Page 189 - Calculus Demystified

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As a result, CHAPTER 6 Transcendental Functions
√ t
5
B(t) = 6000 · √ .
3
At time t =−2 (7:00 p.m.) the number of bacteria was therefore
√ −2
5 3
B(−2) = 6000 · √ = · 6000 = 3600.
3 5

You Try It: A petri dish has 5000 bacteria at 1:00 p.m. on a certain day and 8000
bacteria at 5:00 p.m. that same day. How many bacteria were there at noon?

6.5.3 RADIOACTIVE DECAY

Another natural phenomenon which fits into our theoretical framework is radio-
active decay. Radioactive material, such as C 14 (radioactive carbon), has a half life.
Saying that the half life of a material is h years means that if A grams of material is
present at time t then A/2 grams will be present at time t + h. In other words, half
of the material decays every h years. But this is another way of saying that the rate
at which the radioactive material vanishes is proportional to the amount present.
So equation (.) will apply to problems about radioactive decay.

EXAMPLE 6.32
Five gramsof a certain radioactive isotope decay to three gramsin 100
years. After how many more years will there be just one gram?
SOLUTION
First note that the answer is not “we lose two grams every hundred years
so ….” The rate of decay depends on the amount of material present. That is
the key.
Instead, we let R(t) denote the amount of radioactive material at time t.
Equation (.) guarantees that R has the form

R(t) = P · e Kt .
Letting t = 0 denote the time at which there are 5 grams of isotope, and
measuring time in years, we have
R(0) = 5 and R(100) = 3.
From the first piece of information we learn that

5 = P · e K·0 = P.
Hence P = 5 and
K t
R(t) = 5 · e Kt = 5 · (e ) .