8.4
Growth and Decay
Name____________________________
In this section we look at exponential growth and decay equations. Consider the population of a
region. If there is no immigration or emigration, the rate at which the population is changing is
often proportional to the population. In other words, the larger the population, the faster it is
growing, because there are more people to have babies. If the population at time t is P, and its
continuous growth rate is 2% per unit time, then we know
Rate of growth of population = 2% (Current population)
and we can write this as
dP
 0.02 P
dt
Relative growth rate:
Absolute growth rate:
Every solution to the equation
dP
= kP
dt
kt
Can be written in the form P = P0e ,
Where P0 is the initial value of P, and k  0 represents growth, while k  0 represents decay.
Recall that the doubling time of an exponentially growing quantity is the time required for it to
𝒍𝒏𝟐
double. 𝑻𝟐 = 𝒌
The half-life of an exponentially decaying quantity is the time for half of it to decay.
𝑻𝟏 =
𝟐
𝒍𝒏𝟐
𝒌
Continuously Compounded Interest
Continuous compounding means that, at any time, interest is being accrued at a rate that is a fixed
percentage of the balance at that moment. Thus, the larger the balance the faster interest is being
earned and the faster the balance grows.
Example 1
A bank account earns interest continuously at a rate of 5% of the current balance per year.
Assume that the initial deposit is $1000, and that no other deposits or withdrawals are made.
a)
Write the differential equation satisfied by the balance in the account.
b)
Solve the differential equation and graph the solution.
Difference between Continuous and Annual Percentage Growth Rates
Newton’s Law of Heating and Cooling
Newton proposed that the temperature of a hot object decreases at a rate proportional to the
difference between its temperature and that of its surroundings. Similarly a cold object heats up
at a rate proportional to the temperature difference between the object and its surroundings.
Example 3
When a murder is committed, the body, originally at 37C , cools according to Newton’s Law of
Cooling. Suppose that after two hours the temperature is 35 C, and that the temperature of the
surrounding air is a constant 20 C.
a)
Find the temperature, H of the body as a function of t, the time in hours since the
murder was committed.
b)
Sketch a graph of temperature against time.
c)
What happens to the temperature in the long run? Show this on the graph and
algebraically.
d)
If the body is found at 4 pm at a temperature of 30 C, when was the murder
committed?
Equilibrium Solutions
The graphs below show the temperatures of several objects in a 20 C room. One is initially hotter
than 20C and cools down toward 20C ; another is initially cooler and warms up toward 20C . All
these curves are solutions to the differential equation
dH
= -k  H - 20 
dt

for some fixed k  0 , and all have the form
Where H  20 as t   , because e
 kt

H = 20 + Be -kt
1
 0 as t  
ekt
In other words, in the long run, the temperature of the object always tends toward T , the
temperature of the room. This means that what happens in the long run is independent of the
initial condition.
In special cases where B  0 , we have the equilibrium solution.
H  20
for all t . This means that if the object starts at the temperature 20C , it remains at 20C for all
time. Notice that such a solution can be found directly from the differential equation by solving
dH
 0:
dt
dH
 k  H  20   0 giving H  20
dt
Stable Equilibrium
k 0
Unstable Equilibrium k  0
B
10
20
HW p 588 9, 11, 15, 21,29,43