Transcript Lesson 6-2

Lesson 9-4
Exponential Growth and Decay
Exponential Growth and Decay
Generally these take on the form
P = p0e±kt
Where p0 is the initial condition at time t= 0
population shrinking  decay (negative exponent)
population expanding  growth (positive exponent)
k is the exponential rate of change
Although these equations can be derived like is
demonstrated in your notes, most students will use the
equation above.
d2y
dy
 2  8y  0
2
dt
dt
Example 1 - Growth
The number of bacteria in a rapidly growing culture was
estimated to be 10,000 at noon and 40000 after 2 hours.
How many bacteria will there be at 5 pm?
General Equation:
Equation:
P = p0e±kt
growth k > 0; p0 = 10000
P(2)= 40000 = 10000e2k
4 = e2k
ln 4 = ln e2k = 2k
ln 2² / 2 = k
ln 2 = k
P(5) = 10000e5ln2
= 320,000 bacteria
Use to solve for k
Example 2 - Decay
Carbon 14 is radioactive and decays at a rate
proportional to the amount present. Its half life is 5730
years. If 10 grams were present originally, how much
will be left after 2000 years?
General Equation:
Equation:
P = p0e±kt
decay k < 0; p0 = 10 grams
Use to solve for k
P(5730)= 5 = 10e5730k
0.5 = e5730k
ln 0.5 = ln e5730k = 5730k
ln 2-1 / 5730 = k
-(ln 2)/5730 = k = -0.000121
P(2000) = 10e2000(-0.000121)
= 7.851 grams
dT
 k T  M 
dt
Compounded Interest
• Compounded Interest Formula:
A(t) = A0 (1 + r/n)nt
where A0 is the initial amount
r is the interest rate
n is the number of times compounded per year
t is the time period in years
• Compounded Continuously (n→∞):
A(t) = A0 ert
dy
 2 xy
dx
Example 3a
Suppose Joe put $500 in the bank at 4% interest. How
much will it be worth after 5 years if it is compounded
annually?
Annually yields time increments of years, so n = 1
A(t) = A0(1 + r/n)nt
A(5) = 500 (1 + 0.04/1)(1)5
A(5) = 500(1.04)5
A(5) = $608.33
dy
 2 xy
dx
Example 3b
Suppose Joe put $500 in the bank at 4% interest. How
much will it be worth after 5 years if it is compounded
monthly?
Monthly yields time increments of months, so n = 12
A(t) = A0(1 + r/n)nt
A(5) = 500 (1 + 0.04/12)(12)5
A(5) = 500(1.003333)60
A(5) = $610.50
dy
 2 xy
dx
Example 3c
Suppose Joe put $500 in the bank at 4% interest. How
much will it be worth after 5 years if it is compounded
weekly?
Weekly yields time increments of weeks, so n = 52
A(t) = A0(1 + r/n)nt
A(5) = 500 (1 + 0.04/52)(52)5
A(5) = 500(1.000769231)260
A(5) = $610.65
dy
 2 xy
dx
Example 3d
Suppose Joe put $500 in the bank at 4% interest. How
much will it be worth after 5 years if it is compounded
daily?
Daily yields time increments of days, so n = 365
A(t) = A0(1 + r/n)nt
A(5) = 500 (1 + 0.04/365)(365)5
A(5) = 500(1.000109589)1825
A(5) = $610.69
dy
 2 xy
dx
Example 3e
Suppose Joe put $500 in the bank at 4% interest. How
much will it be worth after 5 years if it is compounded
continuously?
Continuously yields infinite time increments, so n  
A(t) = A0ert
2.718281828
A(5) = 500 (2.718281828)(0.04)5
A(5) = 500(2.718281828)0.2
A(5) = $610.70
So why should we save?
• Time is the key to savings growth and 5 years is just too
short a time period!
• What if we put away $5000 the day our child was born with
just 4% interest, how much would they have at age 65?:
A(65) = 5000 (1 + 0.04/365)(365)65
A(65) = 5000(1.000109589)23725
A(65) = $67,309.10
• If we got 10% interest (average SP500 growth):
A(65) = 5000 (1 + 0.1/365)(365)65
A(65) = 5000(1.000273973)23725
A(65) = $3,322,748.59
Summary & Homework
• Summary:
– Exponential Growth and Decay is a common realworld problem that can be solved using
differential equations
– Interest earning accounts need long periods of
time to earn significant amounts of money
• Homework:
– pg 620 – 621: Day One: 3, 4, 8, 9,
Day Two: 13, 19. 20