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Homework 2
AMATH 383, Autumn 2009
Due: Monday, October 26, after class
1. (6 points) Age of Cave Painting: The caves near Lascaux, France, contain multiple wall paintings
of animals from ancient times. To determine the age charcoal found in the caves have been
investigated using the C14 dating method. The charcoal gave an average count of
R
measured
= 0
.
97
disintegrations
min gram
of C14 in a sample. Fresh wood measurements today give
R
fresh
= 6
.
68
disintegrations
min gram
.
Estimate the age of the charcoal and hence the possible date of the wall paintings.
Solution:
C14 concentration
y
(
t
) =
y
0
e

λt
with
λ
= 1
.
245
×
10

4
1
year
R
(
t
) =

d
y
d
t
=
λy
0
e

λt
R
(0) =
λy
0
!
= 6
.
68
R
(
T
) =
λy
0
e

λT
!
= 0
.
97
⇒
T
=

1
λ
ln
R
(
T
)
R
(0)
= 15498years
2. (1+4+10+10+5+5+1 points) Reduced Thorium Chain: When restricted to the slowest compo
nents the Thorium chain reduces to
Pu244
τ
1
→
U236
τ
2
→
Th232
τ
3
→
Pb208
with halflifes
τ
1
,
2
,
3
around 10
8
years. In order to investigate this chain analytically we make the
simplifying assumptions for the decay constants
λ
i
= ln2
/τ
i
λ
3
=
1
10
λ
1
and
λ
2
= 10
λ
1
(a) Which decay is the fastest and which one the slowest in this chain?
(b) Explain why the system
y
0
1
(
t
) =

λ
1
y
1
(
t
)
y
0
2
(
t
) =

λ
2
y
2
(
t
) +
λ
1
y
1
(
t
)
y
0
3
(
t
) =

λ
3
y
3
(
t
) +
λ
2
y
2
(
t
)
y
0
4
(
t
) = +
λ
3
y
3
(
t
)
describes the above chain. What elements are the function
y
1
,
2
,
3
,
4
refering to?
1
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View Full DocumentHomework, AMATH 383, Autumn 2009
(c) Write the system in matrix form. The general solution of the system is a superposition of
solutions of the form
y
(
t
) =
v
exp(
αt
)
with
α
∈
R
,
v
∈
R
4
and
y
(
t
) = (
y
1
(
t
)
,y
2
(
t
)
,y
3
(
t
)
,y
4
(
t
))
T
. What are the possible values for
α
and for the vectors
v
?
(d) We choose as initial condition the situation where only Pu244 is given (891 units, say) and
no other element, that is
(
y
1
(0)
,y
2
(0)
,y
3
(0)
,y
4
(0)) = (891
,
0
,
0
,
0)
.
Derive the following solution for initial value problem
y
1
(
t
) = 891exp(

λ
1
t
)
y
2
(
t
) = 99exp(

λ
1
t
)

99exp(

10
λ
1
t
)
y
3
(
t
) =

1100exp(

λ
1
t
) + 100exp(

10
λ
1
t
) + 1000exp(

1
10
λ
1
t
)
y
4
(
t
) = 110exp(
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 Spring '08
 WAlker
 Math, Calculus

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