Assignment 10 Solutions

6.18
Problem 6.18 **   If we use polar coordinates (r,f) and write the path in the form f = f(r), then the path length takes the form L = f dr with f = f(f, f, r) = {1 + r2 f2}. Because f/f = 0, the Euler-Lagrange equation is

f

f
= r2 f




1 + r2 f2
= const    whence     f = K

r


r2 - K2
 .
This can be integrated (make the subtitution K/r = cosu) to give

f- fo= arccos(K/r)     or     r = K

cos(f-fo)
.
This is the equation of a straight line perpendicular to the direction f = fo, a distance K from the origin. [To see this, note that r cos(f-fo) is the component of r in the direction f = fo; that this equals the constant K says that r lies on the line indicated.]
6.19
Problem 6.19 **   The area of the surface of revolution is A = 2py  ds = 2 py1y2 y {1 + x2} dy, which we can write as A = 2 pf dy where f = y{1 + x2}. Because f/x = 0, the Euler-Lagrange equation reads

f

x
= yx




1 + x2
= yo    whence     x = yo




y2 - yo2
 ,
where yo is a constant. Using the substitution y/yo = coshu, you can integrate this to give x - xo = yoarccosh(y/yo) (where xo is another constant) or y = yocosh[(x - xo)/yo].
6.27
Problem 6.27 **   The element of path length is ds = {dx2 + dy2 + dz2} = {x2 + y2 + z2}  du. Thus the total path length is L = f  du where f = {x2 + y2 + z2}. There are three Euler-Lagrange equations, which involve the following six derivatives:

f

x
= f

y
= f

z
= 0,
and

f

x
= x




x2 + y2 + z2
,   f

y
= y




x2 + y2 + z2
,   f

z
= z




x2 + y2 + z2
.
Since the first three derivatives are zero, the Euler-Lagrange equations imply simply that each of the last three is constant. This means that the ratios, x:y:z are constant, which implies in turn that as we move along the curve the ratios dx:dy:dz are constant. In other words, the curve is a straight line.

4. So the "action" that needs to be extremized is the total amount of money spent on the trip,
I =
2

1 
t (dD/dt)2 dt   and the effective Lagrangian is   L = t(dD/dt)2
Since D does not occur undifferentiated we immediately have the conservation law
pD = L/

D
 
= t

D
 
= const = k  so   dD/dt = k/t   and   D = k ln(k/t)
where the additive constant in the last integration was incorporated into k' so the ln can have a dimensionless argument. k and k' would be determined by the initial and final conditions (where you start, end, and at what times).

5. The total energy is U = (d2Y/dx2)2 dx. If we let Y = y + ah (Eq 6.8), differentiate with respect to a, and evaluate at a = 0 we get for the extermum condition
dU = 2

d2y

dx2


d2h

dx2

dx = 0
We integrate by parts once (assuming everything necessary is fixed at then endpoints):
dU = -
d

dx

d2y

dx2


d2h

dx2

  and again   = +
d2

dx2

d2y

dx2

hdx = 0
Because h is arbitrary we conclude in the usual way d4y/dx4 = 0. This is solved by
y = ax3 + bx2 + cx + d .
The boundary conditions are y(-1) = -1, y(-1) = 0 = y(1), y(1) = 1, which tell us a = c = 0 and b = 1/2, c = -1/2, thus y(x) = 1/2(x2 - 1) -- a parabola.