On the Stick and Rope Problem - Draft 1
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On The Stick and Rope Problem
Iwan Pranoto
Institute of Technology Bandung
November 19, 2012
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Problem
Isoperimetric Problem in Halfplane?
Finding a smooth function x on I = [0, 1] satisfyingx(0) = x(1) = 0 and the area formed by the graph of x and theinterval I is maximum, with a constrain: the length of the graph ofx must be `, where ` is a positive number given but fixed.
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Stick and Rope
Figure: Free Ends, Left; Fixed Ends, Right
In this paper, we consider the right situation.
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Figure: Half a Bubble
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We assume x(t) ≥ 0 for all t. Therefore, the area A of the regionbounded by the graph of x and I is to be maximized
A[x ] =
∫ 1
0x(t) dt,
and at the same time the length L[x ] = `, where
L[x ] =
∫ 1
0
√1 + (x)2 dt. (1)
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Results
TheoremIf 1 < ` ≤ π
2 , the optimal solution of the problem is a segment of acircle whose center is on the vertical line t = 1
2 .
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Proof
Using Lagrange method, define a new form
A[x , x , λ] =
∫ 1
0
[x(t)− λ
(√1 + (x)2 − `
)]dt.
After substituting this into the Euler-Lagrange equation, we obtain
∂
∂x
[x(t)− λ
(√1 + (x)2 − `
)]
=d
dt
(∂
∂x
[x(t)− λ
(√1 + (x)2 − `
)]). (2)
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This leads to
1 =d
dt
(λx√
1 + (x)2
).
After integrating both sides with respect to t, we obtain
t =λx√
1 + (x)2+ C1
for some constant C1. Thus, x must satisfy
(t − C1)2 =(λx)2
1 + (x)2. (3)
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Since x vanishes only if t = C1, one obtains
1 + (x)2
x2=
λ2
(t − C1)2.
Thus,1
x2=λ2 − (t − C1)2
(t − C1)2,
or
x = ± (t − C1)√λ2 − (t − C1)2
.
Therefore, x must satisfy
x = ±√λ2 − (t − C1)2 + C2
for some constant C2, or
(x − C2)2 = λ2 − (t − C1)2. (4)
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This means, x must satisfy the equation of a circle whose center is(C1,C2)
(t − C1)2 + (x − C2)2 = λ2. (5)
After using the conditions x(0) = x(1) = 0, one obtains C1 = 12 .
Thus, the circle’s center must lie on the vertical line t = 12 .
One may compute the other constants, the center’s ordinate −|C2|and radius |λ|, geometrically.
�
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Special Cases
Figure: Left ` = π/2; Right ` = π/3
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How if ` > π/2?
We consider ` = 2 and a special kind of functions:
x(t) = a
(|t − 1
2|m −
(1
2
)m),
where m is any natural number greater than 1 and a is someconstant to be chosen, depending on m.
Figure: Table, Left; Numerical Results, Right
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How About A Rectangle and A Semicircle?
Figure: The Area = (1− π8 ) ≈ 0.60730091830 · · ·
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The End