Solutions
Problems for Chapter 2 2.1 We obtain directly dr / dz = f(1 + f2 - rr) / (1 + f2)3/2. The equation of the curve is 1 + f2 - rr = 0, from which the result follows. Therefore
r(z) = avl + f(z)2.
Setting f(z) = sinh(¢(z)), we obtain
r(z) = acosh(¢(z));
i.e.,
f
= a¢(z) sinh(¢(z)),
and therefore a¢(z) = 1 and the solution r(z) = acosh((z - zo)/a). This is a particular case of the use of conserved quantities discussed in Chapter 3. 2.2 Lagrange Multipliers We must minimize
(7.68) with the constraints
z(O) = zo,
z(a) =
Zl,
and
i. VI B
+ z(x)2dx = L.
One can transform the problem into min
V=
loa (p,gz + >')Vl + z(x)2dx,
(7.69)
with z(O) = zo, z(a) = Zl. The conserved quantity
(p,gz + >.) = C + z(x)2
VI
(7.70)
168
Solutions
yields z
= sinh cf>(x) , i.e., Z
j.£gZ
+ A = Ccoshcf> with C¢ = j.£g.
The solution is
A + -C cosh (J-Lg = --(x - xo) ) . J-Lg
(7.71)
C
J-Lg
The constants xo, C, and A are fixed by the conditions z(O) = zo, and Joa JI + z(x)2dx = L.
z(a) =
Zl,
2.3 Brachistochrone Energy conservation gives
-I (dS)2 + g(z dt
2
a) = O.
(7.72)
We want to minimize
T-
l
b
I
(
+ Z2
2g(a-z)
a
)
dx
(7.73)
with the constraints z(a) = a, z(b) = {3. The Lagrange function I:- = ylr'cI-+---'-,z2"-j""-2-g-;-(a---z--:-) does not depend on x, and therefore there is conservation of
(7.74) where we introduce a positive constant R. Setting the parametric form Z-Zo
Rcoscf>
= ---, 2
x - Xo
=
z=
tan(cf>j2), we obtain
R(cf> + sincf» 2
'
(7.75)
which is the equation of a cycloid.
2.4 Win a Slalom 1. With this definition of the variable x, we have (z - zo) = (x - xo) sina and the potential energy is V = mg(z - zo) = -mgxsina. 2. The total energy is E = ~m(j;2+1?)-mgxsina. Since energy is conserved, and since it is taken to be zero initially, we have j;2 + iP = 2gx sin a. 3. Therefore dt 2 = (dx 2 + dy2)j(2gx sin a). 4. The total time to get from 0 to A is therefore T
=
fAo dt = v'2gsina I fA JI +x(y')2 dx 0
5. Using the Lagrange-Euler equation, we obtain
o=
-
d
y'
"'t=::;=====;=~
dx ylx(1 + (y')2)
Solutions
169
6. We deduce
where C is a constant. However,
y'
y'x(l + (y')2)
dy y'x(dx 2 + dy2)
if =C xyf2g sin 0: ' (7.76)
and therefore if = Kx with K = Cyf2g sin 0:. 7. The parametric form x(B) = (1- cos2B)/2C 2 = sin 2 B/C 2, y(B) = (2Bsin2B)/2C 2 satisfies the equation (y')2 = C 2x/(1- C 2x); i.e., (dy/dB)2 = (dx/dB)2tan 2 B. From if/x = K, we obtain (dy/dB)(dB/dt)/x = K; i.e., dB/dt = K/2 and B = Kt/2 since, for t = 0, B = O. 8. The curve is a portion of a cycloid. We have dy / dx = tan B and therefore y' » 1 for B rv 7r /2. The trajectory starts vertically (dy / dx = 0 for B = 0) and becomes horizontal if y(A) » x(A), as shown in Figure 7.1.
o
y
A
x
Fig. 7.1. Optimal trajectory from 0 to A.
9. Since point A is fixed, the velocity VA at A is fixed by energy conservation. It is the maximum velocity of the skier. Therefore, the time to get horizontally from y(A) to y(O) is larger than the time (y(A) - y(O))/VA it would take to cover this distance at the maximum velocity. On the other hand, one must start vertically in order to acquire the maximum velocity as quickly as possible. The ideal trajectory comes from an optimization between these two effects.
2.5 Strategy of a Regatta 1. We have by definition x = Vx = V cos B, i = Vz = v sin B, and therefore z' = dz/dx = tanB. 2. We have Vx = vcosB = w/h. This velocity is maximum when h(z') is minimum; i.e., for z' = 1, namely B = 7r/4. We then have Vx = w/2. In fact, it is sufficient to multiply h by a constant to be in the appropriate situation for a given sailboat for which vx,max = )..w. 3. We have dt = dx/v x = h'(z') dx/w(z), and therefore
-l
T-
L
o
dx
h'(z') (). w z
(7.77)
170
Solutions
4. Setting
