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8.3 Applications of Exponential Functions 3/25/2013

8.3 Applications of Exponential Functions 3/25/2013

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Page 1: 8.3 Applications of Exponential Functions 3/25/2013

8.3 Applications of Exponential Functions

3/25/2013

Page 2: 8.3 Applications of Exponential Functions 3/25/2013

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.

Page 3: 8.3 Applications of Exponential Functions 3/25/2013

Compounding Interest Formula

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.

Page 4: 8.3 Applications of Exponential Functions 3/25/2013

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

Calculator: Follow order of Operations: Do what’s in the ( ), then raise it to the exponent, then multiply by 1500.

Page 5: 8.3 Applications of Exponential Functions 3/25/2013

Exponential Growth Formula

Where P(t) = the amount of substance after time tPo = initial/starting amount

b = growth factor = 2 for doubling = 3 for triplingt = time elapsedr = time it takes for growth to occur.

Page 6: 8.3 Applications of Exponential Functions 3/25/2013

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

Calculator: raise 2 to (5÷0.5) then multiply by 50.

Page 7: 8.3 Applications of Exponential Functions 3/25/2013

Half Lifeis the amount of time that the substance's total amount is halved.

Page 8: 8.3 Applications of Exponential Functions 3/25/2013

Exponential Decay Formula (half- life)

Where P(t) = the amount of substance left after time tPo = initial/starting amount

d = decay factor = ½ for half-lifet = time elapsedr = time it takes for decay to occur.

Page 9: 8.3 Applications of Exponential Functions 3/25/2013

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.

Calculator: raise 0.5 to (12÷6 or 2) then multiply by 200.

Page 10: 8.3 Applications of Exponential Functions 3/25/2013

Homework:

WS 8.3 do ALL