Continuous Growth and the
Number e
Lesson 3.4
Compounding Multiple Times Per
Year
• Given the following formula for compounding
 r
B  P  1  
 n
nt
 P = initial investment
 r = yearly rate
 n = number of compounding periods
 t = number of years
Compounding Multiple Times Per
Year
nt
 r
B  P  1  
 n
• What if we invested
$1000 for 5 years
at 4% interest
• Try the formula for different numbers of
compounding periods
 Monthly
 Weekly (n = 52)
 Daily (n = 365)
 Hourly (n = 365 * 25)
What phenomenon do
you notice?
Compounding Multiple Times Per
Year
• You should see that we seem to reach a limit
as to how much multiple compounding
periods increase the final amount
n
 1
lim 1    e
n 
 n
• So we come up with continuous
compounding
B  Pe
k t
Using e As the Base
• We have used
• Consider letting
y = A * Bt
B = ek
• Then by substitution
• Recall
B = (1 + r)
• It turns out that
y = A * (ek)t
(the growth factor)
k r
Continuous Growth
• The constant k is called the continuous
percent growth rate
• For Q = a bt
 k can be found by solving ek = b
• Then Q = a ek*t
• For positive a
 if k > 0 then Q is an increasing function
 if k < 0 then Q is a decreasing function
Continuous Growth
• For Q = a ek*t
• k>0
• k<0
Assume a > 0
Continuous Growth
Q  3 e
0.4t
• For the function
what is the
continuous growth rate?
• The growth rate is the coefficient of t
 Growth rate = 0.4 or 40%
• Graph the function (predict what it looks like)
Converting Between Forms
• Change to the form Q = A*Bt
Q  3 e
0.4t
• We know B = ek
• Change to the form Q = A*ek*t
Q  94.5(1.076)
t
• We will eventually discover that k = ln B
Assignment
• Lesson 3.4
• Page 133
• Exercises 1 – 25 odd