Problem 1 (TM 6-4) The element of distance along the surface is
dx 2 + dy 2 + dz 2
dS =
(1)
In cylindrical coordinates (x,y,z) are related to (ρ,φ,z) by
x = ρ cos φ 

y = ρ sin φ 


z= z

(2)
dx = − ρ sin φ dφ 

dy = ρ cos φ dφ 


dz = dz

(3)
from which
Substituting (3) into (1) and integrating along the entire path, we find
S=
2
∫
ρ 2 dφ 2 + dz 2 =
∫
φ1
1
where z& =
φ2
dz
. If S is to be minimum, f ≡
dφ
ρ 2 + z& 2 dφ
ρ 2 + z& 2 must satisfy the Euler equation:
∂f
∂ ∂f
−
= 0
∂ z ∂ φ ∂ z&
Since
(4)
(5)
∂f
= 0 , the Euler equation becomes
∂z
∂
∂φ
z&
ρ + z& 2
2
= 0
(6)
from which
z&
ρ + z& 2
2
or,
= constant ≡ C
(7)
z& =
C2
ρ
1 − C2
(8)
Since ρ is constant, (8) means
dz
= constant
dφ
and for any point along the path, z and φ change at the same rate. The curve described
by this condition is a helix.
Problem 2 [TM 6-6]. If we use coordinates with the same orientation as in Example 6.2 and if we place the
minimum point of the cycloid at (2a,0) the parametric equations are
x = a ( 1 + cos θ ) 


y = a ( θ + sin θ ) 
(1)
Since the particle starts from rest at the point ( x1 , y1 ) , the velocity at any elevation x is
[cf. Eq. 6.19]
v=
2 g ( x − x1 )
(2)
Then, the time required to reach the point ( x2 , y2 ) is [cf. Eq. 6.20]
t=
x2
∫
x1
 1 + y′ 2 


 2 g ( x − x1 ) 
12
(3)
dx
Using (1) and the derivatives obtained therefrom, (3) can be written as
t=
a
g
θ1
 1 + cos θ 
∫  cosθ − cosθ 1 
θ 2= 0
12
dθ
2
Now, using the trigonometric identity, 1 + cos θ = 2 cos θ 2 , we have
(4)
a
g
t=
a
g
=
cos
θ1
∫
cos 2
0
θ
dθ
2
θ
θ
− cos 2 1
2
2
θ
cos dθ
2
θ
θ
sin 2 1 − sin 2
2
2
θ1
∫
0
(5)
Making the change of variable, z = sin θ 2 , the expression for t becomes
a
t= 2
g
sin
∫
θ
1
2
0
dz
θ
sin 2 1 − z 2
2
(6)
The integral is now in standard form:
∫
dx
 x
= sin − 1  
 a
a2 = x 2
(7)
Evaluating, we find
t= π
a
g
(8)
Thus, the time of transit from ( x1 , y1 ) to the minimum point does not depend on the
position of the starting point.
Problem 3 [TM 6-11} The constraint condition can be found from the relation ds = Rdθ (see the diagram),
where ds is the differential arc length of the path:
(
ds = dx 2 + dy 2
)
12
= Rdθ
(1)
2
which, using y = ax , yields
1 + 4a 2 x 2 dx = Rdθ
If we want the equation of constraint in other than a differential form, (2) can be
integrated to yield
(2)
A + Rθ =
x
2
(
1
ln 2 ax +
4a
4 ax 2 + 1 +
4 a2 x 2 + 1
)
(3)
where A is a constant obtained from the initial conditions. The radius of curvature of a
curve y ( x ) is given by
R=
(1+
y '2
)
3/ 2
| y '' |
2
For parabola, y = ax ,
R=
(
1
2
1 + ( 2 xa )
2a
)
3/ 2
≥
1
2a
The minimum radius is at x = 0 . The condition for the disk to roll with one and only
one point of contact with the parabola is R < r0 ; that is,
1
2a
R<
(4)
Bonus (TM 6.8)
To find the extremum of the following integral
J=
∫ f ( y, x ) dx
we know that we must have from Euler’s equation
∂f
= 0
∂y
This implies that we also have
∂J
=
∂y
∂f
∫ ∂ y dx =
0
giving us a modified form of Euler’s equation. This may be extended to several variables
and to include the imposition of auxiliary conditions similar to the derivation in Sections
6.5 and 6.6. The result is
∂J
+
∂ yi
∑
λ
j
j
( x)
∂ gj
∂ yi
= 0
when there are constraint equations of the form
g j ( yi , x ) = 0
a) The volume of a parallelepiped with sides of lengths a1 , b1 , c1 is given by
V = a1b1c1
(1)
We wish to maximize such a volume under the condition that the parallelepiped is
circumscribed by a sphere of radius R; that is,
a12 + b12 + c12 = 4 R 2
(2)
We consider a1 , b1 , c1 as variables and V is the function that we want to maximize; (2) is
the constraint condition:
g { a1 , b1 , c1 } = 0
(3)

∂g
∂V
+ λ
= 0
∂ a1
∂ a1


∂g
∂V

+ λ
= 0
∂ b1
∂ b1


∂g
∂V
+ λ
= 0

∂ c1
∂ c1
(4)
b1c1 + 2λ a1 = 0 

a1c1 + 2λ b1 = 0 


a1b1 + 2λ c1 = 0 
(5)
Then, the equations for the solution are
from which we obtain
Together with (2), these equations yield
a1 = b1 = c1 =
2
R
3
Thus, the inscribed parallelepiped is a cube with side
(6)
2
R.
3
b) In the same way, if the parallelepiped is now circumscribed by an ellipsoid with
semiaxes a, b, c, the constraint condition is given by
a12
b12
c12
=
=
= 1
4 a 2 4b 2 4 z 2
(7)
where a1 , b1 , c1 are the lengths of the sides of the parallelepiped. Combining (7) with (1)
and (4) gives
a12 b12 c12
=
=
a2 b 2 c 2
(8)
Then,
a1 = a
2
2
2
, b1 = b
, c1 = c
3
3
3
(9)