Page 190 - Calculus Demystified

Transcendental Functions
CHAPTER 6
The second piece of information yields 177
K 100
3 = R(100) = 5 · (e ) .
We conclude that
3
K 100
(e ) =
5
or
1/100
3
e K = .
5
Thus the formula for the amount of isotope present at time t is
t/100
3
R(t) = 5 · .
5
Thus we have complete information about the function R, and we can answer
the original question.
There will be 1 gram of material present when

t/100
3
1 = R(t) = 5 ·
5
or
t/100
1 3
= .
5 5
We solve for t by taking the natural logarithm of both sides:
! "
t/100
3 t
ln(1/5) = ln = · ln(3/5).
5 100
We conclude that there is 1 gram of radioactive material remaining when
ln(1/5)
t = 100 · ≈ 315.066.
ln(3/5)
So at time t = 315.066, or after 215.066 more years, there will be 1 gram of
the isotope remaining.

You Try It: Our analysis of exponential growth and decay is derived from the
hypothesis that the rate of growth is proportional to the amount of matter present.
Suppose instead that we are studying a system in which the rate of decay is propor-
tional to the square of the amount of matter. Let M(t) denote the amount of matter
at time t. Then our physical law is expressed as

dM 2
= C · M .
dt