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Past exam of the mathematics course of the University of Cambridge
/
2025
/
ib
/
Paper 2
/
13C
(Variational Principles)
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Course of the University of Cambridge
Mathematics course of the University of Cambridge
Past exam of the mathematics course of the University of Cambridge
2025
ib
Paper 2
Created
2026-09-24
Updated
2026-09-24
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Solution
13C
Solution
0
0
0
13C
Put
W
=
1
+
ρ
′2
+
ρ
2
ϕ
′2
. The
Euler–Lagrange equations
are
d
z
d
(
W
n
ρ
′
)
−
n
ρ
W
−
W
n
ρ
ϕ
′2
=
0
,
(1)
d
z
d
(
W
n
ρ
2
ϕ
′
)
−
n
ϕ
W
=
0.
(2)
When
n
=
n
(
ρ
)
, the helical
ansatz
ρ
=
R
,
ϕ
=
ϕ
0
+
ω
z
automatically satisfies the
second
equation
. The
first
becomes
n
(
R
)
R
ω
2
+
(
1
+
R
2
ω
2
)
n
′
(
R
)
=
0
,
(3)
so
a
=
1
.
The endpoints have the same cylindrical
angle
exactly when
ω
L
=
2
πk
for some
integer
k
. Thus
a
positive
L
with
L
=
2
πk
/
ω
(
R
)
admits the two special solutions of opposite handedness, using
ω
and
−
ω
.
Ancestors
(9)
Paper 2
ib
2025
Past exam of the mathematics course of the University of Cambridge
Mathematics course of the University of Cambridge
Course of the University of Cambridge
University of Cambridge
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