CK-12-Calculus

http://www.ck12.org Chapter 6. Transcendental Functions

CancelingCfrom both sides,

1

2 =e

kt.

Solving fort,which is the half-life, by taking the natural logarithm on both sides,

ln^12 =lnekt

−ln 2=kt.

Solving fort,and denoting it with new notationt 1 / 2 for half-life (a standard notation in physics),

t 1 / 2 =−ln 2k =−^0 .k^693

This is a famous expression in physics for measuring the half-life of a substance if the decay constantkis known. It
can also be used to computekif the half-lifet 1 / 2 is known.
Example 1:
A radioactive sample contains 2 grams of nobelium. If you know that the half-life of nobelium is 25 seconds, how
much will remain after 3 minutes?
Solution:
Before we compute the mass of nobelium after 3 minutes, we need to first know its decay ratek.Using the half-life
formula,

t 1 / 2 =−ln 2k

k=−tln 2

1 / 2

=− 25 ln 2

=− 0 .028 sec−^1

So the decay rate isk=− 0. 028 /sec.The common unit for the decay rate is the Becquerel(Bq): 1 Bq is equivalent
to 1 decay per sec. Since we foundk, we are now ready to calculate the mass after 3 minutes. We use the radioactive
decay formula. Remember,Crepresents the initial mass,C=2 grams, andt=3 minutes=180 seconds. Thus

y=Cekt

= 2 e(−^0.^028 )(^180 )

= 0 .013 grams.

So after 3 minutes, the mass of the isotope is approximately 0.013 grams.
Population Growth
The same formulay=Cektcan be used for population growth, except thatk> 0 ,since it is an increasing function.