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By — McGraw-Hill Professional
Updated on Aug 31, 2011

Another natural phenomenon which fits into exponential growth and decay is radioactive 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 1

Five grams of a certain radioactive isotope decay to three grams in 100 years. After how many more years will there be just one gram?

Solution 1

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

R ( t ) = 5 · e Kt = 5 · ( e K ) t .

The second piece of information yields

3 = R (100) = 5 · ( e K ) 100 .

We conclude that

or

Thus the formula for the amount of isotope present at time t is

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

or

We solve for t by taking the natural logarithm of both sides:

We conclude that there is 1 gram of radioactive material remaining when

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 proportional 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

Here C is a (negative) constant of proportionality. We apply the method of “separation of variables” described earlier in the section. Thus

so that

Evaluating the integrals, we find that

We have combined the constants from the two integrations. In summary,

For the problem to be realistic, we will require that C < 0 (so that M > 0 for large values of t ) and we see that the population decays like the reciprocal of a linear function when t becomes large.

Re-calculate Example 1 using this new law of exponential decay.

Find practice problems and solutions for these concepts at: Transcendental Functions Practice Test.

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