21a Final Exam Answers (1/24/96)

1)
a) B W(r(5)) is wind velocity, so meters/sec;
b) A The integral measures arc length (speed x time), so meters;
c) C (grad g).(r'(17)) = g_x x_t + g_y y_t + g_z z_t
     Each term is (deg/meter)(meter/sec), so deg/sec;
d) E Flux of a velocity vector field is volume/time,
     so meters^3/sec;
e) F div(grad g) = g_xx + g_yy + g_zz, each term is deg/meter^2;

2)
a) C v // w could be true or false;
b) F since v.w=2=|v||w|cos(theta), we must have cos(theta)>0;
c) T since |v||w|cos(theta)=2 and cos(theta)<=1;
d) F Let S be a level surface of f(x,y,z). Then grad f // grad g,
     so grad g is perpendicular to S and, therefore, to C;
e) F If w tangent to C, since grad g is perpendicular to the tangent
     to C, we would have (grad g).w=0 instead of 2;

3)
a) C The flux of (grad g) through a surface may or may not be 0;
b) C The flux of (curl w) may or may not be 0;
c) T G = (grad g), a gradient field is conservative;
d) T Again, since G is conservative, curl G = 0 everywhere;
e) T div(curl F)=0 for any vector field F;
f) C The divergence of (grad g) may or may not be 0.

4)
a) POSITIVE (xi always points out of the box)
   ZERO (flux in of yi -- not yj! --- cancels flux out)
   ZERO (flux in cancels flux out)
   NEGATIVE (= flux of yi - flux of xi = 0 - positive)
b) Using divergence theorem, div(xi)=1, so the flux out of S
   is the volume of S and the flux out of B is the volume of B.
   That is FLUX OUT OF B > FLUX OUT OF S

5) 
a) NEGATIVE, NONCONSERVATIVE
b) ZERO, CONSERVATIVE
c) NEGATIVE, NONCONSERVATIVE
d) ZERO, NONCONSERVATIVE

6) Use d(x,y,z)=x^2+y^2+z^2 (the square of the distance to the
origin). We want to
	MIN d(x,y,z)
	subject to g(x,y,z)=z-xy=1

   Use Lagrange multipliers:

grad g = 0 => 1=0 (never happens)

grad f = (q) (grad g) => 2x = -qy
                         2y = -qx
                         2z = q
                         z = xy+1

So 4xy = -q^2xy => xy=0 (can't have q^2=-4) => x=0 or y=0
Back to the original equations, we find that (0,0,1) is the
only solution. For geometric reasons, there must be a min, so
(0,0,1) is the point we want.

7) F(x,y,z)=(xi+yj+zk)/(x^2+y^2+z^2)^(3/2); compute div F = 0
(except at the origin).
We can use a sphere T inside the ellipsoid S; by the div theorem, the
flux out of S = flux out of T.
But since the magnitude of F is constant around the sphere T and F is
always normal to T, we have (F.n)=|F|=constant, so

FLUX OUT OF T = |F|(area of sphere) = (1/radius^2)(4Pi radius^2)

FLUX OUT OF S = FLUX OUT OF T = 4Pi.

8)
a) III, I, II, VI
b) III

9)
Volume = Int (0 to 1) (0 to 2Pi) (sqrt(1-r^2) to sqrt(9-r^2)-1)
         r dz dtheta dr +
         Int (1 to 2sqrt(2)) (0 to 2Pi) (0 to sqrt(9-r^2)-1)
         r dz dtheta dr

or

Volume = Int (0 to 1) (0 to 2Pi) (sqrt(1-z^2) to sqrt(9-(z+1)^2))
         r dr dtheta dz +
         Int (1 to 2) (0 to 2Pi) (0 to sqrt(9-(z+1)^2))
         r dr dtheta dz

10) A (L1<L2<L3)
Reason: Green's Theorem says that the circulation of F around a
counterclockwise closed curve C is the double integral of the curl F.
In this case, curl F = x^2+y^2-1 is negative everywhere between C1
and C3. Then:

Circulation around C1-C2 = Double Integral of curl in between < 0
=> L1-L2 < 0 => L1 < L2

Similarly, L2 < L3 using the curve C2-C3.

11)
a) Concentric spheres centered at the origin; indeed, f(x,y,z) fixed
   means g(rho) fixed, which means rho is fixed.
b) grad f is normal to the level surface, i.e., to one of those 
   spheres, therefore parallel to i+2j+2k. An unit vector is
   u = 1/3i + 2/3j + 2/3k
c) Using the chain rule, F = grad f = (dg/drho).(xi+yj+zk)/(rho),
   so |F| = |dg/drho|. From the graph, estimate dg/drho at
   rho = sqrt(1^2+2^2+2^2)=3 as g'(3)=1/3.
   So |F(1,2,2)|=1/3 (approximately).
d) Using b and c (direction and magnitude), we estimate
   F(1,2,2) = 1/9 i + 2/9 j + 2/9 k
e) Similarly, |F(3,0,0)|=g'(3)=1/3 or so and its direction should be
   i, so we estimate
   F(3,0,0)= 1/3 i
f) |F(P)|=|F(Q)| because both are g'(rho(P))=g'(rho(Q)); the directions
   of F are the directions of r(P) and r(Q).

12)
a) Rate = Flux out of drain = Int (F.n) dsigma; but n=-k and z=0 on
   the drain, so F.n = 1 (we assume we want flow DOWN the drain).
   Then Flux = Area of Drain = Pi (cm^3/sec);
b) Compute div F = 0;
c) Using the divergence theorem for the solid region between the drain
   and the hemisphere, we find out that both fluxes must be the same,
   so FLUX OUT OF HEMISPHERE = Pi
d) Take x=sin t, y=cos t, z=0, then G = 1/2 (cos t i - sin t j + k)
   and dr/dt = cos t i - sin t j, so G.dr/dt = 1/2; integrate from
   t=0 to t=2Pi => CIRCULATION = Pi
e) Compute curl G = F;
f) Use Stokes' to show that Circulation of G along C is the
   flux of the curl G=F across drain.