202fk04 - 110.202 Final exam answers [Some students...

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Unformatted text preview: 110.202 Final exam answers [Some students complained on their finals that this exam was kind of sadistic. I apologize if you came away feeling that: it wasn’t meant to be that way. A lot of the calculations are very time-consuming if you attack them head—on, but if you use things like the product formula for divergence and curl, and little basic facts like then things work out reasonably simply. The machinery of vector calculus was worked out so that you don’t have to do the same calculation over and over again: it’s like a higher—order computer language, intended to allow something like parallel processing. These are really three—dimensional problems; you can do them by one—dimensional methods, but that can be very inefficient] I Notation: Let r = (x,y, z) denote the radial vector from the origin in R3 to the point (:13,y, 2), let 7" = M = x/($2+y2+z2) be its length, and let B be the vector field defined on R3 — {O} (i.e. Where r yé 0) by k — 3r‘2zr 7.3 lB(:1:,y, z) = 7 where k = (0,0, 1) is the standard unit vector pointing along the z—axis. [This is what physicists call a dipole field, directed along the z—axis; it is a model (for example) of the Earth’s magnetic field] 1. Calculate the divergence V - B. Solution: We have v - (r‘gk) = V(r_3) - k + r—Bv - k but k is constant, and V(r") = nr"“2r, so this equals —-3r’5r - k = —3r'5z. On the other hand, we have V(r‘5zr) = ~5r“7r - (zr) + 7"”5V — (2r) ; but V- (2r) 2 81(zx)+ 6y(zy) + 6z(z2) = z + z + 22 = 42, so V - B equals ——37'_52 —— 3[—57'_7zr - r + T—5 -4z] Which simplifies to —3r‘5z + 15r'52: — 127*52 = 0. 1 2. Calculate the curl V X 18. Solution: we have V X (r_3k) = V(7‘_3) X k+ T—3 - 0 since k is constant; this then equals —3r‘5r X k. The next thing we have to calculate is then V X (r_5 zr) = —5r_7r X (zr) + r‘EV X (21') . Now the first term in the last sum is essentially just the cross product of r with itself, and such products always vanish. The second term contains V X (21'), which equals V(z) Xr+zV xr, and it is easy to verify that V(z) = k, while V X r is zero. Thus V X (2r) 2 k X r. Plugging this back into the equation above gives V X (7“_5 2r) 2 7'_5k X r , so it follows that V x (r‘3k — 3r’5 2r) 2 —3r‘5r x k — 3(T_5k X r) = 0' //SIBS-dS of the vector field 18% across the surfaces of a sphere S’ of radius R centered at the origin. 3. Calculate the flux Solution: Let’s use the standard parametrization (¢, 0) |-—> R(cos 6 sin gb, sin 0 sin qfi, cos gt) for the sphere around the origin; here 0 g 0 g 27r and 0 S 45 5 71'. To calculate the integral we need to evaluate 1B in terms of this parametrization, and take the dot product of that quantity with the normal vector N = (Rsingb) r given by the cross product of the ¢ and 0 derivatives of the parametrization. [This is a standard calculation; see M 85 T §7 .6 p. 497 (formula 3b) for example] In other words, we need to calculate 271' 7r / le-rRsinqfidgbdQ. o 0 But now 183 - r = (7""3k -— 3r—52r) - r = 7'_3(z — 3z) = —227'_2 , 2 and r = R on the sphere of radius R, while 2 = Rcosd there; so our integral becomes 27r 7r 27r 7r /‘ / (—2Ram¢fl?£-R$n¢d¢d6==—2/i'/ fin¢a$¢d¢d0. 0 0 O 0 Now the theta integral is trivial, and we can integrate with respect to ¢> by using the double—angle formula sin 2¢ = 2 sinqficos ¢>. Thus the integral becomes ~2ar/ sin2¢ dqb = 7r cos2qfi [3; 0 but cos(27r) —— cos(0) = 1 — 1 = 0, so the total flux of B across the sphere is zero. 4. Calculate the gradient Vd), where ¢(r) = zr‘3. Solution: We have Vqfi = V(r—3) z + T’3V(z) = —3r“5zr + r'3k = B . Note that the equipotential surfaces for this function (i.e. the level surfaces 45(r) = C for some constant C) are lemniscates: in polar coordinates, z = 7" cos 45, so r c0s¢ =C<=>r2=Ccosq§. 450‘) = [More properly, the level surfaces are surfaces of revolution, obtained by rotating such a lemniscate, in plane polar coordinates, around the zaxis. The vector field B is perpendicular to this system of level surfaces] 73 Remarks: Physicists interpret a dipole field as resulting from the coalescence of two charged particles of opposite charge; but the story is a little tricky. Suppose we have a particle of charge +1 at the point (0, 0,17) and another particle of charge —1 at (O, 0, —p); then the potential energy of a particle moving in their vicinity is the sum of their potential energies, i.e. 1__;_ h+PM h-pm' It is not hard to see that as p —> 0, this potential energy goes to zero: the particles cancel each other out. However, we can do something more subtle: we can scale up the energy of the particles, as they approach one another, and consider instead the function 11 1 l; [ _ __ Ir + Pk! lr — Pk! to calculate this limit we can rewrite the expression above as ¢(r) := lim iii—>0 p 71.2 um 11" - pkl * I96 + pk] 1H0 19 and use l’Hopital’s rule. It’s straightforward to see that 2+1; 6‘ 1 _ =_ k-1 =__._ aplr+pkl 21r+pl 2(z+p) lr+pk|7 and similarly that Elr— kl—— Z—p ' 3p p Ir-Pkl’ (the minus sign is a little tricky). Now it’s easy to take the limit: we get _—22_2 r _ ¢< > r T , and I threw the factor of two away for simplicity. [Actually it might make better sense to think of the two charges as sitting at ::(p/2)k; that would also eliminate the extra factor of two.] In a more advanced course on these topics we could make sense of the assertion that Vzé) = 47r5(r), where 6 (r) is a subtle gadget called the Dirac delta-function, which equals Zero if 7" 79 0 and 00 at r = 0. [As it stands this is nonsensical; in fact it can be made rigorous, but what’s really going on is that 6 isn’t a ‘function’ in any classical sense, but is something more complicated Anyway, by reasoning as above one can show that the Laplace operator applied to ng (or equivalently, the divergence of B) is proportional to the derivative V2¢(r) = lim z 27,- 3250.) p—>0 2 p in the z—direction of the 6—function. [How you take the derivative of a function which equals 00 at 0 and is zero elsewhere will have to be left for a more advanced math course . . II Notation: Suppose el and e2 are vectors of length one in R3, which are perpendicular to each other: e1 - e2 2 0. Then (7', 0) r——> r(r, 6) = r(e1 cos 0 + e2 sin 0) (for 0 S 0 g 271‘, 0 S 7‘ S R) parametrizes a disk D of radius R centered at 0 in the plane spanned by the vectors 91,62. Note that this plane is perpendicular to the unit vector e1 >< e2, which I Will call n. If we keep 7" fixed, say with constant value R, then this parametrization restricts to define a parametrized circle 0 : r(0) +—> 1(9) 2 R(e1 c050 + 92 sin 0) in the e1,e2-plane: the circle C' 2 6D which bounds the disk. Let V(r) = w x r be the velocity vector field defined by a rotating rigid body with angular speed specified by the constant vector to = (w1,w2,w3). 1. Calculate the circulation per unit area of V around this disk: this is the ratio C(v,R)=;1%/Cv.ds. Then evaluate the limit limRno C(V, R). Solution: We have 27r /V-ds=/ (wxr(0))-r’(6)d6; C 0 but the integrand is a vector triple product, which can be more easily evaluated as w - (r(6) >< r’(0)) . Now r(0) x r’(9) = R(e1 cos 0 + e2 sin 6) X R(——e1 0050 + e2 cos 0) ; but e1 >< el = e2 >< e2 2 07 so this simplifies to 2 R2(e1 >< e2c0s20—e2 x elsin29) = R n. Our integrand thus equals w - (Rzn) 2 R200 - 117 which is constant; so the integral itself equals 27rR2w - n, and the circulation is the ratio 271'sz - n _ 2w n 7TR2 _ ’ which depends on the tilt direction 11 of the disk, but not on its radius. 2. Calculate the similar ratio 6(V,R)=$//D(VXV)-ds, and evaluate its limit as R —> 0. Solution: We know that V X (to X r) 2 2w [because our math teacher told us it would be on the test]. We can plug this into the definition of the surface integral to get f/D(VXV)-dS=/O2Tr/OR2w-(rrxr9)drd0; but the cross product of derivatives equals (e1 cos 0 + e2 sin 6) x r(—e1 sin 6 + e2 cos 0) , 5 which is almost exactly what we calculated in the first part of the problem: to be precise, it equals rn rather than Rzn. It thus remains to Calculate the ratio 1 271' R _ . 0 - WRz f0 f0 2w (7‘11) (11' d , but most of the stuff inside the integral is constant, so we can rewrite the integral as 2w-n 27‘ R Zw-n 1 2 «122/0 /Ordrd6— fiRZ ‘(27T-5R)—2w-n, which is What Stokes’ theorem says it should be. My work here is done . . . ...
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202fk04 - 110.202 Final exam answers [Some students...

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