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Unformatted text preview: Model Answers Mathematical Methods for Mechanical Sciences ASSIGNMENT 1
Problems 2A
2A1: 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) 2A3:
Solve for x the vector equation x + a(b  x) = c, where a, b, c are constant vectors. What happens when ab = —1?[x = c— a(bc)/(1 +ab)]. Solution: To solve
x+a(bx)=c. (1) First ﬁnd b  x. Take the scalar product of b with (1):
bx+(ba)(bx) =boc (b'x)[l+ba] =bc
b  c
m Substitution of (2) into (1) and rearrangement yields
_ a(b  c)
1 + a  b ’ (bX)= 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. 2A5: Solve the simultaneous equations x + y x p = a, y + x x p = b.
[x = {(Pa)p+a—b X p}/(1 +102), y = {(pb)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(py) p2y Then
2 _
xxpp y+p(py) —a><p Elimination of x x p between (2) and (3) results in 3'(1 +p2)=p(pY)+b(a><1_>) To ﬁnd p  y take the scalar product of p and (2):
p  y = p ~ b. Substitute into (4) to get
_ (pb)p+b—aXp 1 + p2
To ﬁnd x, substitute for y in The result will be (pa)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.) 2A7:
Show that (axb)(axc) xd= (ad)(abxc). Solution: (axb)(axc)xd
= (axb)'[—a(cd)+c(ad)] = 0+(ad)axbc {because aXbaEO} (ad)(abxc). {because abxc=axbc} 2 AM 400 Engineering Mathematics Model Answers Mathematical Methods for Mechanical Sciences Problems 2B
2B1: Calculate the gradient of (p = as. Solution: 690 6<p 6:12. 851:. 6:1: W=<5;’5;’az =al+a—yJ+b§k=‘ 233:
Calculate the gradient of (p = r", r = xi + yj + zk, ' [mm2r], Solution: Using the chain rule, and noting that Vr = r/r, _ df __ df r
Vﬂr) — drvr _ drr’
V<p = mm“; — rum—21' 235:
Calculate the gradient of (p = rV(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. 2B7: 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 2B9:
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/ $2 + 112:1) /\/§ (3,4, —5)/5\/§, since M552 + y2 = «9 +16 = 5. Problems 2C
ZCl:
Find the divergence of F = yi + zj + xk Solution: The divergence of F is 8F1 3F2 + 8F3) __ ﬁg 62 6a: __ V'F=(—BTE—+a—y a—z —+'—’ 0. _8:1: 8314—52— 20—3:
Find the divergence of F = (23,312,213) [1 + 23/ + 322]. Solution: The divergence of F is 2 3
ﬁ+9€3+§§)—6ﬁ @+a—z=1+2y+3z2.
8m 8y Oz VIF=( _5;+6y 62 205:
Find the divergence of F = myz(i +j + k) [yz + x2 + my]. Solution: The divergence of F is ﬂ+§§+%)_
6:1: 3g 52 — (a 6 6 g+a—y+a)xyz=yz+$z+$y. VF=( 4 AM 400 Engineering Mathematics Model Answers 2C7: 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)=<pVF+F0V<p. Take (p = (r  a), a scalar, then: 2C9: Vr(ra) ll (ra)Vr+rV(r.a) (ra)( 3(r~a)+r(—— —— — (9:6 (9:1: [email protected][email protected])
0y Oz
(9 0 6
Bx’ay’az 3(ra)+(ra)=4ra. 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.
2011: Prove that div(ng) — div(gi) = fvzg — gV2f. Solution: V 0%) V(9Vf) (fV  Vg+Vg
fV2g9V2f +%gggwm
Bar’ay’az ) (alas + (1231 + (132) {because V7" 2 r/r} {because V  r = 3 and VT 2 r/r} Vf)(gVVf+Vf'Vg) AM 400 Engineering Mathematics Model Answers Mathematical Methods for Mechanical Sciences 2C13:
Prove that V = % fs nV(r2)dS, Where r = xi + yj + zk and V is the volume enclosed by S. Solution: By the divergence theorem
1 1
—f{nV(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. 2C15: 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, fFds = [vde
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 finFdS =%(1)x%(1)(1)= a. 2C17:
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 = fVFdV
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 2E1: Evaluate the line integral fC Fdr: F = (3x4, 3316, 0) where C is the curve: m2 + 3/2 = 4, z = 0
from (2,0,0) to (—2,0,0) [21392]. Solution: The contour is the semicircle 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 2E3: Evaluate the line integral [0 Fdr: F = (ez,e4y/m,ezz/y) where C is r = (t,t2,t3), 0 < t < 1 [%e4 + ie2 + e — 183]. Solution: We have dr = (1, 2t, 3t2)dt and F = (et,e4‘, em), 1 / Fdr = / [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 2E5: Show that the following integral is pathindependent 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. 2E7:
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 2F1: Evaluate fs FdS: 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 Fru x r,, = (—4u—6v). Choose the positive sign and obtain 1 2
/ FdS —/ 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.
2F3:
Evaluate fs FdS: 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 FdS = 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 / FdS
S = /01[vz—iv4+%v5] d1)
1 1 1 1 = l‘§”3‘%”5+1—s”6]o _ 1 1 1 — _§_E 1s _ 59 _ TEE 2F5: Evaluate f3 FdS: 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. 2F7: Evaluate [S FdS: 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% FdS = [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[2an11—2an1q 2pg—ﬂ 271'. II 1‘ Integrate over —1 S 2 g 1 because the shape is constant in 2. In the my plane,
332411234,$20,yZO;thereforeOSySV4—m2,and03wg2. 2F9: Use Stokes’s theorem to evaluate f0 Fdr: 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 $2312 0 In the yz plane, the surface is z = 1 —— y, so the position vector on S is r = (1:,y, 1 — The element of area is d8 = :l:(rz x ry) day dy. Evaluating term by term, rx = (1,0,0), r, = (0,1,—1) and
rm x ry = (0,1,1). (18 = :i:(0,1,1) dxdy. Take the negative sign, because the normal vector must point in the —z direction, and obtain
/V X F ‘ (IS = /(—a;2y,0, 2:1:yz)  (0, 1,1) dxdy
s S
—— / ny2 da: dy
S —/ 2xy(1 — y) dz: dy {because 2 = 1 — y}
S' 1 1
— / da: / 2xy(1 — y) dy {see i for limits of integration}
0 0
1 1 2 3
_ _ E__y_
— 2/0 xdw[2 3L 1 1 1 1 1 1
__ d=___2 =__,
30mg” 3[2$]0 6 ll 1 The region of integration in the xy plane is the square 0 S x S 1 and 0 S y S 1. 2F11: 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 FdS = 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 533; 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. 2F13:
Use Stokes’s theorem to evaluate f0 Fdr: F = (—3y, 3m, 2) where C is m2 +y2 = 4, z = 1, [247r]. Solution: From Stokes’s theorem, f0 Fdr = fS V X F  d8 = f3 nV 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, /nVdeS’ = 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. 2F15:
Use Stokes’s theorem to evaluate fc Fdr: 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 Fdr = fSV x F  d8 = fs nV 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 FdS, (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=?iFdS.
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 % FdS
Sk where divF is evaluated at a point within (Wk. Then k—mo k—)oo / dideV = lim :6Vk(div F), = lim 2 FdS.
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 nonzero
contribution correspond to the surface elements on the boundary S, and the sum is ultimately equalto fSFdSask—>oo. Q. E. D. 15 AM 400 Engineering Mathematics ...
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