EXPONENTIAL
GROWTH AND DECAY
Section 6.4a
Law of Exponential Change
Suppose we are interested in a quantity y that increases or
decreases at a rate proportional to the amount present…
Can you think of any examples???
If we also know the initial amount of y , we can model this
situation with the following initial value problem:
Differential Equation:
Initial Condition: y
dy
 ky
dt
 0  y0
Note: k can be either
positive or negative
 What happens in each
of these instances?
Law of Exponential Change
dy
Let’s solve this differential equation:
 ky
dt
dy
 kdt Separate variables
y
dy
 y   kdt  ln y  kt  C Integrate
ln y
e
e
kt C
y  e e
C
kt
Exponentiate
Laws of Logs/Exps
Law of Exponential Change
dy
Let’s solve this differential equation:
 ky
dt
y  e e
C kt
y  e e Def. of Abs. Value
kt
C
Let
A
=
+
e
–
y  Ae
 k  0
y  0   Ae
 A  y0 Apply the
C
Solution:
y  y0 e
kt
Initial Cond.
kt
Law of Exponential Change
If y changes at a rate proportional to the amount present
(dy/dt = ky) and y = y 0 when t = 0, then
y  y0 e
kt
where k > 0 represents growth and k < 0 represents
decay. The number k is the rate constant of the
equation.
Compounding Interest
Suppose that A 0 dollars are invested at a fixed annual interest
rate r. If interest is added to the account k times a year, the
amount of money present after t years is
 r
A  t   A0  1  
 k
kt
Interest can be compounded monthly (k = 12), weekly (k = 52),
daily (k = 365), etc…
Compounding Interest
What if we compound interest continuously at a rate proportional
to the amount in the account?
We have another initial value problem!!!
Differential Equation:
dA
 rA
dt
Look familiar???
Initial Condition:
Solution:
A  0  A0
A  t   A0e
rt
Interest paid according to this formula is compounded
continuously. The number r is the continuous interest rate.
Radioactivity
Radioactive Decay – the process of a radioactive substance
emitting some of its mass as it changes forms.
Important Point: It has been shown that the rate at which a
radioactive substance decays is approximately proportional to
the number of radioactive nuclei present…
So we can use our
familiar equation!!!
y  y0e
 kt
k 0
Half-Life – the time required for half of the radioactive nuclei
present in a sample to decay.
Guided Practice
Find the solution to the differential equation dy/dt = ky, k a
constant, that satisfies the given conditions.
1. k = – 0.5, y(0) = 200
Solution:
y  t   200e
0.5t
Guided Practice
Find the solution to the differential equation dy/dt = ky, k a
constant, that satisfies the given conditions.
y0  60
2. y(0) = 60, y(10) = 30
y 10  30  60e
10 k
0.5  e
10 k
Solution:
or
ln 0.5  10k  k  0.1ln 0.5
 0.1ln 2
y  t   60e
t 10
y  t   60  2
 0.1ln 2t
Guided Practice
Suppose you deposit $800 in an account that pays 6.3% annual
interest. How much will you have 8 years later if the interest is
(a) compounded continuously? (b) compounded quarterly?
(a)
(b)
A 8  800e
.0638
 $1324.26
 .063 
A  8   800 1 

4 

 4 8
 $1319.07
Guided Practice
Find the half-life of a radioactive substance with the given decay
equation, and show that the half-life depends only on k.
y  y0e
 kt
Need to solve:
1
y0 e  y 0
2
1
 kt  ln
2
 kt
1 1 ln 2
t   ln 
k 2
k
This is always the half-life
of a radioactive substance
with rate constant k (k > 0)!!!
Guided Practice
Scientists who do carbon-14 dating use 5700 years for its halflife. Find the age of the sample in which 10% of the radioactive
nuclei originally present have decayed.
Half-Life =
y0 e
 kt
 kt
ln 2
ln 2
k 
5700 
5700
k
 0.9 y0
e  0.9
kt  ln 0.9
1
t   ln 0.9
k
5700

ln 0.9  866.418
ln 2
The sample is about 866.418 years old
Guided Practice
A colony of bacteria is increasing exponentially with time. At the
end of 3 hours there are 10,000 bacteria. At the end of 5 hours
there are 40,000 bacteria. How many bacteria were present
initially?
10, 000  y0 e
5k
40, 000 y0e

3k
10, 000 y0e
3k
40, 000  y0 e
4e
2k
5k
 k  ln 2
10, 000  y0 e
3ln 2
 y0  1250
There were 1250 bacteria initially
Guided Practice
The number of radioactive atoms remaining after t days in a
sample of polonium-210 that starts with y 0 radioactive atoms
is
0.005t
0
yye
(a) Find the element’s half-life.
Half-life =
ln 2
ln 2

 138.629 days
k
0.005
Guided Practice
The number of radioactive atoms remaining after t days in a
sample of polonium-210 that starts with y 0 radioactive atoms
is
0.005t
0
yye
(b) Your sample is no longer useful after 95% of the initial
radioactive atoms have disintegrated. For about how many
days after the sample arrives will you be able to use the
sample?
0.05 y0  y0 e
0.005t
ln 0.05  0.005t
ln 0.05
t
 599.146 days
0.005