Wednesday, May 11, 2011

Chap 9 Section 3: Continuous Compounding

Hello class. At the beginning of the day, Mr. Cope gave a 5 minute review on sections 9.1 and 9.2. He simply taught exponential growth (9.1) and exponential decay (9.2). He taught the equation y=ab^x. A being the number you begin with, b being the growth factor (growth b>1, decay 0<b<1) and the x being the number of years being used. This was just a quick review of what was taught on Monday and Tuesday.

Next Mr. Cope handed back the tests and homework quizzes we took last week. Unfortunately, Teacher Logic no longer shows the class average so there will be no comparison to period 1 in this blog post. Anyway, we were able to compare our answers to Mr. Copes key and then we moved on to today's lesson.

Today we learned about continuous compounding. We began with a chart showing a bank pays 100% interest for 1 year. So we went through annually, quarterly, monthly, daily, and hourly using the equation P(1+r/n)^nt. The final value (hourly) was very close to e (2.71828) or Euters number which is used to represent exponential growth. So the formal equation for Continuously Compounded Interest is a=Pe^rt. P stands for the original number, e stands for Euters number, r stands for either the growth of decay (growth=if number is positive, decay if number is negative), and t stands for the number of years. We then did an example 1 which was "Suppose $1800 is invested at an annual interest rate of 7% compounded continuously and the money is what happens in 5 years. So the equation would be 1800e^.07(5). You then would solve it on the calculator and your answer should be 2554.32.

Moodle notes:

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