#
Section 9-6

#
Exponential Growth and Decay

##
EXPONENTIAL GROWTH

Exponential growth
models are often used for real-world situations like interest earned on an investment, human or animal population, bacterial culture growth, etc.

The general exponential growth model is

**
***
y
*
=
*
C
*
(1 +
*
r
*
)
^{
t
}
,

where
*
C
*
is the initial amount or number,
*
r
*
is the growth rate (for example, a 2% growth rate means
*
r
*
= 0.02), and
*
t
*
is the time elapsed.

Example:

A population of 32,000 with a 5% annual growth rate would be modeled by the equation:

**
***
y
*
= 32000(1.05
*
*
)
^{
t
}

with
*
t
*
in years.

Sometimes, you may be given a doubling or tripling rate rather than a growth rate in percent. For example, if you are told that the number of cells in a bacterial culture doubles every hour, then the equation to model the situation would be:

**
***
y
*
=
*
C
*
· 2
*
*
^{
t
}

with
*
t
*
in hours.

##
EXPONENTIAL DECAY

Exponential decay
models are also used very commonly, especially for radioactive decay, drug concentration in the bloodstream, of depreciation of value.

##
Radioactive Decay

Radioactive decay problems are often given in terms of half-life of a radioactive element. This is modeled by the equation:

**
**

where
*
N
*
_{
0
}
is the initial amount of the element,
*
N
*
is the amount remaining after
*
t
*
years, and τ is the half-life.

Example:

If you start with a quantity of the unstable element Potassium-40, it takes 1.26 billion years for half of it to decay into Argon-40. So the
**
half-life
**
of Potassium-40 is 1.26 billion years.

Write an exponential decay model to find the number of Potassium-40 atoms remaining after

*
t
*
years, if you start with 2000 Potassium-40 atoms.

Here,
*
N
*
_{
0
}
= 2000 and τ = 1,260,000,000. So the model is:

##
Drug Concentration

For drug concentration problems, you may be given the fraction
*
p
*
of the original amount of the drug left in the bloodstream after a unit of time. In this case, the situation is modeled by the equation

**
***
y
*
=
*
A
*
*
p
*^{
t
}
,

where
*
y
*
is the concentration remaining after time
*
t
*
, and
*
A
*
is the initial amount
*
.
*

Example:

If a person takes
*
A
*
milligrams of a drug at time 0, then
*
y
*
=
*
A
*
(0.8)
*
*^{
t
}
gives the concentration left in the bloodstream after
*
t
*
hours. If the initial dose is 200 mg, what is the concentration of the drug in the bloodstream after 4 hours?

Substitute.

*
***
y
**
**
= 200(0.8)
**^{
4
}

You might want a calculator!

*
***
y
**
**
= 200(0.4096)
**

*
***
y
**
**
= 81.92
**

So there are about 82 milligrams of the drug left in the bloodstream after four hours.

##
Depreciation

If the value of some article (for example, a car), originally $
*
C
*
, depreciates
*
x
*
% per year, then the value after
*
t
*
years is given by the formula:

**
***
y
*
=
*
C
*
(1
–
**
***
x
*
/100)
*
*^{
t
}

Example:

The original value of a car is $28,000. If it depreciates by 15% each year, find its value in 4 years.

Substitute.

*
***
y
**
**
= 28000(1
**
–
**
0.15)
**^{
4
}

*
***
y
**
**
= 28000(0.85)
**^{
4
}

*
***
y
**
**
= 28000(0.52200625)
**

*
***
y
**
**
= 14616.175
**

So after four years, the car is worth about $14,616.