-d)] = 0+(a-d)axb-c {because aXb-aEO} (a-d)(a-bxc). {because a-bxc=axb-c} 2 AM 400 Engineering Mathematics Model Answers Mathematical Methods for Mechanical Sciences Problems 2B 2B-1: Calculate the gradient of (p = as. Solution: 690 6<p 6:12. 851:. 6:1: W=<5;’5;’az =al+a—yJ+b§k=‘- 23-3: Calculate the gradient of (p = r", r = xi + yj + zk, ' [mm-2r], Solution: Using the chain rule, and noting that Vr = r/r, _ df __ df r Vflr) — drvr _ drr’ V<p = mm“; — rum—21' 23-5: Calculate the gradient of (p = r-V(a: + y + z), r = xi + yj + zk. [i +j + k]. Solution: Begin by evaluating (p: <p=r~V(w+y+z) =r-(1,1,1) = (x+y+z), Then: V<p=V(x+y+z)=(1,1,1)=i+j+k. 2B-7: Find the directional derivative in the direction of a of (p = e’ cos y, a = (2,3,0), at x = (2,7r,0) [-2e2/x/13]. Solution: The directional derivative of (p is é - V<p, where 5. is the unit vector in the direction of a: (2, 3, 0) W (Zeac cosy —— 3e$ siny + 0)/\/fi (—2e2 + 0 + 0)/\/fi -—2e2/\/_1§. 51- V90 = - (ex cos y, —ez sin y, 0) 3 AM 400 Engineering Mathematics Model Answers Mathematical Methods for Mechanical Sciences 2B-9: Find the unit normal to the surface 2 = V132 + y2 at x = (3, 4, 5) [(3,4, —5)/5\/§]. Solution: The unit normal to the surface (,0 2 Va? + y2 —- z = 0 is: . W) n=———— (Hr/VIE2 + y2,y/\/w2 + yz, -1) Vang/(:17:2 + y?) + y7/:1:2 + y?) +1 ( (as/vac? + y2,y/ $2 + 112:1) /\/§ (3,4, —5)/5\/§, since M552 + y2 = «9 +16 = 5. Problems 2C ZC-l: Find the divergence of F = yi + zj + xk Solution: The divergence of F is 8F1 3F2 + 8F3) __ fig 62 6a: __ V'F=(—BTE—+a—y a—z —+'—’ 0. _8:1: 8314—52—- 20—3: Find the divergence of F = (23,312,213) [1 + 23/ + 322]. Solution: The divergence of F is 2 3 fi+9€3+§§)—6fi @-+a—z-=1+2y+3z2. 8m 8y Oz VIF=( _5;+6y 62 20-5: Find the divergence of F = myz(i +j + k) [yz + x2 + my]. Solution: The divergence of F is fl+§§+%)_ 6:1: 3g 52 — (a 6 6 g+a—y+a)xyz=yz+$z+$y. V-F=( 4 AM 400 Engineering Mathematics Model Answers 2C-7: Find the divergence of F = r(r - a), r = xi + yj + zk, a = constant. Solution: For a scalar field (p: Mathematical Methods for Mechanical Sciences [4r - a]. V-(<pF)=<pV-F+F0V<p. Take (p = (r - a), a scalar, then: 2C-9: V-r(r-a) ll (r-a)V-r+r-V(r.a) (r-a)( 3(r~a)+r-(—— —— — (9:6 (9:1: +@+@) 0y Oz (9 0 6 Bx’ay’az 3(r-a)+(r-a)=4r-a. Prove that V20”) = n(n + Urn—2, 1' = xi + yj + 2k- Solution: V2(rn) = V - V(r") = V ' (nr"" Expand the divergence: V - (nr""2r) = nr”_2V - r + r - Vnrn‘2 = 37w”—2 + r - — 2)r"_4r) = 3717‘"-2 + n(n — 2)r"’2 = n(n + 1)r"‘2. 20-11: Prove that div(ng) -— div(gi) = fvzg — gV2f. Solution: V- 0%) -V-(9Vf) (fV - Vg+Vg- fV2g-9V2f- +%gggwm Bar’ay’az ) (alas + (1231 + (132) {because V7" 2 r/r} {because V - r = 3 and VT 2 r/r} Vf)-(gV-Vf+Vf'Vg) AM 400 Engineering Mathematics Model Answers Mathematical Methods for Mechanical Sciences 2C-13: Prove that V = % fs n-V(r2)dS, Where r = xi + yj + zk and V is the volume enclosed by S. Solution: By the divergence theorem 1 1 —f{n-V(r2)d3 = —/ V2(r2)dV 6 s 6 V 2 2 2 gal/VF? +i+a—)(x2+y2+22)dV 5? W 622 1 E/VGdV V. 2C-15: Evaluate f3 n - FdS' when F = (m, (1223;, —(II2Z) and S is the surface of the tetrahedron with vertices (0,0,0), (1,0,0), (0,1,0) (0,0,1) [21;]. Solution: Let V denote the interior of the tetrahedron, then by the divergence theorem, fF-ds = [v-de s v _ 3 3 2 -3 2) _ i/v(aza:+aymy azxz dV = /(1+:1:2-—x2)dV V m V which is just the volume of the tetrahedron, one third the height times the area of the triangular base. 1 fin-FdS =%(1)x%(1)(1)= a. 2C-17: Evaluate f5 11 - FdS when F = ami + byj + czk where a, b, c are constants, and S is the unit sphere [xi = 1. [g—fla + b + Solution: If V denotes the interior of the sphere, then by the divergence theorem, /F~ds = fV-FdV S V a 0 8 fv (-a—xax + Egby + Ecz) dV /V(a+b+c)dV 6 AM 400 Engineering Mathematics Model Answers Mathematical Methods for Mechanical Sciences = (a+b+c)/VdV §7r(a + b + c), because the volume integral is over the unit sphere of volume §7r(1)3. Problems 2E 2E-1: Evaluate the line integral fC F-dr: F = (3x4, 3316, 0) where C is the curve: m2 + 3/2 = 4, z = 0 from (2,0,0) to (—2,0,0) [2139-2]. Solution: The contour is the semi-circle r = (2c0s0,23in0,0), OSOS’K dr = (—2sin6,2c030,0)d0 /F~dr = / [3(2cos0)4,3(2sin0)6,0]-[—2sin0,2cos9,o]d0 C 0 71' = 96/ (—cos4flsin0+4sin60cos6’)d6’ 0 c0350 4sin70 W 96[ 5 + 7 —2 192 95 (fl " “5—- 0 ll 2E-3: Evaluate the line integral [0 F-dr: F = (ez,e4y/m,ezz/y) where C is r = (t,t2,t3), 0 < t < 1 [%e4 + i-e2 + e — 183]. Solution: We have dr = (1, 2t, 3t2)dt and F = (et,e4‘, em), 1 / F-dr = / [et + 2te4t +3t2e2t]dt C 0 1 t 1 3t2 3t 3 t ___ 4t ___ _ 2t e+<2 8)e +(2 2+4)e]0 _ 34 32) (_l §> —(e+8e+4e 1 8+4 34 32 13 8e+4e+e 8. 7 AM 400 Engineering Mathematics Model Answers Mathematical Methods for Mechanical Sciences 2E-5: Show that the following integral is path-independent and find the corresponding potential function: fC [ycoswy dart + mcosmy dy — dz], [90 = sinmy — Solution: Path independence is assured because V x [y cos my,:rcos my, —1] = 0. To find the corresponding potential function, let 6 0—: = ycos my, (1) 6 5—: = xcos my, and I (2) W — = —1. 32 (3) From (3), (p = —z + f(x,y). Substitute into (1): a 5—i— = ycos my, f = sinwy+9(y), w = —z + sinmy + g(y)- Substitute into (2): xcosm +d—g — mcosa: y _ y? g = a constant, and we can take (,0 = —z + sinzy. 2E-7: Show that if r = r(t) on C, the length of are between t = a, and t = b is given by fl = |r(t)|dt. Solution: On C, dr = 1"(t) dt. for dt > 0, d3 ldrl = Ii'(t)| dt. b b e = ds= / |i-(t)|dt. a. t=a 8 AM 400 Engineering Mathematics Model Answers Mathematical Methods for Mechanical Sciences 2E—9: r = ((1 cos 0, asin 0, a9 tan a) on a helix, where a, a are constants. Show that the length of arc measured from 0 = 0 is given by 6 = (10 sec (1. Solution: Let 0 < t < 0 and r = (a cos t, asin t, at tan a), then i‘(t) = (——a sin t, acos t, atan 0:) dt. 0 9 0 32/ |r(t)|dt =a/ Vsin2t+coszt+tan20¢ dtzaseca/ dt=a05eca, 0 0 0 where the formulae sin2 t + cos2 t = 1 and 1 + tan2 a = sec2 a have been used. Problems 2F 2F-1: Evaluate fs F-dS: F = (2m, 2y, 0) where S is the surface: 2 = 2:1: + 3y, 0 < av < 2, —1 < y < 1, [-16]. Solution: f5 F'dS = ifs F - ru x rv du dv, where u = a: and v = y, and r = (u, v, 2u + 31)) on S. Then F = (2U, 21), 0). Evaluating term by term, ru = (1,0,2), rv = (0,1,3), and i j k ruxrv = 1 0 2 =(—2,—3,1). 0 1 3 F-ru x r,, = (—4u—6v). Choose the positive sign and obtain 1 2 / F-dS —/ dv/ (4u+61)) du s —1 0 1 2 — / dv [2112 +6uv] -1 0 1 —/ (8 + 12v)dv —1 ll 1 = — [81) + 6112] —1 = —-16. 2F-3: Evaluate fs F-dS: F = (1,$2,myz) where S is z = my, 0 < a: < y, 0 < y < 1, —T5:—0. 9 AM 400 Engineering Mathematics Model Answers Mathematical Methods for Mechanical Sciences Solution: f3 F-dS = ifs F - r“ x rv du dv, where u = a: and v = y, and r = (u,v,uv) on S. Then F = (1,112, 112122). Evaluating term by term, ru = (1,0, 1)), rv = (0, 1,11), and i J k ru X rv = 1 0 'u = (—v, —u, 1) 0 1 '11. F - ru x r,, = (1,112,131)?) (—v, —u, 1) = (—’U — 113 + U202) Choose the positive sign and obtain / F-dS S = /01[-vz—iv4+%v5] d1) 1 1 1 1 = l‘§”3‘%”5+1—s”6]o _ 1 1 1 — _§_E 1s _ 59 _ TEE 2F-5: Evaluate f3 F-dS: F = (2:1:y,:c2,0) where S is r = (coshu,sinhu,v), 0 < u < 2, —3 < v < 3, [2 cosh3 2 — 2]. Solution: fs F.dS = :1: IS F o ru x r1, du d1). Evaluating term by term, ru = (sinh u, cosh u, 0), 1'2: = (0,0, 1), and i j k I‘u X 1‘1; = sinhu coshu 0 = (cosh u, — sinhu,0). 0 0 1 F . ru X rv = (2 coshu sinh u, cosh2 u, 0) - (cosh u, —- sinh'u, 0) = cosh2 usinhu. 10 AM 400 Engineering Mathematics Model Answers Mathematical Methods for Mechanical Sciences Choose the positive Sign and obtain de3 S 2 3 / cosh2 u sinh u du / dz; 0 —3 2 6 / cosh2 u sinhu du 0 2 2 cosh3 11L) 2 cosh3 2 — 2. 2F-7: Evaluate [S F-dS: F = (y2, z2,a:22) where S is the surface bounding the region :32 +y2 g 4, as 2 0, y 2 0, M £1, [27F]- Solution: Let V denote the interior of S, then by the divergence theorem, 3% F-dS = [V V - F dV. With V ~ F = gaiyz + 621122 + %m22 = 3:2, the integral becomes 2 v4—x2 1 / V ' F dV = / d9: / dy / 3:2 dz {see 1‘ below for explanation of integration limits} V 0 0 —1 2 V4—z7 2/ dx/ 2:2 dy 0 0 2 2/ m2x/4—rr2da: 0 2 [—x(4—$2)3/2 mv4—m2 _1 m 2 4 +—§-—+2sm E 0 2[2an-11—-2an-1q 2pg—fl 271'. II 1‘ Integrate over —1 S 2 g 1 because the shape is constant in 2. In the m-y plane, 3324-11234,$20,yZO;thereforeOSySV4—m2,and03wg2. 2F-9: Use Stokes’s theorem to evaluate f0 F-dr: F = :32ij Where C is the quadrilateral with vertices (0,1,0), (1,1,0), (1,0,1), (0,0,1) traversed in this order. [—%]. Solution: From Stokes’s theorem, fc F.dr = [S V x F - dS, where S is the plane surface of the 11 AM 400 Engineering Mathematics Model Answers Mathematical Methods for Mechanical Sciences quadrilateral whose normal is orientated in the positive direction with respect to C, and i j k V X F = 6% 6% 6% = (—:I:2y,0,2zyz). 0 $2312 0 In the y-z plane, the surface is z = 1 —— y, so the position vector on S is r = (1:,y, 1 — The element of area is d8 = :l:(rz x ry) day dy. Evaluating term by term, rx = (1,0,0), r, = (0,1,—1) and rm x ry = (0,1,1). (18 = :i:(0,1,1) dxdy. Take the negative sign, because the normal vector must point in the —z direction, and obtain /V X F ‘ (IS = -/(—-a;2y,0, 2:1:yz) - (0, 1,1) dxdy s S —— / ny2 da: dy S —/ 2xy(1 — y) dz: dy {because 2 = 1 — y} S' 1 1 — / da: / 2xy(1 — y) dy {see i for limits of integration} 0 0 1 1 2 3 _ _ E__y_ — 2/0 xdw[2 3L 1 1 1 1 1 1 __ d=___2 =__, 30mg” 3[2$]0 6 ll 1 The region of integration in the x-y plane is the square 0 S x S 1 and 0 S y S 1. 2F-11: Use Stokes’s theorem to evaluate fc F»dr: F = myzj where C is the triangle with vertices (1,0,0), (0,1,0), (0,0,1), Solution: Horn Stokes’s theorem, f0 F0dr = fs V x F - dS, where S is the plane surface of the triangle whose normal is orientated in the positive direction with respect to C, and .i a i VxF= 3% a; =(rvy,0,yz)- OS’IQJ W 0 myz 12 AM 400 Engineering Mathematics Model Answers Mathematical Methods for Mechanical Sciences The surface is a: + y + z — 1 = 0, so the position vector on S is r = (:r,y, 1 — a: — y). The element of area is (18 = :i:(rx x ry) da: dy. Evaluating term by term, rz = (1,0,—1), ry = (0,1,—1) and r3 x ry = (1,1,1). d3 = :l:(1,1,1)dmdy. Take the positive sign because the normal vector must point in the +2 direction and obtain Av x F-dS = fs(—xy,0,yz) - (1,1,1)d:cdy = /S(—xy + yz) dx dy = /S[—my+y(1——x—y)] dxdy {becausez= l—x—y} 1 1—2 = / dac / [y(1 — a: — y) — my] dy {see § for integration limits} 0 0 1 2 3 yzl—m / dm — rcy2 — 0 y=o [01 Eu — 11:)2 — x(1—- 9:)2 — 311(1- m3] d3; _ ‘1 3 1 412 2314]1 _ [6(1 x)+12(1 :17) 2$+3x 4:1; 0 _ [:1 L1] [1+1]_0 _ 2 3 4 6 12" § The area of integration in the 53-3; plane is the triangle formed by 0 S :1: S 1, 0 S y S 1, and the line y = 1 - :5. Therefore, the limits of integration are 0 g y S (1 — m) and 0 g :3 5 1. 2F-13: Use Stokes’s theorem to evaluate f0 F-dr: F = (—3y, 3m, 2) where C is m2 +y2 = 4, z = 1, [247r]. Solution: From Stokes’s theorem, f0 F-dr = fS V X F - d8 = f3 n-V x FdS, where S is the plane circular surface :32 + y2 S 4, z = 1, and 1 j k V x F = (9—35 3% % = (0,0, 6). {a constant} —3y 31: 2 By inspection, in = (0, 0, 1). Choosing the positive direction, /n-VdeS’ = f(0,0,1)-(0,0,6)ds S S 13 AM 400 Engineering Mathematics Model Answers Mathematical Methods for Mechanical Sciences II II C} 1 c, 33 a. NC!) [I 10 94>. 5‘ because the integral is over the circular area of radius 2. 2F-15: Use Stokes’s theorem to evaluate fc F-dr: F = 2zi + 4xj + 5yk where C is x2 + y2 = 4, z —— x = 4, orientated anticlockwise when viewed from above. [—47r]. Solution: From Stokes’s theorem, fC F-dr = fSV x F - d8 = fs n-V x FdS, where S is the elliptical surface defined by the intersection of the plane 2 = :1: + 4 and the cylinder 2:2 + y2 = 4, whose normal is orientated in the positive direction with respect to C, and i j k V x F = % 0% 53; = (5,2,4). {a constant} 22 4m 5y Unit vector 11 is normal to the plane —w + z — 4 = 0 so (1&1; fl . 11: Therefore, 1 -VdeS —/ —1,0,1 - 5,2,4 (13 [47T\/§] = —47r. 1 1 7278th The integral is over the elliptical area of major axis 4%? and minor axis 4, hence Is ds = «(afixa = 4m. 14 AM 400 Engineering Mathematics Model Answers Mathematical Methods for Mechanical Sciences Essay: Write a short essay (not more than two sides of paper) in which you give the definitions of scalar and vector fields, define in physical terms the divergence of a vector field F(x), and state and prove the divergence theorem. Solution: Single valued, scalar and vector functions defined over a region of space are respectively called scalar and vector fields. Scalar fields include, for example, mass density, temperature, gravitational and electrical potentials; fluid velocity, and the gravitational and electrical force strengths are typical vector fields. The surface integral 3% F - dS over a closed surface S enclosing a volume V (with surface element dS directed out of S) determines the net flux of the vector field F from S. The divergence divF of F is the ‘net flux per unit volume’. This is a scalar function of position whose value at x is defined by . _ . 1 divF _ \l/llgvfi F-dS, (1) where the integration is over the surface S of an infinitesimal volume element V containing x. The divergence theorem states that, for a general closed surface S with interior volume V, /dideV=?iF-dS. V S This is proved by decomposing V into infinitesimal volume elements 6V,c with surfaces Sh. For each element the definition (1) implies that 5V]; (div F)k % F-dS Sk where divF is evaluated at a point within (Wk. Then k—mo k—)oo / dideV = lim :6Vk(div F), = lim 2 F-dS. V k k S]: The surface integrals over the common surface of neighboring, touching volume elements 6V,c are equal and Opposite; the only terms in the final sum on the right hand side that make a non-zero contribution correspond to the surface elements on the boundary S, and the sum is ultimately equalto fSF-dSask—>oo. Q. E. D. 15 AM 400 Engineering Mathematics ...
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This note was uploaded on 05/12/2008 for the course ENG 419 taught by Professor E during the Spring '08 term at BU.

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Assignment#1 - Model Answers Mathematical Methods for Mechanical Sciences ASSIGNMENT 1 Problems 2A 2A-1 Find the value of:1 given that a 2 3i —

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