Exponential Functions Topic 1: Graphs and Equations of Exponential Functions.
4.5 Applications of Exponential Functions 2/8/2013.
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Transcript of 4.5 Applications of Exponential Functions 2/8/2013.
4.5 Applications of Exponential Functions
2/8/2013
Compound InterestInterest 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
Where P(t) = amount of money accumulated after n years, including interest. Po = principal amount (the initial amount you borrow or deposit) r = annual rate of interest (as a decimal)t = number of years the amount is deposited or borrowed for.n = number of times the interest is compounded per year
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?
Po = $1,500
r = .04n = 4 (quarterly = 4 times per year)t = 6 yrs
= = $1,904.60
Depreciation Formula
VWhere V = value of item after time tC = original costr = depreciation rate (as a decimal)t = time
If you purchase a BMW for $45,000, the value of the car depreciates by 16% each year. How much will the car be worth in 8 years?
V = $45,000r = .16t = 8 yrs
V
= = $11,154
Exponential Growth Formula
Where P(t) = the amount after time tPo = initial/starting amount
b = growth factor = 2 for doubling = 3 for triplingt = time elapsedr = 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 after 5 hours?
Po = 50
b = 2 t = 5 hrsr = .5 hrs (30mins)
𝑃 (𝑡 )=𝑃0(𝑏)𝑡𝑟
= =51,200 bacteria
After 5 hrs, there are 51,200 bacteria
Half Lifeis the amount of time that the substance's total amount is halved.
Exponential Decay Formula (half- life)
Where P(t) = the amount left after time tPo = initial/starting amount
d = decay factor = ½ for half-lifet = time elapsedr = 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?
Po = 200 mg
d = ½ t = 12 hrsr = 6 hrs
𝑃 (𝑡 )=𝑃0(𝑑)𝑡𝑟
= = 50mg
After 12 hrs, there’s 50mg left.
Homework:
WS 4.5 do ALL