Suppose you have a bacteria culture and you feed it all it needs and there is no
restriction on its growth due to crowding for example. Then in this case, the rate of
growth is proportional to the amount of bacteria present. This is because the more you
have, the more bacteria there are to divide and make new bacteria. Consider equally
spaced intervals of time such that n of them equal one unit of time where n is
large. The unit might be years, days, etc. Also let A_{k} denote the amount of
whatever is growing at the end of the k^{th} increment of time. In exponential
growth
Ak+1 − Ak ≈ rAk (1∕n ),A0 = P
where r is a proportionality constant called the growth rate and the above difference
equation should give a good description of the amount provided n is large enough. The
symbol ≈ indicates approximately equal. Now this is easy to solve.
( )
Ak+1 ≈ 1 + r- Ak,A0 = P
n
and you look for A_{k} = α^{k} and find α. This is easily seen to be α =
( r)
1+ n
and
so
( )
A ≈ P 1 + r-k
k n
Now let A
(t)
denote the amount at time t. What is it? There are tn time intervals so k
goes up to tn and you get
( )
( r-)nt rt nt
A (t) ≈ P 1+ n ≈ P 1 + nt
the approximation for A
(t)
getting better as n →∞. As will be shown later, this
becomes very close to
A(t) = P exp(rt)
when n is large which is the formula for exponential growth. As an example,
consider P = 10 and r = .01. Then the following is the graph of the function
y = 10e^{.01t}.
PICT
When the rate of change is negative, the process is called exponential decay. This is
the process which governs radioactive substances. It is the same formula which results,
only this time it is of the form
A (t) = P exp(− rt)
where r > 0. Consider the same example only this time consider y = 10e^{−.01t}.
PICT
Exercise 2.7.3Carbon 14 has a half life of 5730 years. This means that if youstart with a given amount of it and wait 5730 years, there will be half as much left.Carbon 14 is assumed to be constantly created by cosmic rays hitting the atmosphereso that the proportion of carbon in a living organism is the same now as it was along time ago. This is of course an assumption and there is evidence it is not truebut this does not concern us here. When the living thing dies, it quits replenishingthe carbon 14 and so that which it has decays according to the above half life. Bymeasuring the amount in the remains of the dead thing and comparing with whatit had when it was alive, one can determine an estimate for how long it has beendead. Suppose then you measure the amount of carbon 14 in some dead wood andfind there is 1/3 the amount there would have been when it was alive. How long agodid the tree from which the wood came die?
Let A
(t)
be the amount of carbon 14 in the sample and let A_{0} be the amount when it
died. Then
A(t) = A0 exp(− rt)
By assumption
.33A0 = A0 exp(− rt)
and cancelling the A_{0} one can solve for t as follows.
ln(.33) = − rt
If I knew what r was, I could then solve for t. The half life is 5730 and so
.5 = exp (− r5730)
and so
ln(.5) = − r(5730)
from properties of ln described above, −ln
(1∕2)
= ln
( −1)
(1∕2)
= ln
(2)
. Therefore,
ln2
r = ---- = 1.2097× 10−4
5730
To get this number, I just used the computer. As mentioned above ln has been
tabulated. Therefore, in the problem of interest,
ln (.33) = − (1.2097 × 10−4)t
and so
---ln(.33)-----
t = − 1.2097 × 10−4 = 9164.8 years.
So how did they find the half life of carbon 14? Did Noah have a sample of
recently dead wood in the ark and make some measurements which he recorded
in the Book of the Law of Noah which were then compared to measurements
made in the twentieth century using chronology determined by Bishop Usher to
determine that exactly 5730 years had passed? Actually, this is not the way it
was done. The half life was also not established by the decree of omniscient
scientists.
Example 2.7.4Find the half lifeof a radioactive substance if after 5 years thereis.999395 of the original amount present.
This says
.999395 = exp (− 5r)
and so
ln(.999395) −4
r = − 5 = 1.21036617× 10
Now to find the half life T, you need to solve the equation
1 = e−(1.21036617×10−4)T
2
Thus ln
(.5)
= −
( −4)
1.21036617× 10
T and so
ln(.5)
T = ----------------−4-= 5726
− (1.21036617 × 10 )
Using the known properties of exponential decay, you can compute the half life without
waiting for over 5000 years.