Transcript 4.5 Applications of Exponential Functions

8.3 Applications of Exponential
Functions
3/25/2013
Compound Interest
Interest that accrues on the initial principal and the accumulated
interest of a principal deposit, loan or debt. Compounding of
interest allows a principal amount to grow at a faster rate than
simple interest, which is calculated as a percentage of only the
principal amount.
Compounding Interest Formula
𝑟 𝑛𝑡
𝑃 𝑡 = 𝑃𝑜 1 +
𝑛
Where P(t) = amount of money accumulated after t years,
including interest.
Po = principal amount (the initial amount you borrow or
deposit)
r = annual rate of interest (as a decimal)
n = number of times the interest is compounded per year
t = number of years the amount is deposited or borrowed
for.
An amount of $1,500.00 is deposited in a bank paying an annual
interest rate of 4%, compounded quarterly. What is the balance
after 6 years?
𝑛𝑡
𝑟
𝑃 𝑡 = 𝑃𝑜 1 +
𝑛
4(6)
.04
Po = $1,500
𝑃 𝑡 = 1500 1 +
4
r = .04
24
=1500
1.01
n = 4 (quarterly = 4 times per year)
= $1,904.60
t = 6 yrs
Calculator: Follow order of Operations: Do what’s in the ( ),
then raise it to the exponent, then multiply by 1500.
Exponential Growth Formula
𝑡
𝑃 𝑡 = 𝑃0 (𝑏) 𝑟
Where P(t) = the amount of substance after time t
Po = initial/starting amount
b = growth factor
= 2 for doubling
= 3 for tripling
t = time elapsed
r = time it takes for growth to occur.
Sarah observes that the number of bacteria in the colony in the
lab doubles every 30mins. If the initial number of bacteria in the
colony is 50, what is the total number of bacteria in the colony
𝑡
after 5 hours?
𝑃 𝑡 = 𝑃 (𝑏) 𝑟
0
5
Po = 50
b=2
t = 5 hrs
r = .5 hrs (30mins)
𝑃 𝑡 = 50(2) .5
=50(2)10
=51,200 bacteria
After 5 hrs, there are 51,200 bacteria
Calculator: raise 2 to (5÷0.5) then multiply by 50.
Half Life
is the amount of time that the substance's
total amount is halved.
Exponential Decay Formula (half- life)
𝑡
𝑃 𝑡 = 𝑃0 (𝑑) 𝑟
Where P(t) = the amount of substance left after
time t
Po = initial/starting amount
d = decay factor
= ½ for half-life
t = time elapsed
r = time it takes for decay to occur.
Technitium-99m is a radioactive substance used to diagnose brain,
thyroid liver and kidney diseases. This radioactive substance has a
half life of 6 hours. If there are 200 mgs of this technetium-99m,
how much will there be in 12 hours? 𝑃 𝑡 = 𝑃 (𝑑)𝑡 𝑟
0
12
Po = 200 mg
d=½
t = 12 hrs
r = 6 hrs
𝑃 𝑡 = 200(½) 6
=200(½)2
= 50mg
After 12 hrs, there’s 50mg left.
Calculator: raise 0.5 to (12÷6 or 2) then multiply by 200.
Homework:
WS 8.3 do ALL