This preview shows pages 1–6. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 timeconsuming 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' //SIBSdS 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
/ lerRsinqﬁdgbdQ.
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 hpm' 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
_ =_ k1 =__._
aplr+pkl 21r+pl 2(z+p) lr+pk7
and similarly that
Elr— kl—— Z—p '
3p p IrPkl’ (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 deltafunction, 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,e2plane: 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
/Vds=/ (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
2wn 27‘ R Zwn 1 2
«122/0 /Ordrd6— ﬁRZ ‘(27T5R)—2wn, which is What Stokes’ theorem says it should be. My work here is done . . . ...
View Full
Document
 Fall '04
 Staff
 Calculus

Click to edit the document details