Exponential growth and decay

A Exponential growth and decay On the same day two friends both buy a foal. Both foals have a weight of 50 kilogram. After one month they compare the weights. Fred says, "My foal grew 10 kg." Andy answers, "My foal grew 20 %. After another month they meet again and compare the weights. Fred: "another 10 kg!"; Andy: "another 20%." A1 What are the weights of the foals two months after they were bought? A2 Compare the weight of Andy's foal after two months to the weight on the day it was bought. What is the percentage increment (percent increase)? Now suppose this growth continues the same way for a number of months: one foal with 10 kg per month, the other one with 20% per month. A3 Complete the following table, but first explain what has already been written: # months after purchase weight of Fred's foal weight of Andy's foal 0 50 50 1 50 + 10 = 60 50 + 0.20 50 = 60 2 60 + 10 = 60 + 0.20 60 = 3 4 5 6 7 8 Part A: exponential growth and decay 1 4/8/2008

The weight of both foals can also be graphed. A4 Use the grid to plot the points you found in the table. A5 Do you think either of these growth patterns is realistic for a longer period of time? A6 What is the most striking difference between the two graphs? weight (in kg) 150 140 130 120 110 100 90 80 70 60 50 1 2 3 4 5 6 7 8 9 10 time t (in months) The kind of growth of Fred's foal is called linear growth. Every month a fixed number (10) is added to the weight of the previous month. We may call the weight of Fred's foal W F and the time after the purchase date t (in months) A7 What function describes the relationship between W F and t? What type of function is it? The growth of Andy's foal deserves a closer look! Looking at the weights in the table, you can find the following relationship: weight this month = 1.2 weight last month A8 Check this ratio for each pair of successive months and explain why this is correct for this kind of growth. This relationship can also be written as: weigh this month = 1.2 weight last month We call the consecutive weights of Andy's foal: W A (t) with t = 0, 1, 2, 3,... Part A: exponential growth and decay 2 4/8/2008

With this function we can write the statement as: W A (1) = 1.2 W A (0) W A (2) = 1.2 W A (1) = 1.2 1.2 W A (0) = 1.2 2 W A (0) W A (3) = 1.2 W A (2) = 1.2 1.2 2 W A (0) = 1.2 3 W A (0) W A (4) = 1.2 W A (3) =... A9 Compute W A (4) (using W A (0)) A10 Explain why the function W A (t) = 50 1.2 t describes the relationship between W A (t) and t A growth process that can be described with a function like G(t) = c g t is called an exponential growth process In this function, c is often called the initial value and g (the base) the growth factor. The name "exponential" comes from the fact that the independent variable (t) is an exponent. Mathematically spoken, the growth factor g is called the base, but for the time being we will call it growth factor because that makes more sense in applications. Although the growth of foals described in the two different ways is not very realistic for the long run, it showed some interesting differences between linear and exponential growth: Linear growth is characterized by: every fixed time step results in a fixed addition Exponential growth is characterized by: every fixed time step results in a fixed multiplication. Exponential growth is a very natural phenomenon. This biological statement shows why: For all living species, each individual contributes to the growth of the whole population So, if in a population 10 individuals are creating one newborn on the average, the growth factor for the whole population will be 1.1. And it is obvious that there will be more newborns when there are more species in the population. Part A: exponential growth and decay 3 4/8/2008

A last question about Andy's foal. Suppose the graph is continuous and it is a smooth curve that goes through the given points. Then it will look like this: A11 Use the graph to make an accurate estimation of the time needed to arrive at a weight of 100 kg. weight (in kg) Now without looking at the graph: how long does it take to arrive at a weight of 120 kg? How can you answer this question without looking at the graph, only using the time needed to arrive at 100 kg? time t (in months) Use the graph to verify the time it takes to grow from 100 kg to 120 kg. Part A: exponential growth and decay 4 4/8/2008

We will now look at two real life applications of exponential growth. Bacteria: Escherichia coli From the internet (http://escherichia coli.sergi5.com/ ), the following information about a bacteria was found: Escherichia coli (E. coli), is one of the main species of bacteria that live in the lower intestines of mammals including us. They are necessary for the proper digestion of food and are part of the intestinal flora. The human body has the same number of human cells and E.coli cells: 10 000 000 000 000.... In real life, E.coli duplicate every two hours. E.coli is important for digestion. Incredible, but true: they double every two hours! Suppose a culture of E.coli is grown in a lab. We put 128 of them in an ideal environment and they duplicate every two hours. A12 How many are there after one day? E. coli magnified to 10,000x. And how many after one week? The number of E.coli is E(t) and t indicates the time (in units of 2 hours) after the start. A13 What formula describes the number of E.coli as a function of time t? Part A: exponential growth and decay 5 4/8/2008

The duplication of E.coli is a discrete process: after every two hours, the number of bacteria is doubled. Therefore the graph of this growth process is a set of isolated points. But what happens during these two hours? The following information gives some insight in the process (see the picture): An individual cell grows in volume during these two hours. At the moment it splits into two new cells, the volume of that one cell is equal to the volume of two original cells. So, if we consider the total volume of E.coli cells over time, this is a continuous process. And we assume that this process of volume growth is also exponential. It means that for every fixed time unit, the volume is multiplied with a constant (fixed) number. The graph of this continuous exponential process, with starting value 128 and growth factor 2 (for the time unit of 2 hours), is shown for the time interval [ 2, 2]. The function that fits this process is: V(t) = 128 2 t with the time (in units of 2 hours) and V the volume of E.coli. At t = 0, the volume of E.coli is 128. At t = 1, the volume (and also the number) is doubled to 256. But between t = 0 and t = 1, the volume is exponentially increasing from 128 to 256. As you can see, the continuous graph shows all values in between 128 and 256. A14 Using the graph, make an accurate estimate of the volume of E.coli after one hour (this is at the moment t = 0.5!) and also after 3 hours (t = 1.5). Dividing both numbers, the result should be exactly 2. Why? Part A: exponential growth and decay 6 4/8/2008

A15 Use your Graphing Calculator to find the values V(0.5) and V(1.5). V (0.5) V (1) V (1.5) V (2) Explain why,, and all should have the same outcome. V (0) V (0.5) V (1) V (1.5) What is the outcome for each ratio? And... what does this outcome mean? What does this outcome have to do with a growth factor of 2 (for two hours)? The given graph did not start at the moment that the lab experiment started (t = 0), but 4 hours before that moment (t = 2). The assumption was that the 128 E.coli bacteria were growing before the experiment started in the same way as they did from t = 0 on. A16 How many were there at t = 1 and t = 2? Do you see this also in the graph of A14? A17 Explain: If an exponential process has a growth factor 2 per time unit going forward in time, you will have to halve the amount when you go back in time for one time unit. In a table, the information of the given graph is: t 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 V(t) 32 64 128 256 512 A18 Use the doubling (forward) and halving (backwards) to calculate the exact values of V for these values of t in the table. The formula for the growth is: V(t) = 128 2 t Using this formula to calculate the value for t = 1 results in: V( 1) = 128 2 1. In this way you can add a new row to the table with the values of V as they are expressed by the formula. Part A: exponential growth and decay 7 4/8/2008

t 2 1 0 1 2 V through doubling and halving 32 64 128 256 512 V as expressed in the formula 128 2 2 128 2 1 128 2 0 128 2 1 128 2 2 On the one hand V( 1) = 64 and on the other hand V( 1) = 128 2 1 A19 Use this information to show that 1 1 2 = 2 Using the same way of calculating values, what should be the meaning of 2 2, 2 3 and 2 0? A20 The growth factor for 2 hours is 2. What will be the growth factor for one hour? Hint: How would you adjust the table above with another entry for one hour? Bank account You start a bank account on January 1 st with an initial investment of $100. The annual interest rate is 5%. If you leave the money in the bank, the amount of money will grow exponentially over the years. After one year, you have $105. The 5 dollars interest is added to the account on the last day of the year. After the second year it will be $105 + 5% of $105 = $110.25. Again: the interest $5.25 is added on the last day of the second year. A21 >> Explain: the growth factor per year is 1.05. >> Which formula describes the amount of money (M) in the bank account over the years (t)? >> How much money is in the bank after 10 years? >> How many years will it take to double the money in the account? Part A: exponential growth and decay 8 4/8/2008

Keep in mind that this is a discrete process. Every last day of the year, the interest of that year is added to the amount. It is also possible to ask the bank to add the interest every half year or every two years. Sam proposes the bank to use the following formulas for these two options. 2 For the half year interest option: M ( t ) = 100 (1.025) t For the two year interest option: ( ) Time t is still in units of one year. 1 2 M t = 100 (1.10) t A22 What interest percentages does Sam use for the half year and for the two year option? Explain why the exponent is changed to 2t and to 1 t for the two options. 2 The mistake Sam makes is that he thinks about addition instead of multiplication. He thinks: 2.5% + 2.5% = 5% and 5% + 5% = 10%. Do you think the bank will accept the half year option? And how about the two year option? Explain. A23 How can you use the growth factor for one year (1.05) to find the growth factor for the half year option and for the two year option? What are the right percentages for the half year and the two year options? (Find the percentages accurate to three decimal places.) Part A: exponential growth and decay 9 4/8/2008

The year can also be split in 8 equal periods. If you want to know the growth factor g for each of these periods, you should realize that repeating this growth 8 times has to result in a total growth with factor 1.05 (the growth factor for a whole year). A visualization may help here: 1.05 g g g g g g g g So, the number you are looking for can be found from g 8 = 1.05 A24 Find this growth factor g accurate to 5 decimal places. Part A: exponential growth and decay 10 4/8/2008

Exponential decay (or: negative growth) All examples of exponential growth so far were about increasing numbers. But exponential processes are also possible with decreasing numbers. The word 'growth' is a little strange for that, so normally we talk about 'decay' when the numbers are decreasing. The following information can be found on wikipedia.org The LZ 129 Hindenburg was the largest airship ever built. It was a brand-new, allduralumin design: 245 meters long (804 ft), 41 meters in diameter (135 ft), containing 200,000 cubic meters (7,000,000 ft³) of gas in 16 bags or cells, powered by four reversible 1,200 horsepower (890 kw) Daimler-Benz diesel engines, giving it a maximum speed of 135 kilometers per hour (84 mph). The Hindenburg was longer than three Boeing 747s placed end-to-end. It had cabins for 50 passengers (upgraded to 72 in 1937) and a crew of 61. For aerodynamic reasons, the passenger quarters were contained within the body rather than in gondolas. On the first of August 1936, the Hindenburg (the pride of Hitler-Germany) was present at the opening ceremonies of the eleventh modern day Olympic Games in Berlin, Germany. Moments before the arrival of Adolf Hitler, the airship crossed over the Olympic stadium trailing the Olympic flag from its tail. During 1936, its first year of commercial operation, the Hindenburg flew 191,583 miles carrying 2,798 passengers and 160 tons of freight and mail. In that year the ship made 17 round trips across the Atlantic Ocean with 10 trips to the US and 7 to Brazil. The picture on the right shows what happened on May 6, 1937. While landing at Lakehurst Naval Air Station in New Jersey, the airship was destroyed by fire. Most of the crew and passengers survived. Of 36 passengers and 61 crew, 13 passengers and 22 crew died. Compared to more modern ones, airships in these days (1930's) were not very well constructed. Through the envelope that covered the hydrogen filled cavity, gas was leaking. A loss of 50% in ten days was normal. So, here we have an example of an exponential process in which there is no growth, but decay. Every 10 days, 50% of the gas disappeared. Part A: exponential growth and decay 11 4/8/2008

A25 Given a completely filled Hindenburg (200,000 m 3 of hydrogen), how much gas was left after 10 days? And how much after another 10 days? Find a formula that describes the amount of gas (G) left in the airship after t units of time (the unit is 10 days) A26 A loss of 50% in 10 days means a loss per day of 6.7%. Explain. So, the formula that describes G as a function of the time t (with time unit 1 day) is: G(t) = 200,000 (0.933) t Plotted with a graphing calculator, with the formula and the chosen WINDOW, the graph looks like this: (the independent variable in the GC is always X, so t should be replaced by X and G(t) is here Y1) Modern airships are quite small (about 3,000 m 3 ) and the loss of gas is much better: only 2% per 10 days. A27 Find a formula that describes the amount of gas left after t days, for a modern airship that is completely filled at t = 0. Part A: exponential growth and decay 12 4/8/2008

When 20% of the gas is lost, an airship cannot stay in the air. A28 How many days can an airship stay up in the air? Answer this question both for an old one like the Hindenburg and for a modern one. Another (well known) example of exponential decay is the half life of medicines, alcohol, radioactive materials and caffeine. (additional information can be found at http://en.wikipedia.org/wiki/caffeine): The half-life, or time it takes for the amount of caffeine in the blood to decrease by 50%, ranges from 3.5 to 10 hours. In adults the half-life is generally around 5 hours. A 7 oz cup filled with Espresso has 100 mg caffeine (source: http://nootropics.com/caffeine/faq.html) I (an adult person) drink an Espresso at 11 am. A29 How many mg of caffeine are still in my body when I go to bed at 11 pm? One day I drink an Espresso at 11 am and another one at 5 pm. A30 How many mg of caffeine are still in my body now when I go to bed at 11 pm? A31 What is the 'growth factor' of caffeine per hour for an adult? Describe the relationship between the amount of caffeine C (in mg) in the body as a function of the time t (in hours) after drinking a 7 oz cup of Espresso. Part A: exponential growth and decay 13 4/8/2008

Summary part A In this paragraph characteristics of linear and exponential growth are explored. It is about change over a fixed time period: from "amount old" to "amount new"; the rule is add a constant number (c) for linear growth and multiply with a constant number (g, the growth factor) for exponential growth (or decay). Given a starting value b (at t = 0) the functions can be found. Schematic: Linear growth Exponential growth Exponential decay rule amount new = amount old + c amount new = amount old g for g > 1 amount new = amount old g for 0 < g < 1 function N(t) = c t + b N(t) = b g t N(t) = b g t graph A visualization of the difference between linear and exponential growth: For linear growth: every fixed time step ( + 1) causes a fixed addition ( + c) amount + c + c + c + c + c + c + c time + 1 + 1 + 1 + 1 + 1 + 1 + 1 For exponential growth: every fixed time step ( + 1) causes a fixed multiplication ( g) amount g g g g g g g time + 1 + 1 + 1 + 1 + 1 + 1 + 1 Another observation: If the time step is doubled ( + 2 instead of + 1), 2c is added to the amount (for linear growth) If the time step is doubled ( + 2 instead of + 1), the amount is multiplied by g 2 (for exponential growth) Some exponential rules were found, like: g 0 1 2 = 1, g = g, g 3 = 1 3 g Part A: exponential growth and decay 14 4/8/2008

Exploration (Homework Exercises) E1 G(t) = ( 1 ) t and F(t) = 2 t 2 Plot both graphs in the coordinate system on the right. ( 4 t 4). What do you notice? Explain! E2 G(t) = ( 1 ) t and F(t) = 2 t 2 Plot both graphs in the coordinate system on the right. ( 4 t 4). What do you notice? Explain! E3 G(t) = 3 2t and F(t) = 9 t The two functions have the same graph. Explain. Part A: exponential growth and decay 15 4/8/2008

E4 A student entered an exponential function in her graphing calculator, which gave the following table: The time unit is 1 day. What is the function? What is the exact value at t = 0.25 (the table only shows a four decimal approximation) Explain: from t = 0.25 to t = 0.75 the growth factor is 2 E5 A growth process starts with 20 (at t = 0) and grows to 200 in 10 days. What formula fits this growth if the process is linear? What formula fits this growth if the process is exponential? The exponential growth can also be written with a growth factor of 10. What is the time unit in this case? What formula describes the process in this case? Part A: exponential growth and decay 16 4/8/2008