EECS+145+HW+2 - 552 VECTOR CALCULUS

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Unformatted text preview: 552 VECTOR CALCULUS _'_________,__—_.—.._————-———-~——— -' 5: . 7-32 1;. I Find div :2 where (a) v(r) = 3x2yi + zj + xlk (b) v{r) = (3x +y)i + (22 + 10} + (z — 2y)k - 32} HF = (nyz + 251' + (3x222 —y223)j + (y:3 # x23)k, calculate div f at the point (—1, 2, 3). everywhere zero, the flow entering any element of the space is exactly balanced by the outflow. This implies that the lines offlow of the field F(r) Where divF : 0 must either form closed curves or finish at boundaries or extend to infinity. Vectors satisfying this condition are sometimes termed solenoidal. F = (2x2);2 + 22H + (Bury3 — xzz)j + (3.30222 + xy)k is soienoidal. :36 ; (Spherical polar coordinates) Using the notation I introduced in Exercise 30, show, working from first principles, that - Vu-divv=ii(rzv)+ 1 3(t) sine) 33 Find V(a - r), (a ' V)r and a(V - r), where a is a _ r2 6,, ’ r Sin 9 39 3 constant vector and, as usual, r is the position vector 1 a r = (16,}: Z} _ + r sin 68¢(v‘9) The vector v is defined by v : rr‘l, where where v _ y u + v u + U u r=(x,y,z)andr=|r|.Showthat _ ’ ’ 9 9 '9 °' WV _ v) E grad div” 3 _23 r V force field F, defined by the inverse square law, r is given by . F = rer 35- Find the value of the constant 1 such that the vector field defined by Show that V- F = 0. 7.3.3 Curl of a vector field It is clear from observations (for example, by watching the movements of marked corks _v on water) that many fluid flows involve rotational motion of the fluid particles. Com- D 1 C 7, plete determination of this motion requires knowledge of the axis of rotation, the rate -03 v3 Z'AZ of rotation and its sense (clockwise or anticlockwise). The measure of rotation is thus a (x’y’ 2) it ' vector quantity, which we shall find by calculating its 2:, y and 2 components separately. A,“ _ _2 gay, _ _,,B Consider the vector field v(r). To find the flow around an axis in the x direction at the Figure 7.10 Flow around a rectangle. point r, we take an elementary rectangle surrounding r perpendicular to the x direction, as shown in Figure 7.10. To measure the circulation around the point r about an axis parallel to the x direction, we calculate the flow around the elementary rectangle ABCD and divide by its area, giving [222(x, y*, z — Az)(2Ay) + 123(x, y + Ay, 2*)(2Az) — v2(x, )7, z + Az)(2Ay) — v3(x, y — Ay, §)(2Az)1/(4AyAz) where y*, )7 E (y — Ay, y + Ay), 2*, E E(z - AZ, 2 + A2) and v I vli + vgj + t/Jk. Rearranging, we obtain —[t}2(x, 3?, z + A2) — v2(x, 32*, z — Az)]/(2Az) + [va(x, y + Ay, 2*) - 123(x, y - Ay, 5)l/(2Ay) 13 DERIVATIUES OF A 1rtECTIJR PUINT FUNCTION 555 Henee when any rigid body is in motion, the our! of its linear velocity at any point is twice its angular velocity in magnitude and has the same direction. Applying this result to the motion of a fluid, we can see by regarding particles of the fluid as miniature bodies that when the curl of the velocity is zero there is no rotation of the particle, and the motion is said to be curl-free or irrotational. When the eurl is non-zero, the motion is rotational. .. .. .. .. .. __ l - - ‘ ‘ “39 .361.- Isa} Find u = eurl r; when U = (3x33, —ya, .1: + 23). 1ft: 2—yi +xj+xysk is the velocity veetor eta fluid! find the local value of the angular velocity at the A vector field is defined by r: 2 (ye, .rsg, xy). Show 9'3“th (1* 3r 2)- that our] v : 0. __ __ _ If the veloeity of a fluid at the point (it, y? a) is given by Show that ifu : (2x +ya,~ 2y + at, 22 + xy) then H = (in? + byfi + (Cl + dylf l : ' f] h th = . ' ' I C”? i" 0: and ndfm SUE at U grfldf find the conditions on the eonstants o, b, r: and d in order that By evaluating eaeh term separately, verify the identity div L! = U, curl I; = 0 V K (jg) Efl‘pr x U) + (19f) K v Verify that in this ease r: = % grit-14:10:11:2 + 2hr}: m nyi) forflr) = .r‘l — y and 120*) z (s, 0, —x). H (Spherieal polar eoordinates) Using the notation Find constants o, t) and e sueh that the veetor field introdueed in Exereise 30, show that defined b y V X U = eurl s! F = (4,132 + 5133):? + (tori + Bali + (6x32 + ey)k is irrotational. With these values of (I? h and e, = 1 i i ‘3 determine a scalar fimetion $(I, y, a) such that F=va 7.3.5 Further properties of the vector operator V ' So. far-we have used ways":- I a. fl") “cases I . F(-_r) a vector field I ; (gi— gliil (%_%lj (Elijah, .- - For) a veetor field ._._l . —.. (d) aurl E H -— _—— Also: gala 3r "- C832 '1 .93 : ""s(gf3d f?” k), a 32‘” 1 33 a (ERIE 132 a 3:2 I "—2 C i j E E. C 3.3:: 3y 0 0 1 33¢ I 2 (3)23: 2 [All grad d) >< k 2 (3:2 mi ?.3 DERIVATIVES OF A HECTOR POINT FUNCTION 559 —12C111‘1(k§ £9) + 3» aurl grad d: :92: k d . . 3; , smaa Cur} grad (9 = 0 by (7.21) it c? a? 3% j 3x352 sirtaa k is a aanstarit vaatar [(aaaaaaxi] 3x By r— —J—~—~— ('33 '32? sa that wa hava “13:1 VXE— a 3: Shaw that ifg is a flirtatiaa af r : (a, y, 3) than Ida]r gradg= Fd—ir Daduaa that if a is a vaatar fiald than div [(u :s: fig] = (r- auri u)g Far Ma, y, a) = xlyzai' and Hart y, a) : rzyr' + xyzaj — yzlk datarrnina (a) 371$ (13) grad divF (a) aurl aurl F Shaw that if a is a aarrstant vaatar and r is tha pasitian vaatar r = (x, y, 3) than div {grad [(r - r)(r - a)]} = 100' - a) 3: )_I[.33¢5 .aip] _ _ I a "J a a dyé‘t 5x3: Varify the identity V21: = grad div v — aurlaurla1 far tha vaatar fiald v : argyflri + yj + 3k). Verify, by aaiaulatirrg aaah tarrrr saparataly, tha idarrtitias div(aaa):v'auria—a*aurla aurl(u K v) = adith —.vdiva + (H‘V)H * (_u*?)v whart at = xyj + Lift? and r: = xyr‘ + yak. If r is tha usual pasitiarr vaatar r = (x, y, z), shaw that 560 VECTOR CALCULUS ________‘__—______.—.—.—-——w—-—-— By evaluating V - (Vf), show that the 'Laplacian in spherical polar coordinates (see Exercise 30) is given by 13 gar l a . 8f v2 =——[ —]+ — s 9—] f r28!" Far rzsin989(m as 1 2 + a .2 fl: r‘sm 98¢ (a) div grad[%) = 0 (b) curl [k x gradEJl/ + grad [k - gradv; = 0 If A is a constant vector and r is the position vector r = (x, y, 2), show that A . r A (A . 1,.) Show that Maxwetl’s equations in free space, namely (a) grad( .) )=—3—3 5 r r‘ r r divH 2 0, divE = 0 AXr _2A 3 _lr9_E f_13ii (b)curl( r3 ]—?+F(AXr)Xr VxH—C t, V><E_ Cat are satisfied by If r is the position vector r = (x, y, z), and a and b are constant vectors, show that H = 1 our] Egg c r 3 er=0 ( ) E 2 curl curlZ (b) (a i V». = a where the Hertzian vector Z satisfies (c) V'X[(a-r)b—(b-r)a1=2(a><b) 7 1322 V'Z = — — (d)V-[(a-r)b—(b-r)a]=0 c :2 __________________,____.._.___—_—-—-——-—-— Figure 7.13 Definite y integral as an area. Topics in integration In the previous sections we saw how the idea of the differentiation of a function of a single variable is generalized to include scalar and vector point functions. We now turn to the inverse process of integration. The fundamental idea of an integral is that of summing all the constituent parts that make a whole. More formally, we define the integral of a function f(x) by a all Age—>0 i=1 whereoizx0 < x1< x2 < < xH < xn=b,Ax[=xi—x,_. andxH —‘-’= 2?, S x,. Geometrically, we can interpret this integral as the area between the graph y = f (x), the x axis and the lines 3: = a and x = b, as illustrated in Figure 7.13. 10 X1 X2 :0 *. =b xi 564 VECTOR CALCULUS In general, W depends on the path chosen. If, however, Hr) is such that F (r) - dr is an exact differential, say *dU, then W: fC — dU : UA — UB, which depends only on A and B and is the same for all paths C joining A and B. Such a force is a conservative force, and U(r) is its potential energy, with F (r) = - grad U. Forces that do not have this prop- erty are said to be dissipative or non—conservative. Similarly, if v(r) represents the velocity field of a fluid then 95C 1) 'dr is the flow around the closed curve C in unit time. This is sometimes termed the net circulation integral of v. If sficu -dr 2 0 then the fluid is curl-free or irrotational, and in this case v has a potential function Mr) such that v = — grad gt). Evaluate fy d5 along the parabola y = 2pc from $61 i Find the work done in moving a particle in the force . 2 2 7 2- 7 _ - z . n A(3, 23) to 8(24, 4‘96). [Recth (37:) = 1+ [3—9 ,1 field F - 3x I + (2x. )0} + k alone (a) the curve defined by )c2 : 4y, 33c3 = 82 from x : 0 to x = 2; E l t BZxd + t 2d- 1 th vauaehl y x (x y) Haong em (b) thestraightlinefrom(0,0,0)to(2,1,3). of the circle x2 + y2 : l in the first quadrant from A“; 0) to B“), 1) Does this mean that F is a conservative force?I Give .; _ reasons for Our answer. _; Evaluate the integral f5 V-dr, where y V: (2yz + 3x2, )22 + 4x2, 222 + 6xy), and C is the .62. curve with parametric equations x : I], y = t3, 2 = r ' " joining the points (0, 0, 0) and (1 , l, 1). Prove that the vector field F = (3x2 7 y, 2}:2 i x, 2y22) is conservative, but not solenoidal. Hence evaluate the scalar line integral fp F ~ dr along any curve Cjoining the point (0, 0, 0) to the lfA = (2y + 3): +x21 + (yzex)k, evaluate ffi-A -dr point (13 2, 3)- along the following paths C: lfF =xyi —zj+x2k and C is the curve 1: = {2,}; = 2!. 2' :13 from t 2 0 to t: 1, evaluate the vector line (a) x=2t2,y=t,z:t3fromt=0tot=l; (b) the straight lines from (0, 0, 0) to (0, 0, 1), then to (0, l. 1) and then to (2, l, l}; (c) the straight line joining (0, 0, 0) to (2, 1. l). 64_ Prove that F 2 (y2 (:08): + 23)i' + (2}: sinx — 4)} + (3ch2 + z)k is a conservative force field. Hence find the work done in moving an object in this field from (0, 1, —1) to (TE/2. fl. 2). 7.4.3 Double integrals integral chx dr. lfA = (3x+y, way—z} and B: (2, —3, 1) evaluate the line integral 95C (A x B) x dr around the circle in the (x, y) plane having centre at the origin and radius 2. traversed in the positive direction. In the introduction to Section 7.4 we defined the definite integral of a function fix) of one variable by the limit l f(x)dx= lim Era-Ma n—>><= ‘ an Ase—>0 ’"1 where a = x0 < x1 < x2 < . . . < x” = b, Ax). = x1. —xH and xH g 3?, g x, This integral is represented by the area between the curve y =f(x) and the x axis and between x 2 a and x = b, as shown in Figure 7.13. ...
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This note was uploaded on 12/13/2010 for the course ELECTRICAL EECS 145A taught by Professor Chinc.lee during the Fall '10 term at UC Irvine.

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EECS+145+HW+2 - 552 VECTOR CALCULUS

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