1 / 10

8.3 Applications of Exponential Functions

8.3 Applications of Exponential Functions. 3/25/2013. Compound Interest.

abril
Télécharger la présentation

8.3 Applications of Exponential Functions

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. 8.3 Applications of Exponential Functions 3/25/2013

  2. 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.

  3. Compounding Interest Formula Where P(t) = amount of money accumulated after tyears, 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.

  4. 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 = .04 n = 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.

  5. Exponential Growth Formula 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.

  6. 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 hrs r = .5 hrs (30mins) = =51,200 bacteria After 5 hrs, there are 51,200 bacteria Calculator: raise 2 to (5÷0.5) then multiply by 50.

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

  8. Exponential Decay Formula (half- life) 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.

  9. 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 hrs r = 6 hrs = = 50mg After 12 hrs, there’s 50mg left. Calculator: raise 0.5 to (12÷6 or 2) then multiply by 200.

  10. Homework: WS 8.3 do ALL

More Related