VIEWS: 9 PAGES: 4 POSTED ON: 7/30/2012 Public Domain
Algebra 2: McDougal Ch 8.1 & 8.2 Objective: Cover basic growth and decay models, asymptotes and shifts. Basic exponential equation: y a b x a = starting amount (in banking this is the principal, or the $ you start with) x = often represents the amount of time that has passed, sometimes a “t” b = rate of growth, AKA “growth constant”, AKA “interest rate” such as doubling, tripling, half life Example: b = 1 means no change b = .95 means decays by 5% each time period b = 1.05 means grows by 5% each time period b = 2 means grows by 100% each time period (doubles) and b = .5 means drops by 50% each time period (half life) b = e this is a rate of change that is found frequently in actual populations = 2.7182…, it is an irrational number like π Show graphs of growth and decay equations on a graphing calculator, changing “b”. Doubling time discussion: Humans In 2000 there were about 6 billion people. Using the surface area formula S 4 r 2 , and a radius of 6400 km, the surface area of Earth is 514,718,540 km. Current population density: 6 billion/514,718,540 km = .086 persons per square km. Note: 1 km2 = .386 mi2 . Our school campus is about 1 square km. Suppose human doubling time is 50 years under ideal conditions. How many people will be on the planet by the year 2500? Figure out how many doubling times: 2500/50 = 50 a = 6 billion , b = 2 (doubling), x = number of doubling time periods = 50 Equation: y = 6*2^50 = 6.755 x 10^15 billions = 6.755 x 10^24 people. Actually the census bureau predicts 9.4 billion in 2500 = 18 people per square kilometer. Why is this number so much less than the prediction? Ans: limiting factors such as space, food, water, pollution, disease. What would the population density be in the year 2500? (6.755 x 10^24 people)/ 514,718,540 km = 13,312,445,000,000,000,000 people/km2 Squeeze all those people on our school campus! This would be about 5,700,000,000,000 people into a classroom (5.7 trillion)! Or about 12,000,000,000 people per square foot! In banking, if earning 7% interest and paid quarterly, then part of that rate is paid each quarter (so divide by 4). .07/4 =.0175 So, if you are paid interest every quarter (4 times per year) how much money do you have at the r end of 10 years (40 time periods), if you start with $100? A P(1 )nt where A = amount after the time, P = n principal, r = interest rate, n= number of times compounded per year, t = number of years. (The “nt” is going to give you how many time periods. Bankers want an “nt”, rather than just an “x”) . So: 100 1 .0175 ) = 40 $200.16 Half life: Is it ever really gone? Suppose a student walks halfway to the wall every minute. The distance is measured from the tip of the student’s nose to the wall. If the room is 48 feet long, then after 1 minute the student is 24 feet from the wall. After 2 minutes the distance is 12 feet. After 3 minutes, 6 ft. After 4 minutes, 3 ft, … When will the student’s nose touch the wall? Never. Talk about ½ life approaching 0. (Asymptotes) Iodine 131 is used medically to treat thyroid cancer. Half life is 8 days. If being treated, then adult patients are supposed to stay away from children (no hugs) because risk of exposure to radioactivity for a month. Suppose a dose is 100 Gy (what ever that unit is). How much is left after 1 month (30 days)? Calculate the number of time units 30/8 = 3.75 3.75 1 y 100 = 7 Gy 2 Other decay models: If you know 78% of something is left after the time period, then b = .78, and 22% decayed away. If you know 35% decayed away, then b = 65%, because you have 65% left. Additional discussion: Shifts in the basic exponential equation: y a b x h k . Name _________________________________ Class Period______ McDougal Chapter 8: Exponential Growth and Decay (Use in place of 8.1-8.2) Basic exponential formulas are of the form y a b x h k . Just like other families of equations, the h and k values shift the graph left/right and up/down. Notice that the x variable is in the exponent. This makes it a little tricky to find the inverse equation, since you currently lack the information for how to solve an equation for a variable in the exponent. Exponential Growth Example: The populations of living Exponential Decay Example: Radioactive elements decay things tend to increase exponentially until their growth is into other particles in a curve that is exponential. If you slowed by environmental factors, such as lack of space or have a barrel of Uranium 238, it will take 4.5 billion years food. If you have money earning interest in a bank, the to decay into a barrel that is ½ Uranium and ½ Lead. amount grows at an exponential rate. Radioactive Carbon decays much faster, ½ of it is decayed within 5734 years. In an exponential equation, y a b x h k a = starting amount (in banking this is the principal, or the $ you start with) x = often represents the amount of time that has passed, sometimes a “t” b: b is the rate that the curve changes. Example: b = 1 means no change b = .95 means decays by 5% each time period b = 1.05 means grows by 5% each time period b = 2 means grows by 100% each time period (doubles) and b = .5 means drops by 50% each time period (half life) h: shifts the graph up or down k: shifts the graph to the left or right Homework: 1. The amount y (in grams) of a sample of radioactive iodine-131 after t days is given by the basic exponential formula y a b x such that y 50(0.92)t . a) How many grams of the substance did you start with? a = ________ b) Look at the formula and find the percent of substance that remains at the end of each day. b = _________ Using the previous value, what percent decays after each day? _________ c) Calculate how many grams of the substance is left after 3 days (t = 3) 2. Miniaturization of computer chips gets more amazing each year. More and more memory can be stored on smaller and smaller computer chips, and as a result cell phones and I-Pods are shrinking. From 1971 to 1995 the number of transistors on a computer chip can be modeled by the exponential equation n 2300(1.59)t . t represents the number of years that have passed since 1971. a) From the equation, identify how many transistors could be put on a chip at the beginning, in 1971. b) What is the growth factor (the rate of growth?) c) Estimate the number of transistors on a computer chip in 2007 (t = how many years since 1971). 3. E. coli is an intestinal bacteria that is also found in contaminated water and sewage. Some strains of E. coli can double in population in only 20 minutes (so 1 time period = 20 minutes). Suppose you start with one bacteria (a = 1) a) Use the basic exponential formula and the Extension: The mass of information above to write a formula for E. coli the Earth = 5.9763 x 1024 growth. grams. b) Graph the equation and estimate how many time The mass of one periods had passed so that the population exceeded bacterium = 10-12 grams. 1000. Let each mark on the y axis = 100 bacteria. How long until the mass c) How many time periods have passed in 1 of bacteria would equal day?____ Calculate how many E-coli there are after the mass of the Earth? 1 day. (It doesn’t take long!) 4. From the years 1982 to 1993 fewer and fewer record albums were sold each year as people replaced their music collections with CD’s. The decline of record sales during that time period can be modeled by the equation A 265 (.39)t , where A is the number of albums in millions, and t is the amount of time since 1982. a) How many millions of record albums were sold at the beginning of the time period, in 1982? b) What percent of the total is still being sold at the end of each year? c) Calculate how many albums were sold in 1988. (Hint: First find t, how many years had passed since 1982). Graph the following: 1. y 2 x 2. y 2 x 4 3. y 2 x 4 4. y 2 x 1 5 x x4 x x 3 1 1 1 1 5. y 6. y 7. y 4 8. y 4 2 2 2 2