Page 31 - Advanced engineering mathematics

1.1 Terminology and Separable Equations 11

of the element. The constants h and k are both uniquely tied to the particular element and
to each other by h =−(1/k)ln(2). Plutonium has one half-life, and radium has a different
half-life.
kt
Now look at the numbers A and k in the expression m(t) = Ae . k is tied to the element’s
half-life. The meaning of A is made clear by observing that
0
m(0) = Ae = A.
A is the mass that is present at some time designated for convenience as time zero (think of this
as starting the clock when the first measurement is made). A is called the initial mass, usually
denoted m 0 . Then
kt
m(t) = m 0 e .
It is sometimes convenient to write this expression in terms of the half-life h. Since
h =−(1/k)ln(2), then k =−(1/h)ln(2),so
kt
m(t) = m 0 e = m 0 e −ln(2)t/h . (1.1)
This expression is the basis for an important technique used to estimate the ages of certain ancient
artifacts. The Earth’s upper atmosphere is bombarded by high-energy cosmic rays, producing
large numbers of neutrons which collide with nitrogen, converting some of it into radioactive
carbon-14, or 14 C. This has a half-life h = 5,730 years. Over the geologically short time in
14
which life has evolved on Earth, the ratio of C to regular carbon in the atmosphere has remained
14
approximately constant. This means that the rate at which a plant or animal ingests C is about
the same now as in the past. When a living organism dies, it ceases its intake of 14 C, which
14
then begins to decay. By measuring the ratio of C to carbon in an artifact, we can estimate the
amount of this decay and hence the time it took, giving an estimate of the last time the organism
lived. This method of estimating the age of an artifact is called carbon dating. Since an artifact
may have been contaminated by exposure to other living organisms, this is a sensitive process.
However, when applied rigorously and combined with other tests and information, carbon dating
has proved a valuable tool in historical and archeological studies.
If we put h = 5730 into equation (1.1) with m 0 = 1, we get
m(t) = e −ln(2)t/5730 ≈ e −0.000120968t .

As a specific example, suppose we have a piece of fossilized wood. Measurements show that the
14
ratio of C to carbon is .37 of the current ratio. To calibrate our clock, say the wood died at time
zero. If T is the time it would take for one gram of the radioactive carbon to decay to .37 of one
gram, then T satisfies the equation

0.37 = e −0.000120968T
from which we obtain
ln(0.37)
T =− ≈ 8,219
0.000120968
years. This is approximately the age of the wood.

EXAMPLE 1.7 Draining a Container
Suppose we have a container or tank that is at least partially filled with a fluid. The container
is drained through an opening. How long will it take the container to empty? This is a simple
enough problem for something like a soda can, but it is not so easy with a large storage tank
(such as the gasoline tank at a service station).

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October 14, 2010 14:9 THM/NEIL Page-11 27410_01_ch01_p01-42