5.6 Differential Equations:
[ Day 2 ]
Growth and Decay
Def.
Radioactive Decay - the process in which
a substance disintegrates by
conversion of its mass into radiation.
an element whose atoms go through this
process spontaneously is called radioactive
Key! the rate of decay is proportional to its mass, y
dy
= ky
dt
...
y = Ce
.
kt
.
Note:
Radioactive Decay - is measured in terms
of half-life - the time required for
half of the atoms in a sample of
radioactive material to decay.
p 363 Short List
Uranium (238U) 4,510,000,000 years
Plutonium (239Pu) 24,360 years
Carbon (14C) 5730 years
Radium (226Ra) 1620 years
Einsteinium (254Es) 270 days
Nobelium (257No) 23 seconds!!
Historical Note:
Application:
Carbon Dating
- determining the age of fossils
(Willard Libby, UCLA, 1950)
1
2)
Radioactive Decay
Living tissue contains two isotopes of carbon, one radioactive
and the other stable. (The ratio of the two being constant). But the radioactive
one decays with a half-life of about 5500 years.
a)
k in y = Cek t.
.
Find
2
Then use this
b)
k to answer the following question.
Determine the age of a fossil in which the radioactive isotope has decayed to
20% of its original amount. (The percentage is determined by comparing the
present ratio of isotopes in the fossil to the known ratio in living tissue).
3
Newton's Law of Cooling
- "the rate of change in
the temperature of an object is proportional to the difference
between the object's temperature and the temperature of the
surrounding medium"
dT k ( T
=
object - Tsurrounding)
medium
dt
.
4
3) (p365 ­ Example 6)
Newton's Law of Cooling
applies the separation of variables technique!
Let T represent the temperature (in F) of an object in a room whose temperature is
kept at a constant 60 . If the object cools from 100 to 90 in 10 minutes, how
much longer will it take for its temperature to decrease to 80 ?
"how much longer?"
(after 10 minutes)
...
it will require about 14.09 more minutes for the
object to cool to a temperature of 80 F.
5
5.6 Exercise 15 (page 366)
6
Assignment
p.367 #33, 35, 39, 41, 42, 43­49 odd,
#53, 57­69 odd, 74
#42.
­ ln 2
k = 5730
t 15,682 years
#74.
k = ln 104
142
T 379.2 F
7