### 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 = ò f dr with f = f(f, f¢, r) = Ö{1 + r2 f¢2}. Because f/f = 0, the Euler-Lagrange equation is

f

f¢
= r2 f¢

 Ö 1 + r2 f¢2
= const    whence     f¢ = K

 r Ö 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 = ò2py  ds = 2 pòy1y2 y Ö{1 + x¢2} dy, which we can write as A = 2 pòf dy where f = yÖ{1 + x¢2}. Because f/x = 0, the Euler-Lagrange equation reads

f

x¢
= yx¢

 Ö 1 + x¢2
= yo    whence     x¢ = yo

 Ö 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 = Ö{dx2 + dy2 + dz2} = Ö{x¢2 + y¢2 + z¢2}  du. Thus the total path length is L = òf  du where f = Ö{x¢2 + y¢2 + z¢2}. There are three Euler-Lagrange equations, which involve the following six derivatives:

 ¶f ¶x = ¶f ¶y = ¶f ¶z = 0,
and

f

x¢
= x¢

 Ö x¢2 + y¢2 + z¢2
,   f

y¢
= y¢

 Ö x¢2 + y¢2 + z¢2
,   f

z¢
= z¢

 Ö x¢2 + y¢2 + z¢2
.
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 = óõ 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 = ò(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 óõ æè d2y dx2 öø æè d2h dx2 öø dx = 0
We integrate by parts once (assuming everything necessary is fixed at then endpoints):
 dU = - óõ d dx æè d2y dx2 öø æè d2h dx2 öø and again   = + óõ d2 dx2 æè d2y dx2 öø 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.